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### 7.3 : Motion Characteristics for Circular Motion

### Example

### Lesson 1: Motion Characteristics for Circular Motion

### Lesson 1: Motion Characteristics for Circular Motion

### Lesson 2

### Now consider the following two problems pertaining to this physical scenario of the car making a turn on a horizontal surface

### Suggested Method of Solving

### Examples

Parts of the Chapter

7.1 Forces n 2D

Equilibrium and

Inclined Planes

Independence of 2D Motion,

Projectiles Horizontally Launched and

Launched at an Angle

7.2 Projectile Motion

7.3

Circular Motion

Centripetal Acceleration,

Uniform Circular Motion, and

Torque

4.11 Equilibrium Application of Newton’s Laws of Motion

Definition of Equilibrium

An object is in equilibrium when it has zero

acceleration.

4.11 Equilibrium Application of Newton’s Laws of Motion

- Reasoning Strategy
- Select an object(s) to which the equations of equilibrium are
- to be applied.
- Draw a free-body diagram for each object chosen above.
- Include only forces acting on the object, not forces the object
- exerts on its environment.
- Choose a set of x, y axes for each object and
- resolve all forces
- in the free-body diagram into components that point
- along these
- axes.
- Apply the equations and solve for the
- unknown quantities.

4.11 Equilibrium Application of Newton’s Laws of Motion

The first equation gives

Substitution into the second gives

4.12 Nonequilibrium Application of Newton’s Laws of Motion

When an object is accelerating, it is not in equilibrium.

4.12 Nonequilibrium Application of Newton’s Laws of Motion

The acceleration is along the x axis so

4.11.1. Consider the following: (i) the book is at rest, (ii) the book is moving at a constant velocity, (iii) the book is moving with a constant acceleration. Under which of these conditions is the book in equilibrium?

a) (i) only

b) (ii) only

c) (iii) only

d) (i) and (ii) only

e) (ii) and (iii) only

4.11.1. Consider the following: (i) the book is at rest, (ii) the book is moving at a constant velocity, (iii) the book is moving with a constant acceleration. Under which of these conditions is the book in equilibrium?

a) (i) only

b) (ii) only

c) (iii) only

d) (i) and (ii) only

e) (ii) and (iii) only

4.11.2. A block of mass M is hung by ropes as shown. The system is in equilibrium. The point O represents the knot, the junction of the three ropes. Which of the following statements is true concerning the magnitudes of the three forces in equilibrium?

a) F1 + F2 = F3

b) F1 = F2 = 0.5×F3

c) F1 = F2 = F3

d) F1 > F3

e) F2 < F3

4.11.2. A block of mass M is hung by ropes as shown. The system is in equilibrium. The point O represents the knot, the junction of the three ropes. Which of the following statements is true concerning the magnitudes of the three forces in equilibrium?

a) F1 + F2 = F3

b) F1 = F2 = 0.5×F3

c) F1 = F2 = F3

d) F1 > F3

e) F2 < F3

4.11.3. A team of dogs pulls a sled of mass 2m with a force . A second sled of mass m is attached by a rope and pulled behind the first sled. The tension in the rope is . Assuming frictional forces are too small to consider, determine the ratio of the magnitudes of the forces and , that is, P/T.

a) 3

b) 2

c) 1

d) 0.5

e) 0.33

4.11.3. A team of dogs pulls a sled of mass 2m with a force . A second sled of mass m is attached by a rope and pulled behind the first sled. The tension in the rope is . Assuming frictional forces are too small to consider, determine the ratio of the magnitudes of the forces and , that is, P/T.

a) 3

b) 2

c) 1

d) 0.5

e) 0.33

Motion along an Inclined Plane

See pp. 152 – 154 in text

3.3 Projectile Motion

Under the influence of gravity alone, an object near the

surface of the Earth will accelerate downwards at 9.80m/s2.

3.3 Projectile Motion

Example 3 A Falling Care Package

The airplane is moving horizontally with a constant velocity of

+115 m/s at an altitude of 1050m. Determine the time required

for the care package to hit the ground.

3.3 Projectile Motion

Example 4 The Velocity of the Care Package

What are the magnitude and direction of the final velocity of

the care package?

3.3 Projectile Motion

Conceptual Example 5 I Shot a Bullet into the Air...

Suppose you are driving a convertible with the top down.

The car is moving to the right at constant velocity. You point

a rifle straight up into the air and fire it. In the absence of air

resistance, where would the bullet land – behind you, ahead

of you, or in the barrel of the rifle?

3.3 Projectile Motion

Example 6 The Height of a Kickoff

A placekicker kicks a football at and angle of 40.0 degrees and

the initial speed of the ball is 22 m/s. Ignoring air resistance,

determine the maximum height that the ball attains.

3.3 Projectile Motion

Example 7 The Time of Flight of a Kickoff

What is the time of flight between kickoff and landing?

3.3 Projectile Motion

Conceptual Example 10 Two Ways to Throw a Stone

From the top of a cliff, a person throws two stones. The stones

have identical initial speeds, but stone 1 is thrown downward

at some angle above the horizontal and stone 2 is thrown at

the same angle below the horizontal. Neglecting air resistance,

which stone, if either, strikes the water with greater velocity?

3.3.1. A football is kicked at an angle 25 with respect to the horizontal. Which one of the following statements best describes the acceleration of the football during this event if air resistance is neglected?

a) The acceleration is zero m/s2 at all times.

b) The acceleration is zero m/s2 when the football has reached the highest point in its trajectory.

c) The acceleration is positive as the football rises, and it is negative as the football falls.

d) The acceleration starts at 9.8 m/s2 and drops to some constant lower value as the ball approaches the ground.

e) The acceleration is 9.8 m/s2 at all times.

