Cohesion and Adhesion in Liquids: Surface Tension and Capillary Action

Children blow soap bubbles and play in the spray of a sprinkler on a hot summer day. (See Figure below ) . A technician draws blood into a small-diameter tube just by touching it to a drop on a pricked finger. A premature infant struggles to inflate her lungs. What is the common thread? All these activities are dominated by the attractive forces between atoms and molecules in liquids—both within a liquid and between the liquid and its surroundings.

Attractive forces between molecules of the same type are called cohesive forces. Liquids can, for example, be held in open containers because cohesive forces hold the molecules together. Attractive forces between molecules of different types are called adhesive forces. Such forces cause liquid drops to cling to window panes, for example. Lets examine effects directly attributable to cohesive and adhesive forces in liquids.

Cohesive Forces
Attractive forces between molecules of the same type are called cohesive forces.

Adhesive Forces
Attractive forces between molecules of different types are called adhesive forces.

The soap bubbles in this photograph are caused by cohesive forces among molecules in liquids. (credit: Steve Ford Elliott)

Surface Tension

Cohesive forces between molecules cause the surface of a liquid to contract to the smallest possible surface area. This general effect is called surface tension. Molecules on the surface are pulled inward by cohesive forces, reducing the surface area. Molecules inside the liquid experience zero net force, since they have neighbors on all sides.

Surface Tension
Cohesive forces between molecules cause the surface of a liquid to contract to the smallest possible surface area. This general effect is called surface tension.

The model of a liquid surface acting like a stretched elastic sheet can effectively explain surface tension effects. For example, some insects can walk on water (as opposed to floating in it) as we would walk on a trampoline—they dent the surface as shown in the below figure (a). The next figure (b) shows another example, where a needle rests on a water surface. The iron needle cannot, and does not, float, because its density is greater than that of water. Rather, its weight is supported by forces in the stretched surface that try to make the surface smaller or flatter. If the needle were placed point down on the surface, its weight acting on a smaller area would break the surface, and it would sink.

Surface tension supporting the weight of an insect and an iron needle, both of which rest on the surface without penetrating it. They are not floating; rather, they are supported by the surface of the liquid. (a) An insect leg dents the water surface.  is a restoring force (surface tension) parallel to the surface. (b) An iron needle similarly dents a water surface until the restoring force (surface tension) grows to equal its weight.

Surface tension is proportional to the strength of the cohesive force, which varies with the type of liquid. Surface tension  is defined to be the force F per unit length  exerted by a stretched liquid membrane

The tables below lists values of  for some liquids.

For the insect of the above figure (a), its weight w is supported by the upward components of the surface tension force: w=γLsinθ, where L is the circumference of the insect’s foot in contact with the water. The figure below shows one way to measure surface tension. The liquid film exerts a force on the movable wire in an attempt to reduce its surface area. The magnitude of this force depends on the surface tension of the liquid and can be measured accurately.

Sliding wire device used for measuring surface tension; the device exerts a force to reduce the film’s surface area. The force needed to hold the wire in place is F=γL=γ(2l), since there are two liquid surfaces attached to the wire. This force remains nearly constant as the film is stretched, until the film approaches its breaking point

Surface tension is the reason why liquids form bubbles and droplets. The inward surface tension force causes bubbles to be approximately spherical and raises the pressure of the gas trapped inside relative to atmospheric pressure outside. It can be shown that the gauge pressure P inside a spherical bubble is given by

where r is the radius of the bubble. Thus the pressure inside a bubble is greatest when the bubble is the smallest. Another bit of evidence for this is illustrated in the figure below . When air is allowed to flow between two balloons of unequal size, the smaller balloon tends to collapse, filling the larger balloon

With the valve closed, two balloons of different sizes are attached to each end of a tube. Upon opening the valve, the smaller balloon decreases in size with the air moving to fill the larger balloon. The pressure in a spherical balloon is inversely proportional to its radius, so that the smaller balloon has a greater internal pressure than the larger balloon, resulting in this flow

Our lungs contain hundreds of millions of mucus-lined sacs called alveoli, which are very similar in size, and about 0.1 mm in diameter. (See figure below.) You can exhale without muscle action by allowing surface tension to contract these sacs. Medical patients whose breathing is aided by a positive pressure respirator have air blown into the lungs, but are generally allowed to exhale on their own. Even if there is paralysis, surface tension in the alveoli will expel air from the lungs. Since pressure increases as the radii of the alveoli decrease, an occasional deep cleansing breath is needed to fully reinflate the alveoli. Respirators are programmed to do this and we find it natural

Bronchial tubes in the lungs branch into ever-smaller structures, finally ending in alveoli. The alveoli act like tiny bubbles. The surface tension of their mucous lining aids in exhalation and can prevent inhalation if too great

The tension in the walls of the alveoli results from the membrane tissue and a liquid on the walls of the alveoli containing a long lipoprotein that acts as a surfactant (a surface-tension reducing substance). The need for the surfactant results from the tendency of small alveoli to collapse and the air to fill into the larger alveoli making them even larger . During inhalation, the lipoprotein molecules are pulled apart and the wall tension increases as the radius increases (increased surface tension). During exhalation, the molecules slide back together and the surface tension decreases, helping to prevent a collapse of the alveoli. The surfactant therefore serves to change the wall tension so that small alveoli don’t collapse and large alveoli are prevented from expanding too much. This tension change is a unique property of these surfactants, and is not shared by detergents (which simply lower surface tension). (See figure below)

Surface tension as a function of surface area. The surface tension for lung surfactant decreases with decreasing area. This ensures that small alveoli don’t collapse and large alveoli are not able to over expand

If water gets into the lungs, the surface tension is too great and you cannot inhale. This is a severe problem in resuscitating drowning victims. A similar problem occurs in newborn infants who are born without this surfactant—their lungs are very difficult to inflate. This condition is known as hyaline membrane disease and is a leading cause of death for infants, particularly in premature births. Some success has been achieved in treating hyaline membrane disease by spraying a surfactant into the infant’s breathing passages. Emphysema produces the opposite problem with alveoli. Alveolar walls of emphysema victims deteriorate, and the sacs combine to form larger sacs. Because pressure produced by surface tension decreases with increasing radius, these larger sacs produce smaller pressure, reducing the ability of emphysema victims to exhale. A common test for emphysema is to measure the pressure and volume of air that can be exhaled.

Adhesion and Capillary Action

Why is it that water beads up on a waxed car but does not on bare paint? The answer is that the adhesive forces between water and wax are much smaller than those between water and paint. Competition between the forces of adhesion and cohesion are important in the macroscopic behavior of liquids. An important factor in studying the roles of these two forces is the angle θ between the tangent to the liquid surface and the surface. (See the figure below .) The contact angle  θ is directly related to the relative strength of the cohesive and adhesive forces. The larger the strength of the cohesive force relative to the adhesive force, the larger θ is, and the more the liquid tends to form a droplet. The smaller θ is, the smaller the relative strength, so that the adhesive force is able to flatten the drop

Contact Angle
The angle θ between the tangent to the liquid surface and the surface is called the contact angle.

In the photograph, water beads on the waxed car paint and flattens on the unwaxed paint. (a) Water forms beads on the waxed surface because the cohesive forces responsible for surface tension are larger than the adhesive forces, which tend to flatten the drop. (b) Water beads on bare paint are flattened considerably because the adhesive forces between water and paint are strong, overcoming surface tension. The contact angle θ is directly related to the relative strengths of the cohesive and adhesive forces. The larger θ is, the larger the ratio of cohesive to adhesive forces. (credit: P. P. Urone)

.  The table below lists contact angles for several combinations of liquids and solids.

One important phenomenon related to the relative strength of cohesive and adhesive forces is capillary action—the tendency of a fluid to be raised or suppressed in a narrow tube, or capillary tube. This action causes blood to be drawn into a small-diameter tube when the tube touches a drop.

Capillary Action
The tendency of a fluid to be raised or suppressed in a narrow tube, or capillary tube, is called capillary action.

If a capillary tube is placed vertically into a liquid, as shown in the figure below , capillary action will raise or suppress the liquid inside the tube depending on the combination of substances. The actual effect depends on the relative strength of the cohesive and adhesive forces and, thus, the contact angle θ given in the table. If θ is less than 90⁰,  then the fluid will be raised; if θ is greater than 90⁰, it will be suppressed. Mercury, for example, has a very large surface tension and a large contact angle with glass. When placed in a tube, the surface of a column of mercury curves downward, somewhat like a drop. The curved surface of a fluid in a tube is called a meniscus. The tendency of surface tension is always to reduce the surface area. Surface tension thus flattens the curved liquid surface in a capillary tube. This results in a downward force in mercury and an upward force in water, as seen in the figure below .

(a) Mercury is suppressed in a glass tube because its contact angle is greater than . Surface tension exerts a downward force as it flattens the mercury, suppressing it in the tube. The dashed line shows the shape the mercury surface would have without the flattening effect of surface tension. (b) Water is raised in a glass tube because its contact angle is nearly . Surface tension therefore exerts an upward force when it flattens the surface to reduce its area.

Capillary action can move liquids horizontally over very large distances, but the height to which it can raise or suppress a liquid in a tube is limited by its weight. It can be shown that this height  is given by

If we look at the different factors in this expression, we might see how it makes good sense. The height is directly proportional to the surface tension , which is its direct cause. Furthermore, the height is inversely proportional to tube radius—the smaller the radius , the higher the fluid can be raised, since a smaller tube holds less mass. The height is also inversely proportional to fluid density , since a larger density means a greater mass in the same volume. (See figure below.)

(a) Capillary action depends on the radius of a tube. The smaller the tube, the greater the height reached. The height is negligible for large-radius tubes. (b) A denser fluid in the same tube rises to a smaller height, all other factors being the same

Thermal Expansion of Solids and Liquids

The expansion of alcohol in a thermometer is one of many commonly encountered examples of thermal expansion, the change in size or volume of a given mass with temperature. Hot air rises because its volume increases, which causes the hot air’s density to be smaller than the density of surrounding air, causing a buoyant (upward) force on the hot air. The same happens in all liquids and gases, driving natural heat transfer upwards in homes, oceans, and weather systems. Solids also undergo thermal expansion. Railroad tracks and bridges, for example, have expansion joints to allow them to freely expand and contract with temperature changes.

Thermal expansion joints like these in the Auckland Harbour Bridge in New Zealand allow bridges to change length without buckling. (credit: Ingolfson, Wikimedia Commons)

First, thermal expansion is clearly related to temperature change. The greater the temperature change, the more a bimetallic strip will bend. Second, it depends on the material. In a thermometer, for example, the expansion of alcohol is much greater than the expansion of the glass containing it.

an increase in temperature implies an increase in the kinetic energy of the individual atoms. In a solid, unlike in a gas, the atoms or molecules are closely packed together, but their kinetic energy (in the form of small, rapid vibrations) pushes neighboring atoms or molecules apart from each other. This neighbor-to-neighbor pushing results in a slightly greater distance, on average, between neighbors, and adds up to a larger size for the whole body. For most substances under ordinary conditions, there is no preferred direction, and an increase in temperature will increase the solid’s size by a certain fraction in each dimension

The change in length ∆ L is proportional to length L. The dependence of thermal expansion on temperature, substance, and length is summarized in the equation

∆ L = α L ∆ T

Where ∆ L  is the change in length , ∆ T  is the change in temperature, and α  is the coefficient of linear expansion, which varies slightly with temperature.

The below table  lists representative values of the coefficient of linear expansion, which may have units of 1/⁰ C or 1/K. Because the size of a kelvin and a degree Celsius are the same, both and can be expressed in units of kelvins or degrees Celsius. The equation ∆ L = α L ∆ T  is accurate for small changes in temperature and can be used for large changes in temperature if an average value of is used.

Thermal Expansion in Two and Three Dimensions

Objects expand in all dimensions, as illustrated in the below figure . That is, their areas and volumes, as well as their lengths, increase with temperature. Holes also get larger with temperature. If you cut a hole in a metal plate, the remaining material will expand exactly as it would if the plug was still in place. The plug would get bigger, and so the hole must get bigger too. (Think of the ring of neighboring atoms or molecules on the wall of the hole as pushing each other farther apart as temperature increases. Obviously, the ring of neighbors must get slightly larger, so the hole gets slightly larger).

For small temperature changes, the change in area ∆ A is given by

∆ A = 2 α A ∆ T

Where ∆ A is the change in area , ∆ T is the change in temperature, and α is the coefficient of linear expansion, which varies slightly with temperature

In general, objects expand in all directions as temperature increases. In these drawings, the original boundaries of the objects are shown with solid lines, and the expanded boundaries with dashed lines. (a) Area increases because both length and width increase. The area of a circular plug also increases. (b) If the plug is removed, the hole it leaves becomes larger with increasing temperature, just as if the expanding plug were still in place. (c) Volume also increases, because all three dimensions increase

In general, objects will expand with increasing temperature. Water is the most important exception to this rule. Water expands with increasing temperature (its density decreases) when it is at temperatures greater than 4^oC (40^oF). However, it expands with decreasing temperature when it is between +4^oC and 0^oC(40^oF 32^oF). Water is densest at +4^oC. (See below graph.) Perhaps the most striking effect of this phenomenon is the freezing of water in a pond. When water near the surface cools down to 4^oC  it is denser than the remaining water and thus will sink to the bottom. This “turnover” results in a layer of warmer water near the surface, which is then cooled.

Eventually the pond has a uniform temperature of 4^oC. If the temperature in the surface layer drops below 4^oC, the water is less dense than the water below, and thus stays near the top. As a result, the pond surface can completely freeze over. The ice on top of liquid water provides an insulating layer from winter’s harsh exterior air temperatures. Fish and other aquatic life can survive in 4^oC water beneath ice, due to this unusual characteristic of water. It also produces circulation of water in the pond that is necessary for a healthy ecosystem of the body of water.

The density of water as a function of temperature. Note that the thermal expansion is actually very small. The maximum density at +4^oC is only 0.0075% greater than the density at 2^oC, and 0.012% greater than that at 0^oC.

Thermal Stress

Thermal stress is created by thermal expansion or contraction . Thermal stress can be destructive, such as when expanding petrol ruptures a tank. It can also be useful, for example, when two parts are joined together by heating one in manufacturing, then slipping it over the other and allowing the combination to cool. Thermal stress can explain many phenomena, such as the weathering of rocks and pavement by the expansion of ice when it freezes.

Forces and pressures created by thermal stress are typically large (See below figure.)

Thermal stress contributes to the formation of potholes. credit: Editor5807, Wikimedia Commons

Power lines sag more in the summer than in the winter, and will snap in cold weather if there is insufficient slack. Cracks open and close in plaster walls as a house warms and cools. Glass cooking pans will crack if cooled rapidly or unevenly, because of differential contraction and the stresses it creates. (Pyrex® is less susceptible because of its small coefficient of thermal expansion.) Nuclear reactor pressure vessels are threatened by overly rapid cooling, and although none have failed, several have been cooled faster than considered desirable. Biological cells are ruptured when foods are frozen, detracting from their taste. Repeated thawing and freezing accentuate the damage. Even the oceans can be affected. A significant portion of the rise in sea level that is resulting from global warming is due to the thermal expansion of sea water.

Metal is regularly used in the human body for hip and knee implants. Most implants need to be replaced over time because, among other things, metal does not bond with bone. Researchers are trying to find better metal coatings that would allow metal-to-bone bonding. One challenge is to find a coating that has an expansion coefficient similar to that of metal. If the expansion coefficients are too different, the thermal stresses during the manufacturing process lead to cracks at the coating-metal interface.

Another example of thermal stress is found in the mouth. Dental fillings can expand differently from tooth enamel. It can give pain when eating ice cream or having a hot drink. Cracks might occur in the filling. Metal fillings (gold, silver, etc.) are being replaced by composite fillings (porcelain), which have smaller coefficients of expansion, and are closer to those of teeth

Heat

 Work is defined as force times distance and in earlier chapters we learned that work done on an object changes its kinetic energy. We also saw in previous lessons that temperature is proportional to the (average) kinetic energy of atoms and molecules. We say that a thermal system has a certain internal energy: its internal energy is higher if the temperature is higher. If two objects at different temperatures are brought in contact with each other, energy is transferred from the hotter to the colder object until equilibrium is reached and the bodies reach thermal equilibrium (i.e., they are at the same temperature). No work is done by either object, because no force acts through a distance. The transfer of energy is caused by the temperature difference, and ceases once the temperatures are equal. These observations lead to the following definition of heat: Heat is the spontaneous transfer of energy due to a temperature difference

Heat is often confused with temperature. For example, we may say the heat was unbearable, when we actually mean that the temperature was high. Heat is a form of energy, whereas temperature is not. The misconception arises because we are sensitive to the flow of heat, rather than the temperature.

Owing to the fact that heat is a form of energy, it has the SI unit of joule (J). The calorie (cal) is a common unit of energy, defined as the energy needed to change the temperature of 1.00 g of water by 1.00^oC —specifically, between 14.5^oC and 15.5^oC , since there is a slight temperature dependence. Perhaps the most common unit of heat is the kilocalorie (kcal), which is the energy needed to change the temperature of 1.00 kg of water by 1.00^oC . Since mass is most often specified in kilograms, kilocalorie is commonly used. Food calories (given the notation Cal, and sometimes called “big calorie”) are actually kilocalories (1 Kilocalorie = 1000 calories), a fact not easily determined from package labeling.

Mechanical Equivalent of Heat

It is also possible to change the temperature of a substance by doing work. Work can transfer energy into or out of a system. This realization helped establish the fact that heat is a form of energy. James Prescott Joule (1818–1889) performed many experiments to establish the mechanical equivalent of heat—the work needed to produce the same effects as heat transfer. In terms of the units used for these two terms, the best modern value for this equivalence is

We consider this equation as the conversion between two different units of energy

Schematic depiction of Joule’s experiment that established the equivalence of heat and work

The figure above shows one of Joule’s most famous experimental setups for demonstrating the mechanical equivalent of heat. It demonstrated that work and heat can produce the same effects, and helped establish the principle of conservation of energy. Gravitational potential energy (PE) (work done by the gravitational force) is converted into kinetic energy (KE), and then randomized by viscosity and turbulence into increased average kinetic energy of atoms and molecules in the system, producing a temperature increase. His contributions to the field of thermodynamics were so significant that the SI unit of energy was named after him.

Heat added or removed from a system changes its internal energy and thus its temperature. Such a temperature increase is observed while cooking. However, adding heat does not necessarily increase the temperature. An example is melting of ice; that is, when a substance changes from one phase to another. Work done on the system or by the system can also change the internal energy of the system. Joule demonstrated that the temperature of a system can be increased by stirring. If an ice cube is rubbed against a rough surface, work is done by the frictional force. A system has a well-defined internal energy, but we cannot say that it has a certain “heat content” or “work content”. We use the phrase “heat transfer” to emphasize its nature.

Temperature Change and Heat Capacity

One of the major effects of heat transfer is temperature change: heating increases the temperature while cooling decreases it. We assume that there is no phase change and that no work is done on or by the system. Experiments show that the transferred heat depends on three factors—the change in temperature, the mass of the system, and the substance and phase of the substance.

 

The heat Q transferred to cause a temperature change depends on the magnitude of the temperature change, the mass of the system, and the substance and phase involved. (a) The amount of heat transferred is directly proportional to the temperature change. To double the temperature change of a mass m, you need to add twice the heat. (b) The amount of heat transferred is also directly proportional to the mass. To cause an equivalent temperature change in a doubled mass, you need to add twice the heat. (c) The amount of heat transferred depends on the substance and its phase. If it takes an amount Q of heat to cause a temperature change ∆T in a given mass of copper, it will take 10.8 times that amount of heat to cause the equivalent temperature change in the same mass of water assuming no phase change in either substance

The dependence on temperature change and mass are easily understood. Owing to the fact that the (average) kinetic energy of an atom or molecule is proportional to the absolute temperature, the internal energy of a system is proportional to the absolute temperature and the number of atoms or molecules. Owing to the fact that the transferred heat is equal to the change in the internal energy, the heat is proportional to the mass of the substance and the temperature change. The transferred heat also depends on the substance so that, for example, the heat necessary to raise the temperature is less for alcohol than for water. For the same substance, the transferred heat also depends on the phase (gas, liquid, or solid).

