College Physics Quiz: Rotational Kinematics

20 questions · exam conditions

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Rotational KinematicsQuestion 1 of 20

A bicycle wheel of radius 0.35 m starts from rest and accelerates uniformly. If the wheel makes 8.0 complete revolutions in the first 4.0 seconds, what is the angular acceleration of the wheel?

π rad/s²

2π rad/s²

3π rad/s²

4π rad/s²

6π rad/s²

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College Physics Quiz

College Physics Quiz: Rotational Kinematics

Practice Rotational Kinematics in College Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Rotational Kinematics, giving you a quick way to practice the rules, question types, and explanations that matter most for College Physics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A bicycle wheel of radius 0.35 m starts from rest and accelerates uniformly. If the wheel makes 8.0 complete revolutions in the first 4.0 seconds, what is the angular acceleration of the wheel?

π rad/s²
2π rad/s² (correct answer)
3π rad/s²
4π rad/s²
6π rad/s²

Explanation: This is a rotational kinematics problem where you need to find angular acceleration from given motion data. When you see questions about rotating objects with uniform acceleration, immediately think of the rotational kinematic equations - they're analogous to linear motion equations but use angular quantities. Start by identifying what you know: the wheel starts from rest (

\omega_0 = 0

), makes 8.0 revolutions in 4.0 seconds, and you need angular acceleration (

\alpha

). First, convert the angular displacement to radians:

\theta = 8.0 \text{ rev} \times 2\pi \text{ rad/rev} = 16\pi \text{ rad}

. Use the kinematic equation:

\theta = \omega_0 t + \frac{1}{2}\alpha t^2

. Since

\omega_0 = 0

, this simplifies to

\theta = \frac{1}{2}\alpha t^2

. Solving for angular acceleration:

\alpha = \frac{2\theta}{t^2} = \frac{2(16\pi)}{(4.0)^2} = \frac{32\pi}{16} = 2\pi \text{ rad/s}^2

. Choice A (

\pi \text{ rad/s}^2

) is half the correct value - you might get this if you forgot the factor of 2 in the kinematic equation. Choice C (

3\pi \text{ rad/s}^2

) could result from calculation errors with the time squared term. Choice D (

4\pi \text{ rad/s}^2

) is double the correct answer, possibly from confusing the kinematic equation or miscalculating the time factor. The correct answer is B (

2\pi \text{ rad/s}^2

). Remember: always convert revolutions to radians first, and double-check that you're using the right kinematic equation for the given information. The factor of

\frac{1}{2}

is crucial when starting from rest.

Question 2

A centrifuge rotor starts from rest and reaches 3000 rpm in 20 seconds with constant angular acceleration. How many revolutions does it complete during this startup period?

250 revolutions
500 revolutions (correct answer)
750 revolutions
1000 revolutions
1500 revolutions

Explanation: When you encounter rotational motion problems involving constant angular acceleration, you're dealing with rotational kinematics - the rotational equivalent of linear motion equations. The key is identifying which rotational kinematic equation to use based on the given information. Here you have initial angular velocity (

\omega_0 = 0

), final angular velocity (

\omega_f = 3000 \text{ rpm}

), and time (

t = 20 \text{ s}

). You need total angular displacement, which you can find using:

\theta = \omega_0 t + \frac{1}{2}\alpha t^2

First, convert 3000 rpm to rad/s:

3000 \times \frac{2\pi}{60} = 314.16 \text{ rad/s}

Find angular acceleration:

\alpha = \frac{\omega_f - \omega_0}{t} = \frac{314.16}{20} = 15.71 \text{ rad/s}^2

Calculate total angular displacement:

\theta = 0 + \frac{1}{2}(15.71)(20)^2 = 3142 \text{ rad}

Convert to revolutions:

\frac{3142}{2\pi} = 500 \text{ revolutions}

Choice A (250 revolutions) represents a common error of forgetting the

\frac{1}{2}

factor in the kinematic equation. Choice C (750 revolutions) might result from incorrectly adding average velocity to the acceleration term. Choice D (1000 revolutions) could come from using the wrong conversion factor or misapplying the equations entirely. Remember: for constant acceleration problems, always check if you're starting from rest - this simplifies the equations significantly by making the initial velocity term zero.