3.3.1. A football is kicked at an angle 25 with respect to the horizontal. Which one of the following statements best describes the acceleration of the football during this event if air resistance is neglected?

a) The acceleration is zero m/s2 at all times.

b) The acceleration is zero m/s2 when the football has reached the highest point in its trajectory.

c) The acceleration is positive as the football rises, and it is negative as the football falls.

d) The acceleration starts at 9.8 m/s2 and drops to some constant lower value as the ball approaches the ground.

e) The acceleration is 9.8 m/s2 at all times.

3.3.2. A baseball is hit upward and travels along a parabolic arc before it strikes the ground. Which one of the following statements is necessarily true?

a) The velocity of the ball is a maximum when the ball is at the highest point in the arc.

b) The x-component of the velocity of the ball is the same throughout the ball\'s flight.

c) The acceleration of the ball decreases as the ball moves upward.

d) The velocity of the ball is zero m/s when the ball is at the highest point in the arc.

e) The acceleration of the ball is zero m/s2 when the ball is at the highest point in the arc.

3.3.2. A baseball is hit upward and travels along a parabolic arc before it strikes the ground. Which one of the following statements is necessarily true?

a) The velocity of the ball is a maximum when the ball is at the highest point in the arc.

b) The x-component of the velocity of the ball is the same throughout the ball\'s flight.

c) The acceleration of the ball decreases as the ball moves upward.

d) The velocity of the ball is zero m/s when the ball is at the highest point in the arc.

e) The acceleration of the ball is zero m/s2 when the ball is at the highest point in the arc.

3.3.3. Two cannons are mounted on a high cliff. Cannon A fires balls with twice the initial velocity of cannon B. Both cannons are aimed horizontally and fired. How does the horizontal range of cannon A compare to that of cannon B?

a) The range for both balls will be the same

b) The range of the cannon ball B is about 0.7 that of cannon ball A.

c) The range of the cannon ball B is about 1.4 times that of cannon

ball A.

d) The range of the cannon ball B is about 2 times that of cannon ball A.

e) The range of the cannon ball B is about 0.5 that of cannon ball A.

3.3.3. Two cannons are mounted on a high cliff. Cannon A fires balls with twice the initial velocity of cannon B. Both cannons are aimed horizontally and fired. How does the horizontal range of cannon A compare to that of cannon B?

a) The range for both balls will be the same

b) The range of the cannon ball B is about 0.7 that of cannon ball A.

c) The range of the cannon ball B is about 1.4 times that of cannon

ball A.

d) The range of the cannon ball B is about 2 times that of cannon ball A.

e) The range of the cannon ball B is about 0.5 that of cannon ball A.

3.3.4. Which one of the following statements concerning the range of a football is true if the football is kicked at an angle with an initial speed v0?

a) The range is independent of initial speed v0.

b) The range is only dependent on the initial speed v0.

c) The range is independent of the angle.

d) The range is only dependent on the angle.

e) The range is dependent on both the initial speed v0 and the angle.

3.3.4. Which one of the following statements concerning the range of a football is true if the football is kicked at an angle with an initial speed v0?

a) The range is independent of initial speed v0.

b) The range is only dependent on the initial speed v0.

c) The range is independent of the angle.

d) The range is only dependent on the angle.

e) The range is dependent on both the initial speed v0 and the angle.

3.3.5. A bullet is aimed at a target on the wall a distance L away from the firing position. Because of gravity, the bullet strikes the wall a distance Δybelow the mark as suggested in the figure. Note: The drawing is not to scale. If the distance L was half as large, and the bullet had the same initial velocity, how would Δybe affected?

a) Δy will double.

b) Δywill be half as large.

c) Δywill be one fourth

as large.

d) Δywill be four times larger.

e) It is not possible to determine unless numerical values are given for the distances.

3.3.1. A bicyclist is riding at a constant speed along a horizontal, straight-line path. The rider throws a ball straight up to a height a few meters above her head. Ignoring air resistance, where will the ball land?

a) in front of the rider

b) behind the rider

c) in the same hand that threw the ball

d) in the opposite hand to the one that threw it

e) This cannot be determined without knowing the speed of the rider and the maximum height of the ball.

3.3.1. A bicyclist is riding at a constant speed along a horizontal, straight-line path. The rider throws a ball straight up to a height a few meters above her head. Ignoring air resistance, where will the ball land?

a) in front of the rider

b) behind the rider

c) in the same hand that threw the ball

d) in the opposite hand to the one that threw it

e) This cannot be determined without knowing the speed of the rider and the maximum height of the ball.

3.3.2. Football A is kicked at a speed v at an angle of with respect to the horizontal direction. If football B is kicked at the same angle, but with a speed 2v, what is the ratio of the range of B to the range of A?

a) 1

b) 2

c) 3

d) 4

e) 9

3.3.2. Football A is kicked at a speed v at an angle of with respect to the horizontal direction. If football B is kicked at the same angle, but with a speed 2v, what is the ratio of the range of B to the range of A?

a) 1

b) 2

c) 3

d) 4

e) 9

3.3.3. Balls A, B, and C are identical. From the top of a tall building, ball A is launched with a velocity of 20 m/s at an angle of 45 above the horizontal direction, ball B is launched with a velocity of 20 m/s in the horizontal direction, and ball C is launched with a velocity of 20 m/s at an angle of 45 below the horizontal direction. Which of the following choices correctly relates the magnitudes of the velocities of the balls just before they hit the ground below? Ignore any effects of air resistance.

a) vA = vC > vB

b) vA = vC = vB

c) vA > vC > vB

d) vA < vC < vB

e) vA > vC < vB

3.3.3. Balls A, B, and C are identical. From the top of a tall building, ball A is launched with a velocity of 20 m/s at an angle of 45 above the horizontal direction, ball B is launched with a velocity of 20 m/s in the horizontal direction, and ball C is launched with a velocity of 20 m/s at an angle of 45 below the horizontal direction. Which of the following choices correctly relates the magnitudes of the velocities of the balls just before they hit the ground below? Ignore any effects of air resistance.