Values of specific heat must generally be looked up in tables, because there is no simple way to calculate them. In general, the specific heat also depends on the temperature. The table below lists representative values of specific heat for various substances. Except for gases, the temperature and volume dependence of the specific heat of most substances is weak. We see from this table that the specific heat of water is five times that of glass and ten times that of iron, which means that it takes five times as much heat to raise the temperature of water the same amount as for glass and ten times as much heat to raise the temperature of water as for iron. In fact, water has one of the largest specific heats of any material, which is important for sustaining life on Earth.

Calorimetry

One technique we can use to measure the amount of heat involved in a physical or chemical process is known as calorimetry. Calorimetry is used to measure amounts of heat transferred to or from a substance. To do so, the heat is exchanged with a calibrated object (calorimeter). The temperature change measured by the calorimeter is used to derive the amount of heat transferred by the process under study. The measurement of heat transfer using this approach requires the definition of a system (the substance or substances undergoing the chemical or physical change) and its surroundings (all other matter, including components of the measurement apparatus, that serve to either provide heat to the system or absorb heat from the system).

A calorimeter is a device used to measure the amount of heat involved in a chemical or physical process. For example, when an exothermic reaction occurs in solution in a calorimeter, the heat produced by the reaction is absorbed by the solution, which increases its temperature. When an endothermic reaction occurs, the heat required is absorbed from the thermal energy of the solution, which decreases its temperature (Figure below). The temperature change, along with the specific heat and mass of the solution, can then be used to calculate the amount of heat involved in either case.

In a calorimetric determination, either (a) an exothermic process occurs and heat, q, is negative, indicating that thermal energy is transferred from the system to its surroundings, or (b) an endothermic process occurs and heat, q, is positive, indicating that thermal energy is transferred from the surroundings to the system

Calorimetry measurements are important in understanding the heat transferred in reactions involving everything from microscopic proteins to massive machines .

Consider a simpler example that illustrates the core idea behind calorimetry. Suppose we initially have a high-temperature substance, such as a hot piece of metal (M), and a low-temperature substance, such as cool water (W). If we place the metal in the water, heat will flow from M to W. The temperature of M will decrease, and the temperature of W will increase, until the two substances have the same temperature—that is, when they reach thermal equilibrium (Figure below ). If this occurs in a calorimeter, ideally all of this heat transfer occurs between the two substances, with no heat gained  or lost by either its external environment

In a simple calorimetry process, (a) heat, q, is transferred from the hot metal, M, to the cool water, W, until (b) both are at the same temperature

Under these ideal circumstances, the net heat change is zero

This relationship can be rearranged to show that the heat gained by substance M is equal to the heat lost by substance W:

The magnitude of the heat (change) is therefore the same for both substances, and the negative sign merely shows that q_{substance M} and q_{substance W} are opposite in direction of heat flow (gain or loss) but does not indicate the arithmetic sign of either q value (that is determined by whether the matter in question gains or loses heat, per definition). In the specific situation described, q_{substance M} is a negative value and q_{substance W} is positive, since heat is transferred from M to W.

What do a falling apple and the orbit of the moon have in common? You will learn in this chapter that each is caused by gravitational force. The motion of all celestial objects, in fact, is determined by the gravitational force, which depends on their mass and separation.

Johannes Kepler discovered three laws of planetary motion that all orbiting planets and moons follow. Years later, Isaac Newton found these laws useful in developing his law of universal gravitation. This law relates gravitational force to the masses of objects and the distance between them.

Kepler’s Laws of Planetary Motion

Hundreds of artificial satellites orbit Earth together with thousands of pieces of debris. The moon’s orbit around Earth has intrigued humans from time immemorial. The orbits of planets, asteroids, meteors, and comets around the sun are no less interesting. If we look farther, we see almost unimaginable numbers of stars, galaxies, and other celestial objects orbiting one another and interacting through gravity.

All these motions are governed by gravitational force. The orbital motions of objects in our own solar system are simple enough to describe with a few fairly simple laws. The orbits of planets and moons satisfy the following two conditions:

• The mass of the orbiting object, m, is small compared to the mass of the object it orbits, M.
• The system is isolated from other massive objects.

Based on the motion of the planets about the sun, Kepler devised a set of three classical laws, called Kepler’s laws of planetary motion, that describe the orbits of all bodies satisfying these two conditions:

1. The orbit of each planet around the sun is an ellipse with the sun at one focus.
2. Each planet moves so that an imaginary line drawn from the sun to the planet sweeps out equal areas in equal times.
3. The ratio of the squares of the periods of any two planets about the sun is equal to the ratio of the cubes of their average distances from the sun.

Kepler’s First Law

The orbit of each planet about the sun is an ellipse with the sun at one focus, as shown in the below figure . The planet’s closest approach to the sun is called aphelion and its farthest distance from the sun is called perihelion

a) An ellipse is a closed curve such that the sum of the distances from a point on the curve to the two foci (f₁ and f₂) is constant.

(b) For any closed orbit, m follows an elliptical path with M at one focus.

(c) The aphelion (ra) is the closest distance between the planet and the sun, while the perihelion (rp) is the farthest distance from the sun

Hence , If you know the aphelion (r_a) and perihelion (r_p) distances, then you can calculate the semi-major axis (a) and semi-minor axis (b).

a = (r_a + r_p)/2

b = √r_ar_p

Kepler’s Second Law

Each planet moves so that an imaginary line drawn from the sun to the planet sweeps out equal areas in equal times, as shown in the below figure

The shaded regions have equal areas. The time for m to go from A to B is the same as the time to go from C to D and from E to F. The mass m moves fastest when it is closest to M. Kepler’s second law was originally devised for planets orbiting the sun, but it has broader validity.

Kepler’s Third Law

The ratio of the periods squared of any two planets around the sun is equal to the ratio of their average distances from the sun cubed. In equation form, this is

where T is the period (time for one orbit) and r is the average distance (also called orbital radius). This equation is valid only for comparing two small masses orbiting a single large mass. Most importantly, this is only a descriptive equation; it gives no information about the cause of the equality.

Newton’s Universal Law of Gravitation :

Newton was the first to propose an exact mathematical form and to use that form to show that the motion of heavenly bodies should be conic sections—circles, ellipses, parabolas, and hyperbolas. This theoretical prediction was a major triumph. It had been known for some time that moons, planets, and comets follow such paths, but no one had been able to propose an explanation of the mechanism that caused them to follow these paths and not others.

The gravitational force between two bodies  is always attractive and depends on the masses involved and the distance between them. Expressed in modern language, Newton’s universal law of gravitation states that every object in the universe attracts every other object with a force that is directed along a line joining them. The force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. This attraction is illustrated by the below figure

Gravitational attraction is along a line joining the centers of mass (CM) of the two bodies. The magnitude of the force on each body is the same, consistent with Newton’s third law (action-reaction).

For two bodies having masses m and M with a distance r between their centers of mass, the equation for Newton’s universal law of gravitation is

where F is the magnitude of the gravitational force and G is a proportionality factor called the gravitational constant. G is a universal constant, meaning that it is thought to be the same everywhere in the universe. It has been measured experimentally to be

If a person has a mass of 60.0 kg, what would be the force of gravitational attraction on him at Earth’s surface? G is given above, Earth’s mass M is 5.97 × 1024 kg, and the radius r of Earth is 6.38 × 106 m. Putting these values into Newton’s universal law of gravitation gives

We can check this result with the relationship:

You may remember that g, the acceleration due to gravity, is another important constant related to gravity. By substituting g for a in the equation for Newton’s second law of motion we get F = mg. Combining this with the equation for universal gravitation gives

Cancelling the mass m on both sides of the equation and filling in the values for the gravitational constant and mass and radius of the Earth, gives the value of g, which may look familiar.

This is a good point to recall the difference between mass and weight. Mass is the amount of matter in an object; weight is the force of attraction between the mass within two objects. Weight can change because g is different on every moon and planet. An object’s mass m does not change but its weight mg can.

We can also derive Kepler’s third law from Newtons’s Universal Law of Gravitation . Applying Newton’s second law of motion to angular motion gives an expression for centripetal force, which can be equated to the expression for force in the universal gravitation equation. This expression can be manipulated to produce the equation for Kepler’s third law. We saw earlier that the expression r3/T2 is a constant for satellites orbiting the same massive object. The derivation of Kepler’s third law from Newton’s law of universal gravitation and Newton’s second law of motion yields that constant:

               r ³ / T² =    GM / 4π²

where M is the mass of the central body about which the satellites orbit (for example, the sun in our solar system).

The universal gravitational constant G is determined experimentally. This definition was first done accurately in 1798 by English scientist Henry Cavendish (1731–1810), more than 100 years after Newton published his universal law of gravitation. The measurement of G is very basic and important because it determines the strength of one of the four forces in nature.

Cavendish’s experiment was very difficult because he measured the tiny gravitational attraction between two ordinary-sized masses (tens of kilograms at most) by using an apparatus like that in the below figure .  Remarkably, his value for G differs by less than 1% from the modern value.

Cavendish used an apparatus like this to measure the gravitational attraction between two suspended spheres (m) and two spheres on a stand (M) by observing the amount of torsion (twisting) created in the fiber. The distance between the masses can be varied to check the dependence of the force on distance. Modern experiments of this type continue to explore gravity

Gravitational Potential Energy

 Gravitational potential energy is the stored energy an object has as a result of its position above Earth’s surface (or another object in space). A roller coaster car at the top of a hill has gravitational potential energy.

Let’s examine how doing work on an object changes the object’s energy. If we apply force to lift a rock off the ground, we increase the rock’s potential energy, PE. If we drop the rock, the force of gravity increases the rock’s kinetic energy as the rock moves downward until it hits the ground.

The force we exert to lift the rock is equal to its weight, w, which is equal to its mass, m, multiplied by acceleration due to gravity, g.

f = w = mg

The work we do on the rock equals the force we exert multiplied by the distance, d, that we lift the rock. The work we do on the rock also equals the rock’s gain in gravitational potential energy, PE_e.

W = PE_e = fmg

Potential energy is particularly useful for forces that change with position, as the gravitational force does over large distances. The change in gravitational potential energy near Earth’s surface is ∆U = mg(y₂ – y₁). This works very well if g does not change significantly between y₁and y₂. We return to the definition of work and potential energy to derive an expression that is correct over larger distances

We know that work (W) is the integral of the dot product between force and distance. Essentially, it is the product of the component of a force along a displacement times that displacement. We define ∆U as the negative of the work done by the force we associate with the potential energy. For clarity, we derive an expression for moving a mass m from distance r₁ from the center of Earth to distance r₂. However, the result can easily be generalized to any two objects changing their separation from one value to another.

Consider the below figure , in which we take m from a distance r₁ from Earth’s center to a distance that is r₂ from the center. Gravity is a conservative force (its magnitude and direction are functions of location only), so we can take any path we wish, and the result for the calculation of work is the same. We take the path shown, as it greatly simplifies the integration. We first move radially outward from distance r₁ to distance r₂ , and then move along the arc of a circle until we reach the final position.

The work integral, which determines the change in potential energy, can be evaluated along the path shown in red.

Note two important items with this definition. First,

The potential energy is zero when the two masses are infinitely far apart. Only the difference in U is important, so the choice of

is merely one of convenience. (Recall that in earlier gravity problems, you were free to take U = 0 at the top or bottom of a building, or anywhere.) Second, note that U becomes increasingly more negative as the masses get closer. That is consistent with what you learned about potential energy in Potential Energy and Conservation of Energy. As the two masses are separated, positive work must be done against the force of gravity, and hence, U increases (becomes less negative). All masses naturally fall together under the influence of gravity, falling from a higher to a lower potential energy.

 Escape velocity

Escape velocity is  defined as the minimum initial velocity of an object that is required to escape the surface of a planet (or any large body like a moon) and never return. As usual, we assume no energy lost to an atmosphere, should there be any.

Consider the case where an object is launched from the surface of a planet with an initial velocity directed away from the planet. With the minimum velocity needed to escape, the object would just come to rest infinitely far away, that is, the object gives up the last of its kinetic energy just as it reaches infinity, where the force of gravity becomes zero.

Since U → 0 as r → ∞, this means the total energy is zero. Thus, we find the escape velocity from the surface of an astronomical body of mass M and radius R by setting the total energy equal to zero. At the surface of the body, the object is located at r₁=R  and it has escape velocity V₁= V𝑒𝑠𝑐. It reaches r2=∞ with velocity v₂=0

The escape velocity is the same for all objects, regardless of mass. Also, we are not restricted to the surface of the planet; R can be any starting point beyond the surface of the planet.

In Potential Energy and Conservation of Energy, we described how to apply conservation of energy for systems with conservative forces. We were able to solve many problems, particularly those involving gravity, more simply using conservation of energy. Those principles and problem-solving strategies apply equally well here. The only change is to place the new expression for potential energy into the conservation of energy equation,

E_tot=K₁+U₁=K₂+U₂

Satellite Orbits and Energy

The Moon orbits Earth. In turn, Earth and the other planets orbit the Sun. The space directly above our atmosphere is filled with artificial satellites in orbit. We examine the simplest of these orbits, the circular orbit, to understand the relationship between the speed and period of planets and satellites in relation to their positions and the bodies that they orbit.

Nicolaus Copernicus first suggested that Earth and all other planets orbit the Sun in circles. He further noted that orbital periods increased with distance from the Sun. Later analysis by Kepler showed that these orbits are actually ellipses, but the orbits of most planets in the solar system are nearly circular. Earth’s orbital distance from the Sun varies a mere 2%. The exception is the eccentric orbit of Mercury, whose orbital distance varies nearly 40%.

Determining the orbital speed and orbital period of a satellite is much easier for circular orbits, so we make that assumption in the derivation that follows. As we described in the previous section, an object with negative total energy is gravitationally bound and therefore is in orbit. Our computation for the special case of circular orbits will confirm this. We focus on objects orbiting Earth, but our results can be generalized for other cases.

Consider a satellite of mass m in a circular orbit about Earth at distance r from the center of Earth (as shown in the below figure) . It has centripetal acceleration directed toward the center of Earth. Earth’s gravity is the only force acting, so Newton’s second law gives

(A satellite of mass m orbiting at radius r from the center of Earth. The gravitational force supplies the centripetal acceleration.)

We solve for the speed of the orbit, noting that m cancels, to get the orbital speed

We see in the next section that this represents Kepler’s third law for the case of circular orbits. It also confirms Copernicus’s observation that the period of a planet increases with increasing distance from the Sun.

Astronauts in orbit appear to be weightless, as if they were free-falling towards Earth. In fact, they are in free fall. Consider the trajectories shown in the below figure. (This figure is based on a drawing by Newton in his Principia and also appeared earlier in Motion in Two and Three Dimensions.) All the trajectories shown that hit the surface of Earth have less than orbital velocity. The astronauts would accelerate toward Earth along the noncircular paths shown and feel weightless. (Astronauts actually train for life in orbit by riding in airplanes that free fall for 30 seconds at a time.) But with the correct orbital velocity, Earth’s surface curves away from them at exactly the same rate as they fall toward Earth. Of course, staying the same distance from the surface is the point of a circular orbit

(A circular orbit is the result of choosing a tangential velocity such that Earth’s surface curves away at the same rate as the object falls toward Earth)

Energy in Circular Orbits

In Gravitational Potential Energy and Total Energy, we argued that objects are gravitationally bound if their total energy is negative. The argument was based on the simple case where the velocity was directly away or toward the planet. We now examine the total energy for a circular orbit and show that indeed, the total energy is negative. As we did earlier, we start with Newton’s second law applied to a circular orbit,

In the last step, we multiplied by r on each side. The right side is just twice the kinetic energy, so we have

K = ½ mv²  = GmM_E / 2r

The total energy is the sum of the kinetic and potential energies, so our final result is

E = K + U

   = GmM_E / 2r  – GmM_E / r

   = – GmM_E / 2r

We can see that the total energy is negative, with the same magnitude as the kinetic energy. For circular orbits, the magnitude of the kinetic energy is exactly one-half the magnitude of the potential energy. Remarkably, this result applies to any two masses in circular orbits about their common center of mass, at a distance r from each other. The proof of this is left as an exercise. We will see in the next section that a very similar expression applies in the case of elliptical orbits

We can see that the total energy is negative, with the same magnitude as the kinetic energy. For circular orbits, the magnitude of the kinetic energy is exactly one-half the magnitude of the potential energy. Remarkably, this result applies to any two masses in circular orbits about their common center of mass, at a distance r from each other. The proof of this is left as an exercise. We will see in the next section that a very similar expression applies in the case of elliptical orbits

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We have learnt about various aspects of motion along a straight line: kinematics (where we learned about displacement, velocity, and acceleration), projectile motion (a special case of two-dimensional kinematics), force, and Newton’s laws of motion. Newton’s first law tells us that objects move along a straight line at constant speed unless a net external force acts on them. Therefore, if an object moves along a circular path, it must be experiencing an external force. In this chapter, we explore both circular motion and rotational motion.

Angle of Rotation and Angular Velocity

What exactly do we mean by circular motion or rotation? Rotational motion is the circular motion of an object about an axis of rotation. We will discuss specifically circular motion and spin. Circular motion is when an object moves in a circular path.

Examples of circular motion include a race car speeding around a circular curve, a toy attached to a string swinging in a circle around your head. Spin is rotation about an axis that goes through the center of mass of the object, such as Earth rotating on its axis, a wheel turning on its axle, the spin of a tornado on its path of destruction, or a figure skater spinning during a performance at the Olympics. Sometimes, objects will be spinning while in circular motion, like the Earth spinning on its axis while revolving around the Sun, but we will focus on these two motions separately.

When solving problems involving rotational motion, we use variables that are similar to linear variables (distance, velocity, acceleration, and force) but take into account the curvature or rotation of the motion. Here, we define the angle of rotation, which is the angular equivalence of distance; and angular velocity, which is the angular equivalence of linear velocity.

When objects rotate about some axis—for example, when the CD in the below figure rotates about its center—each point in the object follows a circular path.

All points on a CD travel in circular paths. The pits (dots) along a line from the center to the edge all move through the same angle Δθ in time Δt

The arc length, , is the distance travelled along a circular path. The radius of curvature, r, is the radius of the circular path. Both are shown in the below figure

The radius (r) of a circle is rotated through an angle Δθ . The arc length, ΔS, is the distance covered along the circumference.

Consider a line from the center of the CD to its edge. In a given time, each pit (used to record information) on this line moves through the same angle. The angle of rotation is the amount of rotation and is the angular analog of distance. The angle of rotation Δθ is the arc length divided by the radius of curvature

Δθ = ΔS / r

The angle of rotation is often measured by using a unit called the radian. (Radians are actually dimensionless, because a radian is defined as the ratio of two distances, radius and arc length.) A revolution is one complete rotation, where every point on the circle returns to its original position. One revolution covers 2π radians (or 360 degrees), and therefore has an angle of rotation of 2π radians, and an arc length that is the same as the circumference of the circle. We can convert between radians, revolutions, and degrees using the relationship

1 revolution = 2π rad = 360°. See the below for the conversion of degrees to radians for some common angles

2 π rad = 360⁰

1 rad = 360⁰ / 2 π

1 rad ≈ 57.3 ⁰

Degree Measures	Radian Measure
300	π/ 6
600	π/ 3
900	π/ 2
1200	2π/ 3
1350	3π/ 4
1800	π

Angular Velocity

How fast is an object rotating? We can answer this question by using the concept of angular velocity. Consider first the angular speed ω is the rate at which the angle of rotation changes. In equation form, the angular speed is

ω = Δθ / Δ t

which means that an angular rotation Δθ occurs in a time, Δt. If an object rotates through a greater angle of rotation in a given time, it has a greater angular speed. The units for angular speed are radians per second (rad/s).

