Angular Momentum

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AP Physics 1 › Angular Momentum

Questions 1 - 10

A satellite is rotating once per minute. It has an moment of inertia of $10,000 kg \cdot m^2$ . Erin, an astronaut, extends the satellite's solar panels, increasing its moment of inertia to $30,000 kg \cdot m^2$ . How quickly is the satellite now rotating?

1 rotation every 3 minutes

3 rotations per minute

1 rotation every 9 minutes

9 rotations per minute

1 rotation per minute

Explanation

The formula for angular momentum is

$L=I\omega$

where

= angular momentum

= moment of inertia

$\omega$ = angular speed

Angular speed is defined as

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

The initial period of the satellite is 1 minute, so:

$\omega = \frac{2\pi}{T} = \frac{2\pi}{60s} = 0.105s^{-1}$

Plugging this in, we can solve for the initial angular momentum:

$L=(10000 kg \cdot m^2)(0.105s^{-1}) = 1050\frac{kg \cdot m^2}{s}$

After Erin extends the the solar panels, momentum is the same (conservation of momentum), but the moment of inertia is now $30,000 kg \cdot m^2$ . Thus,

$1050=(30000 kg \cdot m^2)\omega$

Therefore, $\omega = 0.035$

Plugging this back into the definition, we get

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

$0.035 = \frac{2\pi}{T}$

Therefore,

Thus, the satellite now rotates once every 3 minutes.

I start pushing a merry-go-round with a torque of 10 Newton-meters. It has a moment of inertia of $100 kg \cdot m^2$ . What is its rotational speed after 3 seconds assuming it starts at rest?

$\omega = 0.3s^{-1}$

$\omega = 30s^{-1}$

$\omega = 0.3s$

$\omega = 30s$

$\omega = 3s^{-1}$

Explanation

The angular moment of the merry-go-round after 3 seconds is simply

$L=\tau t=(10 N \cdot m)(3s)=30 \frac{kg \cdot m^2}{s}$

Angular momentum is also given by

$L=I\omega=(100 kg \cdot m^2)\omega$

Plugging in 30 for gives us

$\omega = 0.3s^{-1}$

A blob of clay of mass is dropped on top of a rotating disk's edge of mass spinning at a speed of $\omega _{0}$ . What will the resulting rotational speed $\omega$ be?

$\omega = \frac{1}{2}\omega _{0}$

$\omega = 2\omega _{0}$

$\omega = 4\omega _{0}$

$\omega = \frac{1}{4}\omega _{0}$

$\omega = \omega _{0}$

Explanation

Angular momentum is always conserved.

$L_{initial} = L_{final}$

The equation relating angular momentum, moment of inertia and angular velocity is:

$L = I\omega$

Moment of inertia can also be written as follows:

Plug in given values to find $I_{initial}$ and $I_{final}$

$I_{initial}=I_{disk} = \frac{1}{2}M_{disk}R^{2} = MR^{2}$

$I_{final}=I_{disk} + I_{blob} = \frac{1}{2}M_{disk}R^{2} + M_{blob}R^{2} = 2MR^{2}$

Set the initial angular momentum equal to the final angular momentum and solve.

$\omega_{final} = \frac{I_{initial}}{I_{final}}\omega_{0} = \frac{1}{2}\omega_{0}$

A comet is orbiting a star in a far distant galaxy wth only gravity acting upon it. After a period of time, the sun's radiation partially melts the comet, decreasing the mass to a fourth of what it was. Simultaneously, because of an irregular orbit, its distance to the star doubles. What expression best describes the comet's angular velocity, , compared to its original conditions (before the melting/distance change)?

Explanation

Lets examine the equation , where = angular momentum, = mass of the satellite, = its angular velocity, and = its radius from the center of its orbit. Because of a lack of outside forces (besides the gravity of its star) acting on the satellite, we know that its angular momentum is conserved. Therefore, as its mass is divided by four, and its radius doubled, its angular velocity must also double, making the correct answer .

A bowling ball of mass and radius is traveling down a slope with a vertical height change of . If the ball starts from rest, what is the final linear velocity of the ball as it reaches the bottom of the slope? Neglect air resistance.

$g=10\frac{m}{s^2}$

$13.1\frac{m}{s}$

$15.5\frac{m}{s}$

$11.6\frac{m}{s}$

$8.2\frac{m}{s}$

$6.6\frac{m}{s}$

Explanation

Let's begin with the expression for conservation of energy:

Since the ball begins from rest, we can eliminate initial kinetic energy. Also, if we assume that the bottom of the slope has a height of 0, we can eliminate final potential energy to get:

Plugging in expressions, we get:

$mgh_i=\frac{1}{2}mv_f^2+\frac{1}{2}Iw^2$

Where I is the moment of inertia for a solid sphere:

$I = \frac{2}{5}mr^2$

And w is the angular velocity:

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

Plugging these in, we get:

$mgh_i = \frac{1}{2}mv_f^2+\frac{1}{5}mv_f^2$

Rearranging for final velocity, we get:

$v_f=\sqrt{\left (\frac{10}{7}gh_i \right )}$

Plugging in our values, we get:

$v_f=\sqrt{\frac{10}{7}\left ( 10\frac{m}{s^2}\right )(12m)}$

$v_f = 13.1\frac{m}{s}$

Two identical cars are racing side by side on a circular race track. Which has the greater angular momentum?