a) vA = vC > vB

b) vA = vC = vB

c) vA > vC > vB

d) vA < vC < vB

e) vA > vC < vB

3.3.4. A basketball is launched with an initial speed of 8.5 m/s and follows the trajectory shown. The ball enters the basket 0.92 s after it is launched. What are the distances x and y? Note: The drawing is not to scale.

a) x = 6.0 m, y = 0.88 m

b) x = 5.4 m, y = 0.73 m

c) x = 5.7 m, y = 0.91 m

d) x = 7.6 m, y = 1.1 m

e) x = 6.3 m, y = 0.96 m

3.3.4. A basketball is launched with an initial speed of 8.5 m/s and follows the trajectory shown. The ball enters the basket 0.92 s after it is launched. What are the distances x and y? Note: The drawing is not to scale.

a) x = 6.0 m, y = 0.88 m

b) x = 5.4 m, y = 0.73 m

c) x = 5.7 m, y = 0.91 m

d) x = 7.6 m, y = 1.1 m

e) x = 6.3 m, y = 0.96 m

3.3.5. A physics student standing on the edge of a cliff throws a stone vertically downward with an initial speed of 10.0 m/s. The instant before the stone hits the ground below, it is traveling at a speed of 30.0 m/s. If the physics student were to throw the rock horizontally outward from the cliff instead, with the same initial speed of 10.0 m/s, what is the magnitude of the velocity of the stone just before it hits the ground? Ignore any effects of air resistance.

a) 10.0 m/s

b) 20.0 m/s

c) 30.0 m/s

d) 40.0 m/s

e) The height of the cliff must be specified to answer this question.

3.3.5. A physics student standing on the edge of a cliff throws a stone vertically downward with an initial speed of 10.0 m/s. The instant before the stone hits the ground below, it is traveling at a speed of 30.0 m/s. If the physics student were to throw the rock horizontally outward from the cliff instead, with the same initial speed of 10.0 m/s, what is the magnitude of the velocity of the stone just before it hits the ground? Ignore any effects of air resistance.

a) 10.0 m/s

b) 20.0 m/s

c) 30.0 m/s

d) 40.0 m/s

e) The height of the cliff must be specified to answer this question.

3.3.5. At time t = 0 s, Ball A is thrown vertically upward with an initial speed v0A. Ball B is thrown vertically upward shortly after Ball A at time t. Ball B passes Ball A just as Ball A is reaching the top of its trajectory. What is the initial speed v0B of Ball B in terms of the given parameters? The acceleration due to gravity is g.

a) v0B = v0A (1/2)gt2

b) v0B = v0A (1/2)gt

c)

d)

e) v0B = 2v0Agt

3.3.5. At time t = 0 s, Ball A is thrown vertically upward with an initial speed v0A. Ball B is thrown vertically upward shortly after Ball A at time t. Ball B passes Ball A just as Ball A is reaching the top of its trajectory. What is the initial speed v0B of Ball B in terms of the given parameters? The acceleration due to gravity is g.

a) v0B = v0A (1/2)gt2

b) v0B = v0A (1/2)gt

c)

d)

e) v0B = 2v0Agt

3.3.6. A toy rocket is launched at an angle of 45 with a speed v0. If there is no air resistance, at what point during the time that it is in the air does the speed of the rocketequal 0.5v0?

a) when the rocket is at one half of its maximum height as it is going upward

b) when the rocket is at one half of its maximum height as it is going downward

c) when the rocket is at its maximum height

d) when the rocket is at one fourth of its maximum height as it is going downward

e) at no time during the flight

3.3.6. A toy rocket is launched at an angle of 45 with a speed v0. If there is no air resistance, at what point during the time that it is in the air does the speed of the rocketequal 0.5v0?

a) when the rocket is at one half of its maximum height as it is going upward

b) when the rocket is at one half of its maximum height as it is going downward

c) when the rocket is at its maximum height

d) when the rocket is at one fourth of its maximum height as it is going downward

e) at no time during the flight

3.3.7. During a high school track meet, an athlete performing the long jump runs and leaps at an angle of 25 and lands in a sand pit 8.5 m from his launch point. If the launch point and landing points are at the same height, y = 0 m, with what speed does the athlete land?

a) 6 m/s

b) 8 m/s

c) 10 m/s

d) 2 m/s

e) 4 m/s

3.3.7. During a high school track meet, an athlete performing the long jump runs and leaps at an angle of 25 and lands in a sand pit 8.5 m from his launch point. If the launch point and landing points are at the same height, y = 0 m, with what speed does the athlete land?

a) 6 m/s

b) 8 m/s

c) 10 m/s

d) 2 m/s

e) 4 m/s

3.3.8. An airplane is flying horizontally at a constant velocity when a package is dropped from its cargo bay. Assuming no air resistance, which one of the following statements is correct?

a) The package follows a curved path that lags behind the airplane.

b) The package follows a straight line path that lags behind the airplane.

c) The package follows a straight line path, but it is always vertically below the airplane.

d) The package follows a curved path, but it is always vertically below the airplane.

e) The package follows a curved path, but its horizontal position varies depending on the velocity of the airplane.

3.3.8. An airplane is flying horizontally at a constant velocity when a package is dropped from its cargo bay. Assuming no air resistance, which one of the following statements is correct?

a) The package follows a curved path that lags behind the airplane.

b) The package follows a straight line path that lags behind the airplane.

c) The package follows a straight line path, but it is always vertically below the airplane.

d) The package follows a curved path, but it is always vertically below the airplane.

e) The package follows a curved path, but its horizontal position varies depending on the velocity of the airplane.

