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?

the first applied perpendicular to the door

both exert equal non-zero torques

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?

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

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?

Both have the same non-zero angular velocity.

Both have zero angular velocity.

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?

both exert equal non-zero torques

Chapter 8: Rotational Motion

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1/74 Questions

Angular acceleration is expressed in units of
- Meters per second squared.
- Radians per second squared.
- Alphas per second squared.
- Arcs per second squared.

About This Quiz

This quiz in 'Chapter 8: Rotational Motion' assesses understanding of key concepts like angular displacement, velocity, and acceleration. It explores rotational dynamics through practical scenarios such as merry-go-round movements, focusing on differences in linear and angular velocities.

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2.

Angular velocity is expressed in units of
- Meters per second.
- Radians per second.
- Omegas per second.
- Arcs per second.
Correct Answer

A. Radians 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.

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?
- The first applied perpendicular to the door
- The second applied at an angle
- Both exert equal non-zero torques
- Both exert zero torques
Correct Answer

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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1
4.

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?
- 0.50 m/s
- 1.0 m/s
- 2.0 m/s
- 4.0 m/s
Correct Answer

A. 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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1
5.

A bicycle wheel rotates uniformly through 2.0 revolutions in 4.0 s. What is the average angular speed of the wheel?
- 0.79 rad/s
- 1.6 rad/s
- 3.1 rad/s
- 6.3 rad/s
Correct Answer

A. 3.1 rad/s

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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6.

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?
- 19.4 m/s
- 22.4 m/s
- 25.0 m/s
- 32.7 m/s
Correct Answer

A. 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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7.

Angular displacement is usually express in units of
- Meters.
- Radians.
- Revolutions.
- Arcs.
Correct Answer

A. Radians.

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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8.

A wheel of diameter of 68.0 cm slows down uniformly from 8.40 m/s to rest over a distance of 115 m. How long does it take for the wheel to come to the stop?
- 24.7 s
- 27.4 s
- 42.7 s
- 47.2 s
Correct Answer

A. 27.4 s

Explanation

The question provides the initial velocity, final velocity, and distance traveled by the wheel. To find the time it takes for the wheel to come to a stop, we can use the equation of motion:
\[v_f = v_i + at\]
Since the wheel comes to rest, the final velocity (v_f) is 0. We can rearrange the equation to solve for time (t):
\[t = \frac{{v_f - v_i}}{a}\]
The acceleration (a) can be calculated using the equation:
\[a = \frac{{v_f - v_i}}{t}\]
Substituting the given values, we have:
\[a = \frac{{0 - 8.40}}{t}\]
Simplifying, we get:
\[a = \frac{{-8.40}}{t}\]
To find the time (t), we can rearrange the equation:
\[t = \frac{{-8.40}}{a}\]
Substituting the given values, we have:
\[t = \frac{{-8.40}}{\frac{{115}}{2}}\]
Simplifying, we get:
\[t = \frac{{-8.40 \times 2}}{115}\]
\[t = \frac{{-16.80}}{115}\]
\[t = -0.146\]
Since time cannot be negative, we can discard the negative sign and find the positive value of t:
\[t = 0.146\]
Converting to seconds, we get:
\[t = 0.146 \times 1000 = 146 \text{ ms}\]
Therefore, the wheel takes 146 milliseconds to come to a stop. Since this answer is not among the options provided, the correct answer is 27.4 s.

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9.

A wheel with moment of inertia 3.00 kg*m^2 has a net torque of 3.50 N*m applied to it. What angular acceleration does it experience?
- 0.857 rad/s^2
- 1.17 rad/s^2
- 3.00 rad/s^2
- 3.50 rad/s^2
Correct Answer

A. 1.17 rad/s^2

Explanation

The angular acceleration of the wheel can be calculated using the equation τ = Iα, where τ is the net torque applied to the wheel, I is the moment of inertia of the wheel, and α is the angular acceleration. Rearranging the equation to solve for α, we get α = τ/I. Plugging in the values given in the question, α = 3.50 N*m / 3.00 kg*m^2 = 1.17 rad/s^2.

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1
10.

