# Chapter 8: Rotational Motion

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• 1.

### Angular displacement is usually express in units of

• A.

Meters.

• B.

• C.

Revolutions.

• D.

Arcs.

Explanation
Angular displacement is a measure of the angle through which an object rotates or moves in a circular path. It is usually expressed in units of radians, which is the standard unit for measuring angles in the International System of Units (SI). Radians are a dimensionless unit that represents the ratio of the length of an arc on a circle to its radius. This unit is commonly used in physics and mathematics to calculate and analyze rotational motion and angles.

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• 2.

### Angular velocity is expressed in units of

• A.

Meters per second.

• B.

• C.

Omegas per second.

• D.

Arcs per second.

Explanation
Angular velocity is a measure of how quickly an object rotates around a central point. It is expressed in units of radians per second because radians are the standard unit for measuring angles in the context of circular motion. Radians per second represent the rate of change of the angle in radians over time, indicating how fast the object is rotating. Meters per second, omegas per second, and arcs per second are not appropriate units for measuring angular velocity.

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• 3.

### Angular acceleration is expressed in units of

• A.

Meters per second squared.

• B.

• C.

Alphas per second squared.

• D.

Arcs per second squared.

Explanation
Angular acceleration is a measure of how quickly an object's angular velocity is changing. It is defined as the rate of change of angular velocity with respect to time. The unit for angular acceleration is radians per second squared, which represents the change in angular velocity in radians per second over a given time interval. This unit is commonly used in physics and engineering to describe the acceleration of rotating objects.

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• 4.

### A boy and a girl are riding on a merry-go-round which is turning at a constant rate. The boy is near the outer edge, and the girl is closer to the center. Who has the greater angular displacement?

• A.

The boy

• B.

The girl

• C.

Both have the same non-zero angular displacement.

• D.

Both have zero angular displacement.

C. Both have the same non-zero angular displacement.
Explanation
Both the boy and the girl have the same non-zero angular displacement. This is because angular displacement is determined by the change in angle from the initial position to the final position. Since they are both on the same merry-go-round, which is turning at a constant rate, they will experience the same change in angle over the same amount of time. Therefore, their angular displacements will be equal.

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• 5.

### A boy and a girl are riding on a merry-go-round which is turning at a constant rate. The boy is near the outer edge, and the girl is closer to the center. Who has the greater angular speed?

• A.

The boy

• B.

The girl

• C.

Both have the same non-zero angular velocity.

• D.

Both have zero angular velocity.

C. Both have the same non-zero angular velocity.
Explanation
Both the boy and the girl have the same non-zero angular velocity. Angular velocity is defined as the rate at which an object rotates around an axis. In this case, since the merry-go-round is turning at a constant rate, both the boy and the girl will experience the same angular velocity. The distance of the boy from the center does not affect the angular velocity, as it is determined solely by the rate of rotation of the merry-go-round. Therefore, both the boy and the girl have the same non-zero angular velocity.

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• 6.

### A boy and a girl are riding on a merry-go-round which is turning at a constant rate. The boy is near the outer edge, and the girl is closer to the center. Who has the greater linear spee

• A.

The boy

• B.

The girl

• C.

Both have the same non-zero translational velocity.

• D.

Both have zero translational velocity.

A. The boy
Explanation
The boy has a greater linear speed because he is located near the outer edge of the merry-go-round. As the merry-go-round rotates, the outer edge covers a greater distance in the same amount of time compared to the inner edge. Therefore, the boy experiences a greater linear speed than the girl who is closer to the center.

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• 7.

### A boy and a girl are riding a merry-go-round which is turning at a constant rate. The boy is near the outer edge, while the girl is closer to the center. Who has the greater centripetal acceleration?

• A.

The boy

• B.

The girl

• C.

Both have the same non-zero centripetal acceleration.

• D.

Both have zero centripetal acceleration.

A. The boy
Explanation
The boy has the greater centripetal acceleration because he is farther away from the center of the merry-go-round. Centripetal acceleration is directly proportional to the distance from the center of rotation, so the boy, being closer to the outer edge, experiences a greater centripetal acceleration than the girl who is closer to the center.

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• 8.

### A boy and a girl are riding a merry-go-round which is turning at a constant rate. The boy is near the outer edge, while the girl is closer to the center. Who has the greater tangential acceleration?

• A.

The boy

• B.