3.3.9. In making a movie, a stuntman has to jump from one roof onto another roof, located 2.0 m below. The buildings are separated by a distance of 2.5 m. What is the minimum horizontal speed that the stuntman must have when jumping from the first roof to have a successful jump?

a) 3.9 m/s

b) 2.5 m/s

c) 4.3 m/s

d) 4.5 m/s

e) 3.1 m/s

3.3.9. In making a movie, a stuntman has to jump from one roof onto another roof, located 2.0 m below. The buildings are separated by a distance of 2.5 m. What is the minimum horizontal speed that the stuntman must have when jumping from the first roof to have a successful jump?

a) 3.9 m/s

b) 2.5 m/s

c) 4.3 m/s

d) 4.5 m/s

e) 3.1 m/s

3.3.10. When a projectile is launched at an angle from a height h1 and the projectile lands at the same height, the maximum range, in the absence of air resistance, occurs when = 45. The same projectile is then launched at an angle from a height h1, but it lands at a height h2 that is higher than h1, but less than the maximum height reached by the projectile when = 45. In this case, in the absence of air resistance, does the maximum range still occur for = 45? All angles are measured with respect to the horizontal direction.

a) Yes, = 45 will always have longest range regardless of the height h2.

b) No, depending on the height h2, the longest range may be reached for angles less than 45.

c) No, depending on the height h2, the longest range may be reached for angles greater than 45.

3.3.10. When a projectile is launched at an angle from a height h1 and the projectile lands at the same height, the maximum range, in the absence of air resistance, occurs when = 45. The same projectile is then launched at an angle from a height h1, but it lands at a height h2 that is higher than h1, but less than the maximum height reached by the projectile when = 45. In this case, in the absence of air resistance, does the maximum range still occur for = 45? All angles are measured with respect to the horizontal direction.

a) Yes, = 45 will always have longest range regardless of the height h2.

b) No, depending on the height h2, the longest range may be reached for angles less than 45.

c) No, depending on the height h2, the longest range may be reached for angles greater than 45.

3.3.11. Packages A and B are dropped from the same height simultaneously. Package A is dropped from an airplane that is flying due east at constant speed. Package B is dropped from rest from a helicopter hovering in a stationary position above the ground. Ignoring air friction effects, which of the following statements is true?

a) A and B reach the ground at the same time, but B has a greater velocity in the vertical direction.

b) A and B reach the ground at the same time; and they have the same velocity in the vertical direction.

c) A and B reach the ground at different times because B has a greater velocity in both the horizontal and vertical directions.

d) A and B reach the ground at different times; and they have the same velocity in the vertical direction.

e) A reaches the ground first because it falls straight down, while B has to travel much further than A.

3.3.11. Packages A and B are dropped from the same height simultaneously. Package A is dropped from an airplane that is flying due east at constant speed. Package B is dropped from rest from a helicopter hovering in a stationary position above the ground. Ignoring air friction effects, which of the following statements is true?

a) A and B reach the ground at the same time, but B has a greater velocity in the vertical direction.

b) A and B reach the ground at the same time; and they have the same velocity in the vertical direction.

c) A and B reach the ground at different times because B has a greater velocity in both the horizontal and vertical directions.

d) A and B reach the ground at different times; and they have the same velocity in the vertical direction.

e) A reaches the ground first because it falls straight down, while B has to travel much further than A.

3.3.5. A bullet is aimed at a target on the wall a distance L away from the firing position. Because of gravity, the bullet strikes the wall a distance Δy below the mark as suggested in the figure. Note: The drawing is not to scale. If the distance L was half as large, and the bullet had the same initial velocity, how would Δy be affected?

a) Δy will double.

b) Δy will be half as large.

c) Δy will be one fourth

as large.

d) Δy will be four times larger.

e) It is not possible to determine unless numerical values are given for the distances.

Speed and Velocity

Any moving object can be described using the kinematic concepts discussed in Unit 1. The motion of a moving object can be explained using either Newton\'s Laws (Unit 2) and vector principles (Unit 3) or by means of the Work-Energy Theorem (Ei +Wext =Ef ) . The same concepts and principles used to describe and explain the motion of an object can be used to describe and explain the parabolic motion of a projectile

In this unit, we will see that these same concepts and principles can also be used to describe and explain the motion of objects which either move in circles or can be approximated to be moving in circles. Kinematic concepts and motion principles will be applied to the motion of objects in circles and then extended to analyze the motion of such objects as roller coaster cars, a football player making a circular turn, and a planet orbiting the sun. We will see that the beauty and power of physics lies in the fact that a few simple concepts and principles can be used to explain the mechanics of the entire universe. Lesson 1 of this study will begin with the development of kinematic and dynamic ideas can be used to describe and explain the motion of objects in circles.

Suppose that you were driving a car with the steering wheel turned in such a manner that your car followed the path of a perfect circle with a constant radius. And suppose that as you drove, your speedometer maintained a constant reading of 10 mi/hr. In such a situation as this, the motion of your car would be described to be experiencing uniform circular motion. Uniform circular motion is the motion of an object in a circle with a constant or uniform speed.

Uniform circular motion - circular motion at a constant speed - is one of many forms of circular motion. An object moving in uniform circular motion would cover the same linear distance in each second of time. When moving in a circle, an object traverses a distance around the perimeter of the circle. So if your car were to move in a circle with a constant speed of 5 m/s, then the car would travel 5 meters along the perimeter of the circle in each second of time.

The distance of one complete cycle around the perimeter of a circle is known as the circumference. At a uniform speed of 5 m/s, if the circle had a circumference of 5 meters, then it would take the car 1 second to make a complete cycle around the circle. At this uniform speed of 5 m/s, each cycle around the 5-m circumference circle would require 1 second. At 5 m/s, a circle with a circumference of 20 meters could be made in 4 seconds; and at this uniform speed, every cycle around the 20-m circumference of the circle would take the same time period of 4 seconds. This relationship between the circumference of a circle, the time to complete one cycle around the circle, and the speed of the object is merely an extension of the average speed equation stated in Unit 1.