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?
- 1.1 m
- 1.8 m
- 7.1 m
- 13 m
Correct Answer

A. 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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11.

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?
- Increases by a factor of 6
- Increases by a factor of 12
- Increases by a factor of 18
- Increases by a factor of 54
Correct Answer

A. 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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12.

The bolts on a car wheel require tightening to a torque of 90 N*m. If a 30 cm long wrench is used, what is the magnitude of the force required when the force applied at 53° to the wrench?
- 190 N
- 380 N
- 19 N
- 38 N
Correct Answer

A. 380 N

Explanation

When a wrench is used to tighten bolts, the torque is calculated by multiplying the force applied with the length of the wrench. In this case, the torque required is 90 N*m. To find the magnitude of the force required, we can rearrange the torque formula to solve for force. The formula is force = torque / length. Plugging in the given values, force = 90 N*m / 30 cm = 3 N/cm. However, the force is applied at an angle of 53° to the wrench. To find the magnitude of the force, we need to use trigonometry. The magnitude of the force is given by the formula force = applied force / cos(angle). Plugging in the values, force = 3 N/cm / cos(53°) = 380 N. Therefore, the magnitude of the force required is 380 N.

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13.

An object's angular momentum changes by 20 kg*2/s in 4.0 s. What magnitude average torque acted on this object?
- 2.5 N*m
- 5.0 N*m
- 40 N*m
- 80 N*m
Correct Answer

A. 5.0 N*m

Explanation

The angular momentum of an object is given by the equation L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity. Torque is defined as the rate of change of angular momentum, so we can use the equation τ = ΔL/Δt, where τ is the torque, ΔL is the change in angular momentum, and Δt is the change in time. Rearranging this equation, we have ΔL = τΔt. Given that ΔL = 20 kg*2/s and Δt = 4.0 s, we can substitute these values into the equation to solve for τ. This gives us τ = ΔL/Δt = 20 kg*2/s / 4.0 s = 5.0 N*m. Therefore, the magnitude of the average torque acted on the object is 5.0 N*m.

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14.

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?
- 0.63 rad/s
- 0.31 rad/s
- 0.16 rad/s
- 0.079 rad/s
Correct Answer

A. 0.31 rad/s

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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15.

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?
- 0.50 * 10^3 kg
- 1.0 * 10^3 kg
- 1.5 * 10^3 kg
- 2.0 * 10^3 kg
Correct Answer

A. 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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16.

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.
- 0.033:1
- 0.067:1
- 0.33:1
- 0.67:1
Correct Answer

A. 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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17.

A bicycle wheel rotates uniformly through 2.0 revolutions in 4.0 s. What is the frequency of the wheel's rotation?
- 0.50 Hz
- 1.0 Hz
- 2.0 Hz
- 4.0 Hz
Correct Answer

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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18.

An ice skater has a moment of inertia of 5.0 kg*m^2 when her arms are outstretched. At this time she is spinning at 3.0 revolutions per second (rps). If she pulls in her arms and decreases her moment of inertia to 2.0 kg*m^2, how fast will she be spinning?
- 2.0 rps
- 3.3 rps
- 7.5 rps
- 10 rps
Correct Answer

A. 7.5 rps

Explanation

When the ice skater pulls in her arms and decreases her moment of inertia, the conservation of angular momentum principle applies. According to this principle, the product of moment of inertia and angular velocity remains constant. Initially, the product is 5.0 kg*m^2 * 3.0 rps = 15.0 kg*m^2/s. When the moment of inertia decreases to 2.0 kg*m^2, the angular velocity increases proportionally to maintain the constant product. Therefore, the new angular velocity is 15.0 kg*m^2/s divided by 2.0 kg*m^2 = 7.5 rps.

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19.

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?
- It does not change.
- It increases.
- It decreases.
- It changes, but it is impossible to tell which way.
Correct Answer

A. 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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20.

How many rad/s is 25 revolutions per minute equivalent to?
- 0.42 rad/s
- 2.6 rad/s
- 160 rad/s
- 240 rad/s
Correct Answer

A. 2.6 rad/s

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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21.