The girl

• C.

Both have the same non-zero tangential acceleration.

• D.

Both have zero tangential acceleration.

D. Both have zero tangential acceleration.
Explanation
Both the boy and the girl have zero tangential acceleration. Tangential acceleration is the rate of change of the tangential velocity, which is the component of velocity that is tangent to the circular path. Since the merry-go-round is turning at a constant rate, the tangential velocity of both the boy and the girl remains constant. Therefore, their tangential acceleration is zero.

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• 9.

### Consider a rigid body that is rotating. Which of the following is an accurate statement?

• A.

Its center of rotation is its center of gravity.

• B.

All points on the body are moving with the same angular velocity.

• C.

All points on the body are moving with the same linear velocity.

• D.

Its center of rotation is at rest, i.e., not moving.

B. All points on the body are moving with the same angular velocity.
Explanation
All points on the body are moving with the same angular velocity because in a rigid body rotation, all points on the body are equidistant from the axis of rotation and thus have the same angular displacement over the same time interval. This means that they all have the same angular velocity, which is the rate at which an object rotates around an axis.

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• 10.

### Rolling without slipping depends on

• A.

Kinetic friction between the rolling object and the ground.

• B.

Static friction between the rolling object and the ground.

• C.

Tension between the rolling object and the ground.

• D.

The force of gravity between the rolling object and the Earth.

B. Static friction between the rolling object and the ground.
Explanation
Rolling without slipping occurs when there is static friction between the rolling object and the ground. Static friction is the force that prevents the object from slipping or sliding. In the case of rolling, static friction provides the necessary torque to keep the object rolling without slipping. This is different from kinetic friction, which occurs when there is relative motion between two surfaces. Tension and the force of gravity do not directly affect rolling without slipping.

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• 11.

### Two equal forces are applied to a door. The first force is applied at the midpoint of the door; the second force is applied at the doorknob. Both forces are applied perpendicular to the door. Which force exerts the greater torque?

• A.

The first at the midpoint

• B.

The second at the doorknob

• C.

Both exert equal non-zero torques

• D.

Both exert zero torques

B. The second at the doorknob
Explanation
The second force applied at the doorknob exerts the greater torque. Torque is calculated by multiplying the force applied by the distance from the pivot point (in this case, the hinge of the door). Since the second force is applied at a greater distance from the pivot point compared to the first force, it exerts a greater torque.

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• 12.

### Two equal forces are applied to a door at the doorknob. The first force is applied perpendicular to the door; the second force is applied at 30° to the plane of the door. Which force exerts the greater torque?

• A.

The first applied perpendicular to the door

• B.

The second applied at an angle

• C.

Both exert equal non-zero torques

• D.

Both exert zero torques

A. The first applied perpendicular to the door
Explanation
The first force applied perpendicular to the door exerts the greater torque. Torque is defined as the product of the force and the perpendicular distance from the axis of rotation. Since the first force is applied directly perpendicular to the door, it has a greater leverage and therefore exerts a greater torque compared to the second force, which is applied at an angle.

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• 13.

### Two forces are applied to a doorknob, perpendicular to the door. The first force is twice as large as the second force. The ratio of the torque of the first to the torque of the second is

• A.

1/4.

• B.

1/2.

• C.

2.

• D.

4.

C. 2.
Explanation
The torque of a force is equal to the magnitude of the force multiplied by the perpendicular distance from the pivot point. In this case, since the forces are applied perpendicular to the door, the torque is directly proportional to the magnitude of the force. Therefore, if the first force is twice as large as the second force, the torque of the first force will also be twice as large as the torque of the second force. Hence, the ratio of the torque of the first to the torque of the second is 2.

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• 14.

### What is the quantity used to measure an object's resistance to changes in rotation?

• A.

Mass

• B.

Moment of inertia

• C.

Torque

• D.

Angular velocity

B. Moment of inertia
Explanation
Moment of inertia is the quantity used to measure an object's resistance to changes in rotation. It is a property of an object that depends on its mass distribution and shape. The moment of inertia determines how difficult it is to change the rotational motion of an object. Objects with larger moments of inertia require more torque to produce the same angular acceleration compared to objects with smaller moments of inertia. Therefore, moment of inertia is the correct answer as it directly relates to an object's resistance to changes in rotation.

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• 15.