The circumference of any circle can be computed using from the radius according to the equation

Circumference = 2*pi*Radius

Combining these two equations above will lead to a new equation relating the speed of an object moving in uniform circular motion to the radius of the circle and the time to make one cycle around the circle (period).

where R represents the radius of the circle and T represents the period.

This equation, like all equations, can be used as a algebraic recipe for problem solving. Yet it also can be used to guide our thinking about the variables in the equation relate to each other. For instance, the equation suggests that for objects moving around circles of different radius in the same period, the object traversing the circle of larger radius must be traveling with the greatest speed. In fact, the average speed and the radius of the circle are directly proportional. A twofold increase in radius corresponds to a twofold increase in speed; a threefold increase in radius corresponds to a three--fold increase in speed; and so on.

Objects moving in uniform circular motion will have a constant speed. But does this mean that they will have a constant velocity? Recall from Unit 1 that speed and velocity refer to two distinctly different quantities. Speed is a scalar quantity and velocity is a vector quantity. Velocity, being a vector, has both a magnitude and a direction. The magnitude of the velocity vector is merely the instantaneous speed of the object; the direction of the velocity vector is directed in the same direction which the object moves. Since an object is moving in a circle, its direction is continuously changing. At one moment, the object is moving northward such that the velocity vector is directed northward. One quarter of a cycle later, the object would be moving eastward such that the velocity vector is directed eastward. As the object rounds the circle, the direction of the velocity vector is different than it was the instant before. So while the magnitude of the velocity vector may be constant, the direction of the velocity vector is changing.

The best word that can be used to describe the direction of the velocity vector is the word tangential. The direction of the velocity vector at any instant is in the direction of a tangent line drawn to the circle at the object\'s location. (A tangent line is a line which touches the circle at one point but does not intersect it.) The diagram at the right shows the direction of the velocity vector at four different point for an object moving in a clockwise direction around a circle. While the actual direction of the object (and thus, of the velocity vector) is changing, it\'s direction is always tangent to the circle.

To summarize, an object moving in uniform circular motion is moving around the perimeter of the circle with a constant speed. While the speed of the object is constant, its velocity is changing. Velocity, being a vector, has a constant magnitude but a changing direction. The direction is always directed tangent to the circle and as the object turns the circle, the tangent line is always pointing in a new direction. As we proceed through this unit, we will see that these same principles will have a similar extension to noncircular motion.

Check your understanding

Example 1

- A spiraled tube lies fixed in its horizontal position (i.e., it has been placed upon its side upon a table). When a marble is rolled through it curves around the tube, draw the path of the marble after it exits the tube.

Answer 1

- The ball will move along a path which is tangent to the circle at the point where it exits the tube. At that point, the ball will no longer curve or spiral, but rather travel in a straight line in the tangential direction.

Acceleration

As mentioned earlier in Lesson 1, an object moving in uniform circular motion is moving in a circle with a uniform or constant speed. The velocity vector is constant in magnitude but changing in direction. Because the speed is constant for such a motion, many students have the misconception that there is no acceleration. "After all," they might say, "if I were driving a car in a circle at a constant speed of 20 mi/hr, then the speed is not decreasing or increasing; therefore there must not be an acceleration."

At the heat of this common student misconception is the wrong belief that acceleration has to do with speed and not with velocity. But the fact is that an accelerating object is an object which is changing its velocity. And since velocity is a vector which has both magnitude and direction, a change in either the magnitude or the direction constitutes a change in the velocity. For this reason, it can be boldly declared that an object moving in a circle at constant speed is indeed accelerating. It is accelerating because its velocity is changing its directions.

To understand this at a deeper level, we will have to combine the definition of acceleration with a review of some basic vector principles. Recall from Unit 1 that acceleration as a quantity was defined as the rate at which the velocity of an object changes. As such, it is calculated using the following equation:

where vi represents the initial velocity and vfrepresents the final velocity after some time of t. The numerator of the equation is found by subtracting one vector (vi) from a second vector (vf). But the addition and subtraction of vectors from each other is done in a manner much different than the addition and subtraction of scalar quantities. Consider the case of an object moving in a circle about point C as shown in the diagram below. In a time of t seconds, the object has moved from point A to point B. In this time, the velocity has changed from vi to vf. The process of subtracting vi from vfis shown in the vector diagram; this process yields the change in velocity.

Note in the diagram below that there is a velocity change for an object moving in a circle with a constant speed. Furthermore, note that this velocity change vector is directed towards the center. An object moving in a circle at a constant speed from A to B experiences a velocity change and therefore an acceleration; this acceleration is directed towards point C - the center of the circle.

The acceleration of an object is often measured using a device known as an accelerometer. A simple home- made accelerometer involves a lit candle centered vertically in the middle of a open-air glass. If the glass is held level and at rest (such that there is no acceleration), then the candle flame extends in an upward direction. However, if the glass is held at the end of an outstretched arm as you spin in a circle at a constant rate (such that the flame experiences an acceleration), then the candle flame will no longer extend vertically upwards. Instead the flame deflects from its upright position.

This signifies that there is an acceleration when the flame moves in a circular path at constant speed. The deflection of the flame will be in the direction of the acceleration. This is because the hot gases of the flame are less massive (on a per mL basis) and thus have less inertia than the cooler gases which surround. Subsequently, the hotter and lighter gases of the flame experience the greater acceleration and will lurch towards the direction of the acceleration.

A careful examination of the flame reveals that the flame will point towards the center of the circle, thus indicating that not only is there an acceleration; but that there is an inward acceleration. Objects moving in a circle at a constant speed experience an acceleration which is directed towards the center of the circle.

Cork in water demonstration

- A further demonstrations of this principle was performed in class using a cork accelerometer. A cork was submerged in a sealed flask of water. The flask was then held in an outstretched arm and moved in a circle at a constant rate of turning. Thus, the flask with both the water and the cork were moving in uniform circular motion. Again, the least massive of the two objects will lean in the direction of the acceleration.