Now let’s consider the direction of the angular speed, which means we now must call it the angular velocity. The direction of the angular velocity is along the axis of rotation. For an object rotating clockwise, the angular velocity points away from you along the axis of rotation. For an object rotating counterclockwise, the angular velocity points toward you along the axis of rotation.

Angular velocity (ω) is the angular version of linear velocity v. Tangential velocity is the instantaneous linear velocity of an object in rotational motion. To get the precise relationship between angular velocity and tangential velocity, consider again a pit on the rotating CD. This pit moves through an arc length  Δs  in a short time  Δt so its tangential speed is

V = Δs / Δt

From the definition of the angle of rotation, we know that

Δϑ = Δs / r,

Hence , Δs = r Δϑ

Substituting this into the expression for v gives

V = r Δϑ / Δt

V = r ω

The equation V = r ω says that the tangential speed v is proportional to the distance r from the center of rotation. Consequently, tangential speed is greater for a point on the outer edge of the CD (with larger r) than for a point closer to the center of the CD (with smaller r). This makes sense because a point farther out from the center has to cover a longer arc length in the same amount of time as a point closer to the center. Note that both points will still have the same angular speed, regardless of their distance from the center of rotation

Points 1 and 2 rotate through the same angle (Δθ), but point 2 moves through a greater arc length (Δs₂) because it is farther from the center of rotation.

Now, consider another example: the tire of a moving car . The faster the tire spins, the faster the car moves—large  means large v because . Similarly, a larger-radius tire rotating at the same angular velocity, , will produce a greater linear (tangential) velocity, v, for the car. This is because a larger radius means a longer arc length must contact the road, so the car must move farther in the same amount of time

A car moving at a velocity, v, to the right has a tire rotating with angular velocity ω . The speed of the tread of the tire relative to the axle is v, the same as if the car were jacked up and the wheels spinning without touching the road. Directly below the axle, where the tire touches the road, the tire tread moves backward with respect to the axle with tangential velocity v=rω, where r is the tire radius.

Because the road is stationary with respect to this point of the tire, the car must move forward at the linear velocity v. A larger angular velocity for the tire means a greater linear velocity for the car.

However, there are cases where linear velocity and tangential velocity are not equivalent, such as a car spinning its tires on ice. In this case, the linear velocity will be less than the tangential velocity. Due to the lack of friction under the tires of a car on ice, the arc length through which the tire treads move is greater than the linear distance through which the car moves. It’s similar to running on a treadmill or pedaling a stationary bike; you are literally going nowhere fast

Uniform Circular Motion

In the previous section, we defined circular motion. The simplest case of circular motion is uniform circular motion, where an object travels a circular path at a constant speed. Note that, unlike speed, the linear velocity of an object in circular motion is constantly changing because it is always changing direction. We know from kinematics that acceleration is a change in velocity, either in magnitude or in direction or both. Therefore, an object undergoing uniform circular motion is always accelerating, even though the magnitude of its velocity is constant.

You experience this acceleration yourself every time you ride in a car while it turns a corner. If you hold the steering wheel steady during the turn and move at a constant speed, you are executing uniform circular motion. What you notice is a feeling of sliding (or being flung, depending on the speed) away from the center of the turn. This isn’t an actual force that is acting on you—it only happens because your body wants to continue moving in a straight line (as per Newton’s first law) whereas the car is turning off this straight-line path. Inside the car it appears as if you are forced away from the center of the turn. This fictitious force is known as the centrifugal force. The sharper the curve and the greater your speed, the more noticeable this effect becomes.

Below figure shows an object moving in a circular path at constant speed. The direction of the instantaneous tangential velocity is shown at two points along the path. Acceleration is in the direction of the change in velocity; in this case it points roughly toward the center of rotation. (The center of rotation is at the center of the circular path). If we imagine  becoming smaller and smaller, then the acceleration would point exactly toward the center of rotation, but this case is hard to draw. We call the acceleration of an object moving in uniform circular motion the centripetal acceleration a_c because centripetal means center seeking.

The directions of the velocity of an object at two different points are shown, and the change in velocity  is seen to point approximately toward the center of curvature (see small inset). For an extremely small value of Δs, Δv points exactly toward the center of the circle (but this is hard to draw). Because ac = Δv/Δt, the acceleration is also toward the center, so a_c is called centripetal acceleration.

Now that we know that the direction of centripetal acceleration is toward the center of rotation, let’s discuss the magnitude of centripetal acceleration. For an object traveling at speed v in a circular path with radius r, the magnitude of centripetal acceleration is

a_c  = v² / r

Centripetal acceleration is greater at high speeds and in sharp curves (smaller radius), as you may have noticed when driving a car, because the car actually pushes you toward the center of the turn. But it is a bit surprising that a_c is proportional to the speed squared. This means, for example, that the acceleration is four times greater when you take a curve at 100 km/h than at 50 km/h

We can also express a_c in terms of the magnitude of angular velocity. Substituting v = rω into the equation above, we get

a_c = (rω)² / r = rω²

Centripetal Force

Because an object in uniform circular motion undergoes constant acceleration (by changing direction), we know from Newton’s second law of motion that there must be a constant net external force acting on the object.

Any force or combination of forces can cause a centripetal acceleration. Just a few examples are the tension in the rope on a tether ball, the force of Earth’s gravity on the Moon, the friction between a road and the tires of a car as it goes around a curve, or the normal force of a roller coaster track on the cart during a loop-the-loop.

Any net force causing uniform circular motion is called a centripetal force. The direction of a centripetal force is toward the center of rotation, the same as for centripetal acceleration. According to Newton’s second law of motion, a net force causes the acceleration of mass according to F_net = ma. For uniform circular motion, the acceleration is centripetal acceleration: a = a_c.

Therefore, the magnitude of centripetal force, F_c, is F_c = ma_c

By using the two different forms of the equation for the magnitude of centripetal acceleration, a_c=v²/r and a_c = rω², we get two expressions involving the magnitude of the centripetal force F_c. The first expression is in terms of tangential speed, the second is in terms of angular speed:
F_c = m v²/r and F_c = mω².

Both forms of the equation depend on mass, velocity, and the radius of the circular path. You may use whichever expression for centripetal force is more convenient. Newton’s second law also states that the object will accelerate in the same direction as the net force. By definition, the centripetal force is directed towards the center of rotation, so the object will also accelerate towards the center. A straight line drawn from the circular path to the center of the circle will always be perpendicular to the tangential velocity. Note that, if you solve the first expression for r, you get

r = mv² / F_c

From this expression, we see that, for a given mass and velocity, a large centripetal force causes a small radius of curvature—that is, a tight curve

In this figure, the frictional force f serves as the centripetal force F_c. Centripetal force is perpendicular to tangential velocity and causes uniform circular motion. The larger the centripetal force F_c, the smaller is the radius of curvature r and the sharper is the curve. The lower curve has the same velocity v, but a larger centripetal force F_c produces a smaller radius r‘.

Rotational Motion

In the section on uniform circular motion, we discussed motion in a circle at constant speed and, therefore, constant angular velocity. However, there are times when angular velocity is not constant—rotational motion can speed up, slow down, or reverse directions. Angular velocity is not constant when a spinning skater pulls in her arms, when a child pushes a merry-go-round to make it rotate, or when a CD slows to a halt when switched off. In all these cases, angular acceleration occurs because the angular velocity ω changes. The faster the change occurs, the greater is the angular acceleration. Angular acceleration α is the rate of change of angular velocity. In equation form, angular acceleration is

α = Δω / Δt

where Δω is the change in angular velocity and Δt is the change in time. The units of angular acceleration are (rad/s)/s, or rad/s². If ω increases, then α is positive. If ω decreases, then α is negative. Keep in mind that, by convention, counterclockwise is the positive direction and clockwise is the negative direction. For example, a skater rotates counterclockwise as seen from above, so her angular velocity is positive. Acceleration would be negative, for example, when an object that is rotating counterclockwise slows down. It would be positive when an object that is rotating counterclockwise speeds up.

The relationship between the magnitudes of tangential acceleration, a, and angular acceleration,
α, isa = rαorα = a/r.

These equations mean that the magnitudes of tangential acceleration and angular acceleration are directly proportional to each other. The greater the angular acceleration, the larger the change in tangential acceleration, and vice versa.

Rotational  Linear Relationship

θ	x	θ = x/r
ω	v	ω = v/r
α	a	α = a/r

We can now begin to see how rotational quantities like θ, ω and α are related to each other. For example, if a motorcycle wheel that starts at rest has a large angular acceleration for a fairly long time, it ends up spinning rapidly and rotates through many revolutions. Putting this in terms of the variables, if the wheel’s angular acceleration α is large for a long period of time t, then the final angular velocity ω and angle of rotation θ are large. In the case of linear motion, if an object starts at rest and undergoes a large linear acceleration, then it has a large final velocity and will have traveled a large distance.

The kinematics of rotational motion describes the relationships between the angle of rotation, angular velocity, angular acceleration, and time. It only describes motion—it does not include any forces or masses that may affect rotation (these are part of dynamics). Recall the kinematics equation for linear motion:    v = v₀ + at (constant a)

As in linear kinematics, we assume a is constant, which means that angular acceleration α is also a constant, because
a = r α

The equation for the kinematics relationship between ω, α, and t is
ω = ω₀ + αt (constant α),

where ω₀ is the initial angular velocity. Notice that the equation is identical to the linear version, except with angular analogs of the linear variables. In fact, all of the linear kinematics equations have rotational analogs, which are given in Table 6.3. These equations can be used to solve rotational or linear kinematics problem in which a and αare constant.

Torque

If you have ever spun a bike wheel or pushed a merry-go-round, you know that force is needed to change angular velocity. The farther the force is applied from the pivot point (or fulcrum), the greater the angular acceleration. For example, a door opens slowly if you push too close to its hinge, but opens easily if you push far from the hinges. Furthermore, we know that the more massive the door is, the more slowly it opens; this is because angular acceleration is inversely proportional to mass. These relationships are very similar to the relationships between force, mass, and acceleration from Newton’s second law of motion. Since we have already covered the angular versions of distance, velocity and time, you may wonder what the angular version of force is, and how it relates to linear force.

The angular version of force is torque , which is the turning effectiveness of a force. See the below figure. The equation for the magnitude of torque is

where r is the magnitude of the lever arm, F is the magnitude of the linear force, and θ is the angle between the lever arm and the force. The lever arm is the vector from the point of rotation (pivot point or fulcrum) to the location where force is applied. Since the magnitude of the lever arm is a distance, its units are in meters, and torque has units of N⋅m. Torque is a vector quantity and has the same direction as the angular acceleration that it produces.

The harder the man pushes the merry-go- round in the above figure , the faster it accelerates. Furthermore, the more massive the merry-go-round is, the slower it accelerates for the same torque. If the man wants to maximize the effect of his force on the merry-go-round, he should push as far from the center as possible to get the largest lever arm and, therefore, the greatest torque and angular acceleration. Torque is also maximized when the force is applied perpendicular to the lever arm.

Just as linear forces can balance to produce zero net force and no linear acceleration, the same is true of rotational motion. When two torques of equal magnitude act in opposing directions, there is no net torque and no angular acceleration, as you can see in the following video. If zero net torque acts on a system spinning at a constant angular velocity, the system will continue to spin at the same angular velocity

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Motion itself can be beautiful, such as a dolphin jumping out of the water, the flight of a bird, or the orbit of a satellite. The study of motion is called kinematics, but kinematics describes only the way objects move—their velocity and their acceleration. Dynamics considers the forces that affect the motion of moving objects and systems.

Newton’s laws of motion are the foundation of dynamics. These laws describe the way objects speed up, slow down, stay in motion, and interact with other objects. They are also universal laws: they apply everywhere on Earth as well as in space. A force pushes or pulls an object. The object being moved by a force could be an inanimate object, a table, or an animate object, a person. The pushing or pulling may be done by a person, or even the gravitational pull of Earth. Forces have different magnitudes and directions; this means that some forces are stronger than others and can act in different directions.

For example, a cannon exerts a strong force on the cannonball that is launched into the air. In contrast, a mosquito landing on your arm exerts only a small force on your arm. When multiple forces act on an object, the forces combine. Adding together all of the forces acting on an object gives the total force, or net force.

An external force is a force that acts on an object within the system from outside the system. This type of force is different than an internal force, which acts between two objects that are both within the system. The net external force combines these two definitions; it is the total combined external force.

We discuss further details about net force, external force, and net external force in the coming sections. In mathematical terms, two forces acting in opposite directions have opposite signs (positive or negative). By convention, the negative sign is assigned to any movement to the left or downward. If two forces pushing in opposite directions are added together, the larger force will be somewhat cancelled out by the smaller force pushing in the opposite direction. It is important to be consistent with your chosen coordinate system within a problem; for example, if negative values are assigned to the downward direction for velocity, then distance, force, and acceleration should also be designated as being negative in the downward direction.

Newton’s First Law and Friction :

 Newton’s first law of motion states the following:

1. A body at rest tends to remain at rest.

2. A body in motion tends to remain in motion at a constant velocity unless acted on by a net external force. (Recall that constant velocity means that the body moves in a straight line and at a constant speed.)

At first glance, this law may seem to contradict your everyday experience. You have probably noticed that a moving object will usually slow down and stop unless some effort is made to keep it moving. The key to understanding why, for example, a sliding box slows down (seemingly on its own) is to first understand that a net external force acts on the box to make the box slow down. Without this net external force, the box would continue to slide at a constant velocity (as stated in Newton’s first law of motion). What force acts on the box to slow it down? This force is called friction.

Friction is an external force that acts opposite to the direction of motion (see Figure ). Think of friction as a resistance to motion that slows things down.

Consider an air hockey table. When the air is turned off, the puck slides only a short distance before friction slows it to a stop. However, when the air is turned on, it lifts the puck slightly, so the puck experiences very little friction as it moves over the surface. With friction almost eliminated, the puck glides along with very little change in speed. On a frictionless surface, the puck would experience no net external force (ignoring air resistance, which is also a form of friction). Additionally, if we know enough about friction, we can accurately predict how quickly objects will slow down.

Now let’s think about an example. A man pushes a box across a floor at constant velocity by applying a force of +50 N. (The positive sign indicates that, by convention, the direction of motion is to the right.) What is the force of friction that opposes the motion? The force of friction must be −50 N. Why? According to Newton’s first law of motion, any object moving at constant velocity has no net external force acting upon it, which means that the sum of the forces acting on the object F _net must be zero. The mathematical way to say that no net external force acts on an object is F _net  = 0  or ∑ F = 0 So if the man applies +50 N of force, then the force of friction must be −50 N for the two forces to add up to zero (that is, for the two forces to cancel each other). Whenever you encounter the phrase “at constant velocity” , Newton’s first law tells you that the net external force is zero

The force of friction depends on two factors: the coefficient of friction and the normal force. For any two surfaces that are in contact with one another, the coefficient of friction is a constant that depends on the nature of the surfaces. The normal force is the force exerted by a surface that pushes on an object in response to gravity pulling the object down. In equation form, the force of friction is

f = µ N

Where µ is the coefficient of friction and N is the normal force.

A net external force acts from outside on the object of interest. A more precise definition is that it acts on the system of interest. A system is one or more objects that you choose to study. It is important to define the system at the beginning of a problem to figure out which forces are external and need to be considered, and which are internal and can be ignored.

For example, in the below  Figure , two children push a third child in a wagon at a constant velocity. The system of interest is the wagon plus the small child, as shown in part (b) of the figure. The two children behind the wagon exert external forces on this system (F1, F2). Friction f acting at the axles of the wheels and at the surface where the wheels touch the ground two other external forces acting on the system. Two more external forces act on the system: the weight W of the system pulling down and the normal force N of the ground pushing up. Notice that the wagon is not accelerating vertically, so Newton’s first law tells us that the normal force balances the weight. Because the wagon is moving forward at a constant velocity, the force of friction must have the same strength as the sum of the forces applied by the two children

Figure The wagon and rider form a system that is acted on by external forces. (b) The two children pushing the wagon and child provide two external forces. Friction acting at the wheel axles and on the surface of the tires where they touch the ground provide an external force that act against the direction of motion. The weight W and the normal force N from the ground are two more external forces acting on the system. All external forces are represented in the figure by arrows. All of the external forces acting on the system add together, but because the wagon moves at a constant velocity, all of the forces must add up to zero

Mass and Inertia

Inertia is the tendency for an object at rest to remain at rest, or for a moving object to remain in motion in a straight line with constant speed. This key property of objects was first described by Galileo. Later, Newton incorporated the concept of inertia into his first law, which is often referred to as the law of inertia. As we know from experience, some objects have more inertia than others. For example, changing the motion of a large truck is more difficult than changing the motion of a toy truck. In fact, the inertia of an object is proportional to the mass of the object. Mass is a measure of the amount of matter (or stuff) in an object. The quantity or amount of matter in an object is determined by the number and types of atoms the object contains. Unlike weight (which changes if the gravitational force changes), mass does not depend on gravity. The mass of an object is the same on Earth, in orbit, or on the surface of the moon. In practice, it is very difficult to count and identify all of the atoms and molecules in an object, so mass is usually not determined this way. Instead, the mass of an object is determined by comparing it with the standard kilogram. Mass is therefore expressed in kilograms.

Newton’s Second Law of Motion :

Newton’s first law considered bodies at rest or bodies in motion at a constant velocity. The other state of motion to consider is when an object is moving with a changing velocity, which means a change in the speed and/or the direction of motion. This type of motion is addressed by Newton’s second law of motion, which states how force causes changes in motion. Newton’s second law of motion is used to calculate what happens in situations involving forces and motion, and it shows the mathematical relationship between force, mass, and acceleration. Mathematically, the second law is most often written as

F_net = m a or ∑ F = m a

where F_net  (or ∑F) is the net external force, m is the mass of the system, and a is the acceleration. Note that F_net and ∑F are the same because the net external force is the sum of all the external forces acting on the system.

what do we mean by a change in motion? A change in motion is simply a change in velocity: the speed of an object can become slower or faster, the direction in which the object is moving can change, or both of these variables may change. A change in velocity means, by definition, that an acceleration has occurred.

Newton’s first law says that only a nonzero net external force can cause a change in motion, so a net external force must cause an acceleration. Note that acceleration can refer to slowing down or to speeding up. Acceleration can also refer to a change in the direction of motion with no change in speed, because acceleration is the change in velocity divided by the time it takes for that change to occur, and velocity is defined by speed and direction. From the equation we see that force is directly proportional to both mass and acceleration, which makes sense. To accelerate two objects from rest to the same velocity, you would expect more force to be required to accelerate the more massive object. Likewise, for two objects of the same mass, applying a greater force to one would accelerate it to a greater velocity. Now, let’s rearrange Newton’s second law to solve for acceleration. We get

 a = F_net / m or a = ∑F / m

In this form, we can see that acceleration is directly proportional to force, which we write as

a α F_net

where the symbol means proportional to. This proportionality mathematically states what we just said in words: acceleration is directly proportional to the net external force. When two variables are directly proportional to each other, then if one variable doubles, the other variable must double. Likewise, if one variable is reduced by half, the other variable must also be reduced by half. In general, when one variable is multiplied by a number, the other variable is also multiplied by the same number. It seems reasonable that the acceleration of a system should be directly proportional to and in the same direction as the net external force acting on the system. An object experiences greater acceleration when acted on by a greater force

It is also clear from the equation  a = F_net / m , that acceleration is inversely proportional to mass, which we write as

a α 1/m

Inversely proportional means that if one variable is multiplied by a number, the other variable must be divided by the same number. Now, it also seems reasonable that acceleration should be inversely proportional to the mass of the system. In other words, the larger the mass (the inertia), the smaller the acceleration produced by a given force. This relationship is illustrated in Figure 4.5, which shows that a given net external force applied to a basketball produces a much greater acceleration than when applied to a car.