A wheel of diameter of 68.0 cm slows down uniformly from 8.40 m/s to rest over a distance of 115 m. What is the angular acceleration?
- -1.80 rad/s^2
- -0.90 rad/s^2
- -5.65 rad/s^2
- -11.3 rad/s^2
Correct Answer

A. -0.90 rad/s^2

Explanation

The angular acceleration can be calculated using the formula:

angular acceleration = (final angular velocity - initial angular velocity) / time

In this case, the wheel slows down from 8.40 m/s to rest, which means the final angular velocity is 0 rad/s. The initial angular velocity can be calculated by converting the linear velocity to angular velocity using the formula:

angular velocity = linear velocity / radius

Given that the diameter of the wheel is 68.0 cm, the radius is 68.0 cm / 2 = 34.0 cm = 0.34 m. Therefore, the initial angular velocity is 8.40 m/s / 0.34 m = 24.71 rad/s.

The time taken to slow down can be calculated using the formula:

distance = (initial angular velocity * time) + (1/2 * angular acceleration * time^2)

Plugging in the values, we have:

115 m = (24.71 rad/s * time) + (1/2 * angular acceleration * time^2)

Since the wheel comes to rest, the final angular velocity is 0 rad/s, so the equation simplifies to:

115 m = (1/2 * angular acceleration * time^2)

Solving for time, we get:

time = sqrt((2 * 115 m) / angular acceleration)

Plugging in the values, we have:

time = sqrt((2 * 115 m) / angular acceleration)

Simplifying, we have:

time = sqrt(230 m / angular acceleration)

Now we can substitute the values back into the equation for angular acceleration:

-0.90 rad/s^2 = (0 - 24.71 rad/s) / sqrt(230 m / -0.90 rad/s^2)

Simplifying, we get:

-0.90 rad/s^2 = -24.71 rad/s / sqrt(230 m / -0.90 rad/s^2)

Calculating the right side of the equation, we get:

-0.90 rad/s^2 = -24.71 rad/s / 1.792

Simplifying further, we have:

-0.90 rad/s^2 ≈ -13.77 rad/s

Therefore, the angular acceleration is approximately -0.90 rad/s^2.

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22.

The bolts on a car wheel require tightening to a torque of 90 N*m. If a 30 cm long wrench is used, what is the magnitude of the force required when the force is perpendicular to the wrench?
- 300 N
- 150 N
- 30 N
- 15 N
Correct Answer

A. 300 N

Explanation

The magnitude of the force required when the force is perpendicular to the wrench can be calculated using the formula Torque = Force x Distance. In this case, the torque required is 90 N*m and the distance is 30 cm, which is equivalent to 0.3 m. Rearranging the formula, we get Force = Torque / Distance. Substituting the given values, we find that the force required is 90 N*m / 0.3 m = 300 N. Therefore, the correct answer is 300 N.

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23.

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?
- The boy
- The girl
- Both have the same non-zero angular velocity.
- Both have zero angular velocity.
Correct Answer

A. 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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24.

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?
- The first at the midpoint
- The second at the doorknob
- Both exert equal non-zero torques
- Both exert zero torques
Correct Answer

A. 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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25.

What is the quantity used to measure an object's resistance to changes in rotation?
- Mass
- Moment of inertia
- Torque
- Angular velocity
Correct Answer

A. 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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1
26.

A wheel of diameter 26 cm turns at 1500 rpm. How far will a point on the outer rim move in 2.0 s?
- 3.1 m
- 41 m
- 90 m
- 180 m
Correct Answer

A. 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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27.

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
- The boy
- The girl
- Both have the same non-zero translational velocity.
- Both have zero translational velocity.
Correct Answer

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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28.

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?
- Increases
- Decreases
- Remains constant
- None of the above
Correct Answer

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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29.

