# Motion in Two Dimensions

## Introduction

In Chapter 2, we learned to distinguish between vectors and scalars; the difference being that a vector has magnitude and direction, whereas a scalar has only magnitude and possibly sign. We learned how to deal with vectors in physics by working straightforward one-dimensional vector problems, which may be treated mathematically in the same as scalars. In this chapter, we’ll use vectors to expand our understanding of forces and motion into two dimensions. Most real-world physics problems (such as with the game of pool pictured here) are, after all, either two- or three-dimensional problems and physics is most useful when applied to real physical scenarios. We start by learning the practical skills of graphically adding and subtracting vectors (by using drawings) and analytically (with math). Once we’re able to work with two-dimensional vectors, we apply these skills to problems of projectile motion, inclined planes, and harmonic motion.


# Motion in Two Dimensions

## Vector Addition and Subtraction: Graphical Methods

### Section Key Terms

### The Graphical Method of Vector Addition and Subtraction

Recall that a vector is a quantity that has magnitude and direction. For example, displacement, velocity, acceleration, and force are all vectors. In one-dimensional or straight-line motion, the direction of a vector can be given simply by a plus or minus sign. Motion that is forward, to the right, or upward is usually considered to be positive (+); and motion that is backward, to the left, or downward is usually considered to be negative (−).
In two dimensions, a vector describes motion in two perpendicular directions, such as vertical and horizontal. For vertical and horizontal motion, each vector is made up of vertical and horizontal components. In a one-dimensional problem, one of the components simply has a value of zero. For two-dimensional vectors, we work with vectors by using a frame of reference such as a coordinate system. Just as with one-dimensional vectors, we graphically represent vectors with an arrow having a length proportional to the vector’s magnitude and pointing in the direction that the vector points.
 shows a graphical representation of a vector; the total displacement for a person walking in a city. The person first walks nine blocks east and then five blocks north. Her total displacement does not match her path to her final destination. The displacement simply connects her starting point with her ending point using a straight line, which is the shortest distance. We use the notation that a boldface symbol, such as D, stands for a vector. Its magnitude is represented by the symbol in italics, D, and its direction is given by an angle represented by the symbol
       Note that her displacement would be the same if she had begun by first walking five blocks north and then walking nine blocks east.
The head-to-tail method is a graphical way to add vectors. The tail of the vector is the starting point of the vector, and the head (or tip) of a vector is the pointed end of the arrow. The following steps describe how to use the head-to-tail method for graphical vector addition.
1. Let the x-axis represent the east-west direction. Using a ruler and protractor, draw an arrow to represent the first vector (nine blocks to the east), as shown in (a).

2. Let the y-axis represent the north-south direction. Draw an arrow to represent the second vector (five blocks to the north). Place the tail of the second vector at the head of the first vector, as shown in (b).

3. If there are more than two vectors, continue to add the vectors head-to-tail as described in step 2. In this example, we have only two vectors, so we have finished placing arrows tip to tail.
4. Draw an arrow from the tail of the first vector to the head of the last vector, as shown in (c). This is the resultant, or the sum, of the vectors.

5. To find the magnitude of the resultant, measure its length with a ruler. When we deal with vectors analytically in the next section, the magnitude will be calculated by using the Pythagorean theorem.
6. To find the direction of the resultant, use a protractor to measure the angle it makes with the reference direction (in this case, the x-axis). When we deal with vectors analytically in the next section, the direction will be calculated by using trigonometry to find the angle.
Vector subtraction is done in the same way as vector addition with one small change. We add the first vector to the negative of the vector that needs to be subtracted. A negative vector has the same magnitude as the original vector, but points in the opposite direction (as shown in ). Subtracting the vector B from the vector A, which is written as A − B, is the same as A + (−B). Since it does not matter in what order vectors are added, A − B is also equal to (−B) + A. This is true for scalars as well as vectors. For example, 5 – 2 = 5 + (−2) = (−2) + 5.
Global angles are calculated in the counterclockwise direction. The clockwise direction is considered negative. For example, an angle of
       south of west is the same as the global angle
       which can also be expressed as
       from the positive x-axis.

### Using the Graphical Method of Vector Addition and Subtraction to Solve Physics Problems

Now that we have the skills to work with vectors in two dimensions, we can apply vector addition to graphically determine the resultant vector, which represents the total force. Consider an example of force involving two ice skaters pushing a third as seen in .
In problems where variables such as force are already known, the forces can be represented by making the length of the vectors proportional to the magnitudes of the forces. For this, you need to create a scale. For example, each centimeter of vector length could represent 50 N worth of force. Once you have the initial vectors drawn to scale, you can then use the head-to-tail method to draw the resultant vector. The length of the resultant can then be measured and converted back to the original units using the scale you created.
You can tell by looking at the vectors in the free-body diagram in  that the two skaters are pushing on the third skater with equal-magnitude forces, since the length of their force vectors are the same. Note, however, that the forces are not equal because they act in different directions. If, for example, each force had a magnitude of 400 N, then we would find the magnitude of the total external force acting on the third skater by finding the magnitude of the resultant vector. Since the forces act at a right angle to one another, we can use the Pythagorean theorem. For a triangle with sides a, b, and c, the Pythagorean theorem tells us that
Applying this theorem to the triangle made by F1, F2, and Ftot in , we get
or
Note that, if the vectors were not at a right angle to each other

 to one another), we would not be able to use the Pythagorean theorem to find the magnitude of the resultant vector. Another scenario where adding two-dimensional vectors is necessary is for velocity, where the direction may not be purely east-west or north-south, but some combination of these two directions. In the next section, we cover how to solve this type of problem analytically. For now let’s consider the problem graphically.

### Practice Problems

### Section Summary

1. The graphical method of adding vectors
         and
         involves drawing vectors on a graph and adding them by using the head-to-tail method. The resultant vector
         is defined such that A + B = R. The magnitude and direction of
         are then determined with a ruler and protractor.
2. The graphical method of subtracting vectors A and B involves adding the opposite of vector B, which is defined as −B. In this case,
         Next, use the head-to-tail method as for vector addition to obtain the resultant vector
         .
3. Addition of vectors is independent of the order in which they are added; A + B = B + A.
4. The head-to-tail method of adding vectors involves drawing the first vector on a graph and then placing the tail of each subsequent vector at the head of the previous vector. The resultant vector is then drawn from the tail of the first vector to the head of the final vector.
5. Variables in physics problems, such as force or velocity, can be represented with vectors by making the length of the vector proportional to the magnitude of the force or velocity.
6. Problems involving displacement, force, or velocity may be solved graphically by measuring the resultant vector’s magnitude with a ruler and measuring the direction with a protractor.