Question 3

A disk rotates according to θ(t) = 2.0t³ - 3.0t² + 5.0t radians, where t is in seconds. What is the angular acceleration at t = 2.0 seconds?

6.0 rad/s²
12 rad/s²
18 rad/s² (correct answer)
24 rad/s²
30 rad/s²

Explanation: When you encounter a rotational motion problem with position given as a function of time, you need to find derivatives to get velocity and acceleration. Angular acceleration is the second derivative of angular position with respect to time. Given

θ(t) = 2.0t^3 - 3.0t^2 + 5.0t

, you first find angular velocity by taking the first derivative:

ω(t) = \frac{dθ}{dt} = 6.0t^2 - 6.0t + 5.0

Then find angular acceleration by taking the second derivative:

α(t) = \frac{dω}{dt} = \frac{d^2θ}{dt^2} = 12.0t - 6.0

At t = 2.0 seconds:

α(2.0) = 12.0(2.0) - 6.0 = 24.0 - 6.0 = 18.0 \text{ rad/s}^2

This confirms answer C is correct. Answer A (6.0 rad/s²) represents a common error where students might take only the coefficient from the first derivative or make arithmetic mistakes. Answer B (12 rad/s²) occurs if you forget to substitute t = 2.0 seconds and just use the coefficient of t in the acceleration equation. Answer D (24 rad/s²) happens when you correctly calculate 12.0 × 2.0 but forget to subtract the 6.0 term. Remember the derivative hierarchy: position → velocity → acceleration. For rotational motion, always take two derivatives of the angular position function to find angular acceleration, then carefully substitute your time value into the final expression.

Question 4

A spinning top has an initial angular velocity of 30 rad/s and undergoes constant angular deceleration. If it makes 15 complete revolutions before coming to rest, what is its angular acceleration?

-1.6 rad/s²
-3.2 rad/s²
-4.8 rad/s² (correct answer)
-6.4 rad/s²
-9.5 rad/s²

Explanation: When you encounter rotational motion problems involving constant angular acceleration, you're dealing with rotational kinematics equations—the angular equivalents of linear motion formulas. Here, you have initial angular velocity

\omega_0 = 30 \text{ rad/s}

, final angular velocity

\omega = 0 \text{ rad/s}

(comes to rest), and angular displacement

\theta = 15 \text{ revolutions} = 15 \times 2\pi = 30\pi \text{ rad}

. You need to find angular acceleration

\alpha

. The most direct approach uses the kinematic equation:

\omega^2 = \omega_0^2 + 2\alpha\theta

Substituting the values:

0^2 = (30)^2 + 2\alpha(30\pi)

Solving:

0 = 900 + 60\pi\alpha

, so

\alpha = -\frac{900}{60\pi} = -\frac{15}{\pi} = -4.8 \text{ rad/s}^2

This confirms answer C) -4.8 rad/s². Looking at the wrong answers: A) -1.6 rad/s² might result from incorrectly using 15 revolutions directly without converting to radians. B) -3.2 rad/s² could come from calculation errors or using the wrong kinematic equation. D) -6.4 rad/s² might result from errors in handling the conversion factor or algebraic mistakes. The key strategy for rotational kinematics problems is to identify your known variables, convert units properly (especially revolutions to radians), choose the appropriate kinematic equation, and remember that deceleration yields negative angular acceleration. Always double-check your unit conversions—this is where many students make errors.

Question 5

A motor accelerates from rest with angular acceleration α = 4.0t rad/s², where t is in seconds. What is the angular velocity at t = 3.0 seconds?

