GRAVITATION

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 (ra) and perihelion (rp) distances, then you can calculate the semi-major axis (a) and semi-minor axis (b).

a = (ra + rp)/2

b = √rarp

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, PEe.

W = PEe = 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,

Etot=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  = GmME / 2r

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

E = K + U

   = GmME / 2r  – GmME / r

   = – GmME / 2r

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

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

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Take Quiz

1) Why do some objects fall faster than others near the surface of the earth if all mass is attracted equally by the force of gravity?
a) Some objects fall faster because of air resistance, which acts in the direction of the motion of the object and exerts more force on objects with less surface area.
b) Some objects fall faster because of air resistance, which acts in the direction opposite the motion of the object and exerts more force on objects with less surface area.
c) Some objects fall faster because of air resistance, which acts in the direction of motion of the object and exerts more force on objects with more surface area.
d) Some objects fall faster because of air resistance, which acts in the direction opposite the motion of the object and exerts more force on objects with more surface area.

ANSWER

b) Some objects fall faster because of air resistance, which acts in the direction opposite the motion of the object and exerts more force on objects with less surface area.

2) Would you have the same mass on the moon as you do on Earth? Would you have the same weight?
a) You would weigh more on the moon than on Earth because gravity on the moon is stronger than gravity on Earth.
b) You would weigh less on the moon than on Earth because gravity on the moon is weaker than gravity on Earth.
c) You would weigh less on the moon than on Earth because gravity on the moon is stronger than gravity on Earth.
d) You would weigh more on the moon than on Earth because gravity on the moon is weaker than gravity on Earth.

ANSWER

b) You would weigh less on the moon than on Earth because gravity on the moon is weaker than gravity on Earth.

3) Calculate Earth’s mass given that the acceleration due to gravity at the North Pole is 9.83 m/s2 and the radius of the Earth is 6371 Km from pole to center.
a) 5.94 X 1017 Kg
b) 5.94 X 1024 Kg
c) 9.36 X 1017 Kg
d) 9.36 X 1024 Kg

ANSWER

b) 5.94 X 1024 Kg

4) Comets have very elongated elliptical orbits with the sun at one focus. Using Kepler’s Law, explain why a comet travels much faster near the sun than it does at the other end of the orbit.
a) Because the satellite sweeps out equal areas in equal times
b) Because the satellite sweeps out unequal areas in equal times
c) Because the satellite is at the other focus of the ellipse
d) Because the square of the period of the satellite is proportional to the cube of its average distance from the sun

ANSWER

a) Because the satellite sweeps out equal areas in equal times

5) Titan, with a radius of 2.58 X 106 m , is the largest moon of the planet Saturn. If the mass of Titan is 1.35 X 1023 Kg , what is the acceleration due to gravity on the surface of this moon?
a) 1.35 m/s2
b) 3.49 m/s2
c) 3.49 X 106 m/s2
d) 1.35 X 106 m/s2

ANSWER

a) 1.35 m/s2

6) A moon orbits a planet in an elliptical orbit. The foci of the ellipse are 50, 000 km apart. The closest approach of the moon to the planet is 400, 000 km. What is the length of the major axis of the orbit?
a) 400, 000 km
b) 450, 000, km
c) 800, 000 km
d) 850, 000 km

ANSWER

d) 850, 000 km

7) The closest Earth comes to the sun is 1.47×108 km, and Earth’s farthest distance from the sun is 1.52×108 km. What is the area inside Earth’s orbit?
a) 2.23×1016 km2
b) 6.79×1016 km2
c) 7.02×1016 km2
d) 7.26×1016 km2

ANSWER

c) 7.02×1016 km2

8) An artificial satellite orbits the Earth at a distance of 1.45×l04 km from Earth’s center. The moon orbits the Earth at a distance of 3.84×105 km once every 27.3 days. How long does it take the satellite to orbit the Earth?
a) 0.200 days
b) 3.07 days
c) 243 days
d) 3721 days

ANSWER

a) 0.200 days

9) Earth is 1.496×108 km from the sun, and Venus is 1.08×108 km from the sun. One day on Venus is 243 Earth days long. What best represents the number of Venusian days in a Venusian year?
a) 0.78 days
b) 0.92 days
c) 1.08 days
d) 1.21 days

ANSWER

b) 0.92 days

10) What is the gravitational force between two 60 Kg people sitting 100m apart ?
a) 2.4 X 10 – 11 N
b) 2.4 X 10 – 9 N
c) 3.6 X 10-1 N
d) 3.6 X 101 N

ANSWER

b) 2.4 X 10 – 9 N

11) The masses of Earth and the moon are 5.97×1024 kg and 7.35×l022 kg, respectively. The distance from Earth to the moon is 3.80×105 km. At what point between the Earth and the moon are the opposing gravitational forces equal? (Use subscripts e and m to represent Earth and moon.)

a) 3.42×105 km from the center of Earth
b) 3.80×105 km from the center of Earth
c) 3.42×106 km from the center of Earth
d) 3.10×107 km from the center of Earth

ANSWER

b) 3.80×105 km from the center of Earth

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Morphology of flowering plants – Part II

Structure of Flower

A complete flower typically has four verticils or whorls. They are as follows –

  1. Calyx – The members that make up this whorl are called sepals. Sepals are green leaf-like structures that protect a flower in its early bud stage. Also, it is the outermost whorl of a flower.
  2. Corolla – The members that make up this whorl are called petals. These structures are brightly coloured and aid in pollination by attracting insects. Also, they can be seen in different shapes like tubular, wheel-shaped, funnel-shaped or bell-shaped. The corolla and calyx together make up a non-reproductive structure called the perianth.
  3. Androecium – They are the male reproductive structures composed of stamens. Each stamen consists of a filament or stalk and an anther. Each anther has pollen sacs that aid in the production of pollen grains.
  4. Gynoecium – They are female reproductive structures composed of carpels. A carpel is composed of 3 parts namely ovary, style and stigma

Androecium: 

It is a male reproductive part. It consists of stamens. Each stamen is made up of filament and anthers.

  • The stamens may either be free (polyandrous) or remain fused together. Epipetalous- stamens are attached to petals.

The cohesion of stamens fused conditions are as follows.

  • Monadelphous: Stamens are united into a single group by union of filaments. The anthers remain free.

 (e.g. Hibiscus)

  • Diadelphous: Stamens are united, and arranged in two bundle s.  (e.g. Pisum)

  • Polyadelphous: Stamens are united and grouped onto several bundles. (e.g. Citrus)

  • Synandrous: Stamens are united their whole length. Both filaments and anthers will be fused. (e.g. Cucurbita)

  • Syngenesious: Stamens are united by the joining of anthers. The filaments remain free. (e.g. Tridax)

The stamens can also be found fused with other whorls. This condition is called adhesion. It may be of different types.

1. Epipetalous: Stamens are found attached (wholly or partially) to the corolla, by means of their filaments. Anthers would be free. (e.g. Ixora)

2. Epiphyllous: When stamens are found attached to the perianth of a flower, it is called epiphyllous condition. (e.g. Gloriosa)

3. Gyandrous: Stamens are attached to the carpel (either by whole length, or by the anthers.  (e.g. Calotropis)

Gynoecium :

The gynoecium of a flower is the female reproductive structure that is most essential for sexual reproduction. The gynoecium has ovaries that produce ovules which eventually develop into seeds and fruit.  Carpel is the main component of gynoecium that consists of ovary, style and stigma.

  1. Stigma is a sticky portion at the distal end of a tubular structure called style. It is the primary receptive site where the pollen grains are captured.
  2. Style is the major structure that aids in the process of fertilization. The pollen tube reaches the ovary via this tubular style.
  3. Ovary is located at the swollen base of the carpel. Style is the elongated tube that links the ovary and the stigma. The ovary produces ovules which are present within chambers known as locules. Based on the number of locules, the ovary can also be classified as multilocular, unilocular, bilocular, etc.
  4. pistil : Stigma, ovary and style are together termed as the pistil. A pistil is considered to be a collection of one or more carpels.

Placentation:

It is defined as the arrangement and attachment of ovules within the plant’s ovary.

The part of the ovary where ovules are attached is the placenta.

Types of placentation:

  • Axile placentation: When the placenta is placed axially as in a lemon, then such type of placentation is termed axile placentation.

  • Marginal placentation: When the placenta forms a ridge along the ventral suture as in pea, then this is known as marginal placentation.

  • Free central placentation: When the ovules are born on the central axis, it is called free central as in primrose and Dianthus.

  • Basal placentation: In this type of placentation, a single ovule is attached to the placenta and placed at the base of the ovary. Eg. Marigold.

Parietal placentation: In this type of placentation, the ovules develop on the inner wall or periphery of the ovary. Eg. Sunflower, kiwi, cucumber, cantaloupe.

Superficial: The placenta is formed all around the inner surface of the ovary including the septa.

Nymphaea (waterlily).

  1. Marginal- e.g. pea
  2. Axile- e.g. lemon, china rose
  3. Parietal- e.g. Argemone, mustard
  4. Free central- e.g. Primrose, Dianthus
  5. Basal- e.g. marigold, sunflower

The Flower

A flower has four whorls; calyx, corolla, androecium and gynoecium. These are attached to the swollen terminal of pedicel called the thalamus

Flower symmetry:

  1. Actinomorphic- radially symmetrical flowers, e.g. chilli, datura, mustard
    1. Zygomorphic- when a flower can be divided into two equal parts in only one vertical plane, e.g. Cassia, pea, etc.

