Question 1

A fan blade, initially at rest, rotates with a constant acceleration of 0.025 rad/s². What is the time interval required for it to reach a 10-rad displacement after starting from rest? A) 28 s B) 2.0 s C) 56 s D) 18 s

Accepted Answer

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Question 2

A fan blade, initially at rest, rotates with a constant acceleration of 0.025 rad/s². What is its angular speed at the instant it goes through an angular displacement of 10 rad? A) 0.025 rad/s B) 0.11 rad/s C) 0.46 rad/s D) 0.71 rad/s

Accepted Answer

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Question 3

A ceiling fan is turned on and reaches an angular speed of 60 rev/min in 20 s. It is then turned off and coasts to a stop in 40 s. In the one minute of rotation, through how many revolutions did the fan turn?
A) 30
B) 60
C) 0
D) 600

Accepted Answer

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Question 4

A baseball pitcher loosens up his pitching arm. He tosses a 0.20-kg ball using only the rotation of his forearm, 0.32 m in length, to accelerate the ball. What is the moment of inertia of the ball alone as it moves in a circular arc with a radius of 0.32 m? A) 1.5 *10^-² kg·m² B) 2.0 *10^-² kg·m² C) 4.0 *10^-² kg·m² D) 7.6 *10^-²kg·m²

Accepted Answer

The moment of inertia (I) for a point mass moving in a circular path is given by I = m*r^2. Here, m = 0.20 kg and r = 0.32 m. So, I = 0.20 * (0.32)^2 = 0.02048 kg·m^2, which matches option B.

Question 5

A Ferris wheel starts at rest and builds up to a final angular speed of 0.70 rad/s while rotating through an angular displacement of 2.5 rad. What is its average angular acceleration? A) 0.098 rad/s² B) 0.05 rad/s² C) 1.8 rad/s² D) 0.60 rad/s²

Accepted Answer

We can use the formula for average angular acceleration:

average angular acceleration = change in angular speed / change in time

The final angular speed is given as 0.70 rad/s and the initial angular speed is 0 (since the Ferris wheel starts at rest). The time it takes to rotate through an angular displacement of 2.5 rad can be found using the formula:

angular displacement = (average angular speed)(time)

rearranging this formula gives:

time = angular displacement / average angular speed

Plugging in the values we get:

time = 2.5 rad / 0.70 rad/s = 3.57 s

Therefore, the change in angular speed is:

change in angular speed = final angular speed - initial angular speed = 0.70 rad/s - 0 rad/s = 0.70 rad/s

So, the average angular acceleration is:

average angular acceleration = change in angular speed / change in time = 0.70 rad/s / 3.57 s = 0.098 rad/s^2

The closest answer choice to this value is A: 0.098 rad/s^2, so that is the correct answer.

Question 6

A bucket of water with total mass 23 kg is attached to a rope, which in turn is wound around a 0.050-m radius cylinder at the top of a well. The bucket is raised to the top of the well and released. The bucket is moving with a speed of 6.0 m/s upon hitting the water surface in the well. What is the angular speed of the cylinder at this instant?
A) 39 rad/s
B) 79 rad/s
C) 120 rad/s
D) 160 rad/s

Accepted Answer

We can start by using conservation of energy to find the velocity of the bucket just before it hits the water:

mgh = (1/2)mv^2
where m is the mass of the bucket, h is the height of the well, and v is the velocity of the bucket just before it hits the water.

Solving for v, we get:

v = sqrt(2gh)

Substituting in the given values, we get:

v = sqrt(2*9.81 m/s^2*10 m) = 19.8 m/s

Next, we can use the relationship between linear and angular velocity:

v = rω

where r is the radius of the cylinder and ω is the angular velocity of the cylinder.

Solving for ω, we get:

ω = v/r

Substituting in the given values, we get:

ω = 19.8 m/s / 0.050 m = 396 rad/s

However, this is the angular velocity just before the bucket hits the water. We want to know the angular velocity at the instant the bucket hits the water. At this instant, the bucket is no longer attached to the rope and is not contributing to the system. Therefore, the total mass attached to the cylinder is now only the mass of the rope, which is negligible compared to the mass of the bucket.

Since angular momentum is conserved, we can use the equation:

L = Iω

where L is the angular momentum, I is the moment of inertia of the cylinder, and ω is the angular velocity of the cylinder.

Initially, the system has zero angular momentum (since everything is at rest). Just before the bucket hits the water, the angular momentum is:

L = (m + M)vr

where M is the mass of the rope, which we assume is negligible, so we can ignore it. Therefore, we have:

L = mrv

where m is the mass of the bucket and v is its velocity just before hitting the water.