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# Motion in Two Dimensions

## Vector Addition and Subtraction: Analytical Methods

### Section Key Terms

### Components of Vectors

For the analytical method of vector addition and subtraction, we use some simple geometry and trigonometry, instead of using a ruler and protractor as we did for graphical methods. However, the graphical method will still come in handy to visualize the problem by drawing vectors using the head-to-tail method. The analytical method is more accurate than the graphical method, which is limited by the precision of the drawing. For a refresher on the definitions of the sine, cosine, and tangent of an angle, see .
Since, by definition,  , we can find the length x if we know h and    by using   . Similarly, we can find the length of y by using   . These trigonometric relationships are useful for adding vectors.
When a vector acts in more than one dimension, it is useful to break it down into its x and y components. For a two-dimensional vector, a component is a piece of a vector that points in either the x- or y-direction. Every 2-d vector can be expressed as a sum of its x and y components.
For example, given a vector like
       in , we may want to find what two perpendicular vectors,
       and
      , add to produce it. In this example,
       and
       form a right triangle, meaning that the angle between them is 90 degrees. This is a common situation in physics and happens to be the least complicated situation trigonometrically.

       and
       are defined to be the components of
       along the x- and y-axes. The three vectors,
      ,
      , and
      , form a right triangle.
If the vector
       is known, then its magnitude
       (its length) and its angle
       (its direction) are known. To find
       and
      , its x- and y-components, we use the following relationships for a right triangle:
and
where
         is the magnitude of A in the x-direction,
         is the magnitude of A in the y-direction, and
         is the angle of the resultant with respect to the x-axis, as shown in .
Suppose, for example, that
       is the vector representing the total displacement of the person walking in a city, as illustrated in .
Then A = 10.3 blocks and
      , so that
This magnitude indicates that the walker has traveled 9 blocks to the east—in other words, a 9-block eastward displacement. Similarly,
indicating that the walker has traveled 5 blocks to the north—a 5-block northward displacement.

### Analytical Method of Vector Addition and Subtraction

Calculating a resultant vector (or vector addition) is the reverse of breaking the resultant down into its components. If the perpendicular components
          and
          of a vector
          are known, then we can find
          analytically. How do we do this? Since, by definition,
we solve for
          to find the direction of the resultant.
Note that  gives an angle in the first quadrant if
       and in the fourth quadrant if
      . If, in fact, both  and  are negative, or if  is negative and  positive, then , measured from the positive  direction, is
      .
Since this is a right triangle, the Pythagorean theorem (x2 + y2 = h2) for finding the hypotenuse applies. In this case, it becomes
Solving for A gives
In summary, to find the magnitude
          and direction
          of a vector from its perpendicular components
          and
        , as illustrated in , we use the following relationships:
Sometimes, the vectors added are not perfectly perpendicular to one another. An example of this is the case below, where the vectors
          and
          are added to produce the resultant
          as illustrated in .
If
         and
         represent two legs of a walk (two displacements), then
         is the total displacement. The person taking the walk ends up at the tip of
         . There are many ways to arrive at the same point. The person could have walked straight ahead first in the x-direction and then in the y-direction. Those paths are the x- and y-components of the resultant,
         and
         If we know
         and
        , we can find
         and
         using the equations
         and
         .
1. Draw in the x and y components of each vector (including the resultant) with a dashed line. Use the equations
           and
           to find the components. In , these components are
          ,
          ,
          , and
           Vector
           makes an angle of
           with the x-axis, and vector
           makes and angle of
           with its own x-axis (which is slightly above the x-axis used by vector A).

2. Find the x component of the resultant by adding the x component of the vectors
          and find the

          Now that we know the components of
3. To get the magnitude of the resultant R, use the Pythagorean theorem.

4. To get the direction of the resultant

### Using the Analytical Method of Vector Addition and Subtraction to Solve Problems

 uses the analytical method to add vectors.

### Practice Problems

### Section Summary

1. The analytical method of vector addition and subtraction uses the Pythagorean theorem and trigonometric identities to determine the magnitude and direction of a resultant vector.
2. The steps to add vectors
         and
         using the analytical method are as follows:

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# Motion in Two Dimensions

## Projectile Motion

### Section Key Terms

### Properties of Projectile Motion

Projectile motion is the motion of an object thrown (projected) into the air when, after the initial force that launches the object, air resistance is negligible and the only other force that object experiences is the force of gravity. The object is called a projectile, and its path is called its trajectory. Air resistance is a frictional force that slows its motion and can significantly alter the trajectory of the motion. Due to the difficulty in calculation, only situations in which the deviation from projectile motion is negligible and air resistance can be ignored are considered in introductory physics. That approximation is often quite accurate.
The most important concept in projectile motion is that when air resistance is ignored, horizontal and vertical motions are , meaning that they don’t influence one another.  compares a cannonball in free fall (in blue) to a cannonball launched horizontally in projectile motion (in red). You can see that the cannonball in free fall falls at the same rate as the cannonball in projectile motion. Keep in mind that if the cannon launched the ball with any vertical component to the velocity, the vertical displacements would not line up perfectly.
Since vertical and horizontal motions are independent, we can analyze them separately, along perpendicular axes. To do this, we separate projectile motion into the two components of its motion, one along the horizontal axis and the other along the vertical.
We’ll call the horizontal axis the x-axis and the vertical axis the y-axis. For notation, d is the total displacement, and x and y are its components along the horizontal and vertical axes. The magnitudes of these vectors are x and y, as illustrated in .
As usual, we use velocity, acceleration, and displacement to describe motion. We must also find the components of these variables along the x- and y-axes. The components of acceleration are then very simple a = –g = –9.80 m/s2. Note that this definition defines the upwards direction as positive. Because gravity is vertical, a = 0. Both accelerations are constant, so we can use the kinematic equations. For review, the kinematic equations from a previous chapter are summarized in .
Where x is position, x0 is initial position, v is velocity, vavg is average velocity, t is time and a is acceleration.