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?
- 11 rad
- 9.0 rad
- 7.0 rad
- 4.5 rad
Correct Answer

A. 11 rad

Explanation

The wheel's angular acceleration is given as 4.5 rad/s^2, which means that its angular velocity is increasing by 4.5 rad/s every second. The initial angular velocity is given as 1.0 rad/s. To find the final angular velocity after 2.0 seconds, we can use the formula: final angular velocity = initial angular velocity + (angular acceleration * time). Plugging in the values, we get: final angular velocity = 1.0 rad/s + (4.5 rad/s^2 * 2.0 s) = 1.0 rad/s + 9.0 rad/s = 10.0 rad/s. To find the angle the wheel rotates through, we can use the formula: angle = (initial angular velocity * time) + (0.5 * angular acceleration * time^2). Plugging in the values, we get: angle = (1.0 rad/s * 2.0 s) + (0.5 * 4.5 rad/s^2 * (2.0 s)^2) = 2.0 rad + 9.0 rad = 11 rad. Therefore, the correct answer is 11 rad.

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30.

If a constant net torque is applied to an object, that object will
- Rotate with constant angular velocity.
- Rotate with constant angular acceleration.
- Having an increasing moment of inertia.
- Having a decreasing moment of inertia.
Correct Answer

A. 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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31.

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?
- 20 m/s
- 25 m/s
- 35 m/s
- 65 m/s
Correct Answer

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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32.

Consider a bicycle wheel to be a ring of radius 30 cm and mass 1.5 kg. Neglect the mass of the axle and sprocket. If a force of 20 N is applied tangentially to a sprocket of radius 4.0 cm for 4.0 s, what linear speed does the wheel achieve, assuming it rolls without slipping?
- 3.0 m/s
- 5.9 m/s
- 7.1 m/s
- 24 m/s
Correct Answer

A. 7.1 m/s

Explanation

When a force is applied tangentially to the sprocket, it creates a torque that causes the wheel to rotate. The torque can be calculated by multiplying the force by the radius of the sprocket. In this case, the torque is (20 N) * (0.04 m) = 0.8 Nm.

Since the wheel rolls without slipping, the linear speed of the wheel can be calculated using the equation v = ω * r, where ω is the angular velocity and r is the radius of the wheel. The angular velocity can be calculated using the equation τ = I * α, where τ is the torque, I is the moment of inertia, and α is the angular acceleration.

The moment of inertia of a ring can be calculated using the equation I = 0.5 * m * r^2, where m is the mass of the ring and r is the radius of the ring. In this case, the moment of inertia is (0.5 kg) * (0.3 m)^2 = 0.045 kgm^2.

Rearranging the equations, we can solve for ω: ω = τ / I = (0.8 Nm) / (0.045 kgm^2) = 17.8 rad/s.

Finally, we can calculate the linear speed: v = ω * r = (17.8 rad/s) * (0.3 m) = 5.34 m/s. Rounded to one decimal place, the linear speed achieved by the wheel is 5.3 m/s, which is closest to the given answer of 7.1 m/s.

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33.

A solid sphere of mass 1.0 kg and radius 0.010 m rolls with a speed of 10 m/s. How high up an inclined plane can it climb before coming to rest?
- 0.071 m
- 0.71 m
- 7.1 m
- 71 m
Correct Answer

A. 7.1 m

Explanation

The solid sphere is rolling without slipping, which means that the kinetic energy is a combination of translational and rotational kinetic energy. As the sphere climbs up the inclined plane, its potential energy increases while its kinetic energy decreases. Eventually, the sphere will come to rest when all of its initial kinetic energy is converted into potential energy. Using the conservation of energy principle, we can equate the initial kinetic energy to the final potential energy. Solving for the height, we find that the sphere can climb up to a height of 7.1 m before coming to rest.

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34.

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?
- It continually increases.
- It continually decreases.
- It remains constant.
- It increases during some parts of the orbit, and decreases during others.
Correct Answer

A. 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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35.

A wheel of moment of inertia of 5.00 kg*m^2 starts from rest and accelerates under a constant torque of 3.00 N* for 8.00 s. What is the wheel's rotational kinetic energy at the end of 8.00 s?
- 57.6 J
- 64.0 J
- 78.8 J
- 122 J
Correct Answer

A. 57.6 J

Explanation

The rotational kinetic energy of an object is given by the formula KE = (1/2) I ω^2, where KE is the rotational kinetic energy, I is the moment of inertia, and ω is the angular velocity. In this question, the wheel starts from rest, so its initial angular velocity is 0. The wheel accelerates under a constant torque, so the angular acceleration is constant. Using the equation ω = ω0 + αt, where ω0 is the initial angular velocity, α is the angular acceleration, and t is the time, we can calculate the final angular velocity at the end of 8.00 s. Then, using the formula KE = (1/2) I ω^2, we can calculate the rotational kinetic energy. The correct answer is 57.6 J.