Flowers can be trimerous, tetramerous or pentamerous depending on the multiple of floral appendages present 3, 4 or 5

Types of flowers depending on the presence or absence of bracts (reduced leaf present at the base of pedicel); Bracteate or Ebracteate

Types of flowers based on the position of the ovary:

  1. Hypogynous- gynoecium occupies the highest place, above all the other parts. The ovary is known as superior. The sepals, petals and stamens arise from below the ovary.This is the commonest type of ovary. The flower is technically Hypogynous where hypo means flower parts start ‘below’ the ovary. e.g. brinjal, china rose, mustard

  • Perigynous- gynoecium is present at the same level as the rest of the parts of a flower. The ovary is known as half inferior, The flower parts arise from the rim of a hypanthium – a cup shaped extension of the base which encloses the ovary. Imagine the hypanthium to be a flat ring and you can see that the flower parts actually arise from below the ovary. The flower is technically Perigynous where peri means ‘around’.  e.g. peach, plum, rose

  • Epigynous- thalamus encloses the ovary completely and other parts are present above it. The ovary is known as inferior.  Inferior ovary. The sepals, corolla tube and stamens (filaments plus anthers) arise from above the ovary. The flower is technically Epigynous where epi means ‘above’. e.g. ray florets of a sunflower, guava, cucumber

Aestivation:

The mode of arrangement of sepals or petals in floral bud with respect to the other members of the same whorl is known as aestivation.

The main types of aestivations are valvate, twisted, imbricate and vexillary.

Valvate:

When sepals or petals in a whorl just touch one another at the margin, without overlapping, as in Calotropis, it is said to be valvate.

Twisted:

If one margin of the appendage overlaps that of the next one and so on as in china rose, lady’s finger and cotton, it is called twisted.

Imbricate:

If the margins of sepals or petals overlap one another but not in any particular direction as in Cassia and gulmohur, the aestivation is called imbricate.

Vexillary:

In pea and bean flowers, there are five petals, the largest (standard) overlaps the two lateral petals (wings) which in turn overlap the two smallest anterior petals (keel); this type of aestivation is known as vexillary or papilionaceous.

Families of flowering plants:

Main Difference between Fabaceae, Solanaceae, and Liliaceae

  1. Fabaceae is a legume family, Solanaceae is a potato family, and Liliaceae is the lily family.
  2. Fabaceae contains peas, chickpeas, and soybean, Solanaceae contain potatoes, tomatoes, and bell pepper, while Liliaceae contain onion, lily, and tulip.
  3. Fabaceae is perennials or annuals, Solanaceae is annuals or biennials, while Liliaceae is perennials.
  4. Fabaceae is monocarpellary; Solanaceae is bicarpellary, while Liliaceae is tricarpellary.
  5. Fabaceae contain pods, while Solanaceae and Liliaceae contain berries or capsules.

Refer Flower formula library notes for more information

Fabaceae Solanaceae Liliaceae
It is a large family of trees, shrubs, vines, and herbs. It is a nightshade or potato family of flowering plants These are monocotyledonous, perennial, and herbaceous plants.
Legume family Potato family Lily family
Dicot Dicot Monocot
Peas, chickpeas, and soybean Potato, tomato, and bell pepper. Onion, lily, and tulip
Exhibits marginal placentation. Exhibits axial placentation. Exhibits axial placentation.
Perennial or annuals Perennials, biennials, and annuals. Perennials
Trees, shrubs, vines, and herbs Herbs, shrubs, small trees Herbs and shrubs
Monocarpellary Bicarpellary Tricarpellary
Contain pods Contain berry or capsule Contain berry or capsule
Fix atmospheric nitrogen to improve soil fertility Contain plant varieties with agricultural importance Contain decorative and ornamental plants.

Take Quiz

1. Diadelphous stamens are found in:
a. China rose
b. Citrus
c. Pea
d. China rose and citrus

ANSWER

Diadelphous stamens are found in pea (Fabaceae).

Diadelphous is a condition when stamens are united partially and are present in two brunches, like in Peas. In this condition, filaments of nine separate stamens are fused together forming one unit, and the tenth posterior stamen remains as a stand portion out of the bundle.

2. The sepals, corolla tube and stamens (filaments plus anthers) arise from above the ovary. Based on the position of the ovary it is
a. Hypogyneous
b. Epigyneous
c. Perigyneous

ANSWER

3. The given figure represents a type of aestivation.

The type of aestivation shown in the given figure is known as
a) Valvate
b) Twisted
c) Imbricate
d) Vexillary

ANSWER

4. Axile placentation occurs in
a) Brassicaceae and Solanaceae
b) Asteraceae and Fabaceae
c) Solanaceae and Liliaceae
d) Fabaceae and Liliaceae

ANSWER

5. Gynoecium of Solanaceae is
a) Bicarpellary apocarpous
b) monocarpellary
c) polycarpellary syncarpus
d) bicarpellary syncarpus

ANSWER

In Solanaceae, the gynoecium is bicarpellary, syncarpous, ovary superior, obliquely placed, bilocular, becomes multilocular due to the formation of the false septum, axile placentation is seen.

6. What is the corolla aestivation of the papilionaceous flower which has a large vexillum covering two wings and its keel covered by another wing?
a. Twisted
b. Ascending imbricate
c. Valvate
d. Descending imbricate

ANSWER

Explanation: The corolla of the papilionaceae family has the shape of a butterfly. The butterfly shape is due to the varying size of the petals. The corolla has 5 petals. The vexillum is the large petal on the posterior-most side. The keel which is covered by the wings of the vexillum is made of the innermost fused petals. The large petals followed by small petals kind of overlapping is seen in the descending order in terms of size. Thus, this arrangement is known as descending imbricate.

7. Flowers are Zygomorphic in (NEET 2011 QP)
a) mustard
b) gulmohur
c) tomato
d) Datura

ANSWER

Flowers of gulmohur have bilateral symmetry. So, they are called zygomorphic. Datura, mustard and tomato have actinomorphic flowers.

8. Epipetalous and syngenesious stamens occur in (NEET 1991 QP)
a) solanaceae
b) Brassicaceae
c) Brassicaceae
d) Asteraceae

ANSWER

Syngenesious condition is found in asteraceae. It is the condition when stamens are united by their anthers (filaments free). Epipetalous condition is also seen here.

9. Which of the following flowers exhibit radial symmetry?
a. Pisum
b. Brassica
c. Cassia
d. Trifolium

ANSWER

Explanation. Out of the given options, Brassica is the only flower which exhibits distinct radial symmetry. All the petals of the flower are identical in size and shape. The other type of symmetry is the bilateral symmetry which shows symmetry in only one plane. The radial symmetry in plants is a pollination strategy.

10) Family Fabaceae differs from Solanaceae and Liliaceae. With respect to the stamens, pick out the characteristics specific to family Fabaceae but not found in Solanaceae or Liliaceae. NEET 2023 QP

a) Polyadelphous and epipetalous stamens
b) Monadelphous and Monothecous anthers
c) Epiphyllous and Dithecous anthers
d) Diadelphous and Dithecous anthers

ANSWER

In the family Fabaceae, the stamens are typically arranged in two groups (diadelphous) and each stamen has two distinct chambers (dithecous) in the anther. This arrangement is not present in Solanaceae or Liliaceae.

11. NEET 2023 QP
Axile placentation is observed in

a) China rose, Beans and Lupin
b) Tomato, Dianthus and Pea
c) China rose, Petunia and Lemon
d) Mustard, Cucumber and Primrose

ANSWER

Axile placentation refers to the arrangement of the placenta at the central axis of the ovary. This pattern can be observed in plants like China rose, Petunia, Lemon, and tomato.

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Morphology of flowering plants – Part I

The Root

  • Tap root: Primary root is formed by elongation of the radicle and it bears secondary and tertiary roots, present in dicot plants, e.g. gram, mustard, etc.
  • Fibrous root: found in monocotyledons. A large number of roots originate at the base of the stem, e.g. rice, wheat, etc.
  • Adventitious root: Primary root is not formed from the radicle, e.g. grass, banyan tree, maize, etc.

Modification of the Root

  • For storage: taproots- carrot, turnip; adventitious roots- sweet potato
  • For support: Prop roots of Banyan tree, that arise from branches; stilt root of maize and sugarcane, that comes out of lower stem nodes
  • For aeration: pneumatophores present in mangroves help them in respiration as it grows in swampy areas. These roots grow upwards above the ground, e.g. Rhizophora
  • For nitrogen fixation: root nodules of leguminous plants

The Stem

  • Plumule develops into stem
  • The part of the stem which bears leaves is called a node and the part between two nodes is known as the internode

Modification of the Stem

  1. Underground Stem: They help plants sustain unfavourable conditions for growth
    1. Rhizome- runs parallel to the ground and has nodes, internodes, buds, e.g. ginger, banana
    1. Tubers- the end part gets swollen as in potato
    1. Corm- it grows vertically below the ground, e.g. colocasia, etc.
    1. Bulb- stem is reduced and surrounded by scaly leaves, e.g. garlic, onions
  2. Stem Tendrils: these are a coiled structure that supports tender stem of the plant and help in climbing, e.g. grapes, cucumber, pumpkin
  3. Thorn: axillary bud gets modified into pointed thorns and protects plants from grazing animals, e.g. Bougainvillea, citrus
  4. Subaerial Weak Stem
    1. Offsets- internode of lateral branches decreases resulting in the rosette of leaves, e.g. Eichhornia, Pistia
    1. Suckers- lateral branches arise from the underground portion of the stem, e.g. chrysanthemum, banana, pineapple
    1. Runners- stem run horizontally above the ground and roots arise at nodes, e.g. grasses, strawberry
    1. Stolon- lateral branches arise normally but then bend down and touch the soil where the root grows and the new daughter plant arises, e.g. mint
  5. Aerial modification- The stem is completely metamorphosed for various adaptations, e.g. Phylloclade of xerophytic plants. The stem becomes fleshy and green having photosynthetic pigments to prepare food as leaves are reduced to thorns to check water loss by transpiration, e.g. Euphorbia, Opuntia