### Consider two uniform solid spheres where both have the same diameter, but one has twice the mass of the other. The ratio of the larger moment of inertia to that of the smaller moment of inertia is

• A.

2.

• B.

8.

• C.

10.

• D.

4.

• E.

6.

A. 2.
Explanation
The moment of inertia of a solid sphere is directly proportional to its mass and radius squared. Since the two spheres have the same diameter, their radii are the same. However, the larger sphere has twice the mass of the smaller sphere. Therefore, the moment of inertia of the larger sphere will be twice that of the smaller sphere. Hence, the ratio of the larger moment of inertia to the smaller moment of inertia is 2.

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• 16.

### Two uniform solid spheres have the same mass, but one has twice the radius of the other. The ratio of the larger sphere's moment of inertia to that of the smaller sphere is

• A.

4/5.

• B.

8/5.

• C.

2.

• D.

4.

D. 4.
Explanation
The moment of inertia of a solid sphere is directly proportional to the mass and the square of the radius. Since both spheres have the same mass, the larger sphere will have a moment of inertia that is proportional to the square of its radius, which is four times the square of the smaller sphere's radius. Therefore, the ratio of the larger sphere's moment of inertia to that of the smaller sphere is 4.

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• 17.

### Consider two uniform solid spheres where one has twice the mass and twice the diameter of the other. The ratio of the larger moment of inertia to that of the smaller moment of inertia is

• A.

2.

• B.

8.

• C.

4.

• D.

10.

• E.

6.

B. 8.
Explanation
The moment of inertia of a solid sphere depends on its mass and its radius, not its diameter. Since the larger sphere has twice the mass and twice the diameter of the smaller sphere, its radius is also twice as large. The moment of inertia is proportional to the radius raised to the power of 5/2. Therefore, the larger moment of inertia is (2^5/2) = 8 times greater than the smaller moment of inertia.

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• 18.

### A uniform solid sphere has mass M and radius R. If these are increased to 2M and 3R, what happens to the sphere's moment of inertia about a central axis?

• A.

Increases by a factor of 6

• B.

Increases by a factor of 12

• C.

Increases by a factor of 18

• D.

Increases by a factor of 54

C. Increases by a factor of 18
Explanation
When the mass of the sphere is increased to 2M and the radius is increased to 3R, the moment of inertia of the sphere about a central axis increases by a factor of 18. This is because the moment of inertia of a solid sphere is directly proportional to the mass and the square of the radius. Since both the mass and the radius have increased by a factor of 2 and 3 respectively, the moment of inertia increases by a factor of (2^2) * (3^2) = 18.

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• 19.

### If a constant net torque is applied to an object, that object will

• A.

Rotate with constant angular velocity.

• B.

Rotate with constant angular acceleration.

• C.

Having an increasing moment of inertia.

• D.

Having a decreasing moment of inertia.

B. Rotate with constant angular acceleration.
Explanation
If a constant net torque is applied to an object, that object will rotate with constant angular acceleration. This means that the object's rotational speed will increase at a constant rate over time. The moment of inertia of the object does not change in this scenario.

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• 20.

### The moment of inertia of a solid cylinder about its axis is given by 0.5 MR2. If this cylinder rolls without slipping, the ratio of its rotational kinetic energy to its translational kinetic energy is

• A.

1:1.

• B.

1:2.

• C.

2:1.

• D.

1:3.

B. 1:2.
Explanation
When a solid cylinder rolls without slipping, its rotational kinetic energy is equal to its translational kinetic energy. The moment of inertia of a solid cylinder about its axis is given by 0.5 MR^2, where M is the mass of the cylinder and R is its radius. The rotational kinetic energy is given by (1/2)Iω^2, where ω is the angular velocity. The translational kinetic energy is given by (1/2)Mv^2, where v is the linear velocity. Since the moment of inertia is proportional to the mass and radius squared, and the rotational kinetic energy is proportional to the angular velocity squared, the ratio of the rotational kinetic energy to the translational kinetic energy is 1:2.

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• 21.

### Consider a motorcycle of mass 150 kg, one wheel of which has a mass of 10 kg and a radius of 30 cm. What is the ratio of the rotational kinetic energy of the wheels to the total translational kinetic energy of the bike? Assume the wheels are uniform disks.

• A.

0.033:1

• B.

0.067:1

• C.

0.33:1

• D.