In the case of the cork and the water, the cork is least massive (on a per mL basis) and thus it experiences the greater acceleration. As the cork-water combination spun in the circle, the cork leaned towards the center of the circle. Once more, there is proof that an object moving in circular motion at constant speed experiences an acceleration which directed towards the center of the circle.

So thus far, we have seen a geometric proof and two real-world demonstrations of this inward acceleration. At this point it becomes the decision of the student to believe or not to believe. Is it sensible that an object moving in a circle experiences an acceleration which is directed towards the center of the circle? Can you think of a logical reason to believe in say no acceleration or even an outward acceleration experienced by an object moving in uniform circular motion? In the next part of Lesson 1, additional logical evidence will be presented to support the notion of an inward force for an object moving in circular motion.

1. The initial and final speed of a ball at two different points in time is shown below. The direction of the ball is indicated by the arrow. For each case, indicate if there is an acceleration. Explain why or why not. Indicate the direction of the acceleration.

Answer (a)

- Since the velocity did not change, there is no acceleration.

Answer b

- Since the velocity changes (speeds up) there is acceleration.
- If an object moving rightward speeds up then the acceleration is also rightward.

Answer c

- Since there is a change in velocity there is an acceleration.
- A rightward moving object that slows down has a leftward acceleration.

Answer d

- Since there is a change in velocity (decrease in speed) then there is acceleration.
- A rightward moving object that slows down has a leftward acceleration.

Answer e

- Even though the initial and final speeds are the same, there has been a change in direction, so there is an acceleration.
- The object was moving in a rightward direction until it slowed to 0 m/s then it changed to a leftward movement, so the acceleration is leftward.

2. Explain the connection between your answers to the above questions and the reasoning used to explain why an object moving in a circle at constant speed can be said to experience an acceleration.

Answer 2

- An object that either experiences a change in magnitude or direction of the velocity vector can be said to be accelerating. This explains why an object moving in a circle with constant speed can be said to accelerate—the velocity changes.

Ex. 3

- 3. Dizzy Smith and Hector Vector are still discussing #1e. Dizzy says that the ball is not accelerating because its velocity is not changing. Hector says that since the ball has changed its direction, there is an acceleration. Who do you agree with? Argue a position by explaining the discrepancy in the other student\'s argument.

Answer 3

- Agree with Hector. A change in direction constitutes a change in velocity

4. Identify the three controls on an automobile which allow the car to be accelerated.

Answer 4

- Accelerator allows the car to speed up.
- Brake allows for the car to slow down.
- Steering wheel allows for change in direction.

For questions #5-#8: An object is moving in a clockwise direction around a circle at constant speed. Use your understanding of the concepts of velocity and acceleration to answer the next four questions. Use the diagram shown at the right.

5. Which vector below represents the direction of the velocity vector when the object is located at point B on the circle?

Answer 5

- Answer D
- The velocity vector is directed tangent to the circle that would be downward when at point B.

6. Which vector below represents the direction of the acceleration vector when the object is located at point C on the circle?

Answer 6

- B The acceleration vector would be directed towards the center that would be up and to the right when at point C

7. Which vector below represents the direction of the velocity vector when the object is located at point C on the circle?

Answer 7

- A The velocity vector would be directed tangent to the circle and that would be upwards at point C.

8. Which vector below represents the direction of the acceleration vector when the object is located at point A on the circle?

Answer 8

- D the acceleration vector would be directed towards the center and that would be straight down when the object is at point A

The Centripetal Force Requirement

As mentioned earlier in this lesson, an object moving in a circle is experiencing an acceleration. Even if moving around the perimeter of the circle with a constant speed, there is still a change in velocity and subsequently an acceleration. This acceleration is directed towards the center of the circle. And in accord with Newton\'s second law of motion, an object which experiences an acceleration must also be experiencing a net force; and the direction of the net force is in the same direction as the acceleration.

So for an object moving in a circle, there must be an inward force acting upon it in order to cause its inward acceleration. This is sometimes referred to as the centripetal force requirement. The word "centripetal" (not to be confused with the F-word "centrifugal") means center-seeking. For object\'s moving in circular motion, there is a net force acting towards the center which causes the object to seek the center.

To understand the need for a centripetal force, it is important to have a sturdy understanding of the Newton\'s first law of motion - the law of inertia. The law of inertia states that ...

According to Newton\'s first law of motion, it is the natural tendency of all moving objects to continue in motion in the same direction that they are moving ... unless some form of unbalanced force acts upon the object to deviate the its motion from its straight-line path. Objects will tend to naturally travel in straight lines; an unbalanced force is required to cause it to turn. The presence of the unbalanced force is required for objects to move in circles.

The idea expressed by Newton\'s law of inertia should not be surprising to us. We experience this phenomenon of inertia nearly everyday when we drive our automobile. For example, imagine that your are a passenger in a car at a traffic light. The light turns green and the driver "steps on the gas." The car begins to accelerate forward, yet relative to the seat which you are on, your body begins to lean backwards. Your body being at rest tends to stay at rest. This is one aspect of the law of inertia - "objects at rest tend to stay at rest."

As the wheels of the car spin to generate a forward force upon the car to cause a forward acceleration, your body tends to stay in place. It certainly might seem to you as though your body were experiencing a backwards force causing it to accelerate backwards; yet you would have a difficult time identifying such a backwards force on your body. Indeed there isn\'t one. The feeling of being thrown backwards is merely the tendency of your body to resist the acceleration and to remain in its state of rest. The car is accelerating out from under your body, leaving you with the false feeling of being thrown backwards.