Applying Newton’s Second Law :

Before putting Newton’s second law into action, it is important to consider units. The equation

F_net = m a is used to define the units of force in terms of the three basic units of mass, length, and time (recall that acceleration has units of length divided by time squared). The SI unit of force is called the newton (abbreviated N) and is the force needed to accelerate a 1-kg system at the rate of 1 m/s ². That is, because F_net = m a  we have

1 N = 1 Kg X 1 m/s²  = 1 Kg m s ^-2

One of the most important applications of Newton’s second law is to calculate weight (also known as the gravitational force), which is usually represented mathematically as W. When people talk about gravity, they don’t always realize that it is an acceleration. When an object is dropped, it accelerates toward the center of Earth. Newton’s second law states that the net external force acting on an object is responsible for the acceleration of the object. If air resistance is negligible, the net external force on a falling object is only the gravitational force (i.e., the weight of the object).

 Weight can be represented by a vector because it has a direction. Down is defined as the direction in which gravity pulls, so weight is normally considered a downward force. By using Newton’s second law, we can figure out the equation for weight.

Consider an object with mass m falling toward Earth. It experiences only the force of gravity (i.e., the gravitational force or weight), which is represented by W. Newton’s second law states that  Because the only force acting on the object is the gravitational force, we have

F_net = W 

We know that the acceleration of an object due to gravity is g, so we have

a = g

Substituting these two expressions into Newton’s second law gives

W = mg

This is the equation for weight—the gravitational force on a mass m. On Earth, g = 9.8 m/s²  so the weight (disregarding for now the direction of the weight) of a 1.0-kg object on Earth is

W = mg = 1 X 9.8 m/s² = 9.8 N

When the net external force on an object is its weight, we say that it is in freefall. In this case, the only force acting on the object is the force of gravity.

 On the surface of Earth, when objects fall downward toward Earth, they are never truly in freefall because there is always some upward force due to air resistance that acts on the object (and there is also the buoyancy force of air, which is similar to the buoyancy force in water that keeps boats afloat).

Gravity varies slightly over the surface of Earth, so the weight of an object depends very slightly on its location on Earth. Weight varies dramatically away from Earth’s surface. On the moon, for example, the acceleration due to gravity is only 1.67 m/s2 . Because weight depends on the force of gravity, a 1.0-kg mass weighs 9.8 N on Earth and only about 1.7 N on the moon.

It is important to remember that weight and mass are very different, although they are closely related. Mass is the quantity of matter (how much stuff) in an object and does not vary, but weight is the gravitational force on an object and is proportional to the force of gravity.

It is easy to confuse the two, because our experience is confined to Earth, and the weight of an object is essentially the same no matter where you are on Earth. Adding to the confusion, the terms mass and weight are often used interchangeably in everyday language; for example, our medical records often show our weight in kilograms, but never in the correct unit of newtons.

Momentum, Impulse, and the Impulse-Momentum Theorem :

Linear momentum is the product of a system’s mass and its velocity. In equation form, linear momentum p is

P = m V

You can see from the equation that momentum is directly proportional to the object’s mass (m) and velocity (v). Therefore, the greater an object’s mass or the greater its velocity, the greater its momentum.

A large, fast-moving object has greater momentum than a smaller, slower object. Momentum is a vector and has the same direction as velocity v. Since mass is a scalar, when velocity is in a negative direction (i.e., opposite the direction of motion), the momentum will also be in a negative direction; and when velocity is in a positive direction, momentum will likewise be in a positive direction. The SI unit for momentum is kg m/s.

Momentum is so important for understanding motion that it was called the quantity of motion by physicists such as Newton. Force influences momentum, and we can rearrange Newton’s second law of motion to show the relationship between force and momentum.

Newton actually stated his second law of motion in terms of momentum: The net external force equals the change in momentum of a system divided by the time over which it changes. The change in momentum is the difference between the final and initial values of momentum. In equation form, this law is

F_net = ΔP / Δt

where F_net is the net external force, ΔP is the change in momentum, and Δt  is the change in time. We can solve for ΔP  by rearranging the equation

ΔP = F_net Δt

F_net Δt is known as impulse and this equation is known as the impulse-momentum theorem.

From the equation, we see that the impulse equals the average net external force multiplied by the time this force acts. It is equal to the change in momentum. The effect of a force on an object depends on how long it acts, as well as the strength of the force. Impulse is a useful concept because it quantifies the effect of a force. A very large force acting for a short time can have a great effect on the momentum of an object, such as the force of a racket hitting a tennis ball. A small force could cause the same change in momentum, but it would have to act for a much longer time

Newton’s Second Law in Terms of Momentum

When Newton’s second law is expressed in terms of momentum, it can be used for solving problems where mass varies, since ΔP = Δ(mv)  . In the more traditional form of the law that you are used to working with, mass is assumed to be constant. In fact, this traditional form is a special case of the law, where mass is constant.F_net = ma  is actually derived from the equation:

F_net = ΔP / Δt

The change in momentum ΔP  is given by

ΔP = Δ (mv)

If the mass of the system is constant, then

Δ (mv) = m Δv

By substituting m Δv for ΔP , Newton’s second law of motion becomes

F_net = ΔP / Δt

      = m Δv / Δt for a constant mass

Since Δv / Δt = a , we get

F_net = ma , when the mass of the system is constant

Conservation of Momentum

It is important we realize that momentum is conserved during collisions, explosions, and other events involving objects in motion. To say that a quantity is conserved means that it is constant throughout the event. In the case of conservation of momentum, the total momentum in the system remains the same before and after the collision.

You may have noticed that momentum was not conserved , where forces acting on the objects produced large changes in momentum. Why is this? The systems of interest considered in those problems were not inclusive enough. If the systems were expanded to include more objects, then momentum would in fact be conserved in those sample problems. It is always possible to find a larger system where momentum is conserved, even though momentum changes for individual objects within the system. For example, if a football player runs into the goalpost in the end zone, a force will cause him to bounce backward. His momentum is obviously greatly changed, and considering only the football player, we would find that momentum is not conserved. However, the system can be expanded to contain the entire Earth. Surprisingly, Earth also recoils—conserving momentum—because of the force applied to it through the goalpost. The effect on Earth is not noticeable because it is so much more massive than the player, but the effect is real. Next, consider what happens if the masses of two colliding objects are more similar than the masses of a football player and Earth—in the example shown in Figure 8.4 of one car bumping into another. Both cars are coasting in the same direction when the lead car, labeled m2 , is bumped by the trailing car, labeled m1 . The only unbalanced force on each car is the force of the collision, assuming that the effects due to friction are negligible. Car m1 slows down as a result of the collision, losing some momentum, while car m2 speeds up and gains some momentum. If we choose the system to include both cars and assume that friction is negligible, then the momentum of the two-car system should remain constant. Now we will prove that the total momentum of the two-car system does in fact remain constant, and is therefore conserved

Newton’s Third Law of Motion :

Newton’s third law of motion states that whenever a first object exerts a force on a second object, the first object experiences a force equal in magnitude but opposite in direction to the force that it exerts.

Newton’s third law of motion tells us that forces always occur in pairs, and one object cannot exert a force on another without experiencing the same strength force in return. We sometimes refer to these force pairs as action-reaction pairs, where the force exerted is the action, and the force experienced in return is the reaction (although which is which depends on your point of view).

Newton’s third law is useful for figuring out which forces are external to a system. Recall that identifying external forces is important when setting up a problem, because the external forces must be added together to find the net force.

We can see Newton’s third law at work by looking at how people move about. Consider a swimmer pushing off from the side of a pool. He pushes against the pool wall with his feet and accelerates in the direction opposite to the push. The wall has thus exerted on the swimmer a force of equal magnitude but in the direction opposite that of his push. You might think that two forces of equal magnitude but that act in opposite directions would cancel, but they do not because they act on different systems.

Similarly, a car accelerates because the ground pushes forward on the car’s wheels in reaction to the car’s wheels pushing backward on the ground. You can see evidence of the wheels pushing backward when tires spin on a gravel road and throw rocks backward.

Birds fly by exerting force on air in the direction opposite that in which they wish to fly. For example, the wings of a bird force air downward and backward in order to get lift and move forward

Applying Newton’s Third Law :

The gravitational force (or weight) acts on objects at all times and everywhere on Earth. We know from Newton’s second law that a net force produces an acceleration; so, why is everything not in a constant state of freefall toward the center of Earth? The answer is the normal force. The normal force is the force that a surface applies to an object to support the weight of that object; it acts perpendicular to the surface upon which the object rests. If an object on a flat surface is not accelerating, the net external force is zero, and the normal force has the same magnitude as the weight of the system but acts in the opposite direction. In equation form, we write that

N = mg

Note that this equation is only true for a horizontal surface.

The word tension comes from the Latin word meaning to stretch. Tension is the force along the length of a flexible connector, such as a string, rope, chain, or cable. Regardless of the type of connector attached to the object of interest, one must remember that the connector can only pull (or exert tension) in the direction parallel to its length. Tension is a pull that acts parallel to the connector, and that acts in opposite directions at the two ends of the connector. This is possible because a flexible connector is simply a long series of action-reaction forces, except at the two ends where outside objects provide one member of the actionreaction forces.

Consider a person holding a mass on a rope, as shown in below Figure

Tension in the rope must equal the weight of the supported mass, as we can prove by using Newton’s second law. If the 5 kg mass in the figure is stationary, then its acceleration is zero, so         F_net = 0 .  The only external forces acting on the mass are its weight W and the tension T supplied by the rope. Summing the external forces to find the net force, we obtain

F_net  = T – W = 0

where T and W are the magnitudes of the tension and weight, respectively, and their signs indicate direction, with up being positive. By substituting mg for F_net and rearranging the equation, the tension equals the weight of the supported mass, we get

T = W = mg

For a 5 Kg mass , the we see that

T = mg = 5 Kg X 9.8 m/s² = 49 N

Another example of Newton’s third law in action is thrust. Rockets move forward by expelling gas backward at a high velocity. This means that the rocket exerts a large force backward on the gas in the rocket combustion chamber, and the gas, in turn, exerts a large force forward on the rocket in response. This reaction force is called thrust.

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Work, Power, and the Work–Energy Theorem

In physics, the term work has a very specific definition. Work is application of force, f, to move an object over a distance, d, in the direction that the force is applied. Work, W, is described by the equation

W = fd

Some things that we typically consider to be work are not work in the scientific sense of the term. Let’s consider a few examples. Think about why each of the following statements is true.

• Homework is not work.
• Lifting a rock upwards off the ground is work.
• Carrying a rock in a straight path across the lawn at a constant speed is not work.

The first two examples are fairly simple. Homework is not work because objects are not being moved over a distance. Lifting a rock up off the ground is work because the rock is moving in the direction that force is applied. The last example is less obvious. From the laws of motion , we know that  that force is not required to move an object at constant velocity. Therefore, while some force may be applied to keep the rock up off the ground, no net force is applied to keep the rock moving forward at constant velocity.

Work and energy are closely related. When you do work to move an object, you change the object’s energy. You (or an object) also expend energy to do work. In fact, energy can be defined as the ability to do work. Energy can take a variety of different forms, and one form of energy can transform to another. In this chapter we will be concerned with mechanical energy, which comes in two forms: kinetic energy and potential energy.

• Kinetic energy is also called energy of motion. A moving object has kinetic energy.
• Potential energy, sometimes called stored energy, comes in several forms. Gravitational potential energy is the stored energy an object has as a result of its position above Earth’s surface (or another object in space). A roller coaster car at the top of a hill has gravitational potential energy.

Let’s examine how doing work on an object changes the object’s energy. If we apply force to lift a rock off the ground, we increase the rock’s potential energy, PE. If we drop the rock, the force of gravity increases the rock’s kinetic energy as the rock moves downward until it hits the ground.

The force we exert to lift the rock is equal to its weight, w, which is equal to its mass, m, multiplied by acceleration due to gravity, g.

The work we do on the rock equals the force we exert multiplied by the distance, d, that we lift the rock. The work we do on the rock also equals the rock’s gain in gravitational potential energy, PE_e.

Kinetic energy depends on the mass of an object and its velocity, v.

When we drop the rock the force of gravity causes the rock to fall, giving the rock kinetic energy. When work done on an object increases only its kinetic energy, then the net work equals the change in the value of the quantity 1/2 mv². This is a statement of the work–energy theorem, which is expressed mathematically as

The subscripts ₂ and ₁ indicate the final and initial velocity, respectively. This theorem was proposed and successfully tested by James Joule

The joule (J) is the metric unit of measurement for both work and energy. The measurement of work and energy with the same unit reinforces the idea that work and energy are related and can be converted into one another. 1.0 J = 1.0 N∙m, the units of force multiplied by distance. 1.0 N = 1.0 k∙m/s², so 1.0 J = 1.0 k∙m²/s². Analyzing the units of the term (1/2)mv² will produce the same units for joules.

Calculations Involving Work and Power

In applications that involve work, we are often interested in how fast the work is done. For example, in roller coaster design, the amount of time it takes to lift a roller coaster car to the top of the first hill is an important consideration. Taking a half hour on the ascent will surely irritate riders and decrease ticket sales. Let’s take a look at how to calculate the time it takes to do work.

Recall that a rate can be used to describe a quantity, such as work, over a period of time. Power is the rate at which work is done. In this case, rate means per unit of time. Power is calculated by dividing the work done by the time it took to do the work.

P = W / t

Power can be expressed in units of watts (W). This unit can be used to measure power related to any form of energy or work. You have most likely heard the term used in relation to electrical devices, especially light bulbs. Multiplying power by time gives the amount of energy. Electricity is sold in kilowatt-hours because that equals the amount of electrical energy consumed.

Mechanical Energy and Conservation of Energy

We saw earlier that mechanical energy can be either potential or kinetic. In this section we will see how energy is transformed from one of these forms to the other. We will also see that, in a closed system, the sum of these forms of energy remains constant.

Imagine a roller coaster ride . Quite a bit of potential energy is gained by a roller coaster car and its passengers when they are raised to the top of the first hill. Remember that the potential part of the term means that energy has been stored and can be used at another time. You will see that this stored energy can either be used to do work or can be transformed into kinetic energy. For example, when an object that has gravitational potential energy falls, its energy is converted to kinetic energy. Remember that both work and energy are expressed in joules.

Now, let’s look at the roller coaster in the below figure . Work was done on the roller coaster to get it to the top of the first rise; at this point, the roller coaster has gravitational potential energy. It is moving slowly, so it also has a small amount of kinetic energy. As the car descends the first slope, its PE is converted to KE. At the low point much of the original PE has been transformed to KE, and speed is at a maximum. As the car moves up the next slope, some of the KE is transformed back into PE and the car slows down.

On an actual roller coaster, there are many ups and downs, and each of these is accompanied by transitions between kinetic and potential energy. Assume that no energy is lost to friction. At any point in the ride, the total mechanical energy is the same, and it is equal to the energy the car had at the top of the first rise. This is a result of the law of conservation of energy, which says that, in a closed system, total energy is conserved—that is, it is constant. Using subscripts 1 and 2 to represent initial and final energy, this law is expressed as

Either side equals the total mechanical energy. The phrase in a closed system means we are assuming no energy is lost to the surroundings due to friction and air resistance. If we are making calculations on dense falling objects, this is a good assumption.

Elastic and Inelastic Collisions

When objects collide, they can either stick together or bounce off one another, remaining separate. In this section, we’ll cover these two different types of collisions, first in one dimension and then in two dimensions.

In an elastic collision, the objects separate after impact and don’t lose any of their kinetic energy. Kinetic energy is the energy of motion and is covered in detail elsewhere. The law of conservation of momentum is very useful here, and it can be used whenever the net external force on a system is zero. The below figure shows an elastic collision where momentum is conserved

Perfectly elastic collisions can happen only with subatomic particles. Everyday observable examples of perfectly elastic collisions don’t exist—some kinetic energy is always lost, as it is converted into heat transfer due to friction. However, collisions between everyday objects are almost perfectly elastic when they occur with objects and surfaces that are nearly frictionless, such as with two steel blocks on ice.

Now, to solve problems involving one-dimensional elastic collisions between two objects, we can use the equation for conservation of momentum. First, the equation for conservation of momentum for two objects in a one-dimensional collision is

Substituting the definition of momentum p = mv for each initial and final momentum, we get

where the primes (‘) indicate values after the collision; In some texts, you may see i for initial (before collision) and f for final (after collision). The equation assumes that the mass of each object does not change during the collision

Now, let us turn to the second type of collision. An inelastic collision is one in which objects stick together after impact, and kinetic energy is not conserved. This lack of conservation means that the forces between colliding objects may convert kinetic energy to other forms of energy, such as potential energy or thermal energy. The concepts of energy are discussed more thoroughly elsewhere. For inelastic collisions, kinetic energy may be lost in the form of heat. Below figure shows an example of an inelastic collision. Two objects that have equal masses head toward each other at equal speeds and then stick together. The two objects come to rest after sticking together, conserving momentum but not kinetic energy after they collide. Some of the energy of motion gets converted to thermal energy, or heat.

The figure shows A one-dimensional inelastic collision between two objects. Momentum is conserved, but kinetic energy is not conserved. (a) Two objects of equal mass initially head directly toward each other at the same speed. (b) The objects stick together, creating a perfectly inelastic collision. In the case shown in this figure, the combined objects stop; This is not true for all inelastic collisions.

Since the two objects stick together after colliding, they move together at the same speed. This lets us simplify the conservation of momentum equation from

to

for inelastic collisions, where v′ is the final velocity for both objects as they are stuck together, either in motion or at rest.

But what about collisions, such as those between billiard balls, in which objects scatter to the side? These are two-dimensional collisions, and just as we did with two- dimensional forces, we will solve these problems by first choosing a coordinate system and separating the motion into its x and y components.

We start by assuming that F_net = 0, so that momentum p is conserved. The simplest collision is one in which one of the particles is initially at rest. The best choice for a coordinate system is one with an axis parallel to the velocity of the incoming particle, as shown in below figure . Because momentum is conserved, the components of momentum along the x– and y-axes, displayed as p_x and p_y, will also be conserved. With the chosen coordinate system, p_y is initially zero and p_x is the momentum of the incoming particle.

Now, we will take the conservation of momentum equation, p₁ + p₂ = p′₁ + p′₂ and break it into its x and y components. Along the x-axis, the equation for conservation of momentum is

In terms of masses and velocities, this equation is

But because particle 2 is initially at rest, this equation becomes

The components of the velocities along the x-axis have the form v cos θ . Because particle 1 initially moves along the x-axis, we find v_1x = v₁. Conservation of momentum along the x-axis gives the equation

where θ₁ and θ₂ are as shown in the above Figure

Along the y-axis, the equation for conservation of momentum is

or

But v₁y is zero, because particle 1 initially moves along the x-axis. Because particle 2 is initially at rest, v₂y is also zero. The equation for conservation of momentum along the y-axis becomes

The components of the velocities along the y-axis have the form v sin θ  . Therefore, conservation of momentum along the y-axis gives the following equation:

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There are two major systems of units used in the world: SI units (acronym for the French Le Système International d’Unités, also known as the metric system), and English units (also known as the imperial system). English units were historically used in nations once ruled by the British Empire. Today, the United States is the only country that still uses English units extensively.

Virtually every other country in the world now uses the metric system, which is the standard system agreed upon by scientists and mathematicians.

Some physical quantities are more fundamental than others. In physics, there are seven fundamental physical quantities that are measured in base or physical fundamental units: length, mass, time, electric current temperature, amount of substance, and luminous intensity. Units for other physical quantities (such as force, speed, and electric charge) described by mathematically combining these seven base units. In this course, we will mainly use five of these: length, mass, time, electric current and temperature. The units in which they are measured are the meter, kilogram, second, ampere, kelvin, mole, and candela (Table 1.1). All other units are made by mathematically combining the fundamental units. These are called derived units.