12 rad/s
18 rad/s (correct answer)
24 rad/s
36 rad/s
48 rad/s

Explanation: When you encounter rotational motion problems with time-varying angular acceleration, you need to integrate to find angular velocity and position. The key relationship here is that angular acceleration α is the derivative of angular velocity ω with respect to time. Given that α = 4.0t rad/s² and the motor starts from rest (ω₀ = 0), you can find angular velocity by integrating the angular acceleration:

\omega = \int_0^t \alpha \, dt = \int_0^t 4.0t \, dt = 4.0 \cdot \frac{t^2}{2} = 2.0t^2

At t = 3.0 seconds:

\omega = 2.0(3.0)^2 = 2.0 \times 9.0 = 18 \text{ rad/s}

This confirms answer choice B is correct. Let's examine why the other options are wrong. Choice A (12 rad/s) results from incorrectly using the original acceleration equation α = 4.0t directly at t = 3.0 seconds, giving 4.0 × 3.0 = 12, but this gives acceleration, not velocity. Choice C (24 rad/s) might come from using the formula ω = αt with α = 8.0 rad/s² (perhaps doubling the coefficient incorrectly). Choice D (36 rad/s) could result from calculating 4.0 × 3.0² = 36, which skips the integration step entirely and incorrectly applies the time-dependent acceleration. Remember: when angular acceleration varies with time, you must integrate to find angular velocity. Don't confuse the instantaneous acceleration value with the accumulated velocity change over time.

Question 6

A fan blade starts from rest and reaches an angular velocity of 1200 rpm in 8.0 seconds. Assuming constant angular acceleration, how many revolutions does it complete in the first 4.0 seconds?

20 revolutions (correct answer)
40 revolutions
60 revolutions
80 revolutions
100 revolutions

Explanation: This problem tests rotational kinematics with constant angular acceleration. When you see angular motion problems, think of the rotational analogies to linear motion equations—they follow the same patterns but with angular quantities. Start by converting units and identifying what you know. The fan reaches 1200 rpm in 8.0 seconds, starting from rest. Convert to standard units:

\omega_f = 1200 \text{ rpm} \times \frac{2\pi \text{ rad}}{60 \text{ s}} = 40\pi \text{ rad/s}

. First, find the angular acceleration using

\omega_f = \omega_0 + \alpha t

. Since

\omega_0 = 0

\alpha = \frac{40\pi}{8.0} = 5\pi \text{ rad/s}^2

. Now find the angular displacement in the first 4.0 seconds using

\theta = \omega_0 t + \frac{1}{2}\alpha t^2

\theta = 0 + \frac{1}{2}(5\pi)(4.0)^2 = 40\pi \text{ radians}

Convert to revolutions:

\frac{40\pi \text{ rad}}{2\pi \text{ rad/rev}} = 20 \text{ revolutions}

Answer A (20 revolutions) is correct. Answer B (40 revolutions) likely comes from forgetting to divide by

2\pi

when converting radians to revolutions. Answer C (60 revolutions) might result from using the wrong time value or misapplying the kinematic equation. Answer D (80 revolutions) could come from calculation errors or using the final angular velocity incorrectly. Remember: always convert rpm to rad/s first, use the appropriate kinematic equation for what you're solving, and don't forget the final unit conversion from radians to revolutions.

Question 7

A rotating shaft has angular velocity that varies as ω(t) = 8.0 cos(2.0t) rad/s, where t is in seconds. What is the angular acceleration at t = π/4 seconds?

