Library

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

“Access for free at openstax.org.”

Take Quiz

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

Take Quiz

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.

Take Quiz

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:

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

Take Quiz