The Leaf

Leaves originate from the apical meristem of a shoot

Normally a leaf consists of three parts; leaf base, lamina and petiole

Leaf base attaches to stem and may have two small leaf-like structures known as stipule

Types of venation:

  1. Reticulate- present in dicotyledons, there is a network of veins present, which are irregularly distributed
  2. Parallel- present in monocotyledons, veins are parallel to each other

Types of Leaves:

  1. Simple- lamina is complete and incision doesn’t reach midrib
  2. Compound- incision touches midrib, that divides a leaf into a number of leaflets
    1. Pinnately compound- the leaflets are present on the common axis, i.e. midrib, called the rachis, e.g. Neem
    1. Palmately compound- leaflets are attached at the petiole tip, e.g. silk cotton

Phyllotaxy: pattern of arrangement of leaves around the stem

  1. Alternate phyllotaxy: In this phyllotaxy, a single leaf arises from the node of a branch. Examples are: Sunflower, mustard, and peepal, Hibiscus, Brassica
  2. Opposite phyllotaxy: Opposite phyllotaxy is when two leaves arise from the node of a branch and extend out in opposite directions. Examples are: guava and jamun tree, calotrophis
  3. Whorled phyllotaxy: Whorled phyllotaxy is when 3 or more leaves arise from the node of a branch. An example is alstonia.

Types of Phyllotaxy

Modification of Leaves

  • Tendrils- leaves modified to form a long thread-like structure, it gives support to climbers, e.g. peas

  • Spine- in xerophytic plants to reduce water loss, e.g. cactus, aloe

  • For storage- e.g. garlic, onion
  • Phyllodes- petiole gets modified to form a leaf-like structure and function, e.g. Acacia

  • Pitcher in pitcher plant is a modified leaf which traps insects inside

Inflorescence

  • The arrangement of the flowers around the floral axis
  • The two main types of inflorescence are

1. Racemose: the main axis grows indefinitely, flowers are present laterally in acropetal succession i.e. older flowers at the bottom and the younger ones at the top.

2. Cymose: the main axis terminates in flower and has limited growth. Flowers are borne in a basipetal order i.e. older flowers are at the top and new flowers are at the bottom.

Special types of Inflorescence

3. Verticillaster: sessile flowers arranged in dichasial cyme, e.g. Ocimum, Salvia

4. Cyathium: involucre of bracts form cup shape structure, single female flower is surrounded by numerous male flowers, e.g. Euphorbia

5. Hypenthodium: both male and female flowers are present in a cavity with an apical opening called ostiole

The Flower

  • A flower has four whorls; calyx, corolla, androecium and gynoecium. These are attached to the swollen terminal of pedicel called the thalamus
  • Flower symmetry:

1. Actinomorphic– radially symmetrical flowers, e.g. chilli, datura, mustard

2. Zygomorphic– when a flower can be divided into two equal parts in only one vertical plane, e.g. Cassia, pea, etc.
Flowers can be trimerous, tetramerous or pentamerous depending on the multiple of floral appendages present 3, 4 or 5
Types of flowers depending on the presence or absence of bracts (reduced leaf present at the base of pedicel); Bracteate or Ebracteate

Types of flowers based on the position of the ovary:

  1. Hypogynous- gynoecium occupies the highest place, above all the other parts. The ovary is known as superior, e.g. brinjal, china rose, mustard.
  2. Perigynous- gynoecium is present at the same level as the rest of the parts of a flower. The ovary is known as half inferior, e.g. peach, plum, rose.
  3. Epigynous- thalamus encloses the ovary completely and other parts are present above it. The ovary is known as inferior, e.g. ray florets of a sunflower, guava, cucumber

Parts of a Flower

Calyx: A flower’s outermost whorl is made up of leaf-like structures called sepals

Gamosepalous- sepals united

Polysepalous- sepals free

Corolla: made up of bright coloured petals. Present after sepals

Gamopetalous- petals united

Polypetalous- petals free

[See more about flowers in Morphology of flowering plants part II library notes]

The Fruit

It is a matured and ripened ovary after fertilisation.

  • Parthenocarpic fruit: the fruit formed without fertilisation, it makes seedless fruits, e.g. pineapple
  • When the ovary wall ripens, it changes into a pericarp. Seed and pericarp make a fruit. The fleshy pericarp is made up of three layers; epicarp, mesocarp and endocarp
    • Outermost Layer: Epicarp is the outermost layer of the fruits.
    • Middle Layer: Mesocarp is the middle layer of the fruits.
    • Innermost Layer: Endocarp is the innermost layer of the fruits.

Simple Fruits
Simple Fruits are fruits that are developed from the monocarpellary ovary or multicarpellary syncarpous ovary. Thus, only one fruit is formed in the ovary. The fruit can either be fleshy or dry.

Fleshy Fruits:

  • In fleshy fruits, the fruit wall is differentiated into three layers:
    • Epicarp: The outermost layer.
    • Mesocarp: The middle layer.
    • Endocarp: The innermost layer.
  • Examples of fleshy fruits include:
    • Berry (e.g., grapes)
    • Pepo (e.g., pumpkin)
    • Hesperidium (e.g., orange)
    • Drupe (e.g., plum)

Drupe: is a fleshy fruit that develops from either one or several fused carpels and contains one or many seeds. The seeds are enclosed by the hard protective endocarp (pericarp) of the fruit, e.g., mango. In mango the pericarp is well differentiated into an outer thin epicarp, a middle fleshy edible mesocarp and an inner stony hard endocarp.

Eg. Apricot, Mango, plum

Pome (e.g., apple)
Apple and pear are pome. Pome is a false fruit in which the edible part is thalamus where the true fruit remains embedded.

Dry Fruits

  • The pericarp (fruit wall) of simple dry fruits is usually quite dry and hard.
  • The pericarp (fruit wall) of simple dry fruits is usually quite dry and hard.
  • It is not differentiated into the three layers (epicarp, mesocarp, and endocarp).
  • Some dry fruits dehisce (split open) to release seeds, while others remain indehiscent.
  • Examples of dry fruits include:
    • Dehiscent Fruits: Pericarp breaks down, and seeds are scattered (e.g., pea pods).
    • Schizocarpic Fruits: Pericarp is arranged into one or more seeded segments (e.g., carrot).
    • Indehiscent Fruits: Pericarp remains intact even after maturing (e.g., sunflower seeds).

Aggregate Fruits

When the fruits develop from the multi carpellary apocarpous ovary, they become fruitlets as each carpel is separated from the other. Different types of aggregate fruits include etaerio of follicles, etaerio of achenes, etaerio of berries, etaerio of drupes, etc. 

  • These fruits develop from a single flower with numerous pistils.
  • Each pistil contributes to the aggregate fruit.
  • Examples include:
    • Raspberry: Formed from multiple tiny drupelets.
    • Blackberry: Similar to raspberry but with a different arrangement of drupelets.

Composite Fruits

Composite Fruits consist of many ovaries and other floral parts which combine to form the fruit. It is a false fruit. They are of two types, namely, sorosis and sycosis.

  • Pineapple is a sorosis type of fruit that develops from spike or spadix inflorescence. Here the flowers frise by their succulent tepals and axis bearing the flowers becomes fleshy or woody, thus forming a compact mass. These are composite or multiple fruits. In pineapple fleshy axis, bracts, fused perianth and pericarp are edible.

More on Fruits:

A fruit is a ripened ovary. On the basis of the formation of fruits, they are classified into two types – true fruits and false fruits. True fruits are developed from the ovary only.

Banana is a fleshy fruit – berry. It develops from multicarpellary syncarpous superior or inferior ovary. The pericarp of berries is differentiated into epicarp, mesocarp and endocarp (like drupes) but the endocarp is not stony as is drupes.

Pome is a false fruit in which the edible part is thalamus where the true fruit remains embedded. Fig is a composite fruit. These fruits are the products of the whole inflorescence together with its component parts. Apple and pear are pome.

  • In litchi, aril forms the edible part in fruit. It is a collar like out growth from the base of the ovule forming a kind of third integument. Aril is also found in Asphodelus, Trianthema and Ulmus. Litchi is a nut. In litchi, the epicarp and mesocarp (layers of pericarp) together become leathery and the endocarp is membranous.
  • The juicy hair-like structures in citrus fruits like lemon, commonly known as “pulp hairs,” develop from the endocarp, which is the innermost layer of the pericarp and surrounds the seeds.

The Seed

After the fertilisation, the ovule develops into a seed.

A seed has a seed coat and an embryo. An embryo is made up of radicle, embryonal axis and one or two cotyledons in monocotyledons (maize, wheat) and dicotyledons (pea, gram) respectively.

Structure of monocot seed:

The monocot seeds are mostly endospermic. The seed coat is thin and fuses with pericarp. It contains one cotyledon called scutellum and shoot axis. The lower part of the axis is the radicle, covered by a sheath called coleorhiza. The upper part of an axis is called plumule, covered by a sheath called coleoptile.

During the process of the development of the embryo, the food stored up in the endosperm is continuously drawn up by the developing embryo and thus completely exhausted. Such seeds are known as exalbuminous or non-endospermic. The common examples are : exalbuminous – gram, pea, bean, tamarind, orchid, etc.