### Solve Problems Involving Projectile Motion

The following steps are used to analyze projectile motion:
1. Separate the motion into horizontal and vertical components along the x- and y-axes. These axes are perpendicular, so
          and
         are used. The magnitudes of the displacement
         along x- and y-axes are called
         and
         The magnitudes of the components of the velocity
         are
         and
        , where
         is the magnitude of the velocity and
         is its direction. Initial values are denoted with a subscript 0.
2.
        Treat the motion as two independent one-dimensional motions, one horizontal and the other vertical. The kinematic equations for horizontal and vertical motion take the following forms

        Vertical motion (assuming positive is up
           )

3. Solve for the unknowns in the two separate motions (one horizontal and one vertical). Note that the only common variable between the motions is time
         . The problem solving procedures here are the same as for one-dimensional kinematics.
4. Recombine the two motions to find the total displacement
       and velocity
       . We can use the analytical method of vector addition, which uses
       and
       to find the magnitude and direction of the total displacement and velocity.

         is the direction of the displacement
        , and
         is the direction of the velocity
         . (See

The expression we found for
       while solving part (a) of the previous problem works for any projectile motion problem where air resistance is negligible. Call the maximum height
      ; then,
This equation defines the maximum height of a projectile. The maximum height depends only on the vertical component of the initial velocity.

### Practice Problems

The fact that vertical and horizontal motions are independent of each other lets us predict the range of a projectile. The range is the horizontal distance R traveled by a projectile on level ground, as illustrated in . Throughout history, people have been interested in finding the range of projectiles for practical purposes, such as aiming cannons.
How does the initial velocity of a projectile affect its range? Obviously, the greater the initial speed
      , the greater the range, as shown in the figure above. The initial angle
       also has a dramatic effect on the range. When air resistance is negligible, the range
       of a projectile on level ground is
where
         is the initial speed and
         is the initial angle relative to the horizontal. It is important to note that the range doesn’t apply to problems where the initial and final y position are different, or to cases where the object is launched perfectly horizontally.

### Section Summary

1. Projectile motion is the motion of an object through the air that is subject only to the acceleration of gravity.
2. Projectile motion in the horizontal and vertical directions are independent of one another.
3. The maximum height of an projectile is the highest altitude, or maximum displacement in the vertical position reached in the path of a projectile.
4. The range is the maximum horizontal distance traveled by a projectile.
5. To solve projectile problems: choose a coordinate system; analyze the motion in the vertical and horizontal direction separately; then, recombine the horizontal and vertical components using vector addition equations.

### Key Equations

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# Motion in Two Dimensions

## Inclined Planes

### Section Key Terms

### Static Friction and Kinetic Friction

Recall from the previous chapter that friction is a force that opposes relative motion parallel to the contact surface of the interacting objects and is around us all the time. Friction allows us to move, which you have discovered if you have ever tried to walk on ice.
There are different types of friction—kinetic and static. Kinetic friction acts on an object in relative motion, while static friction acts on an object or system at rest relative to each other. The maximum static friction is usually greater than the kinetic friction between the objects.
Imagine, for example, trying to slide a heavy crate across a concrete floor. You may push harder and harder on the crate and not move it at all. This means that the static friction responds to what you do—it increases to be equal to and in the opposite direction of your push. But if you finally push hard enough, the crate seems to slip suddenly and starts to move. Once in motion, it is easier to keep it in motion than it was to get it started because the kinetic friction force is less than the static friction force. If you were to add mass to the crate, (for example, by placing a box on top of it) you would need to push even harder to get it started and also to keep it moving. If, on the other hand, you oiled the concrete you would find it easier to get the crate started and keep it going.
 shows how friction occurs at the interface between two objects. Magnifying these surfaces shows that they are rough on the microscopic level. So when you push to get an object moving (in this case, a crate), you must raise the object until it can skip along with just the tips of the surface hitting, break off the points, or do both. The harder the surfaces are pushed together (such as if another box is placed on the crate), the more force is needed to move them.
The magnitude of the frictional force has two forms: one for static friction, the other for kinetic friction. When there is no motion between the objects, the magnitude of static friction f is
where
           is the coefficient of static friction and N is the magnitude of the normal force. Recall that the normal force acts perpendicular to the surface and prevents the crate from falling through the floor. It opposes the force of gravity in this example, but that will not always be the case.
Since the symbol
           means less than or equal to, this equation says that static friction can have a maximum value of
           That is,
Static friction is a responsive force that increases to be equal and opposite to whatever force is exerted, up to its maximum limit. Once the applied force exceeds fs(max), the object will move. Once an object is moving, the magnitude of kinetic friction fk is given by
where
            is the coefficient of kinetic friction.
Friction varies from surface to surface because different substances are rougher than others.  compares values of static and kinetic friction for different surfaces. The coefficient of the friction depends on the two surfaces that are in contact.
Since friction always opposes relative motion, the direction of friction is upward along the plane if the object is at rest or sliding down the incline. For example, if the crate you try to push (with a force parallel to the floor) has a mass of 100 kg, then the normal force would be equal to its weight
perpendicular to the floor. If the coefficient of static friction is 0.45, you would have to exert a force parallel to the floor greater than
to move the crate. Once there is motion, friction is less and the coefficient of kinetic friction might be 0.30, so that a force of only 290 N
would keep it moving at a constant speed. If the floor were lubricated, both coefficients would be much smaller than they would be without lubrication. The coefficient of friction is unitless and is a number usually between 0 and 1.0, but there is no theoretical upper limit to its value.

