Angular Momentum - AP Physics 1

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Question

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

Answer

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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Question

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?

Answer

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.

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Question

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?

Answer

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

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

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Question

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?

Answer

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.

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

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Question

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)?

Answer

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 .

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Question

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?

Answer

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.

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

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

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Question

Consider a disk $\left(R_d = 20 cm \right )$ rotating with an angular speed of $12 \frac{rad}{s}$ . At a later time, a ring $\left( R_r = 14 cm \right )$ is suddenly placed on top of the disk. Calculate the new angular speed of the system.

Answer

To solve this problem, we will be using the conservation of angular momentum. Initially we only have a disk rotating. This gives us information about the initial angular momentum of the system. Later, we have both a disk and a ring rotating together at some new angular velocity.

$L_i=L_f\Rightarrow \omega_f = \omega_i \frac{I_i}{I_f}=\omega_i \frac{\frac{1}{2}M_dR_d^2}{\frac{1}{2}M_dR_d^2+M_rR_r^2}$

$\omega_f=12\frac{rad}{s}\frac{\frac{1}{2}30kg(.2m)^2}{\frac{1}{2}30kg(.2m)^2+15kg(.14m)^2}=8.05 \frac{rad}{s}$

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Question

The moment of inertia of two masses separated by distance is $\frac{m_1 m_2}{m_1+m_2}d^2$

Two balls, each of mass are attached by a massless rod of length and spinning at $2\frac{rotations}{second}$ around the point directly on the midpoint of the rod. Then rod then stretches to a length of . Assuming no loss of momentum and that the axis remains constant, determine the new rotational velocity.

Answer

First convert into :

Conservation of angular momentum:

$I_1\omega_1=I_2 \omega_2$

Where, in this case:

$I=\frac{m_1 m_2}{m_1+m_2}d^2$

Combine equations:

$\frac{m_1 m_2}{m_1+m_2}d_1^2\omega_1=\frac{m_1 m_2}{m_1+m_2}d_2^2\omega_2$

Plug in values and solve for the final rotational velocity:

$\frac{6666}{66+66}.15^22=\frac{6666}{66+66}.20^2\omega_2$

$\omega_2=1.125\frac{rotations}{second}$

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Question

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.

Answer

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

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Question

Consider the following system:

Two spherical masses, A and B, are attached to the end of a rigid rod with length l. The rod is attached to a fixed point, p, which is at the midpoint between the masses and is at a height, h, above the ground. The rod spins around the fixed point in a vertical circle that is traced in grey. $\angle c$ is the angle at which the rod makes with the horizontal at any given time ( $c=0^{\circ}$ in the figure).

The rod is spinning, and as it goes through its horizontal position, the total angular momentum, including both masses, is $20 \frac{kg\cdot m^2}{s}$ . What is the instantaneous linear velocity of mass A?

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

Neglect the mass of the rod.

Answer

Since we are given the angular momentum, let's being with that and its formula:

$L = I\cdot w$

Where:

L = angular momentum

I = moment of inertia

w = angular velocity

Since there are two masses on the rod (and we are neglecting the mass of the rod itself), we can expand this expression to:

$L = I_A\cdot w_A + I_B\cdot w_B$

Since the masses are attached to a rigid rod and going in a uniform circle, we can say that:

Since we're asked to determine the velocity of mass A, we'll substitute the angular velocity of mass A for mass B:

$L = I_Aw_A+I_Bw_A = w_A\left ( I_A+I_B\right )$

Rearranging for the angular velocity of A:

$w_A=\frac{L}{I_A+I_B}$

Now we need to determine what the moment of inertia is for each mass. Since they are spheres, we can treat them as point masses and use the following formula:

Thus for each mass, we get:

Where r is half the length of the rod and equal for both masses. Plugging these into the equation, we get:

$w_A=\frac{L}{m_Ar^2+M_Br^2}=\frac{L}{r^2(m_A+m_B)}$

Substituting half the length of the rod in for radius, we get:

$w_A = \frac{L}{\left ( \frac{1}{2}l^2\right )(m_A+m_B)} = \frac{4L}{l^2(m_A+m_B)}$

We know the value of each variable in this equation, so we can solve it. But first, let's check to make sure our units are correct. The units for angular velocity are $\frac{1}{s}$ :

$\frac{\frac{kg\cdot m^2}{s}}{m^2(kg+kg)}=\frac{1}{s}$

It checks out, so time to plug and chug:

$w_A=\frac{4\left ( 20\frac{kg\cdot m^2}{s}\right )}{(4m)^2(5kg+1.5kg)}=0.769\frac{1}{s}$

We can then use the expression for angular velocity to get linear velocity:

$w_A = \frac{v_A}{r}$

$v_A = w_Ar=w_A\left ( \frac{1}{2}l\right )$

$v_A = 0.769\frac{1}{s}\left ( \frac{1}{2}4m\right )=1.5\frac{m}{s}$

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Question

Consider the following system:

Two spherical masses, A and B, are attached to the end of a rigid rod with length l. The rod is attached to a fixed point, p, which is at a height, h, above the ground. The rod spins around the fixed point in a vertical circle that is traced in grey. $\angle c$ is the angle at which the L side of the rod makes with the horizontal at any given time ( $c=0^{\circ}$ in the figure and can be negative if mass A is above the horizontal).

The system is currently at rest and being held so that mass A is below the horizontal and $\angle c = 15^{\circ}$ . The rod is then released and allowed to rotate freely. What is the maximum angular momentum achieved by mass B? Neglect air resistance and internal frictional forces

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

Answer

We can begin with the expression for conservation of energy to solve this problem:

Since we are told that the system is initially at rest, we can eliminate initial kinetic energy to get expression (1):

Expanding each term from left to right to:

$U_i = mgh_i = m_Agh_{A_{i}}+m_Bgh_{B_{i}}$

We are told that mass A is held at an angle of 15 degrees below the horizontal, so we can use the sine function to determine the height of each mass. We will assume that the lowest point of rotation has a height of 0.

$sin(c^{\circ})=\frac{d}{r}$

Where d is the vertical distance below horizontal, and r is the radius from the center of the rod, which is a distance equal to half the length of the rod:

$sin(c^{\circ})=\frac{d}{\frac{1}{2}l}=\frac{2d}{l}$

Rearranging for d:

$d=\frac{sin(c^{\circ})l}{2}=\frac{sin(15^{\circ})12}{2}$

We know that this is the distance below horizontal that mass A begins at and the distance above the horizontal that mass B begins at. Also, horizontal is at a height of half a length rod above our reference height, so we can say:

$h_{A_{i}}=\frac{1}{2}l-d$

$h_{B_{i}}=\frac{1}{2}l+d$

Plugging these into our expanded expression for initial potential energy, we get:

$m_Ag\left ( \frac{1}{2}l-d \right )+m_Bg\left ( \frac{1}{2}l+d \right )$

Moving on to final potential energy:

$U_f = mgh_f = m_Agh_{A_{f}}+m_Bgh_{B_{f}}$

We are asked to find the maximum angular momentum achieved by mass B. This occurs when mass B is traveling at it's highest velocity, which is when the larger of 2 masses, mass A, is traveling through the low point of rotation. This means that mass A is at a height of 0, and thus has no final potential energy. We can then say:

$U_f = m_Bgh_{B_{f}}$

Also, we know that mass B will be at the high point of rotation, which means that its height will be the length of the rod. Therefore, we can say:

Moving on to final kinetic energy:

$K_f = \frac{1}{2}mv_f^2 = \frac{1}{2}v_f^2(m_A+m_B)$

We don't have to expand the velocity term since the two masses are always traveling at the same speed.