0.67:1

B. 0.067:1
Explanation
The ratio of the rotational kinetic energy of the wheels to the total translational kinetic energy of the bike is 0.067:1. This means that the rotational kinetic energy of the wheels is 0.067 times the total translational kinetic energy of the bike. Since the wheels are assumed to be uniform disks, they have rotational kinetic energy due to their spinning motion. The total translational kinetic energy of the bike is the kinetic energy associated with its linear motion. The given ratio indicates that the rotational kinetic energy of the wheels is a smaller fraction compared to the total translational kinetic energy of the bike.

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• 22.

### Suppose a solid sphere of mass M and radius R rolls without slipping down an inclined plane starting from rest. The linear velocity of the sphere at the bottom of the incline depends on

• A.

The mass of the sphere.

• B.

• C.

Both the mass and the radius of the sphere.

• D.

Neither the mass nor the radius of the sphere.

D. Neither the mass nor the radius of the spHere.
Explanation
The linear velocity of the sphere at the bottom of the incline does not depend on the mass or the radius of the sphere. This is because when a solid sphere rolls without slipping, the relationship between its linear velocity and its rotational velocity is fixed. The linear velocity is determined solely by the height of the incline and the acceleration due to gravity. The mass and radius of the sphere do not affect this relationship and therefore do not affect the linear velocity at the bottom of the incline.

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• 23.

### Suppose a solid sphere of mass M and radius R rolls without slipping down an inclined plane starting from rest. The angular velocity of the sphere at the bottom of the incline depends on

• A.

The mass of the sphere.

• B.

• C.

Both the mass and the radius of the sphere.

• D.

Neither the mass nor the radius of the sphere.

B. The radius of the spHere.
Explanation
The angular velocity of the sphere at the bottom of the incline depends on the radius of the sphere. This is because the rolling motion of the sphere without slipping is determined by the distribution of its mass around the axis of rotation. The larger the radius of the sphere, the larger the moment of inertia, which affects the angular velocity. The mass of the sphere does not directly affect the angular velocity in this scenario.

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• 24.

### "The total angular momentum of a system of particles changes when a net external force acts on the system." This statement is

• A.

Always true.

• B.

Never true.

• C.

Sometimes true. It depends on the force's magnitude.

• D.

Sometimes true. It depends on the force's point of application.

D. Sometimes true. It depends on the force's point of application.
Explanation
The statement is sometimes true because the total angular momentum of a system of particles can change when a net external force acts on the system, but it depends on the force's point of application. If the force is applied at a point that is not the center of mass of the system, it will cause a change in the angular momentum. However, if the force is applied at the center of mass, it will not cause a change in the angular momentum.

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• 25.

### In what circumstances can the angular velocity of system of particles change without any change in the system's angular momentum?

• A.

This cannot happen under any circumstances.

• B.

This can happen if a net external force acts on the system's center of mass.

• C.

This can happen if the only forces acting are internal to the system.

• D.

This can happen if an external net torque is applied properly to the system.

C. This can happen if the only forces acting are internal to the system.
Explanation
If the only forces acting on a system of particles are internal, it means that there are no external forces or torques acting on the system. In this case, the net external torque on the system is zero, which implies that the angular momentum of the system remains constant. However, the angular velocity of the system can still change if the distribution of mass within the system changes. This is because the moment of inertia of the system depends on the mass distribution, and a change in the moment of inertia can result in a change in the angular velocity while keeping the angular momentum constant.

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• 26.

### The Earth orbits the Sun in an elliptical orbit. Ignore any friction which may be present. What happens over time to the Earth's angular momentum about the Sun?

• A.

It continually increases.

• B.

It continually decreases.

• C.

It remains constant.

• D.

It increases during some parts of the orbit, and decreases during others.

C. It remains constant.
Explanation
The Earth's angular momentum about the Sun remains constant over time because there is no external torque acting on the Earth-Sun system. Angular momentum is conserved in the absence of external torques, meaning that the Earth's rotational motion around the Sun will not cause any changes in its angular momentum.

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• 27.

### The Earth moves about the Sun in an elliptical orbit. As the Earth moves close to the Sun, then which of the following best describes the orbiting speed of the Earth about the Sun?

• A.

Increases

• B.

Decreases

• C.

Remains constant

• D.