Difference between dicot and monocot seed:

Dicotyledonous seed Monocotyledonous seed
The seed coat is made up of two layers; outer testa and inner tegmen There is a scar present through which the seed was attached to the fruit while developing, it is called the hilum The micropyle is a pore above the hilum The seed coat is fused to the fruit wall and membranous
The embryo consists of an embryonal axis and two cotyledons The embryo consists of one large cotyledon known as scutellum
Cotyledons store food The endosperm is bulky and stores food
Mostly non-endospermic seeds. In castor, an endospermic seed is present- endosperm is formed due to double fertilisation and stores food Mostly endospermic seeds. Some seeds like orchids are non-endospermic. Aleuron (a proteinaceous layer), is the outermost layer of endosperm, which separates the embryo
Radicle and plumule are present at the two ends of the embryonal axis Plumule is enclosed in coleoptile and radicle is enclosed in coleorhiza

Take Quiz

1. Which of the following stem gets modified into the flat and green organ to perform the functions of leaves?
a. Phylloclade
b. Cladodes
c. Scales
d. Phyllodes

ANSWER

a. Phylloclade

Explanation: Phylloclade is the flattened stem that performs the function of a leaf. It is the modified stem. Here, the leaves are modified into spines. Phylloclade becomes thick and fleshy succulent to perform the photosynthesis process. The primary function is to protect the plant from external injury.

2. The edible part of a coconut is known as
a. Pericarp
b. Perisperm
c. Endosperm
d. Cotyledon

ANSWER

Explanation: The edible part of the coconut is the seed. Seed or endosperm is white in mature coconut. This is cellular and opaque. Also, in green coconut water, the endosperm is nuclear. Mesoderm is the fibrous part, and endoderm is the hard or stony layer of the coconut.

3. Which of the following fruit is developed from condensed inflorescence?
a. Etaerio
b. Aggregate fruit
c. Composite fruit
d. Simple fruit

ANSWER

Explanation: Composite fruit is the type of fruit developed from the condensed inflorescence. Both etaerio and aggregate fruits are formed from a single flower’s polycarpellary apocarpous gynoecium. Also, simple fruit is developed from the single ovary of a single flower. Sorosis type of composite fruit is developed from spikes, spades, or catkins inflorescences. Sycosis type of composite fruit is developed from hypanthium inflorescence.

4. Gynoecium with fused carpels is known as-
a. Syncarpous
b. Syngenecium
c. Apocarpous
d. All of the above

ANSWER

Explanation: Gynoecium with fused carpels is known as syncarpous. The term is used for the flower with reference to the fused carpels or pistils in the gynoecium of that flower.

5. Identify the correct set of statements:
(a) The leaflets are modified into pointed hard thorns in Citrus and Bougainvillea
(b) Axillary buds form slender and spirally coiled tendrils in cucumber and pumpkin
(c) Stem is flattened and fleshy on Opuntia and modified to perform the function of leaves
(d) Rhizophora shows vertically upward growing roots that help to get oxygen for respiration
(e) Subaerially growing stems in grasses and strawberry help in vegetative propagation

NEET (UG) – 2022 QP
a. (b) and (c) only
b. (a) and (d) only
c. b), (c), (d) and (e) only
d. (a), (b), (d) and (e) only

ANSWER

6. Which part of the fruit, labelled in the given figure makes it a false fruit?

NEET (UG) – 2022 QP
a) A – Mesocarp
b) B-Endocarp
c) C-Thalamus
d) D-Seed

ANSWER

Solution and Explanation
Fruits are formed from the ovary of a flower after fertilisation. Such fruits are called true fruits. However, in a few species such as apple, strawberry, cashew, etc., some other parts of the flower, such as the thalamus also contribute to fruit formation. Such fruits are called false fruits. The diagram shown here is of an apple. The parts labeled as A is mesocarp, B is endocarp, C is thalamus, and D is seed.

7. The roots that originate from the base of the stem are: (NEET 2020 QP)
a. Fibrous roots
b. Primary roots
c. Prop roots
d. Lateral roots

ANSWER

In most of the monocots, the roots that originate from the base of the stem are called fibrous roots.

8. Inflorescence is racemose in (NEET 2018 QP)

a. Tulip
b. Soyabeen
c. Aloe
d. Brinjal

ANSWER

Racemose inflorescence is also called indefinite and indeterminate type. Growth of the peduncle is indefinite. Here the terminal bud will not modify into a flower. Flowers develop in acropetal succession i.e., mature flowers are towards the base and the younger ones towards the tip of the peduncle. Flowers open in centripetal succession i.e., opening of flowers proceeds from the periphery to the centre of the inflorescence. Peduncle may be unbranched or branched. Soybean belongs to family Fabaceae which has racemose inflorescence.

9. Which one of the following plants shows vexillary aestivation and diadelphous stamens? (NEET 2022 QP)

a. Colchicum autumnale
b. Pisum sativum
c. Allium cepa
d. Solanum nigrum

ANSWER

Pisum, which is also known as pea, exhibits vexillary aestivation and has stamens that are diadelphous. On the other hand, Colchicum autumnale, Solanum nigrum, and Allium cepa exhibit valvate aestivation.

10. Long filamentous threads protruding at the end of a young cob of maize are:
a. anthers
b. Style
c. Ovaries
d. hair

ANSWER

In maize, the styles are the long, thread-like structures that extend from the top of each kernel or ovule. They are part of the female reproductive structure and play a role in pollen capture and fertilization.

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ROTATIONAL MOTION

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 ac 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 ac 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

ac  = 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 ac 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 ac in terms of the magnitude of angular velocity. Substituting v = rω into the equation above, we get

ac = (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 Fnet = ma. For uniform circular motion, the acceleration is centripetal acceleration: a = ac.

Therefore, the magnitude of centripetal force, Fc, is Fc = mac

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

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 Fc. Centripetal force is perpendicular to tangential velocity and causes uniform circular motion. The larger the centripetal force Fc, 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 Fc 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

1) What is the angle of rotation (in degrees) between two hands of a clock, if the radius of the clock is 0.70 m and the arc length separating the two hands is 1.0 m ?

a. 400
b. 800
c. 810
d. 1630

ANSWER

c. 810

2) What is the centripetal force exerted on a 1,600 kg car that rounds a 100 m radius curve at 12 m/s?

a. 192 N
b. 1, 111 N
c. 2, 300 N
d. 13, 333 N

ANSWER

d. 13, 333 N

3) For a given object, what happens to the arc length as the angle of rotation increases?

a. The arc length is directly proportional to the angle of rotation, so it increases with the angle of rotation.
b. The arc length is inversely proportional to the angle of rotation, so it decreases with the angle of rotation.
c. The arc length is directly proportional to the angle of rotation, so it decreases with the angle of rotation.
d. The arc length is inversely proportional to the angle of rotation, so it increases with the angle of rotation

ANSWER

a. The arc length is directly proportional to the angle of rotation, so it increases with the angle of rotation.

4) Which of these quantities is constant in uniform circular motion?

a. Speed
b. Velocity
c. Acceleration
d. Displacement

ANSWER

a. Speed

5) Which of these quantities impact centripetal force?

a. Mass and speed only
b. Mass and radius only
c. Speed and radius only
d. Mass, speed, and radius all impact centripetal force

ANSWER

d. Mass, speed, and radius all impact centripetal force

6) An increase in the magnitude of which of these quantities causes a reduction in centripetal force?

a. Mass
b. Radius of curvature
c. Speed

ANSWER

b. Radius of curvature

7) What happens to centripetal acceleration as the radius of curvature decreases and the speed is constant, and why?

a. It increases, because the centripetal acceleration is inversely proportional to the radius of the curvature.
b. It increases, because the centripetal acceleration is directly proportional to the radius of curvature.
c. It decreases, because the centripetal acceleration is inversely proportional to the radius of the curvature.
d. It decreases, because the centripetal acceleration is directly proportional to the radius of the curvature.

ANSWER

a. It increases, because the centripetal acceleration is inversely proportional to the radius of the curvature.

Read more

LAWS OF MOTION

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

Fnet = m a or ∑ F = m a

where Fnet  (or ∑F) is the net external force, m is the mass of the system, and a is the acceleration. Note that Fnet 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 = Fnet / m or a = ∑F / m

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

a α Fnet

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 = Fnet / 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

Fnet = 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 Fnet = 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

Fnet = 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

Fnet = ΔP / Δt

where Fnet 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 = Fnet Δt

Fnet 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.Fnet = ma  is actually derived from the equation:

Fnet = 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

Fnet = ΔP / Δt

      = m Δv / Δt for a constant mass

Since Δv / Δt = a , we get

Fnet = 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         Fnet = 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

Fnet  = 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 Fnet 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

1. An object is at rest. Two forces, X and Y, are acting on it. Force X has a magnitude of x and acts in the downward direction. What is the magnitude and direction of Y?

a. The magnitude is x and points in the upward direction.
b. The magnitude is 2x and points in the upward direction.
c. The magnitude is x and points in the downward direction.
d. The magnitude is 2x and points in the downward direction

ANSWER

a. The magnitude is x and points in the upward direction.

