First, convert units of the magnitude of velocity (speed):

Convert to meters and plug in values:

Before we even begin thinking about what sort of equations we will be using, we first need to determine at which point mass B is feeling its greatest centripetal force. This will be whenever the mass is traveling at its fastest. Thinking about the law of conservation of energy, the mass will be traveling at its fastest when the potential energy of the system is minimized. There are two different masses in the system that are linked. Therefore, potential energy will be minimized when the larger mass is at its lowest point. This is when the rod is vertical and mass A is at the bottom. Now we can move on, beginning with the expression for centripetal force:

The only value we don't know is velocity. However, we can find it by using the expression for conservation of energy:

If we assume that point p is at a height of 0, we can eliminate initial potential energy to get equation (1):

Now let's expand each term moving from left to right:

Since the velocities of both masses are the same, we can say:

Moving onto final potential:

The final position of the rod is vertical with mass A at the bottom (which is a half length of rod below our reference point) and mass B at the top (half length rod above), so we can rewrite 

Moving on to final kinetic:

Substituting these into equation (1) we get:

To make things look nicer, lets eliminate the fraction from each term:

We are solving for final velocity, so let's begin rearranging:

Note how the masses were switched in the final term since it switched sides. No some more rearranging:

We know all of those values, so time to plug and chug:

Now we go back to our expression for centripetal force:

We need to determine r, which is simply half the length of the rod:

Now that we know all of our values, we can solve the problem:

First we need to determine when at which points mass A will be experiencing the most/least centripetal force. The most will be when the rod is vertical and mass A is at the bottom (initial condition). This is because the potential energy is minimized (thus kinetic is maximized) when the large mass is at its low point. Therefore, the lest centripetal force will be when mass A is at the top (final condition). So now let's begin with the equation for conservation of energy:

Unfortunately we can't eliminate any terms, so let's begin expanding them one at a time:

If we assume that the lowest point of the circle has a height of 0, the initial potential energy of mass A will be 0:

We also know that the height at the top of the circle is exactly one rod length from our reference height, so we can say:

Moving on to initial kinetic:

Since bot velocities are always the same, we can say:

Now final kinetic. We'll just skip ahead to it's final form:

Note how it's the same as before, except that the masses are switched. No final kinetic. We'll skip to it's final form again:

Now plugging all of these back into the expression for conservation of energy, we get:

In order to find the final centripetal force, we are going to need to need the final velocity, so let's begin rearranging for that:

Factoring and dividing, we get:

We know every variable except initial velocity, which we can solve for by using the equation for centripetal force:

Where mass is mass B, v is the initial velocity, and radius is half the length of the rod:

Now we rearrange for initial velocity:

Plugging this back into our big equation along with the rest of the variable:

Now using the expression for centripetal force to find our final answer:

To find the difference in force upwards for liftoff, we need to find how much centrifugal force is generated from the rotation of the Earth.  The equation for centrifugal force is:

Where  is the mass of the object,  is the distance from the center of rotation the object is, and ,  being the period of the rotation.  We know that the earth rotates once per day, so we can find the period of rotation in seconds:

Now we solve for the centrifugal force:

This is all we need to calculate, because we know that the force pulling the shuttle down due to gravity would be identical from the equator to the North Pole on our idealized spherical Earth.

The tension in the rope is equal to the centripetal force on the rocket.

Definition of velocity:

The time necessary to go around one circle is it's period, which is the inverse of the frequency.

Doubling the frequency would double the velocity, which would quadruple the tension.

The bus going around a corner is experiencing centripetal acceleration towards the center of it's turn. Even though the linear velocity is constant (speed), the bus is constantly changing directions, thus its velocity is not constant. Since acceleration is defined as the change in velocity over time, the acceleration is nonzero, which means that due to Newton's second law, the net force on the bus is also nonzero.

To calculate the centripetal acceleration at the point identified in the expression (which is simply ), we need to determine the linear velocity at this point. We can do this by using the period of the bowling ball:

Where both the circumference and velocity are referenced to the outer surface of the bowling ball:

Plugging in our values, we get:

This is the time it takes for any single point of the ball (inside or out, excluding the center) to complete one full rotation. Therefore, we can apply this same expression to the point asked in the question (r=0.05m):

Rearranging for velocity, we get:

Plugging in our values, we get:

The total kinetic energy of a solid sphere can be calculated using the following:

Where I is the moment of inertia for a solid sphere:

And w is the angular velocity:

Plugging these in, we get:

We can obtain an expression for velocity using the expression for centripetal acceleration:

Rearranging:

Plugging this in, we get:

We have all of these values, so let's plug them in:

At the top of the swing, both gravity and the tension of the string will be in the same direction, thus, this is where tension would be at a minimum. 

At the very limit of being taught, 

Thus,

Solving for 

This will be the linear velocity.

The angular velocity will be 

Combining equations

Plugging in values

At the top of the swing, both gravity and the tension of the string will be in the same direction, thus, this is where tension would be at a minimum. 

At the very limit of being taught, 

Thus,

Solving for 

This will be the linear velocity.

The angular velocity will be 

Combining equations

Plugging in values