-16 rad/s² (correct answer)
-8.0 rad/s²
0 rad/s²
8.0 rad/s²
16 rad/s²

Explanation: When you encounter rotational motion problems involving time-varying angular velocity, you need to find angular acceleration by taking the derivative of the angular velocity function with respect to time. Given

\omega(t) = 8.0 \cos(2.0t)

rad/s, the angular acceleration is

\alpha(t) = \frac{d\omega}{dt}

. Using the chain rule to differentiate the cosine function:

\alpha(t) = 8.0 \times (-\sin(2.0t)) \times 2.0 = -16 \sin(2.0t)

rad/s². At

t = \pi/4

seconds:

\alpha(\pi/4) = -16 \sin(2.0 \times \pi/4) = -16 \sin(\pi/2) = -16 \times 1 = -16

rad/s². This confirms answer A is correct. Answer B (-8.0 rad/s²) likely comes from forgetting to apply the chain rule when differentiating

\cos(2.0t)

. If you only considered the derivative of cosine without the factor of 2.0 from the inner function, you'd get

-8.0 \sin(2.0t)

, yielding -8.0 at

t = \pi/4

. Answer C (0 rad/s²) represents a common misconception—perhaps confusing this with the angular velocity at

t = \pi/4

, which is indeed

8.0 \cos(\pi/2) = 0

rad/s. Answer D (8.0 rad/s²) contains a sign error, missing the negative that comes from differentiating cosine to get negative sine. Remember: angular acceleration is always the time derivative of angular velocity. When differentiating trigonometric functions with coefficients inside, don't forget the chain rule—multiply by the derivative of the inner function.

Question 8

A wheel undergoes angular motion described by ω(t) = 10 - 2t rad/s, where t is in seconds. What is the angular displacement between t = 0 and the time when the wheel first reverses direction?

20 rad
25 rad (correct answer)
30 rad
40 rad
50 rad

Explanation: This problem tests rotational kinematics, specifically finding angular displacement when angular velocity changes with time. When you see angular velocity as a function of time, you'll need to integrate to find displacement, and pay special attention to when the motion changes direction. The wheel reverses direction when its angular velocity becomes zero. Setting

\omega(t) = 10 - 2t = 0

, you get

t = 5

seconds. This is when the wheel stops rotating in its initial direction and begins rotating the opposite way. To find angular displacement, integrate the angular velocity function:

\theta = \int_0^5 (10 - 2t) \, dt = [10t - t^2]_0^5 = 10(5) - (5)^2 = 50 - 25 = 25

rad. This confirms answer B is correct. Looking at the wrong answers: A) 20 rad might result from incorrectly calculating

10 \times 5 - 5 \times 5 = 25

but making an arithmetic error. C) 30 rad could come from forgetting to subtract the

t^2

term and calculating just

10 \times 5 - 5 = 45

, then making another error. D) 40 rad might result from incorrectly finding the reversal time or making integration errors. Study tip: For rotational motion problems, always find when direction changes by setting

\omega(t) = 0

, then integrate the velocity function over the appropriate time interval. Remember that angular displacement requires integration of angular velocity, just like linear displacement requires integration of linear velocity.

Question 9

A wheel has an angular velocity that decreases linearly from 20 rad/s to 8.0 rad/s over a time interval of 3.0 seconds. What is the total angular displacement during this interval?

36 rad
42 rad (correct answer)
48 rad
54 rad
60 rad

Explanation: When you encounter rotational motion problems involving changing angular velocity, you're dealing with rotational kinematics. The key insight here is recognizing that "decreases linearly" means constant angular acceleration, so you can use kinematic equations. Since the angular velocity changes linearly from 20 rad/s to 8.0 rad/s over 3.0 seconds, you can find the angular displacement using the average velocity method. The average angular velocity is

\omega_{avg} = \frac{\omega_i + \omega_f}{2} = \frac{20 + 8.0}{2} = 14 \text{ rad/s}

The total angular displacement is:

\theta = \omega_{avg} \times t = 14 \times 3.0 = 42 \text{ rad}

This confirms answer B is correct. Let's examine why the other answers are wrong. Answer A (36 rad) would result from incorrectly using only the final velocity:

8.0 \times 3.0 + 12 = 36

, mixing up the calculation. Answer C (48 rad) comes from a common error of using the sum rather than average of initial and final velocities:

(20 + 8.0) \times 3.0 \div 2 = 42

, but some students mistakenly calculate

(20 + 8.0) + 8.0 = 48

. Answer D (54 rad) results from using only the initial velocity:

20 \times 3.0 - 6 = 54

, another calculation error. Remember: for any motion with constant acceleration (linear or rotational), the displacement equals average velocity times time. When angular velocity changes linearly, always use the average of initial and final values—don't let the changing motion intimidate you into overcomplicating the solution.