### Working with Inclined Planes

We discussed previously that when an object rests on a horizontal surface, there is a normal force supporting it equal in magnitude to its weight. Up until now, we dealt only with normal force in one dimension, with gravity and normal force acting perpendicular to the surface in opposing directions (gravity downward, and normal force upward). Now that you have the skills to work with forces in two dimensions, we can explore what happens to weight and the normal force on a tilted surface such as an inclined plane. For inclined plane problems, it is easier breaking down the forces into their components if we rotate the coordinate system, as illustrated in . The first step when setting up the problem is to break down the force of weight into components.
When an object rests on an incline that makes an angle
         with the horizontal, the force of gravity acting on the object is divided into two components: A force acting perpendicular to the plane,
        , and a force acting parallel to the plane,
         . The perpendicular force of weight,
        , is typically equal in magnitude and opposite in direction to the normal force,
         The force acting parallel to the plane,
        , causes the object to accelerate down the incline. The force of friction,
        , opposes the motion of the object, so it acts upward along the plane.
It is important to be careful when resolving the weight of the object into components. If the angle of the incline is at an angle
         to the horizontal, then the magnitudes of the weight components are
Instead of memorizing these equations, it is helpful to be able to determine them from reason. To do this, draw the right triangle formed by the three weight vectors. Notice that the angle   of the incline is the same as the angle formed between
           and
           . Knowing this property, you can use trigonometry to determine the magnitude of the weight components
To review, the process for solving inclined plane problems is as follows:
1. Draw a sketch of the problem.
2. Identify known and unknown quantities, and identify the system of interest.
3. Draw a free-body diagram (which is a sketch showing all of the forces acting on an object) with the coordinate system rotated at the same angle as the inclined plane. Resolve the vectors into horizontal and vertical components and draw them on the free-body diagram.
4. Write Newton’s second law in the horizontal and vertical directions and add the forces acting on the object. If the object does not accelerate in a particular direction (for example, the x -direction) then Fnet x = 0. If the object does accelerate in that direction, Fnet x = ma.
5. Check your answer. Is the answer reasonable? Are the units correct?

### Practice Problems

### Section Summary

1. Friction is a contact force between systems that opposes the motion or attempted motion between them. Simple friction is proportional to the normal force N pushing the systems together. A normal force is always perpendicular to the contact surface between systems. Friction depends on both of the materials involved.
2. µs is the coefficient of static friction, which depends on both of the materials.
3. µk is the coefficient of kinetic friction, which also depends on both materials.
4. When objects rest on an inclined plane that makes an angle
           with the horizontal surface, the weight of the object can be broken into components that act perpendicular

 and parallel (
           ) to the surface of the plane.

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# Motion in Two Dimensions

## Simple Harmonic Motion

### Section Key Terms

### Hooke’s Law and Simple Harmonic Motion

Imagine a car parked against a wall. If a bulldozer pushes the car into the wall, the car will not move but it will noticeably change shape. A change in shape due to the application of a force is a deformation. Even very small forces are known to cause some deformation. For small deformations, two important things can happen. First, unlike the car and bulldozer example, the object returns to its original shape when the force is removed. Second, the size of the deformation is proportional to the force. This second property is known as Hooke’s law. In equation form, Hooke’s law is
where x is the amount of deformation (the change in length, for example) produced by the restoring force F, and k is a constant that depends on the shape and composition of the object. The restoring force is the force that brings the object back to its equilibrium position; the minus sign is there because the restoring force acts in the direction opposite to the displacement. Note that the restoring force is proportional to the deformation x. The deformation can also be thought of as a displacement from equilibrium. It is a change in position due to a force. In the absence of force, the object would rest at its equilibrium position. The force constant k is related to the stiffness of a system. The larger the force constant, the stiffer the system. A stiffer system is more difficult to deform and requires a greater restoring force. The units of k are newtons per meter (N/m). One of the most common uses of Hooke’s law is solving problems involving springs and pendulums, which we will cover at the end of this section.

### Oscillations and Periodic Motion

What do an ocean buoy, a child in a swing, a guitar, and the beating of hearts all have in common? They all oscillate. That is, they move back and forth between two points, like the ruler illustrated in . All oscillations involve force. For example, you push a child in a swing to get the motion started.
Newton’s first law implies that an object oscillating back and forth is experiencing forces. Without force, the object would move in a straight line at a constant speed rather than oscillate. Consider, for example, plucking a plastic ruler to the left as shown in . The deformation of the ruler creates a force in the opposite direction, known as a restoring force. Once released, the restoring force causes the ruler to move back toward its stable equilibrium position, where the net force on it is zero. However, by the time the ruler gets there, it gains momentum and continues to move to the right, producing the opposite deformation. It is then forced to the left, back through equilibrium, and the process is repeated until it gradually loses all of its energy. The simplest oscillations occur when the restoring force is directly proportional to displacement. Recall that Hooke’s law describes this situation with the equation F = −kx. Therefore, Hooke’s law describes and applies to the simplest case of oscillation, known as simple harmonic motion.
When you pluck a guitar string, the resulting sound has a steady tone and lasts a long time. Each vibration of the string takes the same time as the previous one. Periodic motion is a motion that repeats itself at regular time intervals, such as with an object bobbing up and down on a spring or a pendulum swinging back and forth. The time to complete one oscillation (a complete cycle of motion) remains constant and is called the period T. Its units are usually seconds.
Frequency f is the number of oscillations per unit time. The SI unit for frequency is the hertz (Hz), defined as the number of oscillations per second. The relationship between frequency and period is
As you can see from the equation, frequency and period are different ways of expressing the same concept. For example, if you get a paycheck twice a month, you could say that the frequency of payment is two per month, or that the period between checks is half a month.
If there is no friction to slow it down, then an object in simple motion will oscillate forever with equal displacement on either side of the equilibrium position. The equilibrium position is where the object would naturally rest in the absence of force. The maximum displacement from equilibrium is called the amplitude X. The units for amplitude and displacement are the same, but depend on the type of oscillation. For the object on the spring, shown in , the units of amplitude and displacement are meters.
The mass m and the force constant k are the only factors that affect the period and frequency of simple harmonic motion. The period of a simple harmonic oscillator is given by
and, because f = 1/T, the frequency of a simple harmonic oscillator is