2) Three forces, A, B, and C, are acting on the same object with magnitudes a, b, and c, respectively. Force A acts to the right, force B acts to the left, and force C acts downward. What is a necessary condition for the object to move straight down?

a. The magnitude of force A must be greater than the magnitude of force B, so a > b
b. The magnitude of force A must be equal to the magnitude of force B, so a = b
c. The magnitude of force A must be greater than the magnitude of force C, so A > C.
d. The magnitude of force C must be greater than the magnitude of forces A or B, so A < C > B

ANSWER

d. The magnitude of force C must be greater than the magnitude of forces A or B, so A < C > B

3. Two people push a cart on a horizontal surface by applying forces F1 and F2 in the same direction. Is the magnitude of the net force acting on the cart, Fnet , equal to, greater than, or less than F1 + F2 ? Why?

a. Fnet < F1 + F2 because the net force will not include the frictional force.
b. Fnet = F1 + F2 because the net force will not include the frictional force
c. Fnet  < F1 + F2 because the net force will include the component of frictional force
d. Fnet = F1 + F2 because the net force will include the frictional force

ANSWER

b. Fnet = F1 + F2 because the net force will not include the frictional force

4) True or False: A book placed on a balance scale is balanced by a standard 1-kg iron weight placed on the opposite side of the balance. If these objects are taken to the moon and a similar exercise is performed, the balance is still level because gravity is uniform on the moon’s surface as it is on Earth’s surface.

a. True
b. False

ANSWER

b. False

5) A car weighs 2,000 kg. It moves along a road by applying a force on the road with a parallel component of 560 N. There are two passengers in the car, each weighing 55 kg. If the magnitude of the force of friction experienced by the car is 45 N, what is the acceleration of the car?

a. 0.244 m/s2
b. 0.265 m/s2
c. 4.00 m/s2
d. 4.10 m/s2

ANSWER

a. 0.244 m/s2

6. A person pushes an object of mass 5.0 kg along the floor by applying a force. If the object experiences a friction force of 10 N and accelerates at 18 m/s2 , what is the magnitude of the force exerted by the person?

a. −90 N
b. −80 N
c. 90 N
d. 100 N

ANSWER

d. 100 N

7. A 2,000-kg car is sitting at rest in a parking lot. A bike and rider with a total mass of 60 kg are traveling along a road at 10 km/h. Which system has more inertia? Why?

a. The car has more inertia, as its mass is greater than the mass of the bike.
b. The bike has more inertia, as its mass is greater than the mass of the car.
c. The car has more inertia, as its mass is less than the mass of the bike.
d. The bike has more inertia, as its mass is less than the mass of the car

ANSWER

a. The car has more inertia, as its mass is greater than the mass of the bike.

8. Two people push a 2,000-kg car to get it started. An acceleration of at least 5.0 m/s2 is required to start the car. Assuming both people apply the same magnitude force, how much force will each need to apply if friction between the car and the road is 300 N?

a. 4850 N
b. 5150 N
c. 97000 N
d. 10300 N

ANSWER

b. 5150 N

9. A 55-kg lady stands on a bathroom scale inside an elevator. The scale reads 70 kg. What do you know about the motion of the elevator?

a. The elevator must be accelerating upward.
b. The elevator must be accelerating downward.
c. The elevator must be moving upward with a constant velocity.
d. The elevator must be moving downward with a constant velocity.

ANSWER

a. The elevator must be accelerating upward.

10. Are rockets more efficient in Earth’s atmosphere or in outer space? Why?

a. Rockets are more efficient in Earth’s atmosphere than in outer space because the air in Earth’s atmosphere helps to provide thrust for the rocket, and Earth has more air friction than outer space.
b. Rockets are more efficient in Earth’s atmosphere than in outer space because the air in Earth’s atmosphere helps to provide thrust to the rocket, and Earth has less air friction than the outer space.
c. Rockets are more efficient in outer space than in Earth’s atmosphere because the air in Earth’s atmosphere does not provide thrust but does create more air friction than in outer space.
d. Rockets are more efficient in outer space than in Earth’s atmosphere because the air in Earth’s atmosphere does not provide thrust but does create less air friction than in outer space.

ANSWER

c. Rockets are more efficient in outer space than in Earth’s atmosphere because the air in Earth’s atmosphere does not provide thrust but does create more air friction than in outer space.

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Morphology of flowering plants

Floral Formula

Definition of Floral Formula

A floral formula can defined as the numeric and symbolic expression, which reveals the flower morphological characteristics by employing different symbols, letters, and figures. It is the conventional method accustomed to formulating the structure of the flower. It elucidates the information about the number of whorls and a relative relationship between each other.

Components

The floral formula uses discrete letters, signs and figures to represent the specific feature of the flower.

Letters used in Floral Formula

1.K: This letter denotes the sepals that form an outermost whorl called the calyx.

2. C: This letter represents the group of petals that constitute the second whorl called the corolla.

3. P: It is used to denote the tepals, which indicates the undifferentiated condition of the perianth members (sepals and petals).

4. A: It specifies the male reproduction part or stamens, which includes the filament and anther that all together makes up the third floral whorl known as androecium.

5. G: It denotes the female reproductive part, i.e. carpel, which includes the stigma, style and ovary that colloquially forms the innermost whorl called gynoecium.

6. Br: It represents the bracteate condition of the flower.

A plant having bracts is referred to as bracteate or bracteolate, while one that lacks them is referred to as ebracteate and ebracteolate, without bracts.
bracteate flowers: Flowers with bracts (a reduced leaf at the base of the pedicel) are called bracteate flowers. Bracts are small leaf-like structures found at a flower’s base. China rose, tulip, lily

7. Ebr: It indicates the ebracteate condition, in which a flower lacks bract.

8. Brl: It indicates the presence of bracteoles or represents the bracteolate condition.

9. Epik: It represents the presence of a secondary whorl surrounding the calyx called epicalyx.

10. Ebrl: It is used to indicate the absence of bracteoles or to represent the ebracteolate condition.

Symbols used in Floral Formula

  1. 0: It indicates the absence of a particular member in flower.
  2. : It is generally used when the number of specific flower parts is more than 10.
  3. : It indicates an actinomorphic condition.
  4. %: It denotes a zygomorphic condition.

  • : It represents the bisexuality of flowers.
  • : It represents the unisexual, staminate flower.
  • ♀️: This represents the unisexual, pistillate flower.

Combination of Letters, Symbols and Numbers

  • K5: It shows the aposepalous condition, in which the five sepals are free.

  • K(5): It shows the gamosepalous condition, in which the five sepals are united.

  • C5: It represents the apopetalous condition or the presence of the five free petals.

  • C(5): It represents the gamopetalous condition or the presence of five fused petals.

  • Cx: It indicates corolla cruciform.
  • curve drawn over the letters P and A: It represents the epiphyllous stamens.
  • curve drawn over the letters C and A: It shows the epipetalous stamens.
  • A3: It indicates the presence of three free stamens.
  • A2+2: It indicates the presence of  4 free stamens, two in each whorl.
  • A(9)+1: It represents the presence of diadelphous stamens (10 in number), in which nine stamens are fused in one whorl, and one stamen remains free.

  • A0: Sterile stamen (staminode).
  • G0: Sterile carpel (pistillode).
  • G-: It represents the semi inferior ovary.
  • line over the letter G: It represents the inferior position of an ovary.
  • The line below the letter G: It represents the superior position of an ovary.
  • curve over the letters G and A: It shows the gynostagium condition.

Floral Formula:

Symbol Description / Full form
Br Bracteate condition
Ebr Ebracteate condition
Brl Bracteolate
Epik Epicalyx
Ebrl Ebracteolate
0 (zero) Absence of a particular whorl
Indefinite number of floral parts in a whorl
Actinomorphic condition (radially symmetric) flowers.
% Zygomorphic condition (bilaterally symmetric)
K Calyx (Sepals)
C Corolla (petals)
P Perianth (tepals)
A Androecium (stamens)
G Gynoecium (carpels)
Bisexual flower
Unisexual, staminate flower
♀️ Unisexual, pistillate flower
K5 Five sepals, aposepalous
K(5) Five sepals gamosepalous
C5 Five petals, apopetalous
C(5) Five petals, gamopetalous
Epiphyllous stamens
Epipetalous stamens
A3 Three stamens free
A2+2 Stamens 4, 2 whorls
A10+1 Stamens 10, diadelphous – 9 stamens unite to form one bundle and 1 other bundle
A0 Sterile stamen (staminode)
G0 Sterile carpel (pistillode)
G- Semi inferior ovary
This image has an empty alt attribute; its file name is 3.png Inferior ovary
This image has an empty alt attribute; its file name is 4.png Superior ovary

How to Use Floral Formula?

Firstly, the presence or absence of bract and bracteoles in the flowers of different families is observed and noted by using particular letters.

Then the symmetry of the flower is written according to the position of floral whorls relative to the mother axis. Flowers can be polysymmetric (actinomorphic), dissymmetric, monosymmetric (zygomorphic), asymmetric or spirally arranged. Therefore, the flowers showing different symmetry is indicated by specific symbols in the floral formula.

After that, the sex of the flower is written, i.e. whether a plant is hermaphrodite (bisexual) or unisexual. Bisexual flowers possess the male plus female reproductive parts, while the unisexual flowers either contain stamens or carpels.

Then the number of floral members starting with calyx, corolla, androecium to gynoecium is written. Here each member is denoted by using specific letters like ‘K’ represents calyx, ‘C’ represents corolla, ‘A’ indicates androecium and ‘G’ represents gynoecium. The number of each whorl is written after the following letters starting from K, C, A to G.

The number used after the letter is placed inside bracket ( ), If the particular whorl is united. But, if the floral members remain free, then the bracket is not applied. Adnation between the floral whorls is represented by drawing a curve from the top of the letters.

The position of the ovary can be shown by drawing a horizontal line above, below or in front of the letter ‘G’, which represents the inferior, superior or half inferior position of the ovary.

Examples

The floral formula can be used to describe the flowers of the particular family or the different species of flower. Let us take few examples of the floral formula to study the floral characteristics.

Floral Formula of Fabaceae Family

(%) shows that the symmetry of the flower is monosymmetric or zygomorphic. (⚥) indicates that the flower is perfect or hermaphrodite (includes stamen plus pistil). A letter ‘K’ indicates the outermost whorl, i.e. calyx and (5) shows that the number of sepals is five that are united with each other.