Question 10

A disk starts from rest and rotates with constant angular acceleration. In the third second of motion (from t = 2 s to t = 3 s), it rotates through an angle of 5.0 radians. What is its angular acceleration?

1.0 rad/s²
2.0 rad/s² (correct answer)
2.5 rad/s²
5.0 rad/s²
10 rad/s²

Explanation: This problem tests rotational kinematics with constant angular acceleration, specifically requiring you to work backwards from motion in a specific time interval to find the acceleration. Since the disk starts from rest with constant angular acceleration

\alpha

, you can use the kinematic equation:

\theta = \omega_0 t + \frac{1}{2}\alpha t^2

. With

\omega_0 = 0

, this becomes

\theta = \frac{1}{2}\alpha t^2

. To find the angle rotated during the third second (from t = 2s to t = 3s), calculate the total angle at each time and subtract:

At t = 3s: $\theta_3 = \frac{1}{2}\alpha(3)^2 = 4.5\alpha$
At t = 2s: $\theta_2 = \frac{1}{2}\alpha(2)^2 = 2\alpha$

The angle rotated in the third second is:

\theta_3 - \theta_2 = 4.5\alpha - 2\alpha = 2.5\alpha = 5.0

radians Solving:

\alpha = \frac{5.0}{2.5} = 2.0

rad/s² Answer B (2.0 rad/s²) is correct. Answer A (1.0 rad/s²) would give only 2.5 radians in the third second. Answer C (2.5 rad/s²) incorrectly assumes the 5.0 radians equals the coefficient rather than solving the equation properly. Answer D (5.0 rad/s²) represents the common error of assuming the given angle equals the acceleration directly. Remember: when dealing with motion during a specific time interval (not from the start), always calculate the difference between positions at the interval's endpoints. Don't confuse the given numerical values with the final answer.

Question 11

A pulley starts from rest with angular acceleration α(t) = 6.0t rad/s², where t is in seconds. What is the angular displacement after 2.0 seconds?

4.0 rad
8.0 rad (correct answer)
12 rad
16 rad
24 rad

Explanation: This problem tests rotational kinematics with non-constant angular acceleration. When you see angular acceleration given as a function of time, you'll need to integrate twice to find angular displacement. Given

α(t) = 6.0t

rad/s², you first integrate to find angular velocity. Since the pulley starts from rest,

ω_0 = 0

ω(t) = ∫α(t)dt = ∫6.0t \, dt = 3.0t^2 + C

With

ω(0) = 0

, the constant C = 0, so

ω(t) = 3.0t^2

. Next, integrate angular velocity to find angular displacement. Starting from rest means

θ_0 = 0

θ(t) = ∫ω(t)dt = ∫3.0t^2 \, dt = t^3 + C

With

θ(0) = 0

, again C = 0, so

θ(t) = t^3

. At t = 2.0 seconds:

θ(2) = (2.0)^3 = 8.0

rad. Choice A (4.0 rad) results from incorrectly using

θ = ½αt^2

with α = 6.0, treating the acceleration as constant rather than time-dependent. Choice C (12 rad) comes from errors in the integration process, possibly forgetting to divide by the increased power. Choice D (16 rad) might result from using

θ = ½at^2

with a = 8.0, confusing linear and rotational formulas. Remember: when acceleration depends on time, you must integrate step-by-step rather than using constant-acceleration formulas. Always check your integration constants using initial conditions.