Now, substituting our expanded terms back into expression (1), we get:

$m_Ag\left ( \frac{1}{2}l-d \right )+m_Bg\left ( \frac{1}{2}l+d \right ) = m_Bgl +\frac{1}{2}v_f^2(m_A+m_B)$

In order to determine angular momentum, we are going to need to solve this equation for final velocity, so let's begin rearranging:

$\frac{1}{2}v_f^2(m_A+m_B) = m_Ag\left ( \frac{1}{2}l-d \right )+m_Bg\left ( -\frac{1}{2}l+d \right )$

$\frac{1}{2}v_f^2(m_A+m_B) = m_Ag\left ( \frac{1}{2}l-d \right )-m_Bg\left ( \frac{1}{2}l-d \right )$

$\frac{1}{2}v_f^2(m_A+m_B) = g\left ( \frac{1}{2}l-d \right )(m_A-m_B)$

$v_f^2= \frac{2g\left ( \frac{1}{2}l-d \right )(m_A-m_B)}{(m_A+m_B) }$

$v_f= \sqrt{\frac{g\left (l-2d \right )(m_A-m_B)}{(m_A+m_B) }}$

We have all of these values, so time to plug and chug:

$v_f = \sqrt{\frac{\left ( 10\frac{m}{s^2}\right )(12m-3.11m)(25kg-10kg)}{(25kg+10kg)}}$

$v_f=6.17\frac{m}{s}$

This is the maximum velocity that mass B will experience. We can now use the expression for angular momentum:

$L_f = I\cdot w_f$

where:

$w_f=\frac{v_f}{r}$

Where r is half the length of the rod:

$I = (m_A+m_B)\left ( \frac{1}{2}l\right )^2$

$w_f = \frac{v_f}{\frac{1}{2}l}$

$L_f =(m_A+m_B)\left ( \frac{1}{2}l\right )^2\frac{v_f}{\frac{1}{2}l}$

$L_f = (m_A+m_B)\left ( \frac{1}{2}l\right )v_f$

We have all of these values, so time to plug and chug:

$L_f =(25kg + 10kg)\left ( \frac{1}{2}(12m)\right )\left ( 6.17\frac{m}{s}\right )$

$L_f = 1300\frac{kg\cdot m^2}{s}$

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Question

Moment of inertia of disk:

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

Answer

Finding angular velocity:

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

Using

and

$L=I\omega$

Where is angular momentum

Combining equations and plugging in values:

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

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Question

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

Answer

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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Question

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.

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

Answer

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:

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

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

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Question

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

Answer

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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Question

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?

Answer

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

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Question

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

Answer

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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Question

Consider the following system:

Two spherical masses, A and B, are attached to the end of a rigid rod with length l. The rod is attached to a fixed point, p, which is at a height, h, above the ground. The rod spins around the fixed point in a vertical circle that is traced in grey. $\angle c$ is the angle at which the L side of the rod makes with the horizontal at any given time ( $c=0^{\circ}$ in the figure and can be negative if mass A is above the horizontal).

The system is currently at rest and being held so that mass A is below the horizontal and $\angle c = 15^{\circ}$ . The rod is then released and allowed to rotate freely. What is the maximum angular momentum achieved by mass B? Neglect air resistance and internal frictional forces

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

Answer

We can begin with the expression for conservation of energy to solve this problem:

Since we are told that the system is initially at rest, we can eliminate initial kinetic energy to get expression (1):

Expanding each term from left to right to:

$U_i = mgh_i = m_Agh_{A_{i}}+m_Bgh_{B_{i}}$

We are told that mass A is held at an angle of 15 degrees below the horizontal, so we can use the sine function to determine the height of each mass. We will assume that the lowest point of rotation has a height of 0.