Now imagine that you\'re driving along at constant speed and then suddenly approach a stop sign. The driver steps on the brakes. The wheels of the car lock and begin to skid across the pavement. This causes a backwards force upon the forward moving car and subsequently a backwards acceleration on the car. However, your body being in motion tends to continue in motion while the car is slowing to a stop. It certainly might seem to you as though your body were experiencing a forwards force causing it to accelerate forwards; yet you would once more have a difficult time identifying such a forwards force on your body

Indeed there is no physical object accelerating you forwards. The feeling of being thrown forwards is merely the tendency of your body to resist the deceleration and to remain in its state of forward motion. This is the second aspect of Newton\'s law of inertia - "an object in motion tends to stay in motion with the same speed and in the same direction... ." The unbalanced force acting upon the car causes it to slow down while your body continues in its forward motion.

In each case - the car starting from rest and the moving car braking to a stop - the direction which the passengers lean is opposite the direction of the acceleration. This is merely the result of the passenger\'s inertia - the tendency to resist acceleration. The passengers lean is not an acceleration in itself but rather the tendency to maintain whatever state of motion they have while the car does the acceleration. The tendency of our body to maintain its state of rest or motion while the surroundings (the car) accelerate is often misconstrued as an acceleration. This becomes particularly problematic when we consider the third possible inertia experience as a passenger in a moving automobile - the left hand turn.

Suppose you continue driving and suddenly round a sharp turn to the left at constant speed. During the turn, the car travels in a circular-type path; that is, the car sweeps out one- quarter of a circle. The unbalanced force acting upon the turned wheels of the car cause an unbalanced force upon the car and a subsequent acceleration. The unbalanced force and the acceleration are both directed towards the center of the circle about which the car is turning. Your body however is in motion and tends to stay in motion. It is the inertia of your body - the tendency to resist acceleration - which causes it to continue in its forward motion.

While the car is accelerating inward, you continue in a straight line. If you are sitting on the passenger side of the car, then eventually the outside door of the car will hit you as the car turns inward. This phenomenon might cause you to think that you were being accelerated outwards away from the center of the circle. In reality, you are continuing in your straight-line inertial path tangent to the circle while the car is accelerating out from under you. The sensation of an outward force and an outward acceleration is a false sensation. There is no physical object capable of pushing you outwards. You are merely experiencing the tendency of your body to continue in its path tangent to the circular path along which the car is turning.

Any object moving in a circle (or along a circular path) experiences a centripetal force; that is there must be some physical force pushing or pulling the object towards the center of the circle. This is the centripetal force requirement. The word "centripetal" is merely an adjective used to describe the direction of the force. We are not introducing a new type of force but rather describing the direction of the net force acting upon the object which moves in the circle. Whatever the object, if it moves in a circle, there is some force acting upon it to cause it to deviate from its straight-line path, accelerate inwards and move along a circular path. Three such examples of centripetal force are shown below.

As a car makes a turn, the force of friction acting upon the turned wheels of the car provide the centripetal force required for circular motion.

As a bucket of water is tied to a string and spun in a circle, the force of tension acting upon the bucket provides the centripetal force required for circular motion.

As the moon orbits the Earth, the force of gravity acting upon the moon provides the centripetal force required for circular motion.

Newton’s 2nd Law Revisited

Newton\'s second law states that the acceleration of an object is directly proportional to the net force acting upon the object and inversely proportional to the mass of the object. The law is often expressed in the form of the following two equations.

In Unit 2, Newton\'s second law was used to analyze a variety of physical situations. The idea was that if any given physical situation is analyzed in terms of the individual forces which are acting upon an object, then those individual forces must add up to the net force. Furthermore, the net force must be equal to the mass times the acceleration. Subsequently, the acceleration of an object can be found if the mass of the object and the magnitudes and directions of each individual force are known.

And the magnitude of any individual force can be determined if the mass of the object, the acceleration of the object, and the magnitude of the other individual forces are known. The process of analyzing such physical situations in order to determine unknown information is dependent upon the ability to represent the physical situation by means of a free-body diagram. A free-body diagram is a vector diagram which depicts the relative magnitude and direction of all the individual forces which are acting upon the object.

The three major equations which will be useful are:

- – the equation for net force (Fnet = m * a),
- – the equation for gravitational force (Fgrav = m * g), and
- – the equation for frictional force
- (Ffrict = µ * Fnorm).

In this Lesson, we will use Unit 2 principles (free-body diagrams, Newton\'s second law equation, etc.) and circular motion concepts in order to analyze a variety of physical situations involving the motion of objects in circles or along curved paths. The mathematical equations discussed in Lesson 1 and the concept of a centripetal force requirement will be applied in order to analyze roller coasters and other amusement park rides, various athletic movements, and other real-world phenomenon.

To illustrate how circular motion principles can be combined with Newton\'s second law to analyze a physical situation, consider a car moving in a horizontal circle on a level surface. The diagram below depicts the car on the left side of the circle.

Applying the concept of a centripetal force requirement, we know that the net force acting upon the object is directed inwards. Since the car is positioned on the left side of the circle, the net force is directed rightward. An analysis of the situation would reveal that there are three forces acting upon the object - the force of gravity (acting downwards), the normal force of the pavement (acting upwards), and the force of friction (acting inwards or rightwards). It is the friction force which supplies the centripetal force requirement for the car to move in a horizontal circle.

Without friction, the car would turn its wheels but would not move in a circle (as is the case on an icy surface). This analysis leads to the free-body diagram shown below. Observe that each force is represented by a vector arrow which points in the specific direction which the force acts; also notice that each force is labeled according to type (Ffrict, Fnorm, and Fgrav). Such an analysis is the first step of any problem involving Newton\'s second law and a circular motion.

The two problems will be solved using the same general principles. Yet because the given and requested information is different in each, the solution method will be slightly different.

Sample Problem #1

- A 900-kg car makes a 180-degree turn with a speed of 10.0 m/s. The radius of the circle through which the car is turning is 25.0 m. Determine the force of friction and the coefficient of friction acting upon the car.