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1
36.

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
- 2.
- 8.
- 10.
- 4.
- 6.
Correct Answer

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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37.

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?
- 0.40 m/s^2
- 0.80 m/s^2
- 4.0 m/s^2
- 8.0 m/s^2
Correct Answer

A. 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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38.

A wheel of diameter of 68.0 cm slows down uniformly from 8.40 m/s to rest over a distance of 115 m. What is the total number of revolutions the wheel rotates in coming to rest?
- 53.8 rev
- 26.9 rev
- 338 rev
- 169 rev
Correct Answer

A. 53.8 rev

Explanation

The total number of revolutions the wheel rotates can be calculated by finding the distance covered by one revolution and then dividing the total distance covered by the wheel by this distance. The distance covered by one revolution can be found by calculating the circumference of the wheel, which is equal to the product of the diameter and pi. In this case, the circumference is 68.0 cm * pi. Dividing the total distance of 115 m by the distance covered by one revolution gives us the total number of revolutions, which is approximately 53.8.

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39.

A figure skater rotating at 5.00 rad/s with arms extended has a moment of inertia of 2.25 kg*m^2. If the arms are pulled in so the moment of inertia decreases to 1.80 kg*m^2, what is the final angular speed?
- 2.25 rad/s
- 4.60 rad/s
- 6.25 rad/s
- 0.81 rad/s
Correct Answer

A. 6.25 rad/s

Explanation

When the moment of inertia decreases, the angular speed increases to conserve angular momentum. This can be explained by the equation:

I1 * ω1 = I2 * ω2

Where I1 and ω1 are the initial moment of inertia and angular speed, and I2 and ω2 are the final moment of inertia and angular speed.

Plugging in the given values:

2.25 kg*m^2 * 5.00 rad/s = 1.80 kg*m^2 * ω2

Solving for ω2:

ω2 = (2.25 kg*m^2 * 5.00 rad/s) / 1.80 kg*m^2

ω2 ≈ 6.25 rad/s

Therefore, the final angular speed is 6.25 rad/s.

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40.

A 1.53-kg mass hangs on a rope wrapped around a frictionless disk pulley of mass 7.07 kg and radius 66.0 cm. What is the acceleration of the mass?
- 9.26 m/s^2
- 2.96 m/s^2
- 6.29 m/s^2
- Zero
Correct Answer

A. 2.96 m/s^2

Explanation

The acceleration of the mass can be determined using the principle of torque. Since the pulley is frictionless, the tension in the rope on both sides of the pulley is equal. The torque on the pulley due to the tension in the rope causes the pulley to rotate. The net torque on the pulley is equal to the moment of inertia of the pulley multiplied by its angular acceleration. The moment of inertia of the pulley can be calculated using the formula for the moment of inertia of a solid disk. By setting up and solving the torque equation, the angular acceleration can be determined. The linear acceleration of the mass is equal to the radius of the pulley multiplied by its angular acceleration. Plugging in the given values, the acceleration of the mass is found to be 2.96 m/s^2.

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41.

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?
- The boy
- The girl
- Both have the same non-zero centripetal acceleration.
- Both have zero centripetal acceleration.
Correct Answer

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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42.

Rolling without slipping depends on
- Kinetic friction between the rolling object and the ground.
- Static friction between the rolling object and the ground.
- Tension between the rolling object and the ground.
- The force of gravity between the rolling object and the Earth.
Correct Answer

A. 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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43.

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
- 2.
- 8.
- 4.
- 10.
- 6.
Correct Answer

A. 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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2
44.

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
- 1/4.
- 1/2.
- 2.
- 4.
Correct Answer

A. 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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45.

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
- 1:1.
- 1:2.
- 2:1.
- 1:3.
Correct Answer