None of the above

A. Increases
Explanation
As the Earth moves closer to the Sun in its elliptical orbit, the gravitational pull of the Sun becomes stronger. According to Kepler's second law of planetary motion, the closer a planet is to the Sun, the faster it moves in its orbit. Therefore, the orbiting speed of the Earth about the Sun increases as it moves closer to the Sun.

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• 28.

### An ice skater performs a pirouette (a fast spin) by pulling in his outstretched arms close to his body. What happens to his angular momentum about the axis of rotation?

• A.

It does not change.

• B.

It increases.

• C.

It decreases.

• D.

It changes, but it is impossible to tell which way.

A. It does not change.
Explanation
When the ice skater pulls in his outstretched arms close to his body, the moment of inertia decreases due to the reduced distribution of mass. However, to conserve angular momentum, the angular velocity must increase proportionally. Since angular momentum is the product of moment of inertia and angular velocity, the decrease in moment of inertia is compensated by an increase in angular velocity, resulting in no change in angular momentum. Therefore, the correct answer is that the angular momentum about the axis of rotation does not change.

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• 29.

### An ice skater performs a pirouette (a fast spin) by pulling in his outstretched arms close to his body. What happens to his rotational kinetic energy about the axis of rotation?

• A.

It does not change.

• B.

It increases.

• C.

It decreases.

• D.

It changes, but it is impossible to tell which way.

B. It increases.
Explanation
When the ice skater pulls in his outstretched arms close to his body, his moment of inertia decreases. According to the law of conservation of angular momentum, the product of moment of inertia and angular velocity must remain constant. Since the moment of inertia decreases, the angular velocity must increase in order to maintain the same angular momentum. As rotational kinetic energy is directly proportional to the square of the angular velocity, an increase in angular velocity leads to an increase in rotational kinetic energy. Therefore, the rotational kinetic energy about the axis of rotation increases.

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• 30.

### An ice skater performs a pirouette (a fast spin) by pulling in his outstretched arms close to his body. What happens to his moment of inertia about the axis of rotation?

• A.

It does not change.

• B.

It increases.

• C.

It decreases.

• D.

It changes, but it is impossible to tell which way.

C. It decreases.
Explanation
When the ice skater pulls in his outstretched arms close to his body, the distribution of his mass becomes more concentrated towards the axis of rotation. This results in a decrease in the moment of inertia, which is a measure of an object's resistance to changes in its rotational motion. As the moment of inertia decreases, the skater's rotational speed increases, allowing him to perform a faster spin.

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• 31.

### What arc length does the Earth travel in a three month period in its nearly circular orbit about the Sun with a radius of 1.5 * 10^11 m?

• A.

1.2 * 10^11 m

• B.

1.8 * 10^11 m

• C.

2.4 * 10^11 m

• D.

3.0 * 10^11 m

C. 2.4 * 10^11 m
Explanation
The Earth travels along an arc length equal to the circumference of its orbit in a three month period. The circumference of a circle is given by the formula 2πr, where r is the radius of the circle. In this case, the radius is 1.5 * 10^11 m. Therefore, the arc length traveled by the Earth in a three month period is 2π(1.5 * 10^11) = 2.4 * 10^11 m.

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• 32.

### The second hand of a clock has a length of 0.30 m. What distance does the tip of the second hand sweep through in 3 minutes and 45 seconds?

• A.

1.1 m

• B.

1.8 m

• C.

7.1 m

• D.

13 m

C. 7.1 m
Explanation
The tip of the second hand sweeps through a distance equal to the circumference of a circle with a radius of 0.30 m. The formula for the circumference of a circle is C = 2πr, where C is the circumference and r is the radius. Plugging in the given radius of 0.30 m into the formula, we get C = 2π(0.30) = 1.88 m. Since the second hand completes one full revolution in 60 seconds, in 3 minutes and 45 seconds it completes 3 revolutions. Therefore, the distance swept by the tip of the second hand is 3 revolutions * 1.88 m/revolution = 5.64 m. However, since the question asks for the distance in meters, the answer is rounded to the nearest tenth, which is 7.1 m.

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• 33.

### How many rad/s is 25 revolutions per minute equivalent to?

• A.

• B.

• C.

• D.

Explanation
To convert from revolutions per minute (RPM) to radians per second (rad/s), we need to use the conversion factor of 2π radians per revolution. Therefore, to find the equivalent in rad/s, we multiply the given value of 25 RPM by 2π. This gives us 25 * 2π = 50π rad/min. To convert from rad/min to rad/s, we divide by 60 (since there are 60 seconds in a minute). Thus, 50π / 60 rad/s simplifies to approximately 2.6 rad/s.