The outside car

The inside car

They are the same

Impossible to determine

Explanation

Where is the radius of the circle

is the mass of the object

is the linear velocity of the object

The car on the outside has a larger and a larger , and the same , thus it has a higher angular momentum.

A weight on a rope is swung around at until it comes into contact with a resting brick on flat ground. The brick travels . What is the coefficient of kinetic friction between the the brick and ground?

Explanation

The first part of this problem is a transfer between angular and linear momentum. Conservation of momentum tells us that we can equate the angular momentum of the pendulum to the linear momentum of the brick, or $L = m_{1}\omega r = p = m_{2}v$ , where $m_{1}$ and $m_{2}$ are the masses of the pendulum and brick respectively, $\omega$ is the angular velocity of the pendulum in , is the length of the pendulum rope, and is the resultant velocity of the brick. We find that the resulting velocity of the brick can be written as $v = \frac{m_{1}\omega r}{m_{2}} = 31.4 \frac{m}{s}$

The next part of the problem is identifying that the kinematic equations can be used here to find the acceleration the brick undergoes as it comes to a stop. Here we can use $v^2 = v_{0}^2 + 2a\Delta x$ , where $v_{0}$ is the end velocity of the brick (zero), is the deceleration it undergoes, and $\Delta x$ is the distance the brick travels before coming to a stop. Solving for we have that $a = \frac{v^2}{2\Delta x} = -49.4 \frac{m}{s^2}$ .

Lastly, we know that the kinetic friction force can be written as $f_{k} = -\mu_{k}N = -\mu_{k}m_{2}g$ . Newton's second law says that this can be written as $f_{k} = -\mu_{k}m_{2}g = m_{2}a$ . Thus we find that $\mu_{k} = \frac{a}{-g} = \frac{-49.4\frac{m}{s^2}}{-9.8\frac{m}{s^2}} = 5.0$ .

Pinball paddle problem

A rectangular rod on a horizontal table top is shown in the left side of the diagram. The rod is hinged at one end, as shown. It rotates in such a way that it hits a stationary ball as shown in the diagram's right side. The rod rotates without friction at a rate of $40 \frac{radians}{second}$ until it contacts the ball. As a result of the collision, the rod comes to rest and the ball moves to the right. The mass of the rod is 0.2kg. The mass of the ball is 0.067kg. The length of the rod is 0.15m. The moment of inertia of a rod rotated about the end is: $I=\frac{mL^{2}}{3}$ . What is the ball's speed as a result of the collision?

$6\frac{m}{s}$

$2\frac{m}{s}$

$4\frac{m}{s}$

$8\frac{m}{s}$

$10\frac{m}{s}$

Explanation

This is a conservation of angular momentum problem, so we set the angular momentum of the rod $(L=I\omega )$ to the equivalent angular momentum of the ball

$I\omega=mvr$

$\frac{mL^{2}}{3}\times \omega =mvr$

Note that the here is length, not angular momentum. Be careful not to cancel the since it refers to the rod on the left and the ball on the right. Finally, the on the right is the effective radius of the ball at the moment of impact, so it's simply the length of the rod.

$\frac{0.2kg*(0.15m)^{2}}{3}*40\frac{rad}{s}=0.067kg* v * 0.15m$

$v=6\frac{m}{s}$

Two cars are racing side by side on a perfectly circular race track. The inner car is from the center of the track. The outer car is from center of the track.

If the outer car is moving at $200\frac{km}{hr}$ , determine it's angular momentum.

$.123\frac{rad}{sec}$

$.480\frac{rad}{sec}$

$.220\frac{rad}{sec}$

$.300\frac{rad}{sec}$

$.411\frac{rad}{sec}$

Explanation

$200\frac{km}{hr}*\frac{1000m}{1 km}*\frac{1 radian}{452m}*\frac{1 hr}{3600sec}=\frac{.123 rad}{sec}$

In an isolated system, the moment of inertia of a rotating object is halved. What happens to the angular velocity of the object?

It is doubled.

It is quadrupled.

It is quartered.

It is halved.

It remains the same.

Explanation

In an isolated system, there is no net torque. If there is no net torque on the system, then the total angular momentum of the system remains the same. The angular momentum of a rotating object is equal to the moment of inertia of the object multiplied by the object's angular velocity.

$L=I\omega$

is the symbol for angular momentum, is the moment of inertia, and $\omega$ is the angular velocity.

Therefore, if the moment of inertia, , is halved, then for the angular momentum, , to remain constant, the angular velocity, $\omega$ , must be doubled. This is because $\frac{1}{2}\cdot2=1$ , which is the multiplicative identity. Anything multiplied by one remains the same. So, the final angular velocity would be twice as large as it was originally.

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