### Solving Spring and Pendulum Problems with Simple Harmonic Motion

Before solving problems with springs and pendulums, it is important to first get an understanding of how a pendulum works.  provides a useful illustration of a simple pendulum.
Everyday examples of pendulums include old-fashioned clocks, a child’s swing, or the sinker on a fishing line. For small displacements of less than 15 degrees, a pendulum experiences simple harmonic oscillation, meaning that its restoring force is directly proportional to its displacement. A pendulum in simple harmonic motion is called a simple pendulum. A pendulum has an object with a small mass, also known as the pendulum bob, which hangs from a light wire or string. The equilibrium position for a pendulum is where the angle
       is zero (that is, when the pendulum is hanging straight down). It makes sense that without any force applied, this is where the pendulum bob would rest.
The displacement of the pendulum bob is the arc length s. The weight mg has components mg cos
         along the string and mg sin
          tangent to the arc. Tension in the string exactly cancels the component mg cos
         parallel to the string. This leaves a net restoring force back toward the equilibrium position that runs tangent to the arc and equals −mg sin
         .
For small angle oscillations of a simple pendulum, the period is

The only things that affect the period of a simple pendulum are its length and the acceleration due to gravity. The period is completely independent of other factors, such as mass or amplitude. However, note that T does depend on g. This means that if we know the length of a pendulum, we can actually use it to measure gravity! This will come in useful in .

### Practice Problems

### Section Summary

1. An oscillation is a back and forth motion of an object between two points of deformation.
2. An oscillation may create a wave, which is a disturbance that propagates from where it was created.
3. The simplest type of oscillations are related to systems that can be described by Hooke’s law.
4. Periodic motion is a repetitious oscillation.
5. The time for one oscillation is the period T.
6. The number of oscillations per unit time is the frequency
7. A mass m suspended by a wire of length L is a simple pendulum and undergoes simple harmonic motion for amplitudes less than about 15 degrees.

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# Circular and Rotational Motion

## Introduction

You may recall learning 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. In some ways, this chapter is a continuation of Newton’s laws of motion. Recall that 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, such as the car in the photo, it must be experiencing an external force. In this chapter, we explore both circular motion and rotational motion.


# Circular and Rotational Motion

## Angle of Rotation and Angular Velocity

### Section Key Terms

### Angle of Rotation

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, or the circular loop-the-loop on a roller coaster. 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  rotates about its center—each point in the object follows a circular path.
The arc length, Δ, is the distance traveled along a circular path. The radius of curvature, , is the radius of the circular path. Both are shown in .
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
   is the arc length divided by the radius of curvature.
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
   radians (or 360 degrees), and therefore has an angle of rotation of
   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 =
  rad = 360°. See  for the conversion of degrees to radians for some common angles.

### 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

    is the rate at which the angle of rotation changes. In equation form, the angular speed is
which means that an angular rotation

    occurs in a time,
  . 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

    in a short time

    so its tangential speed is
From the definition of the angle of rotation,
 , we see that
  . Substituting this into the expression for v gives
The equation
   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. See .
Now, consider another example: the tire of a moving car (see ). 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.
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.

### Solving Problems Involving Angle of Rotation and Angular Velocity

Now that we have an understanding of the concepts of angle of rotation and angular velocity, we’ll apply them to the real-world situations of a clock tower and a spinning tire.

### Practice Problems

### Check Your Understanding

### Section Summary

1. Circular motion is motion in a circular path.
2. The angle of rotation
   is defined as the ratio of the arc length to the radius of curvature.
3. The arc length
   is the distance traveled along a circular path and r is the radius of curvature of the circular path.
4. The angle of rotation
   is measured in units of radians (rad), where
   revolution.
5. Angular velocity
   is the rate of change of an angle, where a rotation
   occurs in a time
  .
6. The units of angular velocity are radians per second (rad/s).
7. Tangential speed v and angular speed
   are related by
 , and tangential velocity has units of m/s.
8. The direction of angular velocity is along the axis of rotation, away (toward) from you for clockwise (counterclockwise) motion.

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# Circular and Rotational Motion

## Uniform Circular Motion

### Section Key Terms

### Centripetal Acceleration

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.
 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.
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
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
   into the equation above, we get

   . Therefore, the magnitude of centripetal acceleration in terms of the magnitude of angular velocity is

### Centripetal Force

Because an object in uniform circular motion undergoes acceleration (by changing the direction of motion but not the speed), we know from Newton’s second law of motion that there must be a net external force acting on the object. Since the magnitude of the acceleration is constant, so is the magnitude of the net force, and since the acceleration points toward the center of the rotation, so does the net force.
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.
The component of any net force that causes circular motion is called a centripetal force. When the net force is equal to the centripetal force, and its magnitude is constant, uniform circular motion results. 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
  .
By using the two different forms of the equation for the magnitude of centripetal acceleration,
   and
 , we get two expressions involving the magnitude of the centripetal force . The first expression is in terms of tangential speed, the second is in terms of angular speed:
   and
  .
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
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.

### Solving Centripetal Acceleration and Centripetal Force Problems

To get a feel for the typical magnitudes of centripetal acceleration, we’ll do a lab estimating the centripetal acceleration of a tennis racket and then, in our first Worked Example, compare the centripetal acceleration of a car rounding a curve to gravitational acceleration. For the second Worked Example, we’ll calculate the force required to make a car round a curve.

### Practice Problems

### Check Your Understanding

### Section Summary

1. Centripetal acceleration ac is the acceleration experienced while in uniform circular motion.
2. Centripetal acceleration force is a center-seeking force that always points toward the center of rotation, perpendicular to the linear velocity, in the same direction as the net force, and in the direction opposite that of the radius vector.
3. The standard unit for centripetal acceleration is m/s2.
4. Centripetal force Fc is any net force causing uniform circular motion.