Sample problem #1 provides kinematic information (v and R) and requests the value of an individual force. As such the solution of the problem will demand that the acceleration and the net force first be determined; then the individual force value can be found by use of the free-body diagram.

Known Information:

- m = 900 kg v = 10.0 m/s
- R = 25.0 m
- Requested Information:
- Ffrict = ??? mu = ????
- ("mu" - coefficient of friction)

The mass of the object can be used to determine the force of gravity acting in the downward direction. Use the equation

- Fgrav = m * g
- where g can be approximated as 10 m/s/s. Knowing that there is no vertical acceleration of the car, it can be concluded that the vertical forces balance each other. Thus, Fgrav = Fnorm= 9000 N. This allows us to determine two of the three forces identified in the free-body diagram. Only the friction force remains unknown.

Since the force of friction is the only horizontal force, it must be equal to the net force acting upon the object. So if the net force can be determined, then the friction force is known. To determine the net force, the mass and the kinematic information (speed and radius) must be substituted into the following equation:

Substituting the given values yields a net force of 3600 Newtons. Thus, the force of friction is 3600 N.

- Finally the coefficient of friction ("mu") can be determined using the equation which relates the coefficient of friction to the force of friction and the normal force.

Substituting 3600 N for Ffrict and 9000 N for Fnorm yields a coefficient of friction of 0.400.

Sample Problem 2

- The coefficient of friction acting upon a 900-kg car is 0.850. The car is making a 180-degree turn around a curve with a radius of 35.0 m. Determine the maximum speed with which the car can make the turn.

Sample problem #2 provides information about the individual force values (or at least information which allows for the determination of the individual force values) and requests the value of the maximum speed of the car. As such, its solution will demand that individual force values be used to determine the net force and acceleration; then the acceleration can be used to determine the maximum speed of the car.

Known Information:

- m = 900 kg "mu" = 0.85 (coefficient of friction)
- R = 35.0 m
- Requested Information:
- v = ??? (the minimum speed would be the speed achieved with the given friction coefficient)

The mass of the car can be used to determine the force of gravity acting in the downward direction. Use the equation

- Fgrav = m * g
- where g can be approximated as 10 m/s/s. Knowing that there is no vertical acceleration of the car, it can be concluded that the vertical forces balance each other. Thus, Fgrav = Fnorm= 9000 N. Since the coefficient of friction ("mu") is given, the force of friction can be determined using the following equation:

The net force acting upon any object is the vector sum of all individual forces acting upon that object. So if all individual force values are known (as is the case here), the net force can be calculated. The vertical forces add to 0 N. Since the force of friction is the only horizontal force, it must be equal to the net force acting upon the object. Thus, Fnet = 7650 N.

- Once the net force is determined, the acceleration can be quickly calculated using the following equation.
- Fnet = m*a

Substituting the given values yields an acceleration of 7.65 m/s/s. Finally, the speed at which the car could travel around the turn can be calculated using the equation for centripetal acceleration

Substituting the known values for a and R into this equation and solving algebraically yields a maximum speed of 16.4 m/s.

Circular Motion Problems

Step 1

- From the verbal description of the physical situation, construct a free-body diagram. Represent each force by a vector arrow and label the forces according to type.

Step 2

- Identify the given and the unknown information (express in terms of variables such as m = , a = , v = , etc.).

Step 3

- If any of the individual forces are directed at angles, then use vector principles to resolve such forces into horizontal and vertical components.

Step 4

- Determine the magnitude of any known forces and label on the free-body diagram.(For example, if the mass is given, then the Fgrav can be determined. And as another example, if there is no vertical acceleration, then it is known that the vertical forces or force components balance, allowing for the possible determination of one or more of the individual forces in the vertical direction.)

Step 5

- Use circular motion equations to determine any unknown information.(For example, if the speed and the radius are known, then the acceleration can be determined. And as another example, if the period and radius are known, then the acceleration can be determined.)

Step 6 (2 parts)

- Use the remaining information to solve for the requested information.

a) If the problem requests the value of an individual force, then use the kinematic information (R, T and v) to determine the acceleration and the Fnet ; then use the free-body diagram to solve for the individual force value.

b) If the problem requests the value of the speed or radius, then use the values of the individual forces to determine the net force and acceleration; then use the acceleration to determine the value of the speed or radius.

The method prescribed above will serve you well as you approach circular motion problems. However, one caution is in order. Every physics problem differs from the previous problem. As such, there is no magic formula for solving every one. Using an appropriate approach to solving such problems (which involves constructing a FBD, identifying known information, identifying the requested information, and using available equations) will never eliminate the need to think, analyze and problem-solve. For this reason, make an effort to develop an appropriate approach to every problem; yet always engage your critical analysis skills in the process of the solution. If physics problems were a mere matter of following a foolproof, 5-step formula or using some memorized algorithm, then we wouldn\'t call them "problems."

Check your understanding

Example 1

- 1. A 1.5-kg bucket of water is tied by a rope and whirled in a circle with a radius of 1.0 m. At the top of the circular loop, the speed of the bucket is 4.0 m/s. Determine the acceleration, the net force and the individual force values when the bucket is at the top of the circular loop.

m = 1.5 kg

- a = ________ m/s/s
- Fnet = _________ N

Answer 1

- Fgrav = 15 N
- a = 16 m/s/s
- Fnet = 24 N
- Ftens = 24 N – 15 N = 9N

Example 2

- 2. A 1.5-kg bucket of water is tied by a rope and whirled in a circle with a radius of 1.0 m. At the bottom of the circular loop, the speed of the bucket is 6.0 m/s. Determine the acceleration, the net force and the individual force values when the bucket is at the bottom of the circular loop.

m = 1.5 kg

- a = ________ m/s/s
- Fnet = _________ N

Answer 2

- F grav = 15 N
- a = 36 m/s/s
- F net = 54 N Up
- F tens = 69 N

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