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.

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

Therefore, if the moment of inertia, , is halved, then for the angular momentum, , to remain constant, the angular velocity, , must be doubled. This is because , 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

Angular momentum is also given by

Plugging in 30 for  gives us

The formula for angular momentum is

where

 = angular momentum

 = moment of inertia

 = angular speed

Angular speed is defined as

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

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

After Erin extends the the solar panels, momentum is the same (conservation of momentum), but the moment of inertia is now . Thus,

Therefore, 

Plugging this back into the definition, we get

Therefore,

Thus, the satellite now rotates once every 3 minutes.

Angular momentum is always conserved.

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

Moment of inertia can also be written as follows:

Plug in given values to find  and 

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

This is a conservation of angular momentum problem, so we set the angular momentum of the rod  to the equivalent angular momentum of the ball 

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.

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 .

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.  

First convert  into :

Conservation of angular momentum:

Where, in this case:

Combine equations:

Plug in values and solve for the final rotational velocity:

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

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:

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:

Rearranging for the angular velocity of A:

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:

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

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 :

It checks out, so time to plug and chug:

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

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:

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.

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:

Rearranging for d:

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:

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

Moving on to final potential energy:

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:

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:

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:

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

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

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

where:

Where r is half the length of the rod:

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