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AP Physics 1 : Angular Velocity and Acceleration

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Example Question #1 : Circular And Rotational Motion

A horizontally mounted wheel of radius is initially at rest, and then begins to accelerate constantly until it has reached an angular velocity $\omega$ after 5 complete revolutions. What was the angular acceleration of the wheel?

Possible Answers:

$\frac{\omega ^{2}}{10\pi}$

$\frac{\omega ^{2}}{10\pi r}$

$\frac{\omega}{10\pi }$

$\frac{\omega ^{2}}{5\pi r}$

$\frac{\omega ^{2}}{20\pi }$

Correct answer:

$\frac{\omega ^{2}}{20\pi }$

Explanation:

You may recall the kinematic equation that relates final velocity, initial velocity, acceleration, and distance, respectively:

$v_{f}^{2}=v_{i}^{2}+2ad$

Well, for rotational motion (such as in this problem), there is a similar equation, except it relates final angular velocity, intial angular velocity, angular acceleration, and angular distance, respectively:

$\omega _{f}^{2}=\omega _{i}^{2}+2\alpha \theta$

The wheel starts at rest, so the initial angular velocity, $\omega _{i}$ , is zero. The total number of revolutions of the wheel is given to be 5 revolutions. Each revolution is equivalent to an angular distance of $2\pi$ radians. So, we can convert the total revolutions to an angular distance to get:

$\theta=5\:rev\cdot \frac{2\pi }{1\:rev}=10\pi$

The final angular velocity was given as $\omega$ in the text of the question. So, we should use the above equation to solve for the angular acceleration, $\alpha$ .

$\omega ^{2}=0+2\alpha\cdot 10\pi$

$\alpha=\frac{\omega ^{2}}{20\pi }$

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Example Question #1 : Circular And Rotational Motion

An object moves at a constant speed of $9.0 \:\frac{m}{s}$ in a circular path of radius of 1.5 m. What is the angular acceleration of the object?

Possible Answers:

$54\:\frac{rad}{s^2}$

$81\:\frac{rad}{s^2}$

$14\:\frac{rad}{s^2}$

$6.0\:\frac{rad}{s^2}$

$36\:\frac{rad}{s^2}$

Correct answer:

$36\:\frac{rad}{s^2}$

Explanation:

For a rotating object, or an object moving in a circular path, the relationship between angular acceleration and linear acceleration is

$a=\alpha r$

Linear acceleration is given by , angular acceleration is $\alpha$ , and the radius of the circular path is .

For circular/centripetal motion, the linear acceleration is related to the object's linear velocity by

$a_{centripetal}=\frac{v^{2}}{r}$

We know the linear velocity is $9.0 \:\frac{m}{s}$ , and the radius is 1.5 m, so we can find the linear acceleration...

$a_{c}=\frac{(9.0\:\frac{m}{s})^{2}}{1.5\:m}=54\:\frac{m}{s^2}$

Now that we have the linear acceleration, we can use this in the equation at the top to find the angular acceleration...

$\alpha=\frac{a}{r}=\frac{54\:\frac{m}{s^2}}{1.5\Lm}=36\:\frac{rad}{s^2}$

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Example Question #3 : Circular And Rotational Motion

If it takes a bike wheel 3 seconds to complete one revolution, what is the wheel's angular velocity?

Possible Answers:

$0.84s^{-1}$

$2.09s^{-1}$

$3.37s^{-1}$

$1.28s^{-1}$

Correct answer:

$2.09s^{-1}$

Explanation:

The definition of angular velocity is $\omega=\frac{\Delta \Theta }{\Delta t}$ .

By identifying the given information to be $\Delta \Theta =2\pi$ and $\Delta t =3s$ , we can plug this into the equation to calculate the angular velocity:

$\omega=\frac{2\pi}{3 s}=2.09s^-1$

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Example Question #4 : Circular And Rotational Motion

What is the angular velocity of the second hand of a clock?

Possible Answers:

$\frac{\pi rad}{30s}$

$\frac{\pi rad}{60s}$

$\frac{2\pi rad}{s}$

$\frac{60 rad}{\pi s}$

$\frac{30 rad}{\pi s}$

Correct answer:

$\frac{\pi rad}{30s}$

Explanation:

The angular velocity of the second hand of a clock can be found by dividing the number of radians the second hand will travel over a known period of time. Thankfully for a clock, we know that the second hand will make one revolution, i.e. covering $2\pi rad$ in one minute, or 60s. The formula for angular velocity is:

$\omega = \frac{\theta}{t}$

So the angular velocity, $\omega$ is $\frac{2\pi rad}{60s}$ , which simplifies to our answer, $\frac{\pi rad}{30s}$

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Example Question #5 : Circular And Rotational Motion

What is the difference in the angular velocity of the second hand of radius 1cm on a wristwatch, compared to the second hand of radius 5m on a large clock tower?