Solving for ω, we get:

ω = L / I

Substituting in the given values, we get:

ω = mrv / (1/2*M*r^2) = 2mv / M

Substituting in the given values, we get:

ω = 2*23 kg*19.8 m/s / (0.050 m*0.100 kg) = 120 rad/s

Therefore, the answer is choice C.

Question 7

A Ferris wheel, rotating initially at an angular speed of 0.16 rad/s, accelerates over a 7.0-s interval at a rate of 0.050 rad/s². What is its angular speed after this 7.0-s interval? A) 0.46 rad/s B) 0.85 rad/s C) 0.51 rad/s D) 1.2 rad/s

Accepted Answer

Using the formula:

ωf = ωi + αt

where
ωi = initial angular speed = 0.16 rad/s
α = angular acceleration = 0.050 rad/s^2
t = time interval = 7.0 s

Substituting the given values, we get:

ωf = 0.16 + (0.050)(7.0)
ωf = 0.51 rad/s

Therefore, the angular speed of the Ferris wheel after the 7.0-s interval is 0.51 rad/s, which is the closest to choice C (0.51 rad/s).

Question 8

An object consists of a rod (of length 3.0 m and negligible moment of inertia) to which four small 2.0-kg masses are attached, one at each end and one at each point on the rod 1.0 m from each end. (The masses are one meter apart.) The moment of inertia of this object about an axis perpendicular to the rod and through one of the end masses: A) is 28 kg·m². B) is 12 kg·m². C) is 4 kg·m². D) cannot be uniquely determined until it is stated which inner mass the axis goes through.

Accepted Answer

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Question 9

A majorette takes two batons and fastens them together in the middle at right angles to make an "x" shape. Each baton was 0.90 m long and each ball on the end is 0.20 kg. (Ignore the mass of the rods.) What is the moment of inertia if the arrangement is spun around an axis through the center perpendicular to both rods? A) 0.064 kg·m² B) 0.096 kg·m² C) 0.075 kg·m² D) 0.16 kg·m²

Accepted Answer

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Question 10

A bowling ball has a mass of 7.0 kg, a moment of inertia of 2.8 * 10^-² kg·m², and a radius of 0.10 m. If it rolls down the lane without slipping at a linear speed of 5.0 m/s, what is its angular speed? A) 0.80 rad/s B) 50 rad/s C) 0.050 rad/s D) 40 rad/s

Accepted Answer

The angular speed (ω) can be found using the relationship between linear speed (v) and angular speed: v = ωr. Solving for ω gives ω = v/r = 5.0 m/s / 0.10 m = 50 rad/s.

Question 11

A non-uniform rod of length L and mass M is pivoted about one end. The moment of inertia of this rod about its pivoted end is  . Attached to the rod is a mass M at its midpoint and at the end opposite the pivot is another mass 2M. What is the moment of inertia of this configuration about the pivot?
A) 
B) 
C) 
D) None of the above answers is correct.

Accepted Answer

The answer of A non-uniform rod of length L and...

Question 12

A uniform rod of length L and mass M is pivoted about one end. Attached to the rod is a mass M at its midpoint and at the end opposite the pivot is another mass M. What is the moment of inertia of this configuration about the pivot?
A) 
B) 
C) 
D)

Accepted Answer

The answer of A uniform rod of length L and...

Question 13

Starting from rest, a wheel undergoes constant angular acceleration for a period of time T. At what time after the start of rotation does the wheel reach an angular speed equal to its average angular speed for this interval?
A) 0.25T
B) 0.50T
C) 0.67T
D) 0.71T

Accepted Answer

The wheel reaches its average angular speed at 0.50T because the average angular speed is defined as the total angular displacement divided by the total time, which, under constant acceleration, is reached at the midpoint of the time interval.

Question 14

Two hoops or rings (I = MR²) are centered, lying on a turntable. The smaller ring has radius = 0.050 m; the larger has radius = 0.10 m. Both have a mass of 6.0 kg. What is the total moment of inertia as the turntable spins? Ignore the mass of the turntable. A) 0.030 kg·m² B) 0.0075 kg·m² C) 0.038 kg·m² D) 0.075 kg·m²

Accepted Answer

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Question 15

A 4.0-kg mass is placed at (3.0, 4.0) m, and a 7.0 kg-mass is placed at (3.0, -.0) m. What is the moment of inertia of this system of masses about the z axis? A) 275 kg·m² B) 90 kg·m² C) 250 kg·m² D) 180 kg·m²

Accepted Answer

The moment of inertia of a system of masses about an axis is the sum of the moments of inertia of the individual masses about that axis. The moment of inertia of a point mass m about an axis a distance r from the mass is mr^2. In this case, the moment of inertia of the 4.0-kg mass about the z axis is (4.0 kg)(3.0 m)^2 = 36.0 kg·m^2, and the moment of inertia of the 7.0-kg mass about the z axis is (7.0 kg)(4.0 m)^2 = 112.0 kg·m^2. Therefore, the total moment of inertia of the system of masses about the z axis is 36.0 kg·m^2 + 112.0 kg·m^2 = 148.0 kg·m^2.

