AP Physics 1 - Angular Velocity and Acceleration

Pluto distance to sun: $1.43*10^{9}km$

Determine the translational velocity of Pluto.

$1137\frac{m}{s}$

$1454\frac{m}{s}$

$1009\frac{m}{s}$

$929\frac{m}{s}$

$717\frac{m}{s}$

Explanation

$\frac{d}{t}=v$

$C=2\pi r$

Combine equations:

$\frac{2\pi r}{t}=v$

Convert to meters and seconds and plug in values:

$\frac{2\pi r}{t}=v$

$\frac{2*\pi* 1.43*10^{12}}{7.9*10^9}=v$

$v=1137\frac{m}{s}$

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?

$13 \frac{rad}{s}$

$6.5 \frac{rad}{s}$

$19.5 \frac{rad}{s}$

$10 \frac{rad}{s}$

$6 \frac{rad}{s}$

Explanation

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

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

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

$1 \ \text{day on Pluto}=6.39 \ \text{days on Earth}$

Pluto radius:

Determine the linear velocity of someone standing on the surface of Pluto due to the rotation of the planet.

$13.5\frac{m}{s}$

$4.3\frac{m}{s}$

$15.7\frac{m}{s}$

$9.3\frac{m}{s}$

$2.5\frac{m}{s}$

Explanation

Convert units of time into radians per second:

$\frac{2\pi~radians}{1~pluto~day}*\frac{1~pluto~day}{6.39~earth~days}*\frac{1 earth~ day}{24 hours}*\frac{1 hour}{3600 seconds}=\frac{1.138*10^{-5}radians}{sec}$

Convert to linear distance:

$\frac{1.138*10^{-5}radians}{second}*\frac{1.186*10^6meters}{1 radian}={13.5\frac{m}{s}}$

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

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

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

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

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

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

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

A model train completes a circle of radius in . Determine the angular frequency in $\frac{rad}{s}$ .

$\frac{.897\ rad}{s}$

None of these

$\frac{3.14\ rad}{s}$

$\frac{6.35\ rad}{s}$

$\frac{.223\ rad}{s}$

Explanation

One circle is equal to $2\pi\ rad$ , thus

$\frac{1 circle}{7 s}*\frac{2\pi \ rad}{1 circle}=\frac{.897\ rad}{s}$

A spinning disk is rotating at a rate of $4 \frac{rad}{s}$ in the positive counterclockwise direction. If the disk is speeding up at a rate of $2\frac{rad}{s}^2$ , find the disk's angular velocity in $\frac{rad}{s}$ after four seconds.

$12\frac{rad}{s}$

$8\frac{rad}{s}$

$10\frac{rad}{s}$

$4\frac{rad}{s}$

None of these

Explanation

The angular velocity is given by: $\omega= \omega_{0}+\alpha t\rightarrow \omega=4\frac{rad}{s}+2\frac{rad}{s^2}*4s=\boldsymbol{12 \frac{rad}{s}}$

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?

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

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

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

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

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

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, $\alpha$ .

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

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

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?

No difference

The clocktower second hand has an angular velocity that is 500 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 20 times faster than that of the wristwatch

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

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:

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

Radius of the earth: $6.37*10^{6}m$

A train is traveling directly north at $110\frac{km}{hr}$ . Estimate its angular velocity with respect to the center of the earth.

$4.8*10^{-6} \frac{radians}{second}$

$6.2*10^{-6} \frac{radians}{second}$

$2.5*10^{-6} \frac{radians}{second}$

$7.8*10^{-6} \frac{radians}{second}$

$9.1*10^{-6} \frac{radians}{second}$

Explanation

Convert $\frac{km}{hr}$ to $\frac{m}{s}$

$110\frac{km}{hr}*\frac{1hr}{3600 seconds}*\frac{1000m}{1km}=30.6\frac{m}{s}$

Use the following relationship and plug in known values:

$\omega=\frac{v}{r}$

$\omega=\frac{30.6}{6.37*10^6}$

$\omega=4.8*10^{-6} \frac{radians}{second}$

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 $\frac{rad}{s}$ ?

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.