The expression for angular momentum is:

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

And w is the angular velocity:

Plugging these in, we get:

Plugging in our values, we get:

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:

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

And w is the angular velocity:

Plugging these in, we get:

Rearranging for final velocity, we get:

Plugging in our values, we get:

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.

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 , where  and  are the masses of the pendulum and brick respectively,  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 

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 , where  is the end velocity of the brick (zero),  is the deceleration it undergoes, and  is the distance the brick travels before coming to a stop. Solving for  we have that .

Lastly, we know that the kinetic friction force can be written as . Newton's second law says that this can be written as . Thus we find that .

We are simply asked to find the centripetal acceleration, which is given by:

We were given  in the problem statement (radius will be equal to the length of the string), so we only need to find the velocity of the ball.

We are told that it travels in a circle with radius 1.5m and completes two full rotations per second. The length of each rotation is just the circumference of the circle:

The velocity can be found by multiplying that distance by the frequency:

Now we have all of our variables and can plug into our first equation:

There is a lot going on in this problem and it will take several steps to get to the answer. However, when you boil the question down, we are pretty much asked how fast must the ride spin so that the centripetal force on the student provides enough static friction to keep the student from falling. Let's work through this problem one step at a time.

First, let's figure out what minimal static frictional force is required. This force will be equal to the weight of the student; the student's weight will pull downward, while the friction of the wall pushes upward.

Now we can calculate the normal force required to reach that magnitude of frictional force. Note that the vector for the normal force will be perpendicular to the wal, directed toward the center of the circlel.

This normal force is the minimum centripital force required to keep the student pinned to the wall. We can convert this to centripital acceleration:

We can now convert centripital acceleration to a translational velocity using the equation:

Rearranging for velocity, we get:

This is the velocity that the outer wall of the ride must be spinning at. Since we know the radius of the ride, we can convert this velocity into a maximum period, the final answer:

If you weren't sure how to come about this equation, just think about your units. You know you need to get to units of seconds, and you have a value with units of m/s. Therefore, you need to cancel out the meter and get the second on top.

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

Substituting in our expressions for each variable and removing initial kinetic energy and final potential energy (which will each be zero), we get:

Rearranging for final velocity:

The only variable we don't have at this point is the distance the gumball travels. However, we can calculate it knowing the height of the track and its slope. We can imagine that the spiral track is unwound, creating a right triangle with an angle of 10 degrees and a height of 1.5m. For this triangle, the hypotenuse will be the total distance of the track.

Now that we have all of our variables, we can solve for the final velocity:

We can then use this to calculate the centripetal force on the gumball:

For this problem, we use the centripetal force equation:

We are given the mass, radius or rotation, and the linear velocity. Using these values, we can find the centripetal force.

First, we need to identify at which point in the circle the string is experiencing the most tension. There are two total forces in the system: gravity and tension. It is important to note that the tension isn't only resulting from gravity; it also includes the centripetal force required to keep the ball in circular motion. Thinking practically, we can say that the greatest tension will be when the ball is at its lowest point (gravity and tension are in opposite