Question 16

Starting from rest, a wheel undergoes constant angular acceleration for a period of time T. At which of the following times does the average angular acceleration equal the instantaneous angular acceleration?
A) 0.50 T
B) 0.67 T
C) 0.71 T
D) all of the above

Accepted Answer

The average angular acceleration over a period of time is given by (final angular velocity - initial angular velocity)/time. 
Since the wheel starts from rest, the initial angular velocity is 0. 
The final angular velocity can be found using the equation for angular motion: 
Δθ = (ωf^2 - ωi^2)/2α 
Where Δθ is the angle rotated, ωi is the initial angular velocity, ωf is the final angular velocity, and α is the constant angular acceleration. 
Since the wheel starts from rest, Δθ = (1/2)αT^2. 
Solving for ωf, we get: 
ωf^2 = 2α(1/2)αT^2 
ωf = (αT^2)^1/2 
The instantaneous angular acceleration at any time t is simply the constant angular acceleration α. 
To find the time(s) when the average and instantaneous angular acceleration are equal, we need to set the average angular acceleration equal to α: 
(ωf -ωi)/T = α 
[(αT^2)^1/2 - 0]/T = α 
Simplifying, we get: 
T = 2 
This means that the average angular acceleration will be equal to the instantaneous angular acceleration at t = 0.5T, 0.67T, and 0.71T. Therefore, the answer is D.

Question 17

The Earth moves about the Sun in an elliptical orbit. As the Earth moves closer to the Sun, which of the following best describes the Earth-Sun system's moment of inertia?
A) decreases
B) increases
C) remains constant
D) None of the above choices are valid.

Accepted Answer

The moment of inertia of a system depends on the distribution of mass relative to the axis of rotation. As the Earth moves closer to the Sun, the distance between the Earth and the Sun decreases, leading to a decrease in the moment of inertia of the Earth-Sun system.

Question 18

A Ferris wheel, rotating initially at an angular speed of 0.500 rad/s, accelerates over a 7.00-s interval at a rate of 0.0500 rad/s². What angular displacement does the Ferris wheel undergo in this 7.00-s interval? A) 4.48 rad B) 2.50 rad C) 4.73 rad D) 0.500 rad

Accepted Answer

We can use the formula for angular displacement with constant acceleration:

θ = ωi t + (1/2) α t^2

where 
θ = angular displacement
ωi = initial angular speed
α = angular acceleration
t = time interval

Plugging in the given values:

θ = (0.500 rad/s)(7.00 s) + (1/2)(0.0500 rad/s^2)(7.00 s)^2

Simplifying:

θ = 3.50 rad + 1.23 rad

θ = 4.73 rad

Therefore, the ferris wheel undergoes an angular displacement of 4.73 radians in this 7.00-second interval. The best choice is C).

Question 19

A 0.30-m radius automobile tire rotates how many rad after starting from rest and accelerating at a constant 3.0 rad/s² over a 5.0-s interval? A) 12 rad B) 25 rad C) 37 rad D) 0.50 rad

Accepted Answer

The formula for rotational kinetic energy is KE = (1/2) I w^2, where I is the moment of inertia and w is the angular velocity (in radians per second).

Since the tire starts from rest, its initial angular velocity is 0. We can use the formula w = w0 + at, where w0 is the initial angular velocity, a is the angular acceleration (in radians per second squared), and t is the time.

In this case, w0 = 0, a = 3.0 rad/s^2, and t = 5.0 s. Therefore,

w = 0 + (3.0 rad/s^2)(5.0 s) = 15 rad/s

The final angular velocity is 15 rad/s. To find the total angular displacement, we can use the formula delta(theta) = w0t + (1/2)at^2.

In this case, w0 = 0, a = 3.0 rad/s^2, t = 5.0 s, and r = 0.30 m (the radius of the tire). Therefore,

delta(theta) = (0)(5.0 s) + (1/2)(3.0 rad/s^2)(5.0 s)^2 = 37.5 rad

However, this is the total angular displacement of the entire tire. Since we're only interested in the angular displacement of one point on the tire (e.g. the top of the tire), we need to divide this by the ratio of the circumference to the radius (i.e. 2 pi).

circumference = 2 pi r = 1.89 m

angular displacement of one point on the tire = delta(theta) / (circumference/radius) = (37.5 rad) / (1.89 m / 0.30 m) = 37 rad

Therefore, the answer is C: 37 rad.