Ǫuantity                 Name        Symbol

Length	Meter	m
Mass	Kilogram	kg
Time	Second	s
Electric current	Ampere	a
Temperature	Kelvin	k
Amount of substance	Mole	mol
Luminous intensity	Candela	cd

The Meter

The SI unit for length is the meter (m). The definition of the meter has changed over time to become more accurate and precise. The meter was first defined in 1791 as 1/10,000,000 of the distance from the equator to the North Pole. This measurement was improved in 1889 by redefining the meter to be the distance between two engraved lines on a platinum-iridium bar. (The bar is now housed at the International Bureau of Weights and Meaures, near Paris). By 1960, some distances could be measured more precisely by comparing them to wavelengths of light. The meter was redefined as 1,650,763.73 wavelengths of orange light emitted by krypton atoms. In 1983, the meter was given its present definition as the distance light travels in a vacuum in 1/ 299,792,458 of a second (Figure 1.14).

Figure 1.14 The meter is defined to be the distance light travels in 1/299,792,458 of a second through a vacuum. Distance traveled is speed multiplied by time.

The Kilogram

The SI unit for mass is the kilogram (kg). It is defined to be the mass of a platinum-iridium cylinder, housed at the International Bureau of Weights and Measures near Paris. Exact replicas of the standard kilogram cylinder are kept in numerous locations throughout the world, such as the National Institute of Standards and Technology in Gaithersburg, Maryland. The determination of all other masses can be done by comparing them with one of these standard kilograms.

The Second

The SI unit for time, the second (s) also has a long history. For many years it was defined as 1/86,400 of an average solar day. However, the average solar day is actually very gradually getting longer due to gradual slowing of Earth’s rotation. Accuracy in the fundamental units is essential, since all other measurements are derived from them. Therefore, a new standard was adopted to define the second in terms of a non-varying, or constant, physical phenomenon. One constant phenomenon is the very steady vibration of Cesium atoms, which can be observed and counted. This vibration forms the basis of the cesium atomic clock. In 1967, the second was redefined as the time required for 9,192,631,770 Cesium atom vibrations (Figure 1.15).

The Ampere

Electric current is measured in the ampere (A), named after Andre Ampere. You have probably heard of amperes, or amps, when people discuss electrical currents or electrical devices. Understanding an ampere requires a basic understanding of electricity and magnetism, something that will be explored in depth in later chapters of this book. Basically, two parallel wires with an electric current running through them will produce an attractive force on each other. One ampere is defined as the amount of electric current that will produce an attractive force of 2.7  10–7 newton per meter of separation between the two wires (the newton is the derived unit of force).                               

Kelvins

The SI unit of temperature is the kelvin (or kelvins, but not degrees kelvin). This scale is named after physicist William Thomson, Lord Kelvin, who was the first to call for an absolute temperature scale. The Kelvin scale is based on absolute zero. This is the point at which all thermal energy has been removed from all atoms or molecules in a system. This temperature, 0 K, is equal to −273.15 °C and −459.67 °F. Conveniently, the Kelvin scale actually changes in the same way as the Celsius scale. For example, the freezing point (0 °C) and boiling points of water (100 °C) are 100 degrees apart on the Celsius scale. These two temperatures are also 100 kelvins apart (freezing point = 273.15 K; boiling point = 373.15 K).

Metric Prefixes

Physical objects or phenomena may vary widely. For example, the size of objects varies from something very small (like an atom) to something very large (like a star). Yet the standard metric unit of length is the meter. So, the metric system includes many prefixes that can be attached to a unit. Each prefix is based on factors of 10 (10, 100, 1,000, etc., as well as 0.1, 0.01, 0.001, etc.). Table 1.2 gives the metric prefixes and symbols used to denote the different various factors of 10 in the metric system.

Prefix	Symbol	Value[1]	Example Name	Example Symbol	Example Value	Example Description
exa	E	1018	Exameter	Em	1018 m	Distance light travels in a century
peta	P	1015	Petasecond	Ps	1015 s	30 million years
tera	T	1012	Terawatt	TW	1012 W	Powerful laser output
giga	G	109	Gigahertz	GHz	109 Hz	A microwave frequency
mega	M	106	Megacurie	MCi	106 Ci	High radioactivity
kilo	k	103	Kilometer	km	103 m	About 6/10 mile
hector	h	102	Hectoliter	hL	102 L	26 gallons
deka	da	101	Dekagram	dag	101 g	Teaspoon of butter
deci	d	10–1	Deciliter	dL	10–1 L	Less than half a soda
centi	c	10–2	Centimeter	Cm	10–2 m	Fingertip thickness
milli	m	10–3	Millimeter	Mm	10–3 m	Flea at its shoulder
micro	µ	10–6	Micrometer	µm	10–6 m	Detail in microscope
nano	n	10–9	Nanogram	Ng	10–9 g	Small speck of dust
pico	p	10–12	Picofarad	pF	10–12 F	Small capacitor in radio
femto	f	10–15	Femtometer	Fm	10–15 m	Size of a proton
atto	a	10–18	Attosecond	as	10–18 s	Time light takes to cross an atom

The metric system is convenient because conversions between metric units can be done simply by moving the decimal place of a number. This is because the metric prefixes are sequential powers of 10. There are 100 centimeters in a meter, 1000 meters in a kilometer, and so on. In nonmetric systems, such as U.S. customary units, the relationships are less simple—there are 12 inches in a foot, 5,280 feet in a mile, 4 quarts in a gallon, and so on. Another advantage of the metric system is that the same unit can be used over extremely large ranges of values simply by switching to the most-appropriate metric prefix. For example, distances in meters are suitable for building construction, but kilometers are used to describe road construction. Therefore, with the metric system, there is no need to invent new units when measuring very small or very large objects—you just have to move the decimal point (and use the appropriate prefix).

Known Ranges of Length, Mass, and Time

Table 1.3 lists known lengths, masses, and time measurements. You can see that scientists use a range of measurement units. This wide range demonstrates the vastness and complexity of the universe, as well as the breadth of phenomena physicists study. As you examine this table, note how the metric system allows us to discuss and compare an enormous range of phenomena, using one system of measurement (Figure 1.16 and Figure 1.17).

Length (m)	Phenomenon Measured	Mass (Kg)	Phenomenon Measured[1]	Time (s)	Phenomenon Measured[1]
10–18	Present experimental limit to smallest observable detail	10–30	Mass of an electron (9.11×10–31 kg)	10–23	Time for light to cross a proton
10–15	Diameter of a proton	10–27	Mass of a hydrogen atom (1.67×10–27 kg)	10–22	Mean life of an extremely unstable nucleus
1014	Diameter of a uranium nucleus	10–15	Mass of a bacterium	10–15	Time for one oscillation of a visible light
10–10	Diameter of a hydrogen atom	10–5	Mass of a mosquito	10–13	Time for one vibration of an atom in a solid
10–8	Thickness of membranes in cell of living organism	10–2	Mass of a hummingbird	10–8	Time for one oscillation of an FM radio wave
10–6	Wavelength of visible light	1	Mass of a liter of water (about a quart)	10–3	Duration of a nerve impulse
10–3	Size of a grain of sand	102	Mass of a person	1	Time for one heartbeat
1	Height of a 4-year-old child	103	Mass of a car	105	One day (8.64×104 s)
102	Length of a football field	108	Mass of a large ship	107	One day (3.16×107 s)
104	Greatest ocean depth	1012	Mass of a large iceberg	109	About half the life expectancy of a human
107	Diameter of Earth	1015	Mass of the nucleus of a comet	1011	Recorded history
1011	Distance from Earth to the sun	1023	Mass of the moon (7.35×1022 kg)	1017	Age of Earth
1016	Distance traveled by light in 1 year (a light year)	1025	Mass of Earth (5.97×1024kg)	1018	Age of the universe
1021	Diameter of the Milky Way Galaxy	1030	Mass of the Sun (1.99×1034 kg)
1022	Distance from Earth to the nearest large galaxy (Andromeda)	1042	Mass of the Milky Way galaxy (current upper limit)
1026	Distance from Earth to the edges of the known universe	1053	Mass of the known universe (current upper limit)

Using Scientific Notation with Physical Measurements

Scientific notation is a way of writing numbers that are too large or small to be conveniently written as a decimal. For example, consider the number 840,000,000,000,000. It’s a rather large number to write out. The scientific notation for this number is 8.40 x 10¹⁴. Scientific notation follows this general format

x x 10^y

In this format x is the value of the measurement with all placeholder zeros removed. In the example above, x is 8.4. The x is multiplied by a factor, 10y, which indicates the number of placeholder zeros in the measurement. Placeholder zeros are those at the end of a number that is 10 or greater, and at the beginning of a decimal number that is less than 1. In the example above, the factor is 1014. This tells you that you should move the decimal point 14 positions to the right, filling in placeholder zeros as you go. In this case, moving the decimal point 14 places creates only 13 placeholder zeros, indicating that the actual measurement value is 840,000,000,000,000.

Numbers that are fractions can be indicated by scientific notation as well. Consider the number 0.0000045. Its scientific notation is 4.5 x 10^–6. Its scientific notation has the same format

x x 10^y

Here, x is 4.5. However, the value of y in the 10y factor is negative, which indicates that the measurement is a fraction of 1. Therefore, we move the decimal place to the left, for a negative y. In our example of 4.5 x 10^–6, the decimal point would be moved to the left six times to yield the original number, which would be 0.0000045.

The term order of magnitude refers to the power of 10 when numbers are expressed in scientific notation. Quantities that have the same power of 10 when expressed in scientific notation, or come close to it, are said to be of the same order of magnitude. For example, the number 800 can be written as 8  102, and the number 450 can be written as 4.5  102. Both numbers have the same value for y. Therefore, 800 and 450 are of the same order of magnitude. Similarly, 101 and 99 would be regarded as the same order of magnitude, 102. Order of magnitude can be thought of as a ballpark estimate for the scale of a value. The diameter of an atom is on the order of 10−9 m, while the diameter of the sun is on the order of 109 m. These two values are 18 orders of magnitude apart.

Scientists make frequent use of scientific notation because of the vast range of physical measurements possible in the universe, such as the distance from Earth to the moon (Figure 1.18), or to the nearest star.

Unit Conversion and Dimensional Analysis

It is often necessary to convert from one type of unit to another. For example, if you are reading a European cookbook in the United States, some quantities may be expressed in liters and you need to convert them to cups. A Canadian tourist driving through the United States might want to convert miles to kilometers, to have a sense of how far away his next destination is. A doctor in the United States might convert a patient’s weight in pounds to kilograms.

Let’s consider a simple example of how to convert units within the metric system. How can we want to convert 1 hour to seconds?

Next, we need to determine a conversion factor relating meters to kilometers. A conversion factor is a ratio expressing how many of one unit are equal to another unit. A conversion factor is simply a fraction which equals 1. You can multiply any number by 1 and get the same value. When you multiply a number by a conversion factor, you are simply multiplying it by one. For example, the following are conversion factors: (1 foot)/(12 inches) = 1 to convert inches to feet, (1 meter)/(100 centimeters) = 1 to convert centimeters to meters, (1 minute)/(60 seconds) = 1 to convert seconds to minutes. In this case, we know that there are 1,000 meters in 1 kilometer.

Now we can set up our unit conversion. We will write the units that we have and then multiply them by the conversion factor (1 km/1,000m) = 1, so we are simply multiplying 80m by 1:

When there is a unit in the original number, and a unit in the denominator (bottom) of the conversion factor, the units cancel. In this case, hours and minutes cancel and the value in seconds remains.

You can use this method to convert between any types of unit, including between the U.S. customary system and metric system. Notice also that, although you can multiply and divide units algebraically, you cannot add or subtract different units. An expression like 10 km + 5 kg makes no sense. Even adding two lengths in different units, such as 10 km + 20 m does not make sense. You express both lengths in the same unit. See Appendix C for a more complete list of conversion factors.

Worked Example

Unit Conversions: A Short Drive Home

Suppose that you drive the 10.0 km from your university to home in 20.0 min. Calculate your average speed (a) in kilometers per hour (km/h) and (b) in meters per second (m/s). (Note—Average speed is distance traveled divided by time of travel.)

Strategy

First we calculate the average speed using the given units. Then we can get the average speed into the desired units by picking the correct conversion factor and multiplying by it. The correct conversion factor is the one that cancels the unwanted unit and leaves the desired unit in its place.

Solution for (a)

1.Calculate average speed. Average speed is distance traveled divided by time of travel. (Take this definition as a given for now—average speed and other motion concepts will be covered in a later module.) In equation form,

2. Substitute the given values for distance and time.

3. Convert km/min to km/h: multiply by the conversion factor that will cancel minutes and leave hours. That conversion factor is 60 min/1h . Thus

Discussion for (a)

To check your answer, consider the following:

1. Be sure that you have properly cancelled the units in the unit conversion. If you have written the unit conversion factor upside down, the units will not cancel properly in the equation. If you accidentally get the ratio upside down, then the units will not cancel; rather, they will give you the wrong units as follows

which are obviously not the desired units of km/h.

2. Check that the units of the final answer are the desired units. The problem asked us to solve for average speed in units of km/h and we have indeed obtained these units.

3. Check the significant figures. Because each of the values given in the problem has three significant figures, the answer should also have three significant figures. The answer 30.0 km/h does indeed have three significant figures, so this is appropriate. Note that the significant figures in the conversion factor are not relevant because an hour is defined to be 60 min, so the precision of the conversion factor is perfect.

4. Next, check whether the answer is reasonable. Let us consider some information from the problem—if you travel 10 km in a third of an hour (20 min), you would travel three times that far in an hour. The answer does seem reasonable.

Worked Example

Using Physics to Evaluate Promotional Materials

A commemorative coin that is 2″ in diameter is advertised to be plated with 15 mg of gold. If the density of gold is 19.3 g/cc, and the amount of gold around the edge of the coin can be ignored, what is the thickness of the gold on the top and bottom faces of the coin?

Strategy

To solve this problem, the volume of the gold needs to be determined using the gold’s mass and density. Half of that volume is distributed on each face of the coin, and, for each face, the gold can be represented as a cylinder that is 2″ in diameter with a height equal to the thickness. Use the volume formula for a cylinder to determine the thickness.

Solution

Discussion

The amount of gold used is stated to be 15 mg, which is equivalent to a thickness of about 0.00019 mm. The mass figure may make the amount of gold sound larger, both because the number is much bigger (15 versus 0.00019), and because people may have a more intuitive feel for how much a millimeter is than for how much a milligram is. A simple analysis of this sort can clarify the significance of claims made by advertisers.

Accuracy, Precision and Significant Figures

Science is based on experimentation that requires good measurements. The validity of a measurement can be described in terms of its accuracy and its precision (see Figure 1.19 and Figure 1.20). Accuracy is how close a measurement is to the correct value for that measurement. For example, let us say that you are measuring the length of standard piece of printer paper. The packaging in which you purchased the paper states that it is 11 inches long, and suppose this stated value is correct. You measure the length of the paper three times and obtain the following measurements: 11.1 inches, 11.2 inches, and 10.9 inches. These measurements are quite accurate because they are very close to the correct value of 11.0 inches. In contrast, if you had obtained a measurement of 12 inches, your measurement would not be very accurate. This is why measuring instruments are calibrated based on a known measurement. If the instrument consistently returns the correct value of the known measurement, it is safe for use in finding unknown values.

Figure 1.19 A double-pan mechanical balance is used to compare different masses. Usually an object with unknown mass is placed in one pan and objects of known mass are placed in the other pan. When the bar that connects the two pans is horizontal, then the masses in both pans are equal. The known masses are typically metal cylinders of standard mass such as 1 gram, 10 grams, and 100 grams. (Serge Melki)

Figure 1.20 Whereas a mechanical balance may only read the mass of an object to the nearest tenth of a gram, some digital scales can measure the mass of an object up to the nearest thousandth of a gram. As in other measuring devices, the precision of a scale is limited to the last measured figures. This is the hundredths place in the scale pictured here. (Splarka, Wikimedia Commons)

Precision states how well repeated measurements of something generate the same or similar results. Therefore, the precision of measurements refers to how close together the measurements are when you measure the same thing several times. One way to analyze the precision of measurements would be to determine the range, or difference between the lowest and the highest measured values. In the case of the printer paper measurements, the lowest value was 10.9 inches and the highest value was 11.2 inches. Thus, the measured values deviated from each other by, at most, 0.3 inches. These measurements were reasonably precise because they varied by only a fraction of an inch. However, if the measured values had been 10.9 inches, 11.1 inches, and 11.9 inches, then the measurements would not be very precise because there is a lot of variation from one measurement to another.

The measurements in the paper example are both accurate and precise, but in some cases, measurements are accurate but not precise, or they are precise but not accurate. Let us consider a GPS system that is attempting to locate the position of a restaurant in a city. Think of the restaurant location as existing at the center of a bull’s-eye target. Then think of each GPS attempt to locate the restaurant as a black dot on the bull’s eye.

In Figure 1.21, you can see that the GPS measurements are spread far apart from each other, but they are all relatively close to the actual location of the restaurant at the center of the target. This indicates a low precision, high accuracy measuring system.

However, in Figure 1.22, the GPS measurements are concentrated quite closely to one another, but they are far away from the target location. This indicates a high precision, low accuracy measuring system. Finally, in Figure 1.23, the GPS is both precise and accurate, allowing the restaurant to be located.

Figure 1.21 A GPS system attempts to locate a restaurant at the center of the bull’s-eye. The black dots represent each attempt to pinpoint the location of the restaurant. The dots are spread out quite far apart from one another, indicating low precision, but they are each rather close to the actual location of the restaurant, indicating high accuracy. (Dark Evil)

Figure 1.22 In this figure, the dots are concentrated close to one another, indicating high precision, but they are rather far away from the actual location of the restaurant, indicating low accuracy. (Dark Evil)

Figure 1.23 In this figure, the dots are concentrated close to one another, indicating high precision, but they are rather far away from the actual location of the restaurant, indicating low accuracy. (Dark Evil)

Uncertainty

The accuracy and precision of a measuring system determine the uncertainty of its measurements. Uncertainty is a way to describe how much your measured value deviates from the actual value that the object has. If your measurements are not very accurate or precise, then the uncertainty of your values will be very high. In more general terms, uncertainty can be thought of as a disclaimer for your measured values. For example, if someone asked you to provide the mileage on your car, you might say that it is 45,000 miles, plus or minus 500 miles. The plus or minus amount is the uncertainty in your value. That is, you are indicating that the actual mileage of your car might be as low as 44,500 miles or as high as 45,500 miles, or anywhere in between. All measurements contain some amount of uncertainty. In our example of measuring the length of the paper, we might say that the length of the paper is 11 inches plus or minus 0.2 inches or 11.0 ± 0.2 inches. The uncertainty in a

measurement, A, is often denoted as δA (“delta A“),

The factors contributing to uncertainty in a measurement include the following:

1. Limitations of the measuring device
2. The skill of the person making the measurement
3. Irregularities in the object being measured
4. Any other factors that affect the outcome (highly dependent on the situation)

In the printer paper example uncertainty could be caused by: the fact that the smallest division on the ruler is 0.1 inches, the person using the ruler has bad eyesight, or uncertainty caused by the paper cutting machine (e.g., one side of the paper is slightly longer than the other.) It is good practice to carefully consider all possible sources of uncertainty in a measurement and reduce or eliminate them,

Percent Uncertainty

One method of expressing uncertainty is as a percent of the measured value. If a measurement, A, is expressed with uncertainty,

Uncertainty in Calculations

There is an uncertainty in anything calculated from measured quantities. For example, the area of a floor calculated from measurements of its length and width has an uncertainty because the both the length and width have uncertainties. How big is the uncertainty in something you calculate by multiplication or division? If the measurements in the calculation have small uncertainties (a few percent or less), then the method of adding percents can be used. This method says that the percent uncertainty in a quantity calculated by multiplication or division is the sum of the percent uncertainties in the items used to make the calculation. For example, if a floor has a length of 4.00 m and a width of 3.00 m, with uncertainties of 2 percent and 1 percent, respectively, then the area of the floor is 12.0 m2 and has an uncertainty of 3 percent (expressed as an area this is 0.36 m2, which we round to 0.4 m2 since the area of the floor is given to a tenth of a square meter).