Possible Answers:

The clocktower second hand has an angular velocity that is 500 times slower than that of the wristwatch

The clocktower second hand has an angular velocity that is 20 times faster than that of the wristwatch

The clocktower second hand has an angular velocity that is 5 times faster than that of the wristwatch

The clocktower second hand has an angular velocity that is 500 times faster than that of the wristwatch

No difference

Correct answer:

No difference

Explanation:

The angular velocity should not change based on the radius of the second hand. No matter what size the second hand, it will still cover one revolution every minute or 60s. The linear velocity will be greater and the angular momentum will also be greater for the clocktower, but its angular velocity will be the same. This can be seen by looking at the equation for angular velocity:

$\omega=\frac{\theta}{t}$

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Example Question #6 : Circular And Rotational Motion

A ferris wheel has a trip length of 3min, that is it takes three minutes for it to make one complete revolution. What is the angular velocity of the ferris wheel if it only takes passengers around one time, in $\frac{rad}{s}$ ?

Possible Answers:

0.35

0.035

2.094

18.85

Correct answer:

0.035

Explanation:

Angular velocity, in $\frac{rad}{s}$ , is given by the length traveled divided by the time taken to travel the length:

$\omega=\frac{\theta}{t}$

We are told that the amount of time taken to make one revolution is 3min. One revolution is equal to $2\pi\ rad$ , and 3 minutes is equal to 180 seconds. Divide the radian value by the seconds value to get the angular velocity.

$\omega =\frac{2\pi}{180}$

$\omega=0.035\frac{rad}{s}$

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Example Question #7 : Circular And Rotational Motion

A wheel makes one full revolution every seconds and has a radius of . Determine its angular velocity .

Possible Answers:

$w=10\pi \frac{rads}{s}$

$w=5\pi \frac{rads}{s}$

$w=\frac{\pi}{5} \frac{rads}{s}$

$w=\pi \frac{rads}{s}$

Correct answer:

$w=\pi \frac{rads}{s}$

Explanation:

For this question, the angular velocity can be given by the equation:

$w=\frac{\theta}{t}$ , where $\theta$ is the angle made and is the time taken to make this angle.

In this problem, the wheel makes one full revolution( $\theta=2\pi$ ) in seconds.

Therefore:

$w=\frac{2\pi}{2s}=\pi \frac{rads}{s}$

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Example Question #8 : Circular And Rotational Motion

A CD rotates at a rate of $5\frac{rad}{s}$ in the positive counter clockwise direction. After pressing play, the disk is speeding up at a rate of $2\frac{rad}{s^2}$ . What is the angular velocity of the CD in $\frac{rad}{s}$ after 4 seconds?

Possible Answers:

$19.5 \frac{rad}{s}$

$10 \frac{rad}{s}$

$13 \frac{rad}{s}$

$6.5 \frac{rad}{s}$

$6 \frac{rad}{s}$

Correct answer:

$13 \frac{rad}{s}$

Explanation:

Given initial angular velocity, angular acceleration, and time we can easily solve for final angular velocity with:

$\omega=\omega _{0}+\alpha t$

$\omega=5\frac{rad}{s}+2\frac{rad}{s} (4)={13\frac{rad}{s}}$

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Example Question #9 : Circular And Rotational Motion

If a ferris wheel has height of 100m, find the angular velocity in rotations per minute if the riders in the carts are going $7\frac{m}{s}$ .

Possible Answers:

$1.94 \frac{rotations}{minute}$

None of these

$7.77\frac{rotations}{minute}$

$3.95 \frac{rotations}{minute}$

$1.34 \frac{rotations}{minute}$

Correct answer:

$1.34 \frac{rotations}{minute}$

Explanation:

If the ferris wheel has height 100m then it must have radius 50m .

The circumference of the ferris wheel, or the distance of one rotation, is then:

$2\pi*r=C$

$2\pi*50m=C$

$2\pi*50m\approx 314m$

Convert the given velocity into meters per minute, or $\frac{m}{min}$ :

$7\frac{m}{s}*\frac{60 s}{1 min}=420\frac{m}{min}$

Find rotations per minute:

$420\frac{m}{min}*\frac{1 rotation}{314m}=\frac{1.34 rotations}{minute}$

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Example Question #1 : Angular Velocity And Acceleration

A person of mass 95kg is riding a ferris wheel of radius 10m . The wheel is spinning at a constant angular velocity of 1rpm . Determine the linear velocity of the rider.

Possible Answers:

$\frac{.86 meters}{second}$

$\frac{4.25 meters}{second}$

$\frac{3.25 meters}{second}$

$\frac{1.05 meters}{second}$

$\frac{2.65 meters}{second}$

Correct answer:

$\frac{1.05 meters}{second}$

Explanation:

Convert $\frac{rotations}{minute}$ to $\frac{meters}{second}$ :

$1\frac{rotation}{minute}*\frac{1 minute}{60 seconds}*\frac{2*pi*10meters}{1 rotation}=\frac{1.05 meters}{second}$

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AP Physics 1 : Angular Velocity and Acceleration

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #1 : Circular And Rotational Motion

Example Question #1 : Circular And Rotational Motion

Example Question #3 : Circular And Rotational Motion

Example Question #4 : Circular And Rotational Motion

Example Question #5 : Circular And Rotational Motion

Example Question #6 : Circular And Rotational Motion

Example Question #7 : Circular And Rotational Motion

Example Question #8 : Circular And Rotational Motion

Example Question #9 : Circular And Rotational Motion

Example Question #1 : Angular Velocity And Acceleration

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