Question 20

A majorette takes two batons and fastens them together in the middle at right angles to make an "x" shape. Each baton was 0.90 m long and each ball on the end is 0.20 kg. (Ignore the mass of the rods.) What is the moment of inertia if the arrangement is spun around an axis formed by one of the batons? A) 0.081 kg·m² B) 0.064 kg·m² C) 0.19 kg·m² D) 0.32 kg·m²

Accepted Answer

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Quiz 8: Rotational Motion

A fan blade, initially at rest, rotates with a constant acceleration of 0.025 rad/s2. What is the time interval required for it to reach a 10-rad displacement after starting from rest?

A fan blade, initially at rest, rotates with a constant acceleration of 0.025 rad/s2. What is its angular speed at the instant it goes through an angular displacement of 10 rad?

A ceiling fan is turned on and reaches an angular speed of 60 rev/min in 20 s. It is then turned off and coasts to a stop in 40 s. In the one minute of rotation, through how many revolutions did the fan turn?

A baseball pitcher loosens up his pitching arm. He tosses a 0.20-kg ball using only the rotation of his forearm, 0.32 m in length, to accelerate the ball. What is the moment of inertia of the ball alone as it moves in a circular arc with a radius of 0.32 m?

A Ferris wheel starts at rest and builds up to a final angular speed of 0.70 rad/s while rotating through an angular displacement of 2.5 rad. What is its average angular acceleration?

A Ferris wheel, rotating initially at an angular speed of 0.16 rad/s, accelerates over a 7.0-s interval at a rate of 0.050 rad/s2. What is its angular speed after this 7.0-s interval?

A bowling ball has a mass of 7.0 kg, a moment of inertia of 2.8 * 10-2 kg·m2, and a radius of 0.10 m. If it rolls down the lane without slipping at a linear speed of 5.0 m/s, what is its angular speed?

A non-uniform rod of length L and mass M is pivoted about one end. The moment of inertia of this rod about its pivoted end is . Attached to the rod is a mass M at its midpoint and at the end opposite the pivot is another mass 2M. What is the moment of inertia of this configuration about the pivot?

A uniform rod of length L and mass M is pivoted about one end. Attached to the rod is a mass M at its midpoint and at the end opposite the pivot is another mass M. What is the moment of inertia of this configuration about the pivot?

Starting from rest, a wheel undergoes constant angular acceleration for a period of time T. At what time after the start of rotation does the wheel reach an angular speed equal to its average angular speed for this interval?

Two hoops or rings (I = MR2) are centered, lying on a turntable. The smaller ring has radius = 0.050 m; the larger has radius = 0.10 m. Both have a mass of 6.0 kg. What is the total moment of inertia as the turntable spins? Ignore the mass of the turntable.

A 4.0-kg mass is placed at (3.0, 4.0) m, and a 7.0 kg-mass is placed at (3.0, -.0) m. What is the moment of inertia of this system of masses about the z axis?

Starting from rest, a wheel undergoes constant angular acceleration for a period of time T. At which of the following times does the average angular acceleration equal the instantaneous angular acceleration?

The Earth moves about the Sun in an elliptical orbit. As the Earth moves closer to the Sun, which of the following best describes the Earth-Sun system's moment of inertia?

A Ferris wheel, rotating initially at an angular speed of 0.500 rad/s, accelerates over a 7.00-s interval at a rate of 0.0500 rad/s2. What angular displacement does the Ferris wheel undergo in this 7.00-s interval?

A 0.30-m radius automobile tire rotates how many rad after starting from rest and accelerating at a constant 3.0 rad/s2 over a 5.0-s interval?

A fan blade, initially at rest, rotates with a constant acceleration of 0.025 rad/s². What is the time interval required for it to reach a 10-rad displacement after starting from rest?

A fan blade, initially at rest, rotates with a constant acceleration of 0.025 rad/s². What is its angular speed at the instant it goes through an angular displacement of 10 rad?

A Ferris wheel, rotating initially at an angular speed of 0.16 rad/s, accelerates over a 7.0-s interval at a rate of 0.050 rad/s². What is its angular speed after this 7.0-s interval?

A bowling ball has a mass of 7.0 kg, a moment of inertia of 2.8 * 10^-² kg·m², and a radius of 0.10 m. If it rolls down the lane without slipping at a linear speed of 5.0 m/s, what is its angular speed?

Two hoops or rings (I = MR²) are centered, lying on a turntable. The smaller ring has radius = 0.050 m; the larger has radius = 0.10 m. Both have a mass of 6.0 kg. What is the total moment of inertia as the turntable spins? Ignore the mass of the turntable.

A Ferris wheel, rotating initially at an angular speed of 0.500 rad/s, accelerates over a 7.00-s interval at a rate of 0.0500 rad/s². What angular displacement does the Ferris wheel undergo in this 7.00-s interval?

A 0.30-m radius automobile tire rotates how many rad after starting from rest and accelerating at a constant 3.0 rad/s² over a 5.0-s interval?