For a quick demonstration of the accuracy, precision, and uncertainty of measurements based upon the units of measurement, try this simulation (http://openstax.org/l/28precision) . You will have the opportunity to measure the length and weight of a desk, using milli- versus centi- units. Which do you think will provide greater accuracy, precision and uncertainty when measuring the desk and the notepad in the simulation? Consider how the nature of the hypothesis or research question might influence how precise of a measuring tool you need to collect data.

Precision of Measuring Tools and Significant Figures

An important factor in the accuracy and precision of measurements is the precision of the measuring tool. In general, a precise measuring tool is one that can measure values in very small increments. For example, consider measuring the thickness of a coin. A standard ruler can measure thickness to the nearest millimeter, while a micrometer can measure the thickness to the nearest 0.005 millimeter. The micrometer is a more precise measuring tool because it can measure extremely small differences in thickness. The more precise the measuring tool, the more precise and accurate the measurements can be.

When we express measured values, we can only list as many digits as we initially measured with our measuring tool (such as the rulers shown in Figure 1.24). For example, if you use a standard ruler to measure the length of a stick, you may measure it with a decimeter ruler as 3.6 cm. You could not express this value as 3.65 cm because your measuring tool was not precise enough to measure a hundredth of a centimeter. It should be noted that the last digit in a measured value has been estimated in some way by the person performing the measurement. For example, the person measuring the length of a stick with a ruler notices that the stick length seems to be somewhere in between 36 mm and 37 mm. He or she must estimate the value of the last digit. The rule is that the last digit written down in a measurement is the first digit with some uncertainty. For example, the last measured value 36.5 mm has three digits, or three significant figures. The number of significant figures in a measurement indicates the precision of the measuring tool. The more precise a measuring tool is, the greater the number of significant figures it can report.

Figure 1.24 Three metric rulers are shown. The first ruler is in decimeters and can measure point three decimeters. The second ruler is in centimeters long and can measure three point six centimeters. The last ruler is in millimeters and can measure thirty-six point five millimeters.

Zeros

Special consideration is given to zeros when counting significant figures. For example, the zeros in 0.053 are not significant because they are only placeholders that locate the decimal point. There are two significant figures in 0.053—the 5 and the 3. However, if the zero occurs between other significant figures, the zeros are significant. For example, both zeros in 10.053 are significant, as these zeros were actually measured. Therefore, the 10.053 placeholder has five significant figures. The zeros in 1300 may or may not be significant, depending on the style of writing numbers. They could mean the number is known to the last zero, or the zeros could be placeholders. So 1300 could have two, three, or four significant figures. To avoid this ambiguity, write 1300 in scientific notation as 1.3 × 103. Only significant figures are given in the x factor for a number in scientific notation (in the form x x 10^y). Therefore, we know that 1 and 3 are the only significant digits in this number. In summary, zeros are significant except when they serve only as placeholders. Table 1.4 provides examples of the number of significant figures in various numbers.

Number	Significant Figures	Rationale
1.657	4	There are no zeros and all non-zero numbers are always significant.
0.4578	4	The first zero is only a placeholder for the decimal point.
0.000458	3	The first four zeros are placeholders needed to report the data to the ten-thoudsandths place.
2000.56	6	The three zeros are significant here because they occur between other significant figures.
45,600	3	With no underlines or scientific notation, we assume that the last two zeros are placeholders and are not significant.
15895000	7	The two underlined zeros are significant, while the last zero is not, as it is not underlined.
5.457 x 1013	4	In scientific notation, all numbers reported in front of the multiplication sign are significant
6.520 x 10–23	4	In scientific notation, all numbers reported in front of the multiplication sign are significant, including zeros.

Significant Figures in Calculations

When combining measurements with different degrees of accuracy and precision, the number of significant digits in the final answer can be no greater than the number of significant digits in the least precise measured value. There are two different rules, one for multiplication and division and another rule for addition and subtraction, as discussed below.

1. For multiplication and division: The answer should have the same number of significant figures as the starting value with the fewest significant figures. For example, the area of a circle can be calculated from its radius using A = πr² . Let us see how many significant figures the area will have if the radius has only two significant figures, for example, r = 2.0 m. Then, using a calculator that keeps eight significant figures, you would get

A = πr² = (3.1415927…) x (2.0 m)² = 4.5238934 m²

But because the radius has only two significant figures, the area calculated is meaningful only to two significant figures or

A = 4.5 m²

even though the value of  is meaningful to at least eight digits.

• For addition and subtraction: The answer should have the same number places (e.g. tens place, ones place, tenths place, etc.) as the least-precise starting value. Suppose that you buy 7.56 kg of potatoes in a grocery store as measured with a scale having a precision of 0.01 kg. Then you drop off 6.052 kg of potatoes at your laboratory as measured by a scale with a precision of 0.001 kg. Finally, you go home and add 13.7 kg of potatoes as measured by a bathroom scale with a precision of 0.1 kg. How many kilograms of potatoes do you now have, and how many significant figures are appropriate in the answer? The mass is found by simple addition and subtraction:

The least precise measurement is 13.7 kg. This measurement is expressed to the 0.1 decimal place, so our final answer must also be expressed to the 0.1 decimal place. Thus, the answer should be rounded to the tenths place, giving 15.2 kg. The same is true for non-decimal numbers. For example,

6527.23 + 2 = 6528.23 = 6528

We cannot report the decimal places in the answer because 2 has no decimal places that would be significant. Therefore, we can only report to the ones place.

(You can access this textbook for free in web view or PDF through OpenStax.org, and for a low cost in print)

Take Quiz

Most results in science are presented in scientific journal articles using graphs. Graphs present data in a way that is easy to visualize for humans in general, especially someone unfamiliar with what is being studied. They are also useful for presenting large amounts of data or data with complicated trends in an easily-readable way.

One commonly-used graph in physics and other sciences is the line graph, probably because it is the best graph for showing how one quantity changes in response to the other. Let’s build a line graph based on the data in Table 1.5, which shows the measured distance that a train travels from its station versus time. Our two variables, or things that change along the graph, are time in minutes, and distance from the station, in kilometers. Remember that measured data may not have perfect accuracy.

Time (min)  Distance from Station (km)

0	0
10	24
20	36
30	60
40	84
50	97
60	116
70	140

1. Draw the two axes. The horizontal axis, or x-axis, shows the independent variable, which is the variable that is controlled or manipulated. The vertical axis, or y-axis, shows the dependent variable, the non-manipulated variable that changes with (or is dependent on) the value of the independent variable. In the data above, time is the independent variable and should be plotted on the x-axis. Distance from the station is the dependent variable and should be plotted on the y-axis.

• Label each axes on the graph with the name of each variable, followed by the symbol for its units in parentheses. Be sure to leave room so that you can number each axis. In this example, use Time (min) as the label for the x-axis.
• Next, you must determine the best scale to use for numbering each axis. Because the time values on the x-axis are taken every 10 minutes, we could easily number the x-axis from 0 to 70 minutes with a tick mark every 10 minutes. Likewise, the y-axis scale should start low enough and continue high enough to include all of the distance from station values. A scale from 0 km to 160 km should suffice, perhaps with a tick mark every 10 km.

In general, you want to pick a scale for both axes that 1) shows all of your data, and 2) makes it easy to identify trends in your data. If you make your scale too large, it will be harder to see how your data change. Likewise, the smaller and more fine you make your scale, the more space you will need to make the graph. The number of significant figures in the axis values should be coarser than the number of significant figures in the measurements.

• Now that your axes are ready, you can begin plotting your data. For the first data point, count along the x-axis until you find the 10 min tick mark. Then, count up from that point to the 10 km tick mark on the y-axis, and approximate where 22 km is along the y-axis. Place a dot at this location. Repeat for the other six data points (Figure 1.26).

• Add a title to the top of the graph to state what the graph is describing, such as the y-axis parameter vs. the x-axis parameter. In the graph shown here, the title is train motion. It could also be titled distance of the train from the station vs. time.
• Finally, with data points now on the graph, you should draw a trend line (Figure 1.27). The trend line represents the dependence you think the graph represents, so that the person who looks at your graph can see how close it is to the real data. In the present case, since the data points look like they ought to fall on a straight line, you would draw a straight line as the trend line. Draw it to come closest to all the points. Real data may have some inaccuracies, and the plotted points may not all fall on the trend line. In some cases, none of the data points fall exactly on the trend line.

Analyzing a Graph Using Its Equation

One way to get a quick snapshot of a dataset is to look at the equation of its trend line. If the graph produces a straight line, the equation of the trend line takes the form

The b in the equation is the y-intercept while the m in the equation is the slope. The y-intercept tells you at what y value the line intersects the y-axis. In the case of the graph above, the y-intercept occurs at 0, at the very beginning of the graph. The y-intercept, therefore, lets you know immediately where on the y-axis the plot line begins.

The m in the equation is the slope. This value describes how much the line on the graph moves up or down on the y-axis along the line’s length. The slope is found using the following equation

In order to solve this equation, you need to pick two points on the line (preferably far apart on the line so the slope you calculate describes the line accurately). The quantities Y₂ and Y₁ represent the y-values from the two points on the line (not data points) that you picked, while X₂ and X₁ represent the two x-values of the those points.

What can the slope value tell you about the graph? The slope of a perfectly horizontal line will equal zero, while the slope of a perfectly vertical line will be undefined because you cannot divide by zero. A positive slope indicates that the line moves up the y-axis as the x-value increases while a negative slope means that the line moves down the y-axis. The more negative or positive the slope is, the steeper the line moves up or down, respectively. The slope of our graph in Figure 1.26 is calculated below based on the two endpoints of the line

Equation of line : y = (2.0 km/min) x + 0

Because the x axis is time in minutes, we would actually be more likely to use the time t as the independent (x-axis) variable and write the equation as

y = (2.0 km/min) t + 0

The formula y = mx + b only applies to linear relationships, or ones that produce a straight line. Another common type of line in physics is the quadratic relationship, which occurs when one of the variables is squared. One quadratic relationship in physics is the relation between the speed of an object its centripetal acceleration, which is used to determine the force needed to keep an object moving in a circle. Another common relationship in physics is the inverse relationship, in which one variable decreases whenever the other variable increases. An example in physics is Coulomb’s law. As the distance between two charged objects increases, the electrical force between the two charged objects decreases. Inverse proportionality, such the relation between x and y in the equation

y = k/x,

for some number k, is one particular kind of inverse relationship. A third commonly-seen relationship is the exponential relationship, in which a change in the independent variable produces a proportional change in the dependent variable. As the value of the dependent variable gets larger, its rate of growth also increases. For example, bacteria often reproduce at an exponential rate when grown under ideal conditions. As each generation passes, there are more and more bacteria to reproduce. As a result, the growth rate of the bacterial population increases every generation (Figure 1.28).

Using Logarithmic Scales in Graphing

Sometimes a variable can have a very large range of values. This presents a problem when you’re trying to figure out the best scale to use for your graph’s axes. One option is to use a logarithmic (log) scale. In a logarithmic scale, the value each mark labels

is the previous mark’s value multiplied by some constant. For a log base 10 scale, each mark labels a value that is 10 times the value of the mark before it. Therefore, a base 10 logarithmic scale would be numbered: 0, 10, 100, 1,000, etc. You can see how the logarithmic scale covers a much larger range of values than the corresponding linear scale, in which the marks would label the values 0, 10, 20, 30, and so on.

If you use a logarithmic scale on one axis of the graph and a linear scale on the other axis, you are using a semi-log plot. The Richter scale, which measures the strength of earthquakes, uses a semi-log plot. The degree of ground movement is plotted on a logarithmic scale against the assigned intensity level of the earthquake, which ranges linearly from 1-10 (Figure 1.29 (a)).

If a graph has both axes in a logarithmic scale, then it is referred to as a log-log plot. The relationship between the wavelength and frequency of electromagnetic radiation such as light is usually shown as a log-log plot (Figure 1.29 (b)). Log-log plots are also commonly used to describe exponential functions, such as radioactive decay.

Figure 1.29 (a) The Richter scale uses a log base 10 scale on its y-axis (microns of amplified maximum ground motion). (b) The relationship between the frequency and wavelength of electromagnetic radiation can be plotted as a straight line if a log-log plot is used.

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Energy can be transferred to or from a system, either through a temperature difference between it andanother system (i.e., by heat) or by exerting a force through a distance (work). In these ways, energy can be converted into otherforms of energy in other systems. For example, a car engine burns fuel for heat transfer into a gas. Work is done by the gas as it exerts a force through a distance by pushing a piston outward. This work converts the energy into a variety of other forms—intoan increase in the car’s kinetic or gravitational potential energy; into electrical energy to run the spark plugs, radio, and lights;and back into stored energy in the car’s battery. But most of the thermal energy transferred by heat from the fuel burning in theengine does not do work on the gas. Instead, much of this energy is released into the surroundings at lower temperature (i.e.,lost through heat), which is quite inefficient. Car engines are only about 25 to 30 percent efficient. This inefficiency leads toincreased fuel costs, so there is great interest in improving fuel efficiency. However, it is common knowledge that moderngasoline engines cannot be made much more efficient. The same is true about the conversion to electrical energy in large powerstations, whether they are coal, oil, natural gas, or nuclear powered. Why is this the case?

The answer lies in the nature of heat. Basic physical laws govern how heat transfer for doing work takes place and limit themaximum possible efficiency of the process. This chapter will explore these laws as well their applications to everyday machines.

This chapter will explore these laws as well their applications to everyday machines. These topics are part of thermodynamics—the study of heat and its relationship to doing work.

Zeroth Law of Thermodynamics: Thermal Equilibrium

When two objects (or systems) are in contact with one another, heat will transfer thermal energy from the object at higher temperature to the one at lower temperature until they both reach the same temperature. The objects are then in thermal equilibrium, and no further temperature changes will occur if they are isolated from other systems. The systems interact and change because their temperatures are different, and the changes stop once their temperatures are the same. Thermal equilibrium is established when two bodies are in thermal contact with each other—meaning heat transfer (i.e., the transfer of energy by heat) can occur between them. If two systems cannot freely exchange energy, they will not reach thermal equilibrium. (It is fortunate that empty space stands between Earth and the sun, because a state of thermal equilibrium with the sun would be too toasty for life on this planet!)

If two systems, A and B, are in thermal equilibrium with each another, and B is in thermal equilibrium with a third system, C, then A is also in thermal equilibrium with C. This statement may seem obvious, because all three have the same temperature, but it is basic to thermodynamics. It is called the zeroth law of thermodynamics.

The zeroth law of thermodynamics is very similar to the transitive property of equality in mathematics: If a = b and b = c, then a = c

Pressure, Volume, Temperature, and the Ideal Gas Law

Before covering the first law of thermodynamics, it is first important to understand the relationship between pressure, volume, and temperature. Pressure, P, is defined as

P = F / A

where F is a force applied to an area, A, that is perpendicular to the force. Depending on the area over which it is exerted, a given force can have a significantly different effect

The SI unit for pressure is the pascal, where Pressure is defined for all states of matter but is particularly important when discussing fluids (such as air).

You have probably heard the word pressure being used in relation to blood (high or low blood pressure) and in relation to the weather (high- and low-pressure weather systems). These are only two of many examples of pressures in fluids.

The relationship between the pressure, volume, and temperature for an ideal gas is given by the ideal gas law. A gas is considered ideal at low pressure and fairly high temperature, and forces between its component particles can be ignored.

The ideal gas law states that

PV = NkT

where P is the pressure of a gas, V is the volume it occupies, N is the number of particles (atoms or molecules) in the gas, , and T is its absolute temperature. The constant k is called the Boltzmann constant and has the value ( k = 1.38 X 10^-23 J/K)

It is important for us to notice from the equation that the following are true for a given mass of gas:

• When volume is constant, pressure is directly proportional to temperature.

 • When temperature is constant, pressure is inversely proportional to volume.

• When pressure is constant, volume is directly proportional to temperature.

This last point describes thermal expansion—the change in size or volume of a given mass with temperature. What is the underlying cause of thermal expansion? An increase in temperature means that there’s an increase in the kinetic energy of the individual atoms. Gases are especially affected by thermal expansion, although liquids expand to a lesser extent with similar increases in temperature, and even solids have minor expansions at higher temperatures. This is why railroad tracks and bridges have expansion joints that allow them to freely expand and contract with temperature changes.

To get some idea of how pressure, temperature, and volume of a gas are related to one another, consider what happens when you pump air into a deflated tire. The tire’s volume first increases in direct proportion to the amount of air injected, without much increase in the tire pressure. Once the tire has expanded to nearly its full size, the walls limit volume expansion. If you continue to pump air into tire (which now has a nearly constant volume), the pressure increases with increasing temperature)

(a) When air is pumped into a deflated tire, its volume first increases without much increase in pressure.

(b) When the tire is filled to a certain point, the tire walls resist further expansion, and the pressure increases as more air is added. (

c) Once the tire is inflated fully, its pressure increases with temperature.

The First Law of Thermodynamics :

 Heat (Q) and work (W) are the two ways to add or remove energy from a system. The processes are very different. Heat is driven by temperature differences, while work involves a force exerted through a distance. Nevertheless, heat and work can produce identical results. For example, both can cause a temperature increase. Heat transfers energy into a system, such as when the sun warms the air in a bicycle tire and increases the air’s temperature. Similarly, work can be done on the system, as when the bicyclist pumps air into the tire. Once the temperature increase has occurred, it is impossible to tell whether it was caused by heat or work. Heat and work are both energy in transit—neither is stored as such in a system. However, both can change the internal energy, U, of a system.

Internal energy is the sum of the kinetic and potential energies of a system’s atoms and molecules. It can be divided into many subcategories, such as thermal and chemical energy, and depends only on the state of a system (that is, P, V, and T), not on how the energy enters or leaves the system.

In order to understand the relationship between heat, work, and internal energy, we use the first law of thermodynamics. The first law of thermodynamics applies the conservation of energy principle to systems where heat and work are the methods of transferring energy into and out of the systems. It can also be used to describe how energy transferred by heat is converted and transferred again by work.

The first law of thermodynamics states that the change in internal energy of a closed system equals the net heat transfer into the system minus the net work done by the system. In equation form, the first law of thermodynamics is

ΔU = Q – W

Here, ΔU  is the change in internal energy, U, of the system. As shown in Figure 12.6, Q is the net heat transferred into the system—that is, Q is the sum of all heat transfers into and out of the system. W is the net work done by the system—that is, W is the sum of all work done on or by the system. By convention, if Q is positive, then there is a net heat transfer into the system; if W is positive, then there is net work done by the system. So positive Q adds energy to the system by heat, and positive W takes energy from the system by work. Note that if heat transfers more energy into the system than that which is done by work, the difference is stored as internal energy

The first law of thermodynamics is the conservation of energy principle stated for a system, where heat and work are the methods of transferring energy to and from a system. Q represents the net heat transfer—it is the sum of all transfers of energy by heat into and out of the system. Q is positive for net heat transfer into the system. W_out is the work done by the system, and W_in is the work done on the system. W is the total work done on or by the system. W is positive when more work is done by the system than on it. The change in the internal energy of the system, ΔU , is related to heat and work by the first law of thermodynamics: ΔU = Q – W

It follows also that negative Q indicates that energy is transferred away from the system by heat and so decreases the system’s internal energy, whereas negative W is work done on the system, which increases the internal energy.

Second Law of Thermodynamics: Entropy

It is not even theoretically possible for engines to be 100 percent efficient. This phenomenon is explained by the second law of thermodynamics, which relies on a concept known as entropy. Entropy is a measure of the disorder of a system. Entropy also describes how much energy is not available to do work. The more disordered a system and higher the entropy, the less of a system’s energy is available to do work.

Although all forms of energy can be used to do work, it is not possible to use the entire available energy for work. Consequently, not all energy transferred by heat can be converted into work, and some of it is lost in the form of waste heat—that is, heat that does not go toward doing work. The unavailability of energy is important in thermodynamics; in fact, the field originated from efforts to convert heat to work, as is done by engines.