Question 12

A rotating shaft has an angular velocity ω(t) = 4.0t - 6.0 rad/s, where t is in seconds. At what time does the shaft momentarily come to rest?

t = 0.67 s
t = 1.0 s
t = 1.5 s (correct answer)
t = 2.0 s
t = 2.5 s

Explanation: When you encounter problems involving rotating objects, remember that "momentarily at rest" means the angular velocity equals zero at that instant. This is analogous to finding when a linear object has zero velocity. To find when the shaft comes to rest, set the angular velocity equation equal to zero and solve for time:

\omega(t) = 4.0t - 6.0 = 0

Adding 6.0 to both sides:

4.0t = 6.0

Dividing by 4.0:

t = \frac{6.0}{4.0} = 1.5 \text{ seconds}

Let's examine why the other answers are incorrect: A) t = 0.67 s - This appears to come from incorrectly dividing 4.0 by 6.0 instead of 6.0 by 4.0. At t = 0.67 s,

\omega = 4.0(0.67) - 6.0 = -3.32 \text{ rad/s}

, so the shaft is still rotating. B) t = 1.0 s - This might result from mental rounding or approximation errors. At t = 1.0 s,

\omega = 4.0(1.0) - 6.0 = -2.0 \text{ rad/s}

, which is not zero. D) t = 2.0 s - This could come from setting up the equation incorrectly or arithmetic mistakes. At t = 2.0 s,

\omega = 4.0(2.0) - 6.0 = 2.0 \text{ rad/s}

, so the shaft has passed through rest and is rotating in the positive direction. Study tip: When solving "at rest" problems in rotational motion, always set the velocity function (linear or angular) equal to zero. Double-check your algebra by substituting your answer back into the original equation.

Question 13

A disk has angular velocity ω(t) = 6.0 + 4.0t - t² rad/s, where t is in seconds. At what time does the disk reach its maximum angular velocity?

t = 1.0 s
t = 2.0 s (correct answer)
t = 3.0 s
t = 4.0 s
t = 6.0 s

Explanation: When you encounter angular velocity as a function of time, you're looking at rotational kinematics. To find when angular velocity reaches its maximum, you need to find where its rate of change equals zero - that is, where the angular acceleration becomes zero. Given

ω(t) = 6.0 + 4.0t - t^2

, you find the angular acceleration by taking the derivative:

α(t) = \frac{dω}{dt} = 4.0 - 2t

. Setting this equal to zero:

4.0 - 2t = 0

, which gives

t = 2.0

seconds. At this moment, the disk stops accelerating and begins decelerating, meaning this is where maximum angular velocity occurs. Choice A (t = 1.0 s) represents the time when the disk is still accelerating, since

α(1) = 4.0 - 2(1) = 2.0

rad/s² > 0. The angular velocity is still increasing at this point. Choice C (t = 3.0 s) and Choice D (t = 4.0 s) both occur after the maximum. At these times, the angular acceleration is negative (

α(3) = -2.0

rad/s²,

α(4) = -4.0

rad/s²), meaning the disk is decelerating and angular velocity is decreasing. You can verify:

ω(2) = 6.0 + 4.0(2) - (2)^2 = 10.0

rad/s, which is indeed greater than the angular velocity at any other given time. Remember: for any function representing motion, maximum or minimum values occur where the derivative equals zero. Always take the derivative and set it to zero to find critical points.

Question 14

A bicycle wheel of radius $0.35 \text{ m}$ accelerates from rest to $15 \text{ m/s}$ (linear speed) in $8.0$ seconds. Assuming constant angular acceleration, through what angle does the wheel rotate during this time?

$86 \text{ rad}$
$120 \text{ rad}$
$171 \text{ rad}$ (correct answer)
$343 \text{ rad}$

Explanation: First find the final angular velocity:

\omega_f = \frac{v}{r} = \frac{15}{0.35} = \frac{300}{7} \approx 42.9 \text{ rad/s}

. Since the wheel starts from rest with constant angular acceleration, we can use:

\theta = \frac{\omega_0 + \omega_f}{2} \times t = \frac{0 + 42.9}{2} \times 8.0 = 171 \text{ rad}

. Choice A uses only half the motion time. Choice B represents an error in the radius conversion. Choice D doubles the correct answer.

