Combine equations:

Convert to meters and seconds and plug in values:

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

Our initial state will be when the rod is horizontal, and our final state will be when mass A is at its highest point. If we assume that point p is at a height of 0 and the system is at rest and mass A is at its highest point, we can eliminate initial potential energy and final kinetic energy to get:

Expanding these terms and applying them to both masses, we get:

We don't need to separate the velocity components for each mass since they are always traveling at the same speed. Since the masses are attached to a rigid rod that spins around its midpoint, we know that the heights of the two masses (with respect to point p) will be equal and opposite. In expression form:

Substituting this into our equation, we get:

Rearranging for final height, we get:

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

Now we can use the sine function to determine what the angle c is at this point:

Where the opposite side is the height we just calculated and the hypotenuse is half the length of the rod. Therefore, we get:

Taking the inverse sine of both sides, we get:

Substituting in our values, we get:

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

The problem statement tells us that the sphere is initially at rest, so we can eliminate initial kinetic energy. Also, if we assume that the height at the bottom of the ramp is 0, we can eliminate final potential energy. We then have:

Then we can expand both of these terms. We need to remember that kinetic energy will have both a linear and rotational aspect. We then get equation (1):

Moving from left to right, let's begin substituting in expressions for unknown variables. The first term we don't know is height. However, we can use the length of the slope and the its angle to determine the height at the top of the ramp:

Rearranging for height, we get equation (2):

The next term we don't know is final velocity. We can use the relationship between angular and linear velocity:

Rearranging for linear velocity, we get equation (3):

Moving on, the next term we don't know is moment of inertia. We will use the expression for a sphere to get equation (4):

The last term is final rotational velocity. However, this is what we're solving for, so we'll leave it alone. Now let's substitute equations 2, 3, and 4 back into equation 1:

Multiplying each side the equation by , we get:

Factoring the right side of the equation:

Rearranging for final rotational velocity:

We know each of these values, so time to plug and chug:

The angular velocity is given by:

One circle is equal to , thus

If the cars are racing side by side on a circular track, then they have the same angular velocity, because they complete their circles in the same amount of time.

This is a centripetal force problem. It's important to know that when you draw your free body diagram (FBD), centripetal force () isn't drawn, similar to how net forces () aren't drawn.  is actually the centripetal or circular form of. So when we talk about circular motion such as this, we can set the two equal to each other. Why is this? Well we are trying to translate this object's circular motion into it's linear velocity (our answer).

We can now set, or  equal to the combination of vertical forces in our FBD

But like we said earlier, we need to connect our linear forces to our centripetal forces. Remember this equation?

Now let's put it all together

Time to pull in some algebra now. Remember we want the speed of the bucket, or the velocity of the bucket at the lowest point of the circle, so solve for :

If you plut in everything and solve (remember to use PEMDAS), your answer should be

Centripetal acceleration is as follows:

All we need to do is solve and plug in for 

 = distance traveled and t is the time it took to travel that distance

We know that the yo-yo made 4 revolutions in .

One revolution is the circumference of the yo-yo path: 

So we can either solve for in terms of how far the yo-yo went in one second (I), or how long it took to make one revolution (II); I'll show both, either one is correct if you wish to plug in

or

REMEMBER: Both of these will give you the same answer (check it out if you're good at algebra). Physics is all about comfort in manipulation so choose the way that suits you!

Now plug this into our  equation. Lucky for you, the radius was 

Note: I put in the units so you can see how they cancel out algebraically 

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