‘C’ indicates the second floral whorl, i.e. corolla or petals and the number after this, i.e. 1+2+(2), represents vexillary aestivation. ‘A’ means the male reproductive part or androecium, and the number after it (9)+1 shows the presence of diadelphous stamens (includes 10 stamens).

In 10 stamens, 9 filaments unite to constitute one bundle, and the remaining one filament makes up another one. The line below the letter ‘G’ indicates the superior position of an ovary, and digit 1 shows the monocarpellary ovary.

Floral Formula of Liliaceae Family

Through this floral formula, we can identify many floral characteristics. This formula indicates that the flower is bracteate, bisexual or hermaphrodite. A letter ‘P’ indicates that the sepals and petals are undifferentiated, i.e. there are three tepals in every two whorls.

Letter ‘A’ denotes androecium, and ‘3+3’ suggests there are three free stamens in every two whorls. A line below a letter ‘G’ specifies the hypogynous condition of an ovary, and the number ‘(3)’ represents the presence of three fused carpels.

Floral Formula of Solanacae Family

The general floral formula of Solanaceae family is as follows:

Here the symbols represent:

Actinomorphic (radial symmetry)
Bisexual
K(5) Calyx – 5 sepals, gamosepalous (united)
C(5) Corolla – 5 petals, gamopetalous
A5 Androecium – 5 stamens, polyandrous (free), epipetalous (attached to petals)
G(2) Gynoecium – bicarpellary, syncarpous (united), superior ovary

  • Calyx (K): Composed of 5 sepals, which are united (gamosepalous).
  • Corolla ©: Consists of 5 petals, which are also united (gamopetalous).
  • Androecium (A): Contains 5 stamens, which are free (polyandrous) and attached to the petals (epipetalous).
  • Gynoecium (G): Bicarpellary (two carpels), syncarpous (united), and has a superior ovary.

The Solanaceae family includes a variety of plants, such as potatoes, tomatoes, bell peppers, eggplants, and tobacco. These plants exhibit diverse characteristics and are widely distributed across tropical, subtropical, and temperate zones.

Fabaceae floral formula

Take Quiz

1. NEET 2021 QP
Match Column – I with Column – II

Select the correct answer from the options given below.

  1. (a)-(iv), (b)-(ii), (c)-(i), (d)-(iii)
  2. (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)
  3. (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
  4. (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)

ANSWER

b) (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)

Fabaceae: The flowers in Fabaceae exhibit zygomorphic symmetry and are bisexual. The calyx is composed of five fused sepals, while the corolla is papilionaceous, consisting of one standard petal, two wing petals, and two fused keel petals. The stamens are diadelphous, and the gynoecium is monocarpellary with a superior ovary.

Solanaceae: Solanaceae flowers are actinomorphic and bisexual. The calyx features five fused sepals, and the corolla consists of five fused petals. There are five epipetalous stamens, and the gynoecium is bicarpellary and apocarpous with a superior ovary.
Liliaceae: Liliaceae flowers are actinomorphic and bisexual. The perianth comprises six fused tepals, arranged in two rows. There are six epipetalous stamens, and the gynoecium is tricarpellary and syncarpous with a superior ovary.

Brassicaceae: Flowers in Brassicaceae exhibit actinomorphic symmetry and are bisexual. The calyx consists of four sepals arranged in two rows (polysepalous), while the corolla comprises four petals (polypetalous). There are six stamens, arranged in two rows (tetradynamous), and the gynoecium is bicarpellary and syncarpous with a superior ovary.

So, the correct option is (B): (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)

2) Floral formula below belongs to which family?

  1. Solanaceae
  2. Liliaceae
  3. Fabaceae
  4. Brassicaceae

ANSWER

c) Fabaceae

(%) shows that the symmetry of the flower is monosymmetric or zygomorphic. (⚥) indicates that the flower is perfect or hermaphrodite (includes stamen plus pistil). A letter ‘K’ indicates the outermost whorl, i.e. calyx and (5) shows that the number of sepals is five that are united with each other.

‘C’ indicates the second floral whorl, i.e. corolla or petals and the number after this, i.e. 1+2+(2), represents vexillary aestivation. ‘A’ means the male reproductive part or androecium, and the number after it (9)+1 shows the presence of diadelphous stamens (includes 10 stamens).

In 10 stamens, 9 filaments unite to constitute one bundle, and the remaining one filament makes up another one. The line below the letter ‘G’ indicates the superior position of an ovary, and digit 1 shows the monocarpellary ovary.

3) Which of the following is false about floral formula – 

  1. Indefinite number of floral parts is shown with the help of infinity symbol ‘∞’
  2. Calyx is denoted with the letter ‘C’
  3. K5 means there are five petals, aposepalous
  4. A2+2 means 4 free stamens, two in each whorl

ANSWER

b) Calyx is denoted with the letter ‘C’

 In the floral formula, calyx (sepals) is denoted as ‘K’ and not ‘C’. it is one of the non-essential organs of the flower, it is the outermost and the first whorl of the flower. In the bud stage of the flower the calyx forms a protective layer around the bud. The ‘C’ in the floral formula is written for corolla.
K5: It shows the aposepalous condition, in which the five sepals are free.
K(5): It shows the gamosepalous condition, in which the five sepals are united.
C5: It represents the apopetalous condition or the presence of the five free petals.
C(5): It represents the gamopetalous condition or the presence of five fused petals.

4) The correct floral formula of soyabean is (NEET 2010 QP)

ANSWER

The plants belonging to the family fabaceae such as soyabean, pea, moong, gram, etc have the floral formula

5) NEET 2016 QP
Tricarpellary, syncarpous gynoecium is found in the flowers of
(a) Liliaceae
(b) Poaceae
(c) Fabaceae
(d) Solanaceae

ANSWER

a) Tricarpellary, syncarpous gynoecium is a unique reproductive structure found in the flowers of Liliaceae family. It is formed by the fusion of three carpels, resulting in a single, compound ovary with three chambers. Each chamber has its own style and stigma, which are designed to facilitate pollination and fertilization.

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Work, Energy, Power and Collision

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, PEe.

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 Fnet = 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 px and py, will also be conserved. With the chosen coordinate system, py is initially zero and px is the momentum of the incoming particle.

Now, we will take the conservation of momentum equation, p1 + p2 = p1 + p2 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 v1x = v1. Conservation of momentum along the x-axis gives the equation

where θ1 and θ2 are as shown in the above Figure

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

or

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

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

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Take Quiz

1) Describe the transformation between forms of mechanical energy that is happening to a falling skydiver before his parachute opens.
a. Kinetic energy is being transformed into potential energy.
b. Potential energy is being transformed into kinetic energy.
c. Work is being transformed into kinetic energy.
d. Kinetic energy is being transformed into work.

ANSWER

b. Potential energy is being transformed into kinetic energy.

2) True or false—If a rock is thrown into the air, the increase in the height would increase the rock’s kinetic energy, and then the increase in the velocity as it falls to the ground would increase its potential energy.
a. True
b. False

ANSWER

b. False

3) A 0.2 kg apple on an apple tree has a potential energy of 10 J. It falls to the ground, converting all of its PE to kinetic energy. What is the velocity of the apple just before it hits the ground?
a. 0 m/s
b. 2 m/s
c. 10 m/s
d. 50 m/s

ANSWER

c. 10 m/s

4) Is it possible for the sum of kinetic energy and potential energy of an object to change without work having been done on the object? Explain.
A) No, because the work-energy theorem states that work done on an object is equal to the change in kinetic energy, and change in KE requires a change in velocity. It is assumed that mass is constant.
b) No, because the work-energy theorem states that work done on an object is equal to the sum of kinetic energy, and the change in KE requires a change in displacement. It is assumed that mass is constant.
C) Yes, because the work-energy theorem states that work done on an object is equal to the change in kinetic energy, and change in KE requires a change in velocity. It is assumed that mass is constant.
D)Yes, because the work-energy theorem states that work done on an object is equal to the sum of kinetic energy, and the change in KE requires a change in displacement. It is assumed that mass is constant

ANSWER

D) Yes, because the work-energy theorem states that work done on an object is equal to the sum of kinetic energy, and the change in KE requires a change in displacement. It is assumed that mass is constant

5) The starting line of a cross country foot race is at the bottom of a hill. Which form(s) of mechanical energy of the runners will change when the starting gun is fired?
a. Kinetic energy only
b. Potential energy only
c. Both kinetic and potential energy
d. Neither kinetic nor potential energy

ANSWER

c. Both kinetic and potential energy

6) True or false—A cyclist coasts down one hill and up another hill until she comes to a stop. The point at which the bicycle stops is lower than the point at which it started coasting because part of the original potential energy has been converted to a quantity of heat and this makes the tires of the bicycle warm.
a. True
b. False

ANSWER

b. False

7) A boy pushes his little sister on a sled. The sled accelerates from 0 to 3.2 m/s . If the combined mass of his sister and the sled is 40.0 kg and 18 W of power were generated, how long did the boy push the sled?
a. 205 s
b. 128 s
c. 23 s
d. 11 s

ANSWER

c. 23 s

8) What is the kinetic energy of a bullet traveling at a velocity of ?
a. 3.5 L
b. 7J
c. 2.45 x 103 J
d. 2.45 x 105 J

ANSWER

c. 2.45 x 103 J

9) A marble rolling across a flat, hard surface at rolls up a ramp. Assuming that and no energy is lost to friction, what will be the vertical height of the marble when it comes to a stop before rolling back down? Ignore effects due to the rotational kinetic energy.
a. 0.1 m
b. 0.2 m
c. 0.4 m
d. 2 m

ANSWER

b. 0.2 m

10) The work–energy theorem states that the change in the kinetic energy of an object is equal to what?
a. The work done on the object
b. The force applied to the object
c. The loss of the object’s potential energy
d. The object’s total mechanical energy minus its kinetic energy

ANSWER

a. The work done on the object

11) A runner at the start of a race generates 250 W of power as he accelerates to 5 m/s . If the runner has a mass of 60 kg, how long did it take him to reach that speed?
a. 0.33 s
b. 0.83 s
c. 1.2 s
d. 3.0 s

ANSWER

d. 3.0 s

12) A car’s engine generates 100,000 W of power as it exerts a force of 10,000 N. How long does it take the car to travel 100 m?
a. 0.001 s
b. 0.01 s
c. 10 s
d. 1,000 s

ANSWER

c. 10 s

13) Why is this expression for kinetic energy incorrect?
KE = (m) (v) 2
a. The constant is missing.
b. The term should not be squared.
c. The expression should be divided by 2 .
d. The energy lost to friction has not been subtracted

ANSWER

c. The expression should be divided by 2 .