Question 15

A rotating disk initially at rest undergoes constant angular acceleration of $2.5 \text{ rad/s}^2$ for $4.0$ seconds, then rotates at constant angular velocity for $6.0$ seconds, and finally decelerates uniformly to rest in $3.0$ seconds. What is the total angular displacement of the disk during the entire motion?

$85 \text{ rad}$
$120 \text{ rad}$
$135 \text{ rad}$ (correct answer)
$150 \text{ rad}$

Explanation: This problem requires analyzing three phases of motion. Phase 1 (acceleration):

\omega_f = \alpha t = 2.5 \times 4.0 = 10 \text{ rad/s}

\theta_1 = \frac{1}{2}\alpha t^2 = \frac{1}{2}(2.5)(4.0)^2 = 20 \text{ rad}

. Phase 2 (constant velocity):

\theta_2 = \omega t = 10 \times 6.0 = 60 \text{ rad}

. Phase 3 (deceleration):

\alpha = -\frac{10}{3.0} = -\frac{10}{3} \text{ rad/s}^2

\theta_3 = \omega_0 t + \frac{1}{2}\alpha t^2 = 10(3.0) + \frac{1}{2}(-\frac{10}{3})(3.0)^2 = 30 - 15 = 15 \text{ rad}

. Total:

20 + 60 + 15 = 135 \text{ rad}

. Choice A uses only the first two phases. Choice B neglects the deceleration phase properly. Choice D assumes constant acceleration throughout.

Question 16

A wheel starts from rest and rotates with constant angular acceleration. After $8.0$ revolutions, its angular velocity is $12 \text{ rad/s}$ . What was the angular velocity after $2.0$ revolutions?

$3.0 \text{ rad/s}$
$4.0 \text{ rad/s}$
$6.0 \text{ rad/s}$ (correct answer)
$8.0 \text{ rad/s}$

Explanation: Using

\omega^2 = \omega_0^2 + 2\alpha\theta

with

\omega_0 = 0

. For 8.0 revolutions:

\theta = 8.0 \times 2\pi = 16\pi \text{ rad}

, so

(12)^2 = 2\alpha(16\pi)

, giving

\alpha = \frac{144}{32\pi} = \frac{9}{2\pi} \text{ rad/s}^2

. For 2.0 revolutions:

\theta = 4\pi \text{ rad}

, so

\omega^2 = 2 \times \frac{9}{2\pi} \times 4\pi = 36

, therefore

\omega = 6.0 \text{ rad/s}

. Choice A assumes linear relationship with revolutions. Choice B uses incorrect proportionality. Choice D assumes direct proportionality with square root of revolutions.

Question 17

A flywheel rotating at $300 \text{ rpm}$ is brought to rest by a constant angular deceleration in $25$ seconds. How many complete revolutions does the flywheel make during this time?

$52$ revolutions
$62$ revolutions (correct answer)
$104$ revolutions
$125$ revolutions

Explanation: First convert initial angular velocity:

\omega_0 = 300 \text{ rpm} = 300 \times \frac{2\pi}{60} = 10\pi \text{ rad/s}

. Using

\theta = \omega_0 t + \frac{1}{2}\alpha t^2

with

\omega_f = 0

, we find

\alpha = -\frac{\omega_0}{t} = -\frac{10\pi}{25} = -\frac{2\pi}{5} \text{ rad/s}^2

. Therefore:

\theta = 10\pi(25) + \frac{1}{2}(-\frac{2\pi}{5})(25)^2 = 250\pi - 125\pi = 125\pi \text{ rad}

. Converting to revolutions:

\frac{125\pi}{2\pi} = 62.5

revolutions, so 62 complete revolutions. Choice A uses average angular velocity incorrectly. Choice C forgets the factor of 1/2. Choice D uses only the initial velocity term.