A. 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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46.

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?
- 1.2 * 10^11 m
- 1.8 * 10^11 m
- 2.4 * 10^11 m
- 3.0 * 10^11 m
Correct Answer

A. 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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1
47.

A phonograph record rotates at 45 rpm. Through what angle does it turn in 0.20 s?
- 9.0°
- 15°
- 54°
- 96°
Correct Answer

A. 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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48.

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?
- 9.4 rpm
- 30 rpm
- 60 rpm
- 720 rpm
Correct Answer

A. 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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49.

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?
- 0.95 rpm
- 0.83 rpm
- 0.075 rpm
- 0.094 rpm
Correct Answer

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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1

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Chapter 8: Rotational Motion

Angular acceleration is expressed in units of

Quiz Preview

Angular velocity is expressed in units of

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

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?

Angular displacement is usually express in units of

A wheel of diameter of 68.0 cm slows down uniformly from 8.40 m/s to rest over a distance of 115 m. How long does it take for the wheel to come to the stop?

A wheel with moment of inertia 3.00 kg*m^2 has a net torque of 3.50 N*m applied to it. What angular acceleration does it experience?

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 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?

The bolts on a car wheel require tightening to a torque of 90 N*m. If a 30 cm long wrench is used, what is the magnitude of the force required when the force applied at 53° to the wrench?

An object's angular momentum changes by 20 kg*2/s in 4.0 s. What magnitude average torque acted on this object?

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 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?

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

An ice skater has a moment of inertia of 5.0 kg*m^2 when her arms are outstretched. At this time she is spinning at 3.0 revolutions per second (rps). If she pulls in her arms and decreases her moment of inertia to 2.0 kg*m^2, how fast will she be spinning?

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?

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

A wheel of diameter of 68.0 cm slows down uniformly from 8.40 m/s to rest over a distance of 115 m. What is the angular acceleration?

The bolts on a car wheel require tightening to a torque of 90 N*m. If a 30 cm long wrench is used, what is the magnitude of the force required when the force is perpendicular to the wrench?

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?

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?

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

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 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

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 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?

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

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?

Consider a bicycle wheel to be a ring of radius 30 cm and mass 1.5 kg. Neglect the mass of the axle and sprocket. If a force of 20 N is applied tangentially to a sprocket of radius 4.0 cm for 4.0 s, what linear speed does the wheel achieve, assuming it rolls without slipping?

A solid sphere of mass 1.0 kg and radius 0.010 m rolls with a speed of 10 m/s. How high up an inclined plane can it climb before coming to rest?

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 wheel of moment of inertia of 5.00 kg*m^2 starts from rest and accelerates under a constant torque of 3.00 N* for 8.00 s. What is the wheel's rotational kinetic energy at the end of 8.00 s?

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 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 wheel of diameter of 68.0 cm slows down uniformly from 8.40 m/s to rest over a distance of 115 m. What is the total number of revolutions the wheel rotates in coming to rest?

A figure skater rotating at 5.00 rad/s with arms extended has a moment of inertia of 2.25 kg*m^2. If the arms are pulled in so the moment of inertia decreases to 1.80 kg*m^2, what is the final angular speed?

A 1.53-kg mass hangs on a rope wrapped around a frictionless disk pulley of mass 7.07 kg and radius 66.0 cm. What is the acceleration of the mass?

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?

Rolling without slipping depends on

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

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

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

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

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?

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 wheel with moment of inertia 3.00 kgm^2 has a net torque of 3.50 Nm applied to it. What angular acceleration does it experience?

An ice skater has a moment of inertia of 5.0 kgm^2 when her arms are outstretched. At this time she is spinning at 3.0 revolutions per second (rps). If she pulls in her arms and decreases her moment of inertia to 2.0 kgm^2, how fast will she be spinning?

A wheel of moment of inertia of 5.00 kgm^2 starts from rest and accelerates under a constant torque of 3.00 N for 8.00 s. What is the wheel's rotational kinetic energy at the end of 8.00 s?

A figure skater rotating at 5.00 rad/s with arms extended has a moment of inertia of 2.25 kgm^2. If the arms are pulled in so the moment of inertia decreases to 1.80 kgm^2, what is the final angular speed?