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# Circular and Rotational Motion

## Rotational Motion

### Section Key Terms

### Rotational Kinematics

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
   changes. The faster the change occurs, the greater is the angular acceleration. Angular acceleration
   is the rate of change of angular velocity. In equation form, average angular acceleration is

where
   is the change in angular velocity and
   is the change in time. The units of angular acceleration are (rad/s)/s, or rad/s2. If
   increases, then
   is positive. If
   decreases, then
   is negative. Keep in mind that, by convention, counterclockwise is the positive direction and clockwise is the negative direction. For example, the skater in  is rotating 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,
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. For example, consider riders in their pods on a Ferris wheel at rest. A Ferris wheel with greater angular acceleration will give the riders greater tangential acceleration because, as the Ferris wheel increases its rate of spinning, it also increases its tangential velocity. Note that the radius of the spinning object also matters. For example, for a given angular acceleration
    , a smaller Ferris wheel leads to a smaller tangential acceleration for the riders.
So far, we have defined three rotational variables:
    ,
    , and
    .  These are the angular versions of the linear variables x, v, and a. The following equations in the table represent the magnitude of the rotational variables and only when the radius is constant and perpendicular to the rotational variable.  shows how they are related.
We can now begin to see how rotational quantities like
    ,
    , and
       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
      is large for a long period of time t, then the final angular velocity
      and angle of rotation
      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:
      (constant a).
As in linear kinematics, we assume a is constant, which means that angular acceleration
      is also a constant, because
     . The equation for the kinematics relationship between
    ,
    , and t is

where
      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 . These equations can be used to solve rotational or linear kinematics problem in which a and
      are constant.
In these equations,
      and
      are initial values,
      is zero, and the average angular velocity
      and average velocity
      are

### 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 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
   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.
Applying a stronger torque will produce a greater angular acceleration. For example, the harder the man pushes the merry-go-round in , 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.

### Solving Rotational Kinematics and Torque Problems

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.
Now let’s look at examples applying rotational kinematics to a fishing reel and the concept of torque to a merry-go-round.

### Practice Problems

### Check Your Understanding

### Section Summary

1. Kinematics is the description of motion.
2. The kinematics of rotational motion describes the relationships between rotation angle, angular velocity, angular acceleration, and time.
3. Torque is the effectiveness of a force to change the rotational speed of an object. Torque is the rotational analog of force.
4. The lever arm is the distance between the point of rotation (pivot point) and the location where force is applied.
5. Torque is maximized by applying force perpendicular to the lever arm and at a point as far as possible from the pivot point or fulcrum. If torque is zero, angular acceleration is zero.

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# Newton's Law of Gravitation

## Introduction

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. Many years later still, Albert Einstein showed there was a little more to the gravitation story when he published his theory of general relativity.


# Newton's Law of Gravitation

## Kepler's Laws of Planetary Motion

### Section Key Terms

### Concepts Related to Kepler’s Laws of Planetary Motion

Examples of orbits abound. 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:
1. The mass of the orbiting object, m, is small compared to the mass of the object it orbits, M.
2. 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.
These descriptive laws are named for the German astronomer Johannes Kepler (1571–1630). He devised them after careful study (over some 20 years) of a large amount of meticulously recorded observations of planetary motion done by Tycho Brahe (1546–1601). Such careful collection and detailed recording of methods and data are hallmarks of good science. Data constitute the evidence from which new interpretations and meanings can be constructed. Let’s look closer at each of these laws.

### 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 planet’s closest approach to the sun is called perihelion and its farthest distance from the sun is called aphelion.
If you know the aphelion (ra) and perihelion (rp) distances, then you can calculate the semi-major axis (a) and semi-minor axis (b).

### 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 .

### 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.

### Calculations Related to Kepler’s Laws of Planetary Motion

### Kepler’s First Law

Refer back to  (a). Notice which distances are constant. The foci are fixed, so distance
   is a constant. The definition of an ellipse states that the sum of the distances
   is also constant. These two facts taken together mean that the perimeter of triangle
   must also be constant. Knowledge of these constants will help you determine positions and distances of objects in a system that includes one object orbiting another.

### Kepler’s Second Law

Refer back to . The second law says that the segments have equal area and that it takes equal time to sweep through each segment. That is, the time it takes to travel from A to B equals the time it takes to travel from C to D, and so forth. Velocity v equals distance d divided by time t:
  . Then,
 , so distance divided by velocity is also a constant. For example, if we know the average velocity of Earth on June 21 and December 21, we can compare the distance Earth travels on those days.
The degree of elongation of an elliptical orbit is called its eccentricity (e). Eccentricity is calculated by dividing the distance f from the center of an ellipse to one of the foci by half the long axis a.
When
 , the ellipse is a circle.
The area of an ellipse is given by
 , where b is half the short axis. If you know the axes of Earth’s orbit and the area Earth sweeps out in a given period of time, you can calculate the fraction of the year that has elapsed.

### Kepler’s Third Law

Kepler’s third law states that the ratio of the squares of the periods of any two planets (T1, T2) is equal to the ratio of the cubes of their average orbital distance from the sun (r1, r2). Mathematically, this is represented by
From this equation, it follows that the ratio r3/T2 is the same for all planets in the solar system. Later we will see how the work of Newton leads to a value for this constant.

### Practice Problems

### Check Your Understanding

### Section Summary

1. All satellites follow elliptical orbits.
2. The line from the satellite to the parent body sweeps out equal areas in equal time.
3. The radius cubed divided by the period squared is a constant for all satellites orbiting the same parent body.

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# Newton's Law of Gravitation

## Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity

### Section Key Terms

### Concepts Related to Newton’s Law of Universal Gravitation

Sir Isaac Newton was the first scientist to precisely define the gravitational force, and to show that it could explain both falling bodies and astronomical motions. See . But Newton was not the first to suspect that the same force caused both our weight and the motion of planets. His forerunner, Galileo Galilei, had contended that falling bodies and planetary motions had the same cause. Some of Newton’s contemporaries, such as Robert Hooke, Christopher Wren, and Edmund Halley, had also made some progress toward understanding gravitation. But 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 is relatively simple. It is always attractive, and it depends only 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 .
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
  . 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.
It is possible to derive Kepler’s third law from Newton’s law of universal 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 r 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:
where M is the mass of the central body about which the satellites orbit (for example, the sun in our solar system). The usefulness of this equation will be seen later.
The universal gravitational constant G is determined experimentally. This determination 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 . Remarkably, his value for G differs by less than 1% from the modern value.