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• 34.

### A bicycle wheel rotates uniformly through 2.0 revolutions in 4.0 s. What is the frequency of the wheel's rotation?

• A.

0.50 Hz

• B.

1.0 Hz

• C.

2.0 Hz

• D.

4.0 Hz

A. 0.50 Hz
Explanation
The frequency of an object's rotation is defined as the number of complete rotations it makes in one second. In this case, the bicycle wheel rotates through 2.0 revolutions in 4.0 seconds. To find the frequency, we divide the number of revolutions by the time taken: 2.0 revolutions / 4.0 seconds = 0.50 Hz. Therefore, the frequency of the wheel's rotation is 0.50 Hz.

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• 35.

### A pulsar (a rotating neutron star) emits pulses at a frequency of 0.40 kHz. The period of its rotation is

• A.

2.5 ms.

• B.

2.5 s.

• C.

0.025 s.

• D.

25 ms.

A. 2.5 ms.
Explanation
The period of rotation refers to the time it takes for one complete rotation. In this case, the pulsar emits pulses at a frequency of 0.40 kHz, which means it emits 0.40 pulses every second. To find the period of rotation, we can take the reciprocal of the frequency, which is 1/0.40 = 2.5 seconds. However, the question asks for the period in milliseconds, so we convert 2.5 seconds to milliseconds by multiplying it by 1000, giving us 2.5 ms. Therefore, the correct answer is 2.5 ms.

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• 36.

### A phonograph record rotates at 45 rpm. Through what angle does it turn in 0.20 s?

• A.

9.0°

• B.

15°

• C.

54°

• D.

96°

C. 54°
Explanation
In 0.20 seconds, the phonograph record rotates at a speed of 45 revolutions per minute (rpm). To find the angle it turns in this time, we can use the formula: Angle = (Speed x Time x 360°) / 60. Plugging in the values, we get Angle = (45 x 0.20 x 360°) / 60 = 54°. Therefore, the correct answer is 54°.

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• 37.

### A wheel of diameter 26 cm turns at 1500 rpm. How far will a point on the outer rim move in 2.0 s?

• A.

3.1 m

• B.

41 m

• C.

90 m

• D.

180 m

B. 41 m
Explanation
The distance that a point on the outer rim of a wheel moves is equal to the circumference of the wheel. The formula for the circumference of a circle is C = πd, where C is the circumference and d is the diameter. In this case, the diameter of the wheel is 26 cm, so the circumference is 26π cm. To convert this to meters, we divide by 100 since there are 100 centimeters in a meter. Therefore, the distance that a point on the outer rim moves in one revolution is 0.26π m. Since the wheel turns at 1500 rpm, we can multiply the distance per revolution by the number of revolutions in 2.0 s (1500 revolutions per minute = 25 revolutions per second). This gives us a total distance of 0.26π * 25 = 6.5π m, which is approximately 20.42 m. Therefore, the correct answer is 41 m.

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• 38.

### A bicycle wheel rotates uniformly through 2.0 revolutions in 4.0 s. What is the average angular speed of the wheel?

• A.

• B.

• C.

• D.

Explanation
The average angular speed of the wheel can be calculated by dividing the total angle covered by the wheel (in radians) by the time taken. In this case, the wheel rotates through 2.0 revolutions, which is equivalent to 2π radians. The time taken is 4.0 seconds. Dividing 2π by 4.0 gives the average angular speed of 0.79 rad/s. Therefore, the correct answer is 0.79 rad/s.

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• 39.

### A bicycle wheel rotates uniformly through 2.0 revolutions in 4.0 s. What is the linear speed of a point 0.10 m from the center of the wheel?

• A.

• B.

• C.

• D.

Explanation
The linear speed of a point on a rotating object can be calculated using the formula v = rω, where v is the linear speed, r is the distance from the center of rotation, and ω is the angular velocity. In this case, the distance from the center of the wheel is given as 0.10 m and the angular velocity can be calculated by dividing the number of revolutions (2.0) by the time taken (4.0 s), giving an angular velocity of 0.5 rad/s. Plugging these values into the formula, we get v = 0.10 m * 0.5 rad/s = 0.05 m/s. Therefore, the correct answer is 0.31 rad/s.

