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AP Physics 1 : Angular Momentum

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Example Questions

Example Question #31 : Circular And Rotational Motion

Two 1500kg cars are racing side by side on a perfectly circular race track. The inner car is 450m from the center of the track. The outer car is 452m from center of the track.

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

1 radian=452m

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

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Example Question #32 : Circular And Rotational Motion

Moment of inertia of disk: .5MR^2

A car with wheels of mass 25kg and wheels of radius .25m is traveling at $50\frac{m}{s}$ . Treating the wheels as disks of uniform mass density, calculate the angular momentum of one wheel.

Possible Answers:

None of these

$156.25\frac{kg*m^2}{s}$

$755.25\frac{kg*m^2}{s}$

$75.25\frac{kg*m^2}{s}$

$625.25\frac{kg*m^2}{s}$

Correct answer:

$156.25\frac{kg*m^2}{s}$

Explanation:

Finding angular velocity:

$50\frac{m}{s}*\frac{1 rotation}{2*\pi*.25m}*\frac{2\pi radians}{rotation}=200\frac{radians}{second}$

Using

I=.5MR^2

and

$L=I\omega$

Where is angular momentum

Combining equations and plugging in values:

L=.5*25*.25^2*200

$L=156.25\frac{kg*m^2}{s}$

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Example Question #33 : Circular And Rotational Motion

A solid bowling ball of mass 6kg and radius r=0.15m is traveling at a linear velocity of $v=4\frac{m}{s}$ . What is the angular momentum of the ball?

Possible Answers:

$1.8\frac{kg\cdot m^2}{s}$

$2.4\frac{kg\cdot m^2}{s}$

$3.6\frac{kg\cdot m^2}{s}$

$1.4\frac{kg\cdot m^2}{s}$

$1.2\frac{kg\cdot m^2}{s}$

Correct answer:

$1.4\frac{kg\cdot m^2}{s}$

Explanation:

The expression for angular momentum is:

$L=I\cdot w$

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:

$L = \frac{2}{5}mvr$

Plugging in our values, we get:

$L = \frac{2}{5}(6kg)\left ( 4\frac{m}{s}\right )(0.15m)$

$1.4\frac{kg\cdot m^2}{s}$

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Example Question #11 : Angular Momentum

A bowling ball of mass 4kg and radius r=0.12m is traveling down a slope with a vertical height change of 12m . 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.

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

Possible Answers:

$8.2\frac{m}{s}$

$11.6\frac{m}{s}$

$13.1\frac{m}{s}$

$6.6\frac{m}{s}$

$15.5\frac{m}{s}$

Correct answer:

$13.1\frac{m}{s}$

Explanation:

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

E=U_i+K_i=U_f+K_f

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:

U_i=K_f

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:

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

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

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Example Question #11 : Angular Momentum

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

Possible Answers:

The inside car

The outside car

Impossible to determine

They are the same

Correct answer:

The outside car

Explanation:

L=rmv

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.

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Example Question #11 : Angular Momentum

A 5kg weight on a rope is swung around at 1 revolution/second until it comes into contact with a resting 1kg brick on flat ground. The brick travels 10m . What is the coefficient of kinetic friction between the the brick and ground?

Possible Answers:

0.5

5.0

3.5

1.5

0.8

Correct answer:

5.0

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 rad/s , 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 $\mu_{k} = \frac{a}{-g} = \frac{-49.4\frac{m}{s^2}}{-9.8\frac{m}{s^2}} = 5.0$ .

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AP Physics 1 : Angular Momentum

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #31 : Circular And Rotational Motion

Example Question #32 : Circular And Rotational Motion

Example Question #33 : Circular And Rotational Motion

Example Question #11 : Angular Momentum

Example Question #11 : Angular Momentum

Example Question #11 : Angular Momentum

All AP Physics 1 Resources

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