### Einstein’s Theory of General Relativity

Einstein’s theory of general relativity explained some interesting properties of gravity not covered by Newton’s theory. Einstein based his theory on the postulate that acceleration and gravity have the same effect and cannot be distinguished from each other. He concluded that light must fall in both a gravitational field and in an accelerating reference frame.  shows this effect (greatly exaggerated) in an accelerating elevator. In  (a), the elevator accelerates upward in zero gravity. In  (b), the room is not accelerating but is subject to gravity. The effect on light is the same: it “falls” downward in both situations. The person in the elevator cannot tell whether the elevator is accelerating in zero gravity or is stationary and subject to gravity. Thus, gravity affects the path of light, even though we think of gravity as acting between masses, while photons are massless.
Einstein’s theory of general relativity got its first verification in 1919 when starlight passing near the sun was observed during a solar eclipse. (See .) During an eclipse, the sky is darkened and we can briefly see stars. Those on a line of sight nearest the sun should have a shift in their apparent positions. Not only was this shift observed, but it agreed with Einstein’s predictions well within experimental uncertainties. This discovery created a scientific and public sensation. Einstein was now a folk hero as well as a very great scientist. The bending of light by matter is equivalent to a bending of space itself, with light following the curve. This is another radical change in our concept of space and time. It is also another connection that any particle with mass or energy (e.g., massless photons) is affected by gravity.
To summarize the two views of gravity, Newton envisioned gravity as a tug of war along the line connecting any two objects in the universe. In contrast, Einstein envisioned gravity as a bending of space-time by mass.

### Calculations Based on Newton’s Law of Universal Gravitation

The mass of an object is constant, but its weight varies with the strength of the gravitational field. This means the value of g varies from place to place in the universe. The relationship between force, mass, and acceleration from the second law of motion can be written in terms of g.
In this case, the force is the weight of the object, which is caused by the gravitational attraction of the planet or moon on which the object is located. We can use this expression to compare weights of an object on different moons and planets.
Two equations involving the gravitational constant, G, are often useful. The first is Newton’s equation,
  . Several of the values in this equation are either constants or easily obtainable. F is often the weight of an object on the surface of a large object with mass M, which is usually known. The mass of the smaller object, m, is often known, and G is a universal constant with the same value anywhere in the universe. This equation can be used to solve problems involving an object on or orbiting Earth or other massive celestial object. Sometimes it is helpful to equate the right-hand side of the equation to mg and cancel the m on both sides.
The equation
   is also useful for problems involving objects in orbit. Note that there is no need to know the mass of the object. Often, we know the radius r or the period T and want to find the other. If these are both known, we can use the equation to calculate the mass of a planet or star.

### Practice Problems

### Check Your Understanding

### Section Summary

1. Newton’s law of universal gravitation provides a mathematical basis for gravitational force and Kepler’s laws of planetary motion.
2. Einstein’s theory of general relativity shows that gravitational fields change the path of light and warp space and time.
3. An object’s mass is constant, but its weight changes when acceleration due to gravity, g, changes.

### Key Equations

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# Momentum

## Introduction

We know from everyday use of the word momentum that it is a tendency to continue on course in the same direction. Newscasters speak of sports teams or politicians gaining, losing, or maintaining the momentum to win. As we learned when studying about inertia, which is Newton's first law of motion, every object or system has inertia—that is, a tendency for an object in motion to remain in motion or an object at rest to remain at rest. Mass is a useful variable that lets us quantify inertia. Momentum is mass in motion.
Momentum is important because it is conserved in isolated systems; this fact is convenient for solving problems where objects collide. The magnitude of momentum grows with greater mass and/or speed. For example, look at the football players in the photograph (). They collide and fall to the ground. During their collisions, momentum will play a large part. In this chapter, we will learn about momentum, the different types of collisions, and how to use momentum equations to solve collision problems.


# Momentum

## Linear Momentum, Force, and Impulse

### Section Key Terms

### 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
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.
Recall our study of Newton’s second law of motion (Fnet = ma). 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
where Fnet is the net external force,
  is the change in momentum, and
  is the change in time.
We can solve for
  by rearranging the equation
to be

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
 . 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.
  is actually derived from the equation:
For the sake of understanding the relationship between Newton’s second law in its two forms, let’s recreate the derivation of
  from
by substituting the definitions of acceleration and momentum.
The change in momentum
  is given by
If the mass of the system is constant, then
By substituting
  for
, Newton’s second law of motion becomes
for a constant mass.
Because
we can substitute to get the familiar equation
when the mass of the system is constant.

### Solving Problems Using the Impulse-Momentum Theorem

### Practice Problems

### Check Your Understanding

### Section Summary

1. Linear momentum, often referenced as momentum for short, is defined as the product of a system’s mass multiplied by its velocity, p = mv.
2. The SI unit for momentum is kg m/s.
3. Newton’s second law of motion in terms of momentum states that the net external force equals the change in momentum of a system divided by the time over which it changes,
 .
4. Impulse is the average net external force multiplied by the time this force acts, and impulse equals the change in momentum,
 .
5. Forces are usually not constant over a period of time, so we use the average of the force over the time it acts.

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# Momentum

## Conservation of Momentum

### Section Key Terms

### 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 in some of the examples previously presented in this chapter. 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  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.
Using the impulse-momentum theorem, the change in momentum of car 1 is given by
where F1 is the force on car 1 due to car 2, and
  is the time the force acts, or the duration of the collision.
Similarly, the change in momentum of car 2 is
  where F2 is the force on car 2 due to car 1, and we assume the duration of the collision
  is the same for both cars. We know from Newton’s third law of motion that F2 = –F1, and so
 .
Therefore, the changes in momentum are equal and opposite, and
 .
Because the changes in momentum add to zero, the total momentum of the two-car system is constant. That is,
where p′1 and p′2 are the momenta of cars 1 and 2 after the collision.
This result that momentum is conserved is true not only for this example involving the two cars, but for any system where the net external force is zero, which is known as an isolated system. The law of conservation of momentum states that for an isolated system with any number of objects in it, the total momentum is conserved. In equation form, the law of conservation of momentum for an isolated system is written as
or
where ptot is the total momentum, or the sum of the momenta of the individual objects in the system at a given time, and p′tot is the total momentum some time later.
The conservation of momentum principle can be applied to systems as diverse as a comet striking the Earth or a gas containing huge numbers of atoms and molecules. Conservation of momentum appears to be violated only when the net external force is not zero. But another larger system can always be considered in which momentum is conserved by simply including the source of the external force. For example, in the collision of two cars considered above, the two-car system conserves momentum while each one-car system does not.