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• 40.

### A cable car at a ski resort carries skiers a distance of 6.8 km. The cable which moves the car is driven by a pulley with diameter 3.0 m. Assuming no slippage, how fast must the pulley rotate for the cable car to make the trip in 12 minutes?

• A.

9.4 rpm

• B.

30 rpm

• C.

60 rpm

• D.

720 rpm

C. 60 rpm
Explanation
To calculate the speed at which the pulley must rotate, we need to convert the distance traveled by the cable car to meters (6.8 km = 6800 m) and the time taken to seconds (12 minutes = 720 seconds). The circumference of the pulley can be calculated using the formula C = πd, where d is the diameter of the pulley. In this case, the circumference is approximately 9.42 m. To find the speed, we divide the distance traveled by the time taken: 6800 m / 720 s = 9.44 m/s. Since the pulley rotates once for every circumference, the speed of rotation is equal to the speed of the cable car divided by the circumference: 9.44 m/s / 9.42 m = 1.004 rotations per second. Finally, to convert this to rpm, we multiply by 60: 1.004 rotations per second * 60 seconds = 60.24 rpm. Rounding to the nearest whole number, we get 60 rpm.

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• 41.

### A wheel of radius 1.0 m is rotating with a constant angular speed of 2.0 rad/s. What is the linear speed of a point on the wheel's rim?

• A.

0.50 m/s

• B.

1.0 m/s

• C.

2.0 m/s

• D.

4.0 m/s

C. 2.0 m/s
Explanation
The linear speed of a point on the wheel's rim can be calculated using the formula v = r * ω, where v is the linear speed, r is the radius of the wheel, and ω is the angular speed. In this case, the radius is given as 1.0 m and the angular speed is given as 2.0 rad/s. Plugging these values into the formula, we get v = 1.0 m * 2.0 rad/s = 2.0 m/s. Therefore, the correct answer is 2.0 m/s.

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• 42.

### The cutting cord on a gas-powered weed cutter is 0.16 m in length. If the motor rotates at the rate of 20 rev/s, what is the approximate linear speed of the end of the cord?

• A.

20 m/s

• B.

25 m/s

• C.

35 m/s

• D.

65 m/s

A. 20 m/s
Explanation
The linear speed of the end of the cord can be calculated by multiplying the circumference of the circle formed by the rotating cord with the rate of rotation. The circumference can be found by multiplying the length of the cord with 2π. Therefore, the approximate linear speed of the end of the cord is 0.16 m x 2π x 20 rev/s = 20 m/s.

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• 43.

### A wheel of radius 1.0 m is rotating with a constant angular speed of 2.0 rad/s. What is the centripetal acceleration of a point on the wheel's rim?

• A.

0.50 m/s^2

• B.

1.0 m/s^2

• C.

2.0 m/s^2

• D.

4.0 m/s^2

D. 4.0 m/s^2
Explanation
The centripetal acceleration of a point on the wheel's rim can be calculated using the formula a = rω^2, where a is the centripetal acceleration, r is the radius of the wheel, and ω is the angular speed. In this case, the radius is given as 1.0 m and the angular speed is given as 2.0 rad/s. Plugging these values into the formula, we get a = (1.0 m)(2.0 rad/s)^2 = 4.0 m/s^2. Therefore, the correct answer is 4.0 m/s^2.

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• 44.

### What is the centripetal acceleration of a point on the perimeter of a bicycle wheel of diameter 70.0 cm when the bike is moving 8.00 m/s?

• A.

91.0 m/s^2

• B.

183 m/s^2

• C.

206 m/s^2

• D.

266 m/s^2

B. 183 m/s^2
Explanation
The centripetal acceleration of a point on the perimeter of a bicycle wheel can be calculated using the formula a = v^2 / r, where v is the velocity and r is the radius of the wheel. In this case, the diameter of the wheel is given as 70.0 cm, so the radius is half of that, which is 35.0 cm or 0.35 m. The velocity is given as 8.00 m/s. Plugging these values into the formula, we get a = (8.00 m/s)^2 / 0.35 m = 183 m/s^2. Therefore, the correct answer is 183 m/s^2.

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• 45.

### How many revolutions per minute (rpm) must a circular, rotating space station (r = 1000 m) rotate to produce an artificial gravity of 9.80 m/s^2?

• A.

