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.

\(\displaystyle L=I\omega\)

\(\displaystyle L\) is the symbol for angular momentum, \(\displaystyle I\) is the moment of inertia, and \(\displaystyle \omega\) is the angular velocity.

Therefore, if the moment of inertia, \(\displaystyle I\), is halved, then for the angular momentum, \(\displaystyle L\), to remain constant, the angular velocity, \(\displaystyle \omega\), must be doubled. This is because \(\displaystyle \frac{1}{2}\cdot2=1\), 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.

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

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

Angular momentum is also given by

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

Plugging in 30 for \(\displaystyle L\) gives us

\(\displaystyle \omega = 0.3s^{-1}\)

The formula for angular momentum is

\(\displaystyle L=I\omega\)

where

\(\displaystyle L\) = angular momentum

\(\displaystyle I\) = moment of inertia

\(\displaystyle \omega\) = angular speed

Angular speed is defined as

\(\displaystyle \omega = \frac{2\pi}{T}\)

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

\(\displaystyle \omega = \frac{2\pi}{T} = \frac{2\pi}{60s} = 0.105s^{-1}\)

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

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

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

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

Therefore, \(\displaystyle \omega = 0.035\)

Plugging this back into the definition, we get

\(\displaystyle \omega = \frac{2\pi}{T}\)

\(\displaystyle 0.035 = \frac{2\pi}{T}\)

Therefore,

\(\displaystyle T=180s\)

Thus, the satellite now rotates once every 3 minutes.

Angular momentum is always conserved.

\(\displaystyle L_{initial} = L_{final}\)

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

\(\displaystyle L = I\omega\) 

Moment of inertia can also be written as follows:

\(\displaystyle I=mr^2\)

Plug in given values to find \(\displaystyle I_{initial}\) and \(\displaystyle I_{final}\)

\(\displaystyle I_{initial}=I_{disk} = \frac{1}{2}M_{disk}R^{2} = MR^{2}\) 

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

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

\(\displaystyle \omega_{final} = \frac{I_{initial}}{I_{final}}\omega_{0} = \frac{1}{2}\omega_{0}\)

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

\(\displaystyle I\omega=mvr\)

\(\displaystyle \frac{mL^{2}}{3}\times \omega =mvr\)

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

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

\(\displaystyle v=6\frac{m}{s}\)

Lets examine the equation \(\displaystyle L = mvr\), where \(\displaystyle L\) = angular momentum, \(\displaystyle m\) = mass of the satellite, \(\displaystyle v\) = its angular velocity, and \(\displaystyle r\) = 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 \(\displaystyle 2v\).

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.  

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

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

First convert \(\displaystyle cm\) into \(\displaystyle m\):

\(\displaystyle 15cm=.15m\)

\(\displaystyle 20cm=.20m\)

Conservation of angular momentum:

\(\displaystyle I_1\omega_1=I_2 \omega_2\)

Where, in this case:

\(\displaystyle I=\frac{m_1 m_2}{m_1+m_2}d^2\)

Combine equations:

\(\displaystyle \frac{m_1 m_2}{m_1+m_2}d_1^2*\omega_1=\frac{m_1 m_2}{m_1+m_2}d_2^2*\omega_2\)

Plug in values and solve for the final rotational velocity:

\(\displaystyle \frac{66*66}{66+66}*.15^2*2=\frac{66*66}{66+66}*.20^2*\omega_2\)

\(\displaystyle \omega_2=1.125\frac{rotations}{second}\)

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

\(\displaystyle L = I\cdot w\)

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:

\(\displaystyle L = I_A\cdot w_A + I_B\cdot w_B\)

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

\(\displaystyle w_A = w_B\)

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

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

Rearranging for the angular velocity of A:

\(\displaystyle w_A=\frac{L}{I_A+I_B}\)

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:

\(\displaystyle I=mr^2\)

Thus for each mass, we get:

\(\displaystyle I_A=m_Ar^2\)

\(\displaystyle I_B=m_Br^2\)

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

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

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

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

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 \(\displaystyle \frac{1}{s}\):

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

It checks out, so time to plug and chug:

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

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

\(\displaystyle w_A = \frac{v_A}{r}\)

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

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

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

\(\displaystyle E = U_i +K_i = U_f +K_f\)

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

\(\displaystyle U_i = U_f +K_f\)

Expanding each term from left to right to:

\(\displaystyle U_i = mgh_i = m_Agh_{A_{i}}+m_Bgh_{B_{i}}\)

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.

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

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:

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

Rearranging for d:

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

\(\displaystyle d=1.55m\)

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:

\(\displaystyle h_{A_{i}}=\frac{1}{2}l-d\)

\(\displaystyle h_{B_{i}}=\frac{1}{2}l+d\)

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

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

Moving on to final potential energy:

\(\displaystyle U_f = mgh_f = m_Agh_{A_{f}}+m_Bgh_{B_{f}}\)

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:

\(\displaystyle U_f = m_Bgh_{B_{f}}\)

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:

\(\displaystyle U_f = m_Bgl\)

Moving on to final kinetic energy:

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

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:

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

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

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

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

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

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

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

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

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

\(\displaystyle v_f=6.17\frac{m}{s}\)

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

\(\displaystyle L_f = I\cdot w_f\)

where:

\(\displaystyle I = mr^2\)

\(\displaystyle w_f=\frac{v_f}{r}\)

Where r is half the length of the rod:

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

\(\displaystyle w_f = \frac{v_f}{\frac{1}{2}l}\)

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

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

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

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

\(\displaystyle L_f = 1300\frac{kg\cdot m^2}{s}\)