The equation for the change in entropy, ΔS  , is ΔS = Q/T

where Q is the heat that transfers energy during a process, and T is the absolute temperature at which the process takes place. Q is positive for energy transferred into the system by heat and negative for energy transferred out of the system by heat. In SI, entropy is expressed in units of joules per kelvin (J/K). If temperature changes during the process, then it is usually a good approximation (for small changes in temperature) to take T to be the average temperature

The second law of thermodynamics states that the total entropy of a system either increases or remains constant in any spontaneous process; it never decreases.

An important implication of this law is that heat transfers energy spontaneously from higher- to lower-temperature objects, but never spontaneously in the reverse direction. This is because entropy increases for heat transfer of energy from hot to cold. Because the change in entropy is Q/T, there is a larger change in at lower temperatures (smaller T). The decrease in entropy of the hot (larger T) object is therefore less than the increase in entropy of the cold (smaller T) object, producing an overall increase in entropy for the system

The ice in this drink is slowly melting. Eventually, the components of the liquid will reach thermal equilibrium, as predicted by the second law of thermodynamics—that is, after heat transfers energy from the warmer liquid to the colder ice

Another way of thinking about this is that it is impossible for any process to have, as its sole result, heat transferring energy from a cooler to a hotter object. Heat cannot transfer energy spontaneously from colder to hotter, because the entropy of the overall system would decrease.

Suppose we mix equal masses of water that are originally at two different temperatures, say and . The result will be water at an intermediate temperature of 30⁰ C . Three outcomes have resulted: entropy has increased, some energy has become unavailable to do work, and the system has become less orderly.

Let us think about each of these results. First, why has entropy increased? Mixing the two bodies of water has the same effect as the heat transfer of energy from the higher-temperature substance to the lower-temperature substance. The mixing decreases the entropy of the hotter water but increases the entropy of the colder water by a greater amount, producing an overall increase in entropy.

Second, once the two masses of water are mixed, there is no more temperature difference left to drive energy transfer by heat and therefore to do work. The energy is still in the water, but it is now unavailable to do work.

Third, the mixture is less orderly, or to use another term, less structured. Rather than having two masses at different temperatures and with different distributions of molecular speeds, we now have a single mass with a broad distribution of molecular speeds, the average of which yields an intermediate temperature. These three results—entropy, unavailability of energy, and disorder—not only are related but are, in fact, essentially equivalent. Heat transfer of energy from hot to cold is related to the tendency in nature for systems to become disordered and for less energy to be available for use as work.

Based on this law, what cannot happen? A cold object in contact with a hot one never spontaneously transfers energy by heat to the hot object, getting colder while the hot object gets hotter. Nor does a hot, stationary automobile ever spontaneously cool off and start moving

We’ve explained that heat never transfers energy spontaneously from a colder to a hotter object. The key word here is spontaneously. If we do work on a system, it is possible to transfer energy by heat from a colder to hotter object. We’ll learn more about this in the next section, covering refrigerators as one of the applications of the laws of thermodynamics. Sometimes people misunderstand the second law of thermodynamics, thinking that based on this law, it is impossible for entropy to decrease at any particular location. But, it actually is possible for the entropy of one part of the universe to decrease, as long as the total change in entropy of the universe increases. In equation form, we can write this as

ΔS_tot = ΔS_system + ΔS_environment  > 0

Based on this equation, we see that ΔS_system can be negative as long as ΔS_environment    is positive and greater in magnitude.

How is it possible for the entropy of a system to decrease? Energy transfer is necessary.

• If you pick up marbles that are scattered about the room and put them into a cup, your work has decreased the entropy of that system.
• If you gather iron ore from the ground and convert it into steel and build a bridge, your work has decreased the entropy of that system. . In this case , although you made the system of the bridge and steel more structured, you did so at the expense of the universe. Altogether, the entropy of the universe is increased by the disorder created by digging up the ore and converting it to stee
• Energy coming from the sun can decrease the entropy of local systems on Earth—that is, is negative. But the overall entropy of the rest of the universe increases by a greater amount—that is, is positive and greater in magnitude

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Kinematics is defined as the study of motion without considering its causes. The word “kinematics” comes from a Greek term meaning motion and is related to other English words such as “cinema” (movies) and “kinesiology” (the study of human motion). In one-dimensional kinematics we will study only the motion of a ball, for example, without worrying about what forces cause or change its motion. Such considerations come in other chapters. In this chapter, we examine the simplest type of motion—namely, motion along a straight line, or one-dimensional motion.

Position

In order to describe the motion of an object, you must first be able to describe its position—where it is at any particular time. More precisely, you need to specify its position relative to a convenient reference frame. Earth is often used as a reference frame, and we often describe the position of an object as it relates to stationary objects in that reference frame. For example, a rocket launch would be described in terms of the positionof the rocket with respect to the Earth as a whole, while a professor’s positioncould be described in terms of where she is in relation to the nearby white board.  In other cases, we use reference frames that are not stationary but are in motionrelative to the Earth. To describe the position of a person in an airplane, for example, we use the airplane, not the Earth, as the reference frame.  

Displacement

If an object moves relative to a reference frame (for example, if a professor moves to the right relative to a white board or a passenger moves toward the rear of an airplane), then the object’s position changes. This change in position is known as displacement. The word “displacement” implies that an object has moved, or has been displaced.

Definition: 

DISPLACEMENT

Displacement is the change in position of an object:

Δx=xf−x0,

where Δx is displacement , x_f is the final position and x ₀ is the initial position

In this text the upper case Greek letter Δ always means “change in” whatever quantity follows it; thus, Δx means change in position

We can find displacement by subtracting initial position X₀ from the final position X_f

Note that the SI unit for displacement is the meter (m), but sometimes kilometres, miles, feet, and other units of length are used. Keep in mind that when units other than the meter are used in a problem, you may need to convert them into meters to complete the calculation.

A professor paces left and right while lecturing. Her position relative to Earth is given by x. The +2.0 m displacement of the professor relative to Earth is represented by an arrow pointing to the right.

A passenger moves from his seat to the back of the plane. His location relative to the airplane is given by x. The −4.0 m displacement of the passenger relative to the plane is represented by an arrow toward the rear of the plane. Notice that the arrow representing his displacement is twice as long as the arrow representing the displacement of the professor (he moves twice as far)  

Note that displacement has a direction as well as a magnitude. The professor’s displacement is 2.0 m to the right, and the airline passenger’s displacement is 4.0 m toward the rear. In one-dimensional motion, direction can be specified with a plus or minus sign. When you begin a problem, you should select which direction is positive (usually that will be to the right or up, but you are free to select positive as being any direction).

The professor’s initial position is X₀ =1.5 m and her final position is X_f =3.5 m . Thus her displacement is

Δx=Xf−X0=3.5 m−1.5 m=+2.0 m.

In this coordinate system, motion to the right is positive, whereas motion to the left is negative.

Similarly, the airplane passenger’s initial position is X0=6.0 m and his final position is Xf=2.0m and hence his displacement is

Δx=Xf−X0=2.0 m−6.0 m=−4.0 m.

His displacement is negative because his motion is toward the rear of the plane, or in the negative x direction in our coordinate system.

Distance

Although displacement is described in terms of direction, distance is not. 

Distance is defined to be the magnitude or size of displacement between two positions. Note that the distance between two positions is not the same as the 

distance travelled between them. 

Distance travelled is the total length of the path traveled between two positions. 

Distancehas no direction and, thus, no sign. For example, the distance

 the professor walks is 2.0 m. The distance the airplane passenger walks is 4.0 m.

MISCONCEPTION ALERT: 

DISTANCE TRAVELED VS. MAGNITUDE OF DISPLACEMENT

It is important to note that the distance traveled, however, can be greater than the magnitude of the displacement (by magnitude, we mean just the size of the displacement without regard to its direction; that is, just a number with a unit). For example, the professor could pace back and forth many times, perhaps walking a distance of 150 m during a lecture, yet still end up only 2.0 m to the right of her starting point. In this case her displacement would be +2.0 m, the magnitude of her displacement would be 2.0 m, but the distance she travelled would be 150 m.

In kinematics we nearly always deal with displacement and magnitude of displacement, and almost never with distance travelled. One way to think about this is to assume you marked the start of the motion and the end of the motion. The 

Displacement is simply the difference in the position of the two marks and is independent of the path taken in traveling between the two marks. The 

distance travelled, however, is the total length of the path taken between the two marks.

Exercise 

A cyclist rides 3 km west and then turns around and rides 2 km east. (a) What is her displacement? (b) What distance does she ride? (c) What is the magnitude of her displacement?

Answer

The rider’s displacement is .

a) Δx=Xf−X0 = 2 – 3 = -1  (The displacement is negative because we take east to be positive and west to be negative.)

 (b) The distance traveled is 3 km + 2 km = 5 km.

(c) The magnitude of the displacement is 1 km

Vectors, Scalars, and Coordinate Systems

What is the difference between distance and displacement?

Whereas displacement is defined by both direction and magnitude, distance is defined only by magnitude. Displacement is an example of a vector quantity. Distance is an example of a scalar quantity.

A vector is any quantity with both magnitude and direction. Other examples of vectors include a velocity of 90 km/h east and a force of 500 newtons straight down. The direction of a vector in one-dimensional motion is given simply by a plus (+) or minus (−) sign. Vectors are represented graphically by arrows. An arrow used to represent a vector has a length proportional to the vector’s magnitude (e.g., the larger the magnitude, the longer the length of the vector) and points in the same direction as the vector. Some physical quantities, like distance, either have no direction or none is specified.

A scalar is any quantity that has a magnitude, but no direction. For example, a 20ºC temperature, the 250 kilocalories (250 Calories) of energy in a candy bar, a 90 km/h speed limit, a person’s 1.8 m height, and a distance of 2.0 m are all scalars—quantities with no specified direction. Note, however, that a scalar can be negative, such as a −20ºC temperature. In this case, the minus sign indicates a point on a scale rather than a direction. Scalars are never represented by arrows.

Coordinate Systems for One-Dimensional Motion

In order to describe the direction of a vector quantity, you must designate a coordinate system within the reference frame. For one dimensional motion, this is a simple coordinate system consisting of a one-dimensional coordinate line. In general, when describing horizontal motion, motion to the right is usually considered positive, and motion to the left is considered negative. With vertical motion, motion up is usually positive and motion down is negative. In some cases, however, as with the jet in Figure , it can be more convenient to switch the positive and negative directions. For example, if you are analyzing the motion of falling objects, it can be useful to define downwards as the positive direction. If people in a race are running to the left, it is useful to define left as the positive direction. It does not matter as long as the system is clear and consistent. Once you assign a positive direction and start solving a problem, you cannot change it. Figure :

It is usually convenient to consider motion upward or to the right as positive (+) and motion downward or to the left as negative (−).

Time, Velocity, and Speed :

There is more to motion than distance and displacement. Questions such as, “How long does a foot race take?” and “What was the runner’s speed?” cannot be answered without an understanding of other concepts. In this section we add definitions of time, velocity, and speed to expand our description of motion.

Time

The most fundamental physical quantities are defined by how they are measured. This is the case with time. Every measurement of time involves measuring a change in some physical quantity. It may be a number on a digital clock, a heartbeat, or the position of the Sun in the sky. In physics, the definition of time is simple—time is change, or the interval over which change occurs. It is impossible to know that time has passed unless something changes. The amount of time or change is calibrated by comparison with a standard. The SI unit for time is the second, abbreviated s. We might, for example, observe that a certain pendulum makes one full swing every 0.75 s. We could then use the pendulum to measure time by counting its swings or, of course, by connecting the pendulum to a clock mechanism that registers time on a dial. This allows us to not only measure the amount of time, but also to determine a sequence of events.

How does time relate to motion? We are usually interested in elapsed time for a particular motion, such as how long it takes an airplane passenger to get from his seat to the back of the plane. To find elapsed time, we note the time at the beginning and end of the motion and subtract the two. For example, a lecture may start at 11:00 A.M. and end at 11:50 A.M., so that the elapsed time would be 50 min.

Elapsed time Δt is the difference between the ending time and beginning time,

Δt = t_f – t₀ , where

Where Δt is the change in time or elapsed time, is the time at the end of the motion, and t₀ is the time at the beginning of the motion.

Velocity

Your notion of velocity is probably the same as its scientific definition. You know that if you have a large displacement in a small amount of time you have a large velocity, and that velocity has units of distance divided by time, such as miles per hour or kilometres per hour.

Average velocity is displacement (change in position) divided by the time of travel,

ṽ =  Δx / Δt  = (X_f −X₀) /  (t_f −t₀)

where ṽ  is the average (indicated by the bar over the v) velocity, Δx is the change in position (or displacement), and X_f  and X₀ are the final and beginning positions at times t_f and t₀  respectively

Notice that this definition indicates that velocity is a vector because displacement is a vector. It has both magnitude and direction. The SI unit for velocity is meters per second or m/s, but many other units, such as km/h, mi/h (also written as mph), and cm/s, are in common use.

Suppose, for example, an airplane passenger took 5 seconds to move −4 m (the negative sign indicates that displacement is toward the back of the plane). His average velocity would be

ṽ =  Δx / Δt  = (-4m)/5s = -0.8 m/s

The average velocity of an object does not tell us anything about what happens to it between the starting point and ending point, however. For example, we cannot tell from average velocity whether the airplane passenger stops momentarily or backs up before he goes to the back of the plane. To get more details, we must consider smaller segments of the trip over smaller time intervals.

The smaller the time intervals considered in a motion, the more detailed the information. When we carry this process to its logical conclusion, we are left with an infinitesimally small interval. Over such an interval, the average velocity becomes the instantaneous velocity or the velocity at a specific instant. A car’s speedometer, for example, shows the magnitude (but not the direction) of the instantaneous velocityof the car. (Police give tickets based on instantaneous velocity, but when calculating how long it will take to get from one place to another on a road trip, you need to use average velocity.) 

Instantaneous velocity v is the average velocity at a specific instant in time (or over an infinitesimally small timeinterval).

Speed

In everyday language, most people use the terms “speed” and “velocity” interchangeably. In physics, however, they do not have the same meaning and they are distinct concepts. One major difference is that speed has no direction. Thus speed is a scalar . Just as we need to distinguish between instantaneous velocity and average velocity, we also need to distinguish between instantaneous speed and average speed

Instantaneous speed is the magnitude of instantaneous velocity . For example, suppose the airplane passenger at one instant had an instantaneous velocity  of −3.0 m/s (the minus meaning toward the rear of the plane). At that same time  his instantaneous speed was 3.0 m/s. Or suppose that at one time during a shopping trip your instantaneous velocity is 40 km/h due north. Your instantaneous speed at that instant would be 40 km/h—the same magnitude but without a direction. 

Average speed , however, is very different from average velocity.  Average speed  is the distance travelled  divided by elapsed time .

We have noted that distance travelled  can be greater than displacement. So average speedcan be greater than average velocity , which is displacement  divided by time . For example, if you drive to a store and return home in half an hour, and your car’s odometer shows the total distance travelled  was 6 km, then your average speed  was 12 km/h. Your average velocity , however, was zero, because your displacement  for the round trip is zero. (Displacement is change in position  and, thus, is zero for a round trip.) Thus average speed is not simply the magnitude of average velocity

Another way of visualizing the motion  of an object is to use a graph. A plot of position or of velocity as a function of time  can be very useful. For example, for this trip to the store, the position , velocity, and speed-vs.-time graphs are displayed (Note that these graphs depict a very simplified model of the trip. We are assuming that speed is constant during the trip, which is unrealistic given that we’ll probably stop at the store. But for simplicity’s sake, we will model it with no stops or changes in speed. We are also assuming that the route between the store and the house is a perfectly straight line.)

.

Position vs. time , velocity vs. time, and speed vs. time on a trip. Note that the velocity for the return trip is negative.

Acceleration :

In everyday conversation, to accelerate means to speed up. The accelerator in a car can in fact cause it to speed up. The greater the acceleration, the greater the change in velocity over a given time . The formal definition of 

Acceleration is consistent with these notions, but more inclusive.

AVERAGE ACCELERATION

Average Acceleration is the rate at which velocity changes,

ặ=Δv / Δt=(V_f−V₀) / (T_f−T₀)

where ặ  is average acceleration ,V is velocity, and T is time

Because acceleration is velocity in m/s divided by time  in s, the SI units for acceleration are m/s ²   (meters per second  squared)

Recall that velocity is a vector —it has both magnitude and direction. This means that a change in velocity can be a change in magnitude (or speed), but it can also be a change in direction. For example, if a car turns a corner at constant speed, it is accelerating because its direction is changing. The quicker you turn, the greater the acceleration. So there is an Acceleration when velocity changes either in magnitude (an increase or decrease in speed) or in direction, or both.

ACCELERATION AS A VECTOR

Acceleration is a  vector in the same direction as the change in velocity, Δv. Since velocity is a vector , it can change either in magnitude or in direction. Acceleration  is therefore a change in either speed or direction, or both. Keep in mind that although acceleration  is in the direction of the change in velocity, it is not always in the direction of motion

. If acceleration  is in a direction opposite to the direction of motion , the object slows down.

Problem :

A racehorse coming out of the gate accelerates from rest to a velocity of 15.0 m/s due west in 1.80 s. What is its average acceleration ?

First we draw a sketch and assign a coordinate system to the problem. This is a simple problem, but it always helps to visualize it. Notice that we assign east as positive and west as negative. Thus, in this case, we have negative velocity.

We can solve this problem by identifying Δv and Δt from the given information and then calculating the average acceleration directly from the equation

ặ=Δv / Δt=(V_f−V₀) / (T_f−T₀)

We know that

V₀=0, V_f =−15.0 m/s (the negative sign indicates direction toward the west), Δt=1.80 s

Since the horse is going from zero to −15.0 m/s, its change in velocity equals its final velocity: Δv=v_f=−15.0 m/s

Substituting the values , we  get

ặ  = (−15.0 m/s) / 1.80 s= −8.33 m/s²

Falling Objects

Falling objects form an interesting class of motion problems. For example, we can estimate the depth of a vertical mine shaft by dropping a rock into it and listening for the rock to hit the bottom. By applying the kinematics developed so far to falling objects, we can examine some interesting situations and learn much about gravity in the process.

Gravity

The most remarkable and unexpected fact about falling objects is that, if air resistance and friction are negligible, then in a given location all objects fall toward the center of Earth with the same constant acceleration, independent of their mass. This experimentally determined fact is unexpected, because we are so accustomed to the effects of air resistance and friction that we expect light objects to fall slower than heavy ones.

In the real world, air resistance can cause a lighter object to fall slower than a heavier object of the same size. A tennis ball will reach the ground after a hard baseball dropped at the same time . (It might be difficult to observe the difference if the height is not large.) Air resistance opposes the motion of an object through the air, while friction  between objects—such as between clothes and a laundry chute or between a stone and a pool into which it is dropped—also opposes motion between them. For the ideal situations of these first few chapters, an object falling without air resistance or friction  is defined to be in free-fall . The forceof gravity causes objects to fall toward the center of Earth. The acceleration of free-falling objects is therefore called the acceleration due to gravity . The acceleration due to gravity is constant, which means we can apply the kinematics equations to any falling object where air resistance and friction are negligible. This opens a broad class of interesting situations to us. The acceleration due to gravity is so important that its magnitude is given its own symbol, g. It is constant at any given location on Earth and has the average value

g=9.80 m/s²

The direction of the acceleration due to gravity  is downward (towards the center of Earth). In fact, its direction defines what we call vertical. Note that whether the acceleration a in the kinematic equations has the Value +g or −g depends on how we define our coordinate system . If we define the upward direction as positive, then a=−g=−9.80 m/s² , and if we define the downward direction as positive, then a=g=9.80 m/s²

KINEMATIC EQUATIONS FOR OBJECTS IN FREE FALL :

V=V₀−gt

y=y₀+v₀ t−1/2(gt²)

v²=(v₀)²−2g(y−y₀)

Example problem :

A person standing on the edge of a high cliff throws a rock straight up with an initial velocity of 13.0 m/s. The rock misses the edge of the cliff as it falls back to earth. Calculate the position  and velocity of the rock 1.00 s, 2.00 s, and 3.00 s after it is thrown, neglecting the effects of air resistance .