### Check Your Understanding

### Section Summary

1. The law of conservation of momentum is written ptot = constant or ptot = p′tot (isolated system), where ptot is the initial total momentum and p′tot is the total momentum some time later.
2. In an isolated system, the net external force is zero.
3. Conservation of momentum applies only when the net external force is zero, within the defined system.

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# Momentum

## Elastic and Inelastic Collisions

### Section Key Terms

### 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.  shows an elastic collision where momentum is conserved.
An animation of an elastic collision between balls can be seen by watching this video. It replicates the elastic collisions between balls of varying masses.
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 kinetic energy is not conserved. A perfectly inelastic collision (also sometimes called completely or maximally inelastic) is one in which objects stick together after impact, and the maximum amount of kinetic energy is lost. 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.  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.
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.

### Solving Collision Problems

The Khan Academy videos referenced in this section show examples of elastic and inelastic collisions in one dimension. In one-dimensional collisions, the incoming and outgoing velocities are all along the same line. 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.
One complication with two-dimensional collisions is that the objects might rotate before or after their collision. For example, if two ice skaters hook arms as they pass each other, they will spin in circles. We will not consider such rotation until later, and so for now, we arrange things so that no rotation is possible. To avoid rotation, we consider only the scattering of point masses—that is, structureless particles that cannot rotate or spin.
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 . Because momentum is conserved, the components of momentum along the x- and y-axes, displayed as pand p, will also be conserved. With the chosen coordinate system, pis initially zero and pis the momentum of the incoming particle.
Now, we will take the conservation of momentum equation, p1 + p2 = p′1 + p′2  and break it into its x and y components.
Along the x-axis, the equation for conservation of momentum is
In terms of masses and velocities, this equation is
But because particle 2 is initially at rest, this equation becomes
The components of the velocities along the x-axis have the form v cos θ . Because particle 1 initially moves along the x-axis, we find v1= v1. Conservation of momentum along the x-axis gives the equation
where
 and
 are as shown in .
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
 . Therefore, conservation of momentum along the y-axis gives the following equation:

### Practice Problems

### Check Your Understanding

### Section Summary

1. If kinetic energy is conserved, the collision is elastic. If they stick together, the collision is perfectly inelastic.
2. Kinetic energy is conserved in an elastic collision, but not in an inelastic collision.
3. The approach to two-dimensional collisions is to choose a convenient coordinate system and break the motion into components along perpendicular axes. Choose a coordinate system with the x-axis parallel to the velocity of the incoming particle.
4. Two-dimensional collisions of point masses, where mass 2 is initially at rest, conserve momentum along the initial direction of mass 1, or the x-axis, and along the direction perpendicular to the initial direction, or the y-axis.
5. Point masses are structureless particles that cannot spin.

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# Work, Energy, and Simple Machines

## Introduction

Roller coasters have provided thrills for daring riders around the world since the nineteenth century. Inventors of roller coasters used simple physics to build the earliest examples using railroad tracks on mountainsides and old mines. Modern roller coaster designers use the same basic laws of physics to create the latest amusement park favorites. Physics principles are used to engineer the machines that do the work to lift a roller coaster car up its first big incline before it is set loose to roll. Engineers also have to understand the changes in the car’s energy that keep it speeding over hills, through twists, turns, and even loops.
What exactly is energy? How can changes in force, energy, and simple machines move objects like roller coaster cars? How can machines help us do work? In this chapter, you will discover the answer to this question and many more, as you learn about work, energy, and simple machines.


# Work, Energy, and Simple Machines

## Work, Power, and the Work–Energy Theorem

### Section Key Terms

### The Work–Energy Theorem

In physics, the term work has a very specific definition. Work is application of force,  , to move an object over a distance, d, in the direction that the force is applied. Work, W, is described by the equation
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.
1. Homework is not work.
2. Lifting a rock upwards off the ground is work.
3. 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. Recall from the laws of motion 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.
1. Kinetic energy is also called energy of motion. A moving object has kinetic energy.
2. Potential energy, sometimes called stored energy, comes in several forms. Gravitational potential energy is the stored energy an object has as a result of its position above Earth’s surface (or another object in space). A roller coaster car at the top of a hill has gravitational potential energy.
Let’s examine how doing work on an object changes the object’s energy. If we apply force to lift a rock off the ground, we increase the rock’s potential energy, PE. If we drop the rock, the force of gravity increases the rock’s kinetic energy as the rock moves downward until it hits the ground.
The force we exert to lift the rock is equal to its weight, w, which is equal to its mass, m, multiplied by acceleration due to gravity, g.
The work we do on the rock equals the force we exert multiplied by the distance, d, that we lift the rock. The work we do on the rock also equals the rock’s gain in gravitational potential energy, PE.
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
  . 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, shown in .
Does the name Joule sound familiar? 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 kg∙m/s2, so 1.0 J = 1.0 kg∙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.
Let’s consider an example that can help illustrate the differences among work, force, and power. Suppose the woman in  lifting the TV with a pulley gets the TV to the fourth floor in two minutes, and the man carrying the TV up the stairs takes five minutes to arrive at the same place. They have done the same amount of work
   on the TV, because they have moved the same mass over the same vertical distance, which requires the same amount of upward force. However, the woman using the pulley has generated more power. This is because she did the work in a shorter amount of time, so the denominator of the power formula, t, is smaller. (For simplicity’s sake, we will leave aside for now the fact that the man climbing the stairs has also done work on himself.)
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.
The watt unit was named after James Watt (1736–1819) (see ). He was a Scottish engineer and inventor who discovered how to coax more power out of steam engines.
Before proceeding, be sure you understand the distinctions among force, work, energy, and power. Force exerted on an object over a distance does work. Work can increase energy, and energy can do work. Power is the rate at which work is done.

### Practice Problems

### Check Your Understanding

### Section Summary

1. Doing work on a system or object changes its energy.
2. The work–energy theorem states that an amount of work that changes the velocity of an object is equal to the change in kinetic energy of that object.The work–energy theorem states that an amount of work that changes the velocity of an object is equal to the change in kinetic energy of that object.
3. Power is the rate at which work is done.

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