$sin(c^{\circ})=\frac{d}{r}$

Where d is the vertical distance below horizontal, and r is the radius from the center of the rod, which is a distance equal to half the length of the rod:

$sin(c^{\circ})=\frac{d}{\frac{1}{2}l}=\frac{2d}{l}$

Rearranging for d:

$d=\frac{sin(c^{\circ})l}{2}=\frac{sin(15^{\circ})12}{2}$

We know that this is the distance below horizontal that mass A begins at and the distance above the horizontal that mass B begins at. Also, horizontal is at a height of half a length rod above our reference height, so we can say:

$h_{A_{i}}=\frac{1}{2}l-d$

$h_{B_{i}}=\frac{1}{2}l+d$

Plugging these into our expanded expression for initial potential energy, we get:

$m_Ag\left ( \frac{1}{2}l-d \right )+m_Bg\left ( \frac{1}{2}l+d \right )$

Moving on to final potential energy:

$U_f = mgh_f = m_Agh_{A_{f}}+m_Bgh_{B_{f}}$

We are asked to find the maximum angular momentum achieved by mass B. This occurs when mass B is traveling at it's highest velocity, which is when the larger of 2 masses, mass A, is traveling through the low point of rotation. This means that mass A is at a height of 0, and thus has no final potential energy. We can then say:

$U_f = m_Bgh_{B_{f}}$

Also, we know that mass B will be at the high point of rotation, which means that its height will be the length of the rod. Therefore, we can say:

Moving on to final kinetic energy:

$K_f = \frac{1}{2}mv_f^2 = \frac{1}{2}v_f^2(m_A+m_B)$

We don't have to expand the velocity term since the two masses are always traveling at the same speed.

Now, substituting our expanded terms back into expression (1), we get:

$m_Ag\left ( \frac{1}{2}l-d \right )+m_Bg\left ( \frac{1}{2}l+d \right ) = m_Bgl +\frac{1}{2}v_f^2(m_A+m_B)$

In order to determine angular momentum, we are going to need to solve this equation for final velocity, so let's begin rearranging:

$\frac{1}{2}v_f^2(m_A+m_B) = m_Ag\left ( \frac{1}{2}l-d \right )+m_Bg\left ( -\frac{1}{2}l+d \right )$

$\frac{1}{2}v_f^2(m_A+m_B) = m_Ag\left ( \frac{1}{2}l-d \right )-m_Bg\left ( \frac{1}{2}l-d \right )$

$\frac{1}{2}v_f^2(m_A+m_B) = g\left ( \frac{1}{2}l-d \right )(m_A-m_B)$

$v_f^2= \frac{2g\left ( \frac{1}{2}l-d \right )(m_A-m_B)}{(m_A+m_B) }$

$v_f= \sqrt{\frac{g\left (l-2d \right )(m_A-m_B)}{(m_A+m_B) }}$

We have all of these values, so time to plug and chug:

$v_f = \sqrt{\frac{\left ( 10\frac{m}{s^2}\right )(12m-3.11m)(25kg-10kg)}{(25kg+10kg)}}$

$v_f=6.17\frac{m}{s}$

This is the maximum velocity that mass B will experience. We can now use the expression for angular momentum:

$L_f = I\cdot w_f$

where:

$w_f=\frac{v_f}{r}$

Where r is half the length of the rod:

$I = (m_A+m_B)\left ( \frac{1}{2}l\right )^2$

$w_f = \frac{v_f}{\frac{1}{2}l}$

$L_f =(m_A+m_B)\left ( \frac{1}{2}l\right )^2\frac{v_f}{\frac{1}{2}l}$

$L_f = (m_A+m_B)\left ( \frac{1}{2}l\right )v_f$

We have all of these values, so time to plug and chug:

$L_f =(25kg + 10kg)\left ( \frac{1}{2}(12m)\right )\left ( 6.17\frac{m}{s}\right )$

$L_f = 1300\frac{kg\cdot m^2}{s}$

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Question

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?

Answer

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.

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Question

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?

Answer

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

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

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