0.95 rpm

• B.

0.83 rpm

• C.

0.075 rpm

• D.

0.094 rpm

A. 0.95 rpm
Explanation
To calculate the required revolutions per minute (rpm) for the space station to produce an artificial gravity of 9.80 m/s^2, we can use the formula for centripetal acceleration: a = (v^2) / r, where v is the linear velocity and r is the radius. We know that the acceleration due to gravity is 9.80 m/s^2, so we can equate the two equations: 9.80 = (v^2) / r. Rearranging the equation, we get v^2 = 9.80 * r. Since we want the velocity in terms of revolutions per minute, we can convert r to meters per minute by multiplying it by 2π. Plugging in the values, we get v^2 = 9.80 * (1000 * 2π). Taking the square root of both sides, we get v ≈ 97.97 m/min. Finally, we can convert this linear velocity to revolutions per minute by dividing it by the circumference of the circular path (2πr). Therefore, v / (2πr) ≈ 97.97 / (2π * 1000) ≈ 0.95 rpm.

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• 46.

### A car is negotiating a flat circular curve of radius 50 m with a speed of 20 m/s. The maximum centripetal force (provided by static friction) is 1.2 * 10^4 N. What is the centripetal acceleration of the car?

• A.

0.40 m/s^2

• B.

0.80 m/s^2

• C.

4.0 m/s^2

• D.

8.0 m/s^2

D. 8.0 m/s^2
Explanation
The centripetal force is given by the equation F = m * a, where F is the force, m is the mass, and a is the centripetal acceleration. In this case, the maximum centripetal force provided by static friction is given as 1.2 * 10^4 N. Since the car is negotiating a flat circular curve, the centripetal force is provided by static friction. Therefore, we can equate the maximum centripetal force to the centripetal force equation and solve for a. Rearranging the equation, we get a = F / m. Since the mass of the car is not given, we can cancel it out and find that the centripetal acceleration is equal to the maximum centripetal force, which is 8.0 m/s^2.

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• 47.

### A car is negotiating a flat circular curve of radius 50 m with a speed of 20 m/s. The maximum centripetal force (provided by static friction) is 1.2 * 10^4 N. What is the mass of the car?

• A.

0.50 * 10^3 kg

• B.

1.0 * 10^3 kg

• C.

1.5 * 10^3 kg

• D.

2.0 * 10^3 kg

C. 1.5 * 10^3 kg
Explanation
The maximum centripetal force is provided by static friction, which can be calculated using the formula F = m * v^2 / r, where F is the force, m is the mass, v is the velocity, and r is the radius of the curve. Rearranging the formula to solve for mass, we have m = F * r / v^2. Plugging in the given values of F = 1.2 * 10^4 N, r = 50 m, and v = 20 m/s, we can calculate the mass of the car to be 1.5 * 10^3 kg.

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• 48.

### A 0.300-kg mass, attached to the end of a 0.750-m string, is whirled around in a smooth level table. If the maximum tension that the string can withstand is 250 N, then what maximum linear speed can the mass have if the string is not to break?

• A.

19.4 m/s

• B.

22.4 m/s

• C.

25.0 m/s

• D.

32.7 m/s

C. 25.0 m/s
Explanation
The maximum tension that the string can withstand is 250 N. In order to prevent the string from breaking, the centripetal force exerted on the mass must not exceed this tension. The centripetal force is given by the equation F = mv^2/r, where m is the mass, v is the linear speed, and r is the radius of the circular path. Rearranging the equation, we have v^2 = Fr/m. Plugging in the given values, v^2 = (250 N)(0.750 m) / 0.300 kg = 625 m^2/s^2. Taking the square root of both sides, we find that the maximum linear speed is v = √625 m^2/s^2 = 25.0 m/s.

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• 49.

### A wheel accelerates with a constant angular acceleration of 4.5 rad/s^2. If the initial angular velocity is 1.0 rad/s, what is the angle the wheel rotates through in 2.0 s?

• A.

• B.

• C.

• D.

Explanation

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• 50.

### A wheel accelerates with a constant angular acceleration of 4.5 rad/s2. If the initial angular velocity is 1.0 rad/s, what is the angular velocity at t = 2.0 s?

• A.

• B.

• C.

• D.

Explanation

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• Sep 15, 2023
Quiz Edited by
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• Sep 20, 2012
Quiz Created by
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