Solution :

Draw a sketch

We are asked to determine the position y at various times. It is reasonable to take the initial position y₀ to be zero. This problem involves one-dimensional Motion  in the vertical direction. We use plus and minus signs to indicate direction, with up being positive and down negative. Since up is positive, and the rock is thrown upward, the initial velocity must be positive too. The acceleration due to gravity  is downward, so a is negative. It is crucial that the initial velocity and the acceleration due to gravity  have opposite signs. Opposite signs indicate that the acceleration due to gravity  opposes the nitial motion and will slow and eventually reverse it.

Since we are asked for values of position  and velocity at three times, we will refer to these as y₁ and v₁ ; y₂ and v₂; and y ₃ and v₃

Solution for Position y₁ :

We know that y₀=0; v₀=13.0 m/s; a=−g=−9.80 m/s² and t=1.00 s.

We need to Identify the best equation to use. We will use 

y=y₀+v₀ t−1/2(gt²)

which is the value we want to find.

Substituting the known values and solve for y1

y1=0+(13.0 m/s)(1.00 s) – 1/ 2(−9.80 m/s2)(1.00 s)² =8.10 m

Solution for Velocity V₁ :

Given that 

y₀=0;  v₀=13.0 m/s;  a=−g=−9.80 m/s²

We also know from the solution above that y1=8.10 m

The best equation to use is v=v₀−gt

 Where (−g )= gravitational acceleration

Substituting the values , we get  

V₁=v₀ −gt=13.0 m/s−(9.80 m/s²)(1.00 s)=3.20 m/s

Solution for Remaining Times

The procedures for calculating the position and velocity att = 2.00 s and 3.00 s are the same as those above. The results are summarized in the Table below :

Time , t	Position , y	Velocity, v
1.00 s	8.10 m	3.20 m
2.00 s	6.40 m	−6.60 m
3.00 s	−5.10m	−16.4 m

Centripetal Acceleration :

We defined acceleration  as a change in velocity, either in its magnitude or in its direction, or both. When an object moves along a circular path, the direction of its velocity changes constantly, so there is always an associated acceleration, even if the speed of the object is constant. You experience this acceleration  yourself when you turn a corner in your car. What you notice is a sideways acceleration  because you and the car are changing direction. The sharper the curve and the greater your speed, the more noticeable this acceleration  will become. In this section we briefly examine the direction and magnitude of that acceleration.

The below image shows an object moving in a circular path at constant speed, called uniform circular motion . The direction of the instantaneous velocity  is shown at two points along the path. Acceleration  is in the direction of the change in velocity, which points directly toward the center of rotation (the center of the circular path). This pointing is shown with the vector diagram in the figure. We call the acceleration of an object moving in uniform circular motion  (resulting from a net external force) the centripetal acceleration (a_c ) – centripetal means “toward the center” or “center seeking.”

The directions of the velocity of an object at two different points are shown, and the change in velocity Δv is seen to point directly toward the center of curvature. (See small inset.) Because ac=Δv/Δt , the acceleration is also toward the center; ac is called centripetal acceleration. (For small time differences, Δθ is very small, and the arc length Δs is approximately equal to the chord length Δr)

The direction of centripetal acceleration is toward the center of curvature, but what is its magnitude? If we use the geometry shown in the above Figure along with some kinematics equations, we can obtain

ac=v² / r

which is the acceleration of an object in a circle of radius r at a speed v.

We see in the above Equation that centripetal acceleration is greater at high speeds and in sharp curves (smaller radius), as you have noticed when driving a car. But it is a bit surprising that ac is proportional to speed squared, implying, for example, that it is four times as hard to take a curve at 100 km/h than at 50 km/h. A sharp corner has a small radius, so that a_c is greater for tighter turns .

The unit of centripetal acceleration is m/s²

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Introduction

Physics is a fascinating branch of science that seeks to understand the fundamental principles governing the behaviour of the universe. One important aspect of physics is the study of physical quantities and their dimensions. In this lesson, we will explore the concept of dimensions, delve into dimensional analysis, and discover its practical application

Dimensions of Physical Quantities

1.1 Definition:
Physical quantities are measurable properties or characteristics of objects or phenomena, such as length, time, mass, and Velocity

Dimensions represent the nature of these quantities and provide a framework for their measurement
1.2 Fundamental Dimensions :

There are seven fundamental dimensions in the International System of Units (SI): length (L), mass (M), time (T), electric current (I), temperature (Θ), amount of substance (N), and luminous intensity (J)

All other physical quantities can be derived from these fundamental dimensions

Quantity	Unit	Symbol	Definition
Length	Meter	m	The meter is the distance traveled by light in a vacuum in 1/299,792,458th of a second.
Mass	Kilogram	kg	The kilogram is defined as the mass of the International Prototype of the Kilogram, a platinum-iridium cylinder kept at the International Bureau of Weights and Measures.
Time	Second	s	The second is the duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the cesium-133 atom.
Electric Current	Ampere	A	The ampere is the constant current that, if maintained in two parallel conductors of infinite length and negligible cross-sectional area, placed 1 meter apart in a vacuum, would produce a force between the conductors of exactly    2 x 10-7 newton per meter of length.
Temperature	Kelvin	K	The kelvin is the fraction 1/273.16 of the thermodynamic temperature of the triple point of water.
Amount of Substance	Mole	mol	The mole is the amount of a substance that contains as many elementary entities (such as atoms, molecules, ions, or particles) as there are atoms in exactly 0.012 kilograms of carbon-12.
Luminous Intensity	Candela	cd	The candela is the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency 540 x 1012 hertz and that has a radiant intensity in that direction of 1/683 watt per steradian.

These fundamental units provide a standardized and universally accepted system of measurement for various physical quantities, allowing for consistency and accuracy in scientific calculations and communications.

Advantages of SI Units

The SI (International System of Units) unit system offers several advantages:

1. Universality: The SI unit system is widely recognized and accepted globally. It provides a consistent and unified framework for measurements across different countries and scientific disciplines. This universality promotes international collaboration, communication, and understanding in various fields of study.
2. Standardization: SI units have well-defined definitions and are based on fundamental constants of nature. This standardization ensures accuracy, reliability, and reproducibility in measurements. It allows for precise comparisons and consistent results in scientific experiments and observations.
3. Coherence: SI units are designed to be coherent, meaning they are interrelated in a logical and consistent manner. The coherent system of units simplifies calculations and conversions between different quantities, as they are based on consistent mathematical relationships.
4. Scalability: The SI unit system is scalable, meaning it provides prefixes that can be used to express measurements across a wide range of magnitudes. This scalability makes it convenient to work with both extremely large and small values without the need for excessive zeros or complex conversions.
5. Compatibility with Scientific Laws: SI units are compatible with the laws of physics and other scientific principles. They align with the fundamental theories and principles that govern the behavior of the physical world, allowing for the formulation and testing of scientific theories, laws, and equations.
6. Practicality: The SI unit system is designed to be practical and user-friendly. The units are intuitive and easily understandable, facilitating effective communication and comprehension of measurements and quantities among scientists, engineers, and the general public.
7. International Recognition: The SI unit system is officially recognized by international organizations such as the International Bureau of Weights and Measures (BIPM), ensuring consistent usage and adoption in scientific research, education, industry, and trade worldwide.
   These qualities make it an indispensable tool for precise and accurate measurements in scientific and technological advancements.

1.3 Derived Dimensions

Derived dimensions are obtained by combining the fundamental dimensions using multiplication, division, or exponentiation
For example, velocity has dimensions of [L][T]⁻¹, where [L] represents length and [T]⁻¹ represents the inverse of time
Here is a table of 22 derived SI units along with the basic units involved in their derivation:

Derived SI Unit	Quantity	Definition	Basic Units Involved
Newton (N)	Force	1 N = 1 kg * m/s2	kilogram (kg), meter (m), second (s)
Pascal (Pa)	Pressure	1 Pa = 1 N/m2	Newton (N), meter (m)
Joule (J)	Energy	1 J = 1 N * m = 1 kg * m2/s2	Newton (N), meter (m), second (s)
Watt (W)	Power	1 W = 1 J/s = 1 kg * m2/s3	Joule (J), second (s)
Coulomb (C)	Electric charge	1 C = 1 A * s	Ampere (A), second (s)
Volt (V)	Electric potential	1 V = 1 W/A = 1 kg * m2/(A * s3)	Watt (W), Ampere (A), second (s)
Ohm (Ω)	Electrical resistance	1 Ω = 1 V/A = 1 kg * m2/(A2 * s3)	Volt (V), Ampere (A), second (s)
Farad (F)	Capacitance	1 F = 1 C/V = 1 A*s/V =            1 kg-1 * m-2 * s4 * A2	Coulomb (C), Volt (V), Ampere (A), second (s), kilogram (kg), meter (m)
Tesla (T)	Magnetic flux density	1 T = 1 Wb/m^2 = 1 kg/(A * s2)	Weber (Wb), meter (m), Ampere (A)
Henry (H)	Inductance	1 H = 1 V * s/A = 1 kg * m2/(A2 * s2)	Volt (V), second (s), Ampere (A)
Hertz (Hz)	Frequency	1 Hz = 1/s	1/second (1/s)
Siemens (S)	Electrical conductance	1 S = 1 A/V = 1 s3 * A2/(kg * m2)	Ampere (A), Volt (V), second (s), kilogram (kg), meter (m)
Weber (Wb)	Magnetic flux	1 Wb = 1 V * s = 1 kg * m2/(A * s2)	Volt (V), second (s), kilogram (kg), meter (m), Ampere (A)
Lux (lx)	Illuminance	1 lx = 1 lm/m2 = 1 cd * sr/m2	Lumen (lm), meter (m), candela (cd), steradian (sr)
Becquerel (Bq)	Radioactivity	1 Bq = 1/s	1/second (1/s)
Gray (Gy)	Absorbed dose	1 Gy = 1 J/kg	Joule (J), kilogram (kg)
Sievert (Sv)	Equivalent dose	1 Sv = 1 J/kg	Joule (J), kilogram (kg)
Weber per square meter	Magnetic field strength	1 Wb/m2 = 1 T	Weber (Wb), meter (m)
Radian (rad)	Plane angle	1 rad = 1 m/m	Meter (m)
Steradian (sr)	Solid angle	1 sr = 1 m2/m2	Meter (m)
Celsius (°C)	Temperature	°C = (K – 273.15)	Kelvin (K)

Prefixes and Multiples of SI Units

Since the magnitude of the SI units vary over a wide range , multiples and sub multiples are used to explain the units more precisely

Here is a table of the standard prefixes for SI units, including both multiples and submultiples

Prefix	Symbol	Multiplication Factor
Yotta	Y	1024
Zetta	Z	1021
Exa	E	1018
Peta	P	1015
Tera	T	1012
Giga	G	109
Mega	M	106
Kilo	K	103
Hecto	H	102
Deca	da	101
–	–	100 (Base Unit)
Deci	D	10-1
Centi	C	10-2
Milli	M	10-3
Micro	µ	10-6
Nano	N	10-9
Pico	P	10-12
Femto	F	10-15
Atto	A	10-18
Zepto	z	10-21
Yocto	y	10-24

Additional Practical Units of the SI system

Here are some practical units of the SI system

Parsec: The parsec (pc) is a unit of length used in astronomy to measure vast distances to stars and galaxies. It is approximately equal to 3.09 x 10¹⁶ meters.

Light Year: The light-year (ly) is another unit of astronomical distance. It represents the distance light travels in one year, and it is about 9.46 x 10¹⁵ meters.

Astronomical Unit: The astronomical unit (AU) is a unit of length used in astronomy to describe distances within our solar system. It is approximately the mean distance from the Earth to the Sun and is about 1.5 x 10¹¹ meters.

Micron: The micron (μm) is a unit of length equal to one millionth of a meter. It is often used to measure small distances, such as the width of a human hair or microscopic objects.

Angstrom: The angstrom (Å) is a unit of length used to measure atomic and molecular scales. It is equal to 0.1 nanometers or 10^-10 meters.

Fermi: The fermi (fm) is a unit of length used in nuclear and particle physics to measure atomic and subatomic scales. It is equal to 10^-15 meters.

X-ray Unit: The X-ray unit (XU) is a unit used to measure the intensity of X-rays. It is a non-SI unit and is defined as the X-ray exposure that produces 2.58 x 10^-4 coulombs per kilogram of air.

Atomic Mass Unit (amu): The atomic mass unit (amu) is a unit used to express the mass of atoms and molecules on a scale relative to the mass of a carbon-12 atom. 1 amu is approximately 1.66 x 10^-27 kilograms.

These units are used in various scientific and practical contexts, especially in astronomy, physics, and other fields dealing with very large or very small distances and masses.

Methods of Measurement of Physical Quantities

The direct method and indirect method are two approaches used to measure physical quantities. Here’s a brief explanation of each method:

Direct Method: In the direct method of measurement, the physical quantity of interest is measured directly using appropriate instruments or techniques. The measurement is made by comparing the quantity being measured to a known standard or using a calibrated instrument. This method provides a straightforward and accurate measurement of the desired quantity. For example, measuring the length of an object using a ruler or measuring the temperature using a thermometer are examples of direct measurements.

Indirect Method: In the indirect method of measurement, the physical quantity of interest is derived or calculated by measuring other related quantities and using mathematical relationships or equations. This method involves measuring multiple quantities and utilizing mathematical models or formulas to determine the desired quantity indirectly. The indirect method is used when direct measurement of the quantity is not feasible or when it is more practical to measure other related quantities. For example, determining the velocity of an object by measuring its displacement and time, or calculating the area of an irregular shape by measuring its dimensions and applying appropriate formulas are examples of indirect measurements.

Both direct and indirect methods have their advantages and limitations. The choice of method depends on the nature of the physical quantity being measured, the available instruments or techniques, and the accuracy required for the measurement. In many cases, a combination of direct and indirect methods may be employed to obtain accurate and reliable measurements of complex physical quantities

Let us consider the measurement of the Length of an object . If the object is measureable by placing a scale , then its called the Direct Method of measurement . However this method is not feasible if we measure the height of a hill , for which we need to adapt indirect methods

Various methods of measurement of Length

Echo Method of Length Measurement :

The echo method of length measurement is a technique used to determine the distance to an object or surface by measuring the time it takes for a sound wave or pulse to travel to the object and back. This method is commonly employed in various applications, such as sonar systems, ultrasound imaging, and distance measurement devices.

Here’s how the echo method works:

A sound wave or pulse is generated and emitted towards the target object or surface.

The sound wave travels through the medium (such as air or water) until it reaches the object.

Upon reaching the object, the sound wave reflects or echoes back towards the source.

A receiver detects the returning sound wave or echo.

The time elapsed between the emission of the sound wave and the reception of the echo is measured.

The distance to the object is then calculated using the known speed of sound in the medium and the measured time.

Since the speed of sound in a given medium is relatively constant, knowing the time it takes for the sound wave to travel allows for the calculation of the distance based on the equation:

Distance = (Speed of Sound × Time) / 2

The division by 2 is necessary because the measured time accounts for the round trip of the sound wave.

The echo method can be used to measure distances accurately over a wide range, from short distances in medical ultrasound imaging to large distances in sonar systems for navigation or oceanographic research. It is a non-contact measurement method that is often preferred when direct physical contact with the object being measured is not possible or practical.

RADAR method of Length measurement :

The radar method of length measurement is a technique that utilizes radar (radio detection and ranging) to determine the distance to an object or surface. It is widely used in applications such as navigation, weather monitoring, and remote sensing.

Here’s how the radar method works:

A radar system emits short pulses of electromagnetic waves, typically radio waves or microwaves.

These waves travel at the speed of light and are directed towards the target object or surface.

Upon encountering the object, a portion of the electromagnetic waves is reflected back towards the radar system.

The radar system’s receiver detects the reflected waves, commonly referred to as radar echoes.

The time delay between the emission of the pulse and the reception of the echo is measured.

The distance to the object is then calculated using the known speed of light and the measured time delay.

The speed of light is approximately 299,792,458 meters per second (in a vacuum). By multiplying the speed of light by the measured time delay and dividing it by 2 (since the time accounts for the round trip), the radar system can accurately determine the distance to the target.

Radar systems can provide precise and reliable distance measurements over various ranges, from short distances to long distances. Additionally, radar can also provide additional information about the target, such as its velocity or size, by analyzing the Doppler shift or the strength of the radar echoes.

The radar method is widely used in applications such as air traffic control, maritime navigation, meteorology, and even in automotive systems like radar-based distance sensors for collision avoidance. It allows for non-contact measurement and provides valuable information in a wide range of scenarios.

SONAR method of Length measurement :

The sonar method of length measurement is a technique that uses sound waves to measure the distance between an object or surface and the source of the sound. Sonar (Sound Navigation And Ranging) is commonly used underwater to determine distances, map the ocean floor, and detect underwater objects. Here’s an overview of how the sonar method works:

Sound Wave Generation: A sonar system emits sound waves, usually in the form of short pulses, into the water or another medium. The sound waves can be produced by specialized sonar transducers

Sound Wave Propagation: The emitted sound waves travel through the medium, such as water, until they encounter an object or surface. The sound waves propagate through the medium at a known speed, which is typically slower than the speed of light.

Reflection or Echo: When the sound waves encounter an object or surface, a portion of the sound energy is reflected back towards the sonar system. This reflection is referred to as an echo.

Detection: The sonar system’s receiver detects the returning echoes. The receiver is designed to capture and analyze the reflected sound waves.

Time Measurement: The time delay between the emission of the sound wave and the reception of the echo is measured precisely. This time delay is often referred to as the “echo time” or “round-trip time.”

Distance Calculation: The distance between the sonar system and the object or surface is then calculated using the known speed of sound in the medium and the measured echo time. The distance is determined by multiplying half of the echo time by the speed of sound:

Distance = (Speed of Sound × Echo Time) / 2

Sonar is widely used in various applications, including underwater navigation, bathymetry (measurement of water depth), fish finding, underwater imaging, and locating underwater structures or objects. It allows for non-contact measurement over long distances underwater, providing valuable information about the environment beneath the surface

LASER method of Length measurement :

The laser method of length measurement utilizes laser technology to accurately measure distances with a high degree of precision. This method is widely employed in various fields, including engineering, surveying, manufacturing, and scientific research. Here’s an overview of how the laser method works:

Laser Emission: A laser emits a highly focused beam of light, which travels in a straight line.

Target Reflection: The laser beam is directed towards the target object or surface. When the laser beam encounters the target, a portion of the light is reflected back towards the laser source.

Detection: The reflected laser light is detected using a sensor or detector. The detector captures the returning light and converts it into an electrical signal.

Time Measurement: The time taken for the laser light to travel to the target and back is precisely measured using electronic timing equipment. This time measurement is usually accomplished using high-frequency electronic circuits or time-of-flight measurement techniques.

Speed of Light: The speed of light in air or a vacuum is a well-known constant (approximately 299,792,458 meters per second). By multiplying the speed of light by the measured time and dividing it by 2 (since the time accounts for the round trip), the distance to the target can be calculated.

Distance = (Speed of Light × Time) / 2

Data Processing: Multiple measurements are often taken to improve accuracy and account for any variations or errors. Advanced data processing techniques, including statistical analysis and error correction algorithms, may be employed to enhance the precision of the measurements.

The laser method of length measurement offers numerous advantages, including non-contact measurement, high accuracy, and rapid data acquisition. It is commonly used for tasks such as distance ranging, 3D scanning, alignment, surface profiling, and dimensional inspection. Additionally, the laser method can be applied in both short-range and long-range measurement scenarios, depending on the specific laser technology and setup used.

So we see that multiple indirect measurement methods are available for all practical purposes.

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