Question 18

A gear system has three connected gears with radii $r_1 = 5.0 \text{ cm}$ , $r_2 = 12.0 \text{ cm}$ , and $r_3 = 8.0 \text{ cm}$ . Gear 1 is connected to gear 2, and gear 2 is connected to gear 3. If gear 1 rotates at $240 \text{ rpm}$ , what is the angular velocity of gear 3?

$50 \text{ rpm}$
$80 \text{ rpm}$
$100 \text{ rpm}$
$150 \text{ rpm}$ (correct answer)

Explanation: For meshing gears, the linear velocity at the contact point must be equal. From gear 1 to gear 2:

r_1\omega_1 = r_2\omega_2

, so

\omega_2 = \frac{r_1}{r_2}\omega_1 = \frac{5.0}{12.0} \times 240 = 100 \text{ rpm}

. From gear 2 to gear 3:

r_2\omega_2 = r_3\omega_3

, so

\omega_3 = \frac{r_2}{r_3}\omega_2 = \frac{12.0}{8.0} \times 100 = 150 \text{ rpm}

. Choice A uses the wrong overall ratio. Choice B miscalculates the intermediate step. Choice C stops at gear 2's angular velocity.

Question 19

A motor starts from rest and reaches $1800 \text{ rpm}$ in $12$ seconds with constant angular acceleration. It then maintains this speed for $20$ seconds before decelerating uniformly to $600 \text{ rpm}$ in $8.0$ seconds. What is the average angular velocity during the entire $40$ -second interval?

$120 \text{ rad/s}$
$135 \text{ rad/s}$
$150 \text{ rad/s}$ (correct answer)
$165 \text{ rad/s}$

Explanation: Convert angular velocities:

1800 \text{ rpm} = 60\pi \text{ rad/s}

and

600 \text{ rpm} = 20\pi \text{ rad/s}

. Calculate angular displacement for each phase: Phase 1 (0-12s):

\theta_1 = \frac{1}{2}(0 + 60\pi) \times 12 = 360\pi \text{ rad}

. Phase 2 (12-32s):

\theta_2 = 60\pi \times 20 = 1200\pi \text{ rad}

. Phase 3 (32-40s):

\theta_3 = \frac{1}{2}(60\pi + 20\pi) \times 8 = 320\pi \text{ rad}

. Total displacement:

\theta_{total} = 1880\pi \text{ rad}

. Average angular velocity:

\omega_{avg} = \frac{1880\pi}{40} = 47\pi \approx 148 \text{ rad/s}

. Choice A neglects part of the motion. Choice B uses arithmetic mean of endpoints. Choice D overestimates the constant-speed contribution.

Question 20

Two wheels, A and B, are connected by a belt that does not slip. Wheel A has radius $0.20 \text{ m}$ and wheel B has radius $0.50 \text{ m}$ . If wheel A rotates at $150 \text{ rpm}$ and increases its angular velocity at $2.0 \text{ rad/s}^2$ , what is the angular acceleration of wheel B?

$0.80 \text{ rad/s}^2$ (correct answer)
$1.2 \text{ rad/s}^2$
$2.0 \text{ rad/s}^2$
$5.0 \text{ rad/s}^2$

Explanation: For a belt connection without slipping, the linear speeds of the belt at both wheel edges must be equal:

v = r_A\omega_A = r_B\omega_B

. Taking the time derivative:

r_A\alpha_A = r_B\alpha_B

. Therefore:

\alpha_B = \frac{r_A}{r_B}\alpha_A = \frac{0.20}{0.50} \times 2.0 = 0.80 \text{ rad/s}^2

. The initial rpm of wheel A is irrelevant for finding acceleration. Choice B uses the wrong radius ratio. Choice C assumes equal angular accelerations. Choice D uses the inverse ratio incorrectly.