14) What is the kinetic energy of a 10kg object moving at 2.0 m/s ?
a. 10 J
b. 20 J
c. 40 J
d. 100 J

ANSWER

b. 20 J

15) A boulder rolls from the top of a mountain, travels across a valley below, and rolls part way up the ridge on the opposite side. Describe all the energy transformations taking place during these events and identify when they happen.
a. As the boulder rolls down the mountainside, KE is converted to PE. As the boulder rolls up the opposite slope, PE is converted to KE. The boulder rolls only partway up the ridge because some of the PE has been converted to thermal energy due to friction.
b. As the boulder rolls down the mountainside, KE is converted to PE. As the boulder rolls up the opposite slope, KE is converted to PE. The boulder rolls only partway up the ridge because some of the PE has been converted to thermal energy due to friction.
c. As the boulder rolls down the mountainside, PE is converted to KE. As the boulder rolls up the opposite slope, PE is converted to KE. The boulder rolls only partway up the ridge because some of the PE has been converted to thermal energy due to friction.
d. As the boulder rolls down the mountainside, PE is converted to KE. As the boulder rolls up the opposite slope, KE is converted to PE. The boulder rolls only partway up the ridge because some of the PE has been converted to thermal energy due to friction.

ANSWER

d. As the boulder rolls down the mountainside, PE is converted to KE. As the boulder rolls up the opposite slope, KE is converted to PE. The boulder rolls only partway up the ridge because some of the PE has been converted to thermal energy due to friction.

Read more

SI Units: Fundamental and Derived Units

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

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

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

Ǫuantity                 Name        Symbol

Length Meter m
Mass Kilogram kg
Time Second s
Electric current Ampere a
Temperature Kelvin k
Amount of substance Mole mol
Luminous intensity Candela cd
Table 1.1 SI Base Units

The Meter

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

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

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

The Kilogram

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

The Second

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

Figure 1.15 An atomic clock such as this one uses the vibrations of cesium atoms to keep time to a precision of one microsecond per year. The fundamental unit of time, the second, is based on such clocks. This image is looking down from the top of an atomic clock. (Steve Jurvetson/Flickr)

The Ampere

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

Kelvins

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

Metric Prefixes

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

Prefix  Symbol Value[1] Example Name Example Symbol Example Value Example Description
exa E 1018 Exameter Em 1018 m Distance light travels in a century
peta P 1015 Petasecond Ps 1015 s 30 million years
tera T 1012 Terawatt TW 1012 W Powerful laser output
giga G 109 Gigahertz GHz 109 Hz A microwave frequency
mega M 106 Megacurie MCi 106 Ci High radioactivity
kilo k 103 Kilometer km 103 m About 6/10 mile
hector h 102 Hectoliter hL 102 L 26 gallons
deka da 101 Dekagram dag 101 g Teaspoon of butter
deci d 10–1 Deciliter dL 10–1 L Less than half a soda
centi c 10–2 Centimeter Cm 10–2 m Fingertip thickness
milli m 10–3 Millimeter Mm 10–3 m Flea at its shoulder
micro µ 10–6 Micrometer µm 10–6 m Detail in microscope
nano n 10–9 Nanogram Ng 10–9 g Small speck of dust
pico p 10–12 Picofarad pF 10–12 F Small capacitor in radio
femto f 10–15 Femtometer Fm 10–15 m Size of a proton
  atto   a  10–18   Attosecond   as   10–18 s Time light takes to cross an atom
Table 1.2 Metric Prefixes for Powers of 10 and Their Symbols [1]See Appendix A for a discussion of powers of 10. Note—Some examples are approximate

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

Known Ranges of Length, Mass, and Time

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

Length (m) Phenomenon Measured Mass
(Kg)
Phenomenon Measured[1] Time (s) Phenomenon Measured[1]
  10–18 Present experimental limit to smallest observable detail 10–30 Mass of an electron (9.11×10–31 kg) 10–23 Time for light to cross a proton
  10–15 Diameter of a proton 10–27   Mass of a hydrogen atom (1.67×10–27 kg) 10–22 Mean life of an extremely unstable nucleus
  1014 Diameter of a uranium nucleus 10–15   Mass of a bacterium 10–15 Time for one oscillation of a visible light
  10–10 Diameter of a hydrogen atom  10–5   Mass of a mosquito 10–13 Time for one vibration of an atom in a solid
  10–8 Thickness of membranes in cell of living organism  10–2   Mass of a hummingbird 10–8 Time for one oscillation of an FM radio wave
  10–6 Wavelength of visible light   1 Mass of a liter of water (about a quart) 10–3 Duration of a nerve impulse
10–3 Size of a grain of sand 102 Mass of a person 1 Time for one heartbeat
1 Height of a 4-year-old child 103 Mass of a car 105 One day (8.64×104 s)
102 Length of a football field 108 Mass of a large ship 107 One day (3.16×107 s)
104 Greatest ocean depth 1012 Mass of a large iceberg 109 About half the life expectancy of a human
107 Diameter of Earth 1015 Mass of the nucleus of a comet 1011 Recorded history
1011 Distance from Earth to the sun 1023 Mass of the moon (7.35×1022 kg) 1017 Age of Earth
1016 Distance traveled by light in 1 year (a light year) 1025 Mass of Earth (5.97×1024kg) 1018 Age of the universe
1021 Diameter of the Milky Way Galaxy 1030 Mass of the Sun (1.99×1034 kg)
1022 Distance from Earth to the nearest large galaxy (Andromeda) 1042 Mass of the Milky Way galaxy (current upper limit)
1026 Distance from Earth to the edges of the known universe 1053 Mass of the known universe (current upper limit)
Table 1.3 Approximate Values of Length, Mass, and Time [1] More precise values are in parentheses.

Figure 1.16 Tiny phytoplankton float among crystals of ice in the Antarctic Sea. They range from a few micrometers to as much as 2 millimeters in length. (Prof. Gordon T. Taylor, Stony Brook University; NOAA Corps Collections)

Figure 1.17 Galaxies collide 2.4 billion light years away from Earth. The tremendous range of observable phenomena in nature challenges the imagination. (NASA/CXC/UVic./A. Mahdavi et al. Optical/lensing: CFHT/UVic./H. Hoekstra et al.)

Using Scientific Notation with Physical Measurements

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

x x 10y

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

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

x x 10y

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

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

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

Figure 1.18 The distance from Earth to the moon may seem immense, but it is just a tiny fraction of the distance from Earth to our closest neighboring star. (NASA)

Unit Conversion and Dimensional Analysis

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

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

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

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

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

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

Worked Example

Unit Conversions: A Short Drive Home

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

Strategy

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

Solution for (a)

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

2. Substitute the given values for distance and time.

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

Discussion for (a)

To check your answer, consider the following:

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

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

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

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

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

Worked Example

Using Physics to Evaluate Promotional Materials

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

Strategy

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

Solution

Discussion

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

Accuracy, Precision and Significant Figures

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

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

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

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

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

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

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

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

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

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

Uncertainty

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

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

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

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

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

Percent Uncertainty

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

Uncertainty in Calculations

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

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

Precision of Measuring Tools and Significant Figures

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

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

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

Zeros

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

Number  Significant Figures Rationale
1.657 4 There are no zeros and all non-zero numbers are always significant.
0.4578 4 The first zero is only a placeholder for the decimal point.
0.000458   3 The first four zeros are placeholders needed to report the data to the ten-thoudsandths place.
2000.56 6 The three zeros are significant here because they occur between other significant figures.
45,600   3 With no underlines or scientific notation, we assume that the last two zeros are placeholders and are not significant.
15895000 7 The two underlined zeros are significant, while the last zero is not, as it is not underlined.
5.457 x 1013   4 In scientific notation, all numbers reported in front of the multiplication sign are significant
6.520 x 10–23   4 In scientific notation, all numbers reported in front of the multiplication sign are significant, including zeros.
Table 1.4

Significant Figures in Calculations

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

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

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

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

A = 4.5 m2

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

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

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

6527.23 + 2 = 6528.23 = 6528

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

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Take Quiz

1. A car engine moves a piston with a circular cross section of  7.500 ± 0.002 cm diameter a distance of 3.250±0.001 cm to compress the gas in the cylinder. By what amount did the gas decrease in volume in cubic centimeters? Find the uncertainty in this volume

  1. 143.6 ± 0.002 cm3
  2. 143.6 ± 0.003 cm3
  3. 143.6 ± 0.005 cm3
  4. 143.6 ± 0.1 cm3

ANSWER

d) 143.6 ± 0.1 cm3

2) The sides of a small rectangular box are measured 1.80 cm and 2.50cm long and 3.1 cm high. Calculate its volume and uncertainty in cubic centimeters. Assume the measuring device is accurate to ± 0.05 cm

  1. 11.4 ± 0.1 cm3
  2. 11.4 ± 0.6 cm3
  3. 11.4 ± 0.8 cm3
  4. 11.4 ± 0.10 cm3

ANSWER

b) 11.4 ± 0.6 cm3

3) Calculate the approximate number of atoms in a bacterium. Assume that the average mass of an atom in the bacterium is ten times the mass of a hydrogen atom. (Hint—The mass of a hydrogen atom is on the order of 10−27 kg and the mass of a bacterium is on the order of 10−15 kg .)

  1. 1010 atoms
  2. 1011 atoms
  3. 1012 atoms
  4. 1013 atoms

ANSWER

b) 1011 atoms

4) Suppose that a bathroom scale reads a person’s mass as 65 kg with a 3 percent uncertainty. What is the uncertainty in their mass in kilograms?

a. 2 kg
b. 98 kg
c. 5 kg
d. 0

ANSWER

a) 2 kg

5) The speed of sound is measured at 342 m/s on a certain day. What is this in km/h ? Report your answer in scientific notation

  1. 1.23 X 10 4  Kmph
  2. 1.23 X 10 3  Kmph
  3. 9.5 X 10 1  Kmph
  4. 2.05 X 10 -1  Kmph

ANSWER

b) 1.23 X 10 3  Kmph

6) Tectonic plates are large segments of the Earth’s crust that move slowly . Suppose one such plate has an average speed of 4.0 cm / year . What distance does it move in 1 second at this speed ?

  1. 1.3 X 10-9 m
  2. 1.3 X 10-6 m
  3. 1.3 X 10– 10  m
  4. 1.3 X 10– 4 m

ANSWER

c) 1.3 X 10– 10  m

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Graphing in Physics

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

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

Time (min)  Distance from Station (km)

0 0
10 24
20 36
30 60
40 84
50 97
60 116
70 140
Table 1.5

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

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

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

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

Figure 1.26 The graph of the train’s distance from the station versus time from the exercise above.

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

Figure 1.27 The completed graph with the trend line included.

Analyzing a Graph Using Its Equation

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

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

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

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

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

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

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

y = (2.0 km/min) t + 0

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

y = k/x,

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

Figure 1.28 Examples of (a) linear, (b) quadratic, (c) inverse, and (d) exponential relationship graphs.

Using Logarithmic Scales in Graphing

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

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

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

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

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

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

Take Quiz

1. How does the independent variable in a graph differ from the dependent variable?

a. The dependent variable varies linearly with the independent variable.
b. The dependent variable depends on the scale of the axis chosen while independent variable does not.
c. The independent variable is directly manipulated or controlled by the person doing the experiment, while dependent variable is the one that changes as a result.
d. The dependent and independent variables are fixed by a convention and hence they are the same.

ANSWER

c) The independent variable is directly manipulated or controlled by the person doing the experiment, while dependent

2) Velocity, or speed, is measured using the following formula: where v is velocity, d is the distance travelled, and t is the time the object took to travel the distance. If the velocity-time data are plotted on a graph, which variable will be on which axis? Why?

a. Time would be on the x-axis and velocity on the y- axis, because time is an independent variable and velocity is a dependent variable.
b. Velocity would be on the x-axis and time on the y- axis, because time is the independent variable and velocity is the dependent variable.
c. Time would be on the x-axis and velocity on the y- axis, because time is a dependent variable and velocity is a independent variable.
d. Velocity would be on x-axis and time on the y-axis, because time is a dependent variable and velocity is a independent variable.

ANSWER

a) Time would be on the x-axis and velocity on the y- axis, because time is an independent variable and velocity is a dependent variable.

3) The graph shows the temperature change over time of a heated cup of water.

What is the slope of the graph between the time period 2 min and 5 min?
a. –15 ºC/min
b. –0.07 ºC/min
c. 0.07 ºC/min
d. 15 ºC/min

ANSWER

d) 15 ºC/min

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Thermodynamics

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

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

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

Zeroth Law of Thermodynamics: Thermal Equilibrium

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

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

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

Pressure, Volume, Temperature, and the Ideal Gas Law

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

P = F / A

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

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

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

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

The ideal gas law states that

PV = NkT

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

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

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

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

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

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

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

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

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

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

The First Law of Thermodynamics :

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

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

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

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

ΔU = Q – W

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

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

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

Second Law of Thermodynamics: Entropy

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

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

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

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

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

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

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

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

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

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

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

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

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

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

ΔStot = ΔSsystem + ΔSenvironment  > 0

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

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

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

Take Quiz

1. What is the net heat out of the system when 25 J is transferred by heat into the system and 45 J  is transferred out of it ?
a) -70 J
b) – 20 J
c) 20 J
d) 70 J

ANSWER

c) 20 J

2) What is internal energy?
a. It is the sum of the kinetic energies of a system’s atoms and molecules.
b. It is the sum of the potential energies of a system’s atoms and molecules.
c. It is the sum of the kinetic and potential energies of a system’s atoms and molecules.
d. It is the difference between the magnitudes of the kinetic and potential energies of a system’s atoms and molecules.

ANSWER

c) It is the sum of the kinetic and potential energies of a system’s atoms and molecules.

3) What is the increase in entropy when 3 Kg of ice at 0 0 celsius  melt to form water at 0 0 degree celsius ?
a . 1.84 X 10 3 J / K
b. 3.67 X 10 3 J / K
c. 1.84 X 10 8 J / K
d. 3.67 X 10 8 J / K

ANSWER

b) 3.67 X 10 3 J / K

4) For heat transferring energy from a high to a low temperature, what usually happens to the entropy of the whole system?
a. It decreases.
b. It must remain constant.
c. The entropy of the system cannot be predicted without specific values for the temperatures.
d. It increases.

ANSWER

d) It increases.

5) A heat engine is given 120 J by heat and releases 20 J by heat to the environment. What is the amount of work done by the system?
a. – 100 J
b. -60 J
c. 60 J
d. 100 J

ANSWER

d) 100 J

6) Eighty joules are added by heat to a system, while it does 20 J of work. Later 30 J, are added by heat to the system, and it does 40 J of work. What is the change in the system’s internal energy ?
a. 30 J
b. 60 J
c. 50 J
d. 110 J

ANSWER

c) 50 J

 7) A coal power station functions at 40.0 percent efficiency. What is the amount of work it does if it takes in 1.20 ×10 12 J by heat?
a. 3 X 1010  J
b. 4.8 X 1011  J
c. 3 X 1012  J
d. 4.8 X 1018  J

ANSWER

b) 4.8 X 1011  J

8) When does a real gas behave like an ideal gas ?
a. A real gas behaves like an ideal gas at high temperature and low pressure.
b. A real gas behaves like an ideal gas at high temperature and high pressure.
c. A real gas behaves like an ideal gas at low temperature and low pressure.
d. A real gas behaves like an ideal gas at low temperature and high pressure.

ANSWER

a) A real gas behaves like an ideal gas at high temperature and low pressure.

9) If Q is 120 J and T  is 350 K , what is the change in entropy?
a. 1.51 J / K
b. 0.343 J / K
c.2.92 J / K
d.34.3 J / K

ANSWER

b) 0.343 J / K

10) 310 joules of heat enter a system, after which the system does 120 J of work. What is the change in its internal energy? Would this amount change if  30 J of  energy ransferred by heat were added after the work was done instead of before?
a. – 190 J ; this would change if heat added energyafter the work was done
b.  190 J ; this would change if heat added energy after the work was done
c. ; -190 Jthis would not change even if heat added energy after the work was done
d. 190 J ; this would not change even if heat added energy after the work was done

ANSWER

b) 190 J ; this would change if heat added energy after the work was done

11) Why does entropy increase when ice melts into water?
a. Melting converts the highly ordered solid structure into a disorderly liquid, thereby increasing entropy.
b. Melting converts the highly ordered liquid into a disorderly solid structure, thereby increasing entropy.
c. Melting converts the highly ordered solid structure into a disorderly solid structure, thereby increasing entropy.
d. Melting converts the highly ordered liquid into a disorderly liquid, thereby increasing entropy.

ANSWER

a) Melting converts the highly ordered solid structure into a disorderly liquid, thereby increasing entropy.

12) Two distinct systems have the same amount of stored internal energy. 500 joules are added by heat to the first system, and 300 J are added by heat to the second system. What will be the change in internal energy of the first system if it does 200 J of work? How much work will the second system have to do in order to have the same internal energy?
a. 700 J; 0 J
b. 300 J; 300 J
c. 700 J; 300 J
d. 300 J; 0 J

ANSWER

d) 300 J; 0 J

13) A coal power station transfers 3.0×1012 J by heat from burning coal and transfers 1.5×1012 J by heat into the environment. What is the efficiency of the power station?
a.0.33
b. 0.5
c. 0.66
d.1

ANSWER

b) 0.5

14) By convention, if Q is positive, what is the direction in which heat transfers energy with regard to the system?
a. The direction of the heat transfer of energy depends on the changes in W, regardless of the sign of Q.
b. The direction of Q cannot be determined from just the sign of Q.
c. The direction of net heat transfer of energy will be out of the system.
d. The direction of net heat transfer of energy will be into the system

ANSWER

d) The direction of net heat transfer of energy will be into the system

15) What is the change in entropy caused by melting 5.00 kg of ice at 0 °C ?
a.0 J / K
b. 6.11 X 10 3 J / K
c. 6.11 X 10 4 J / K
d. Infinite

ANSWER

b) 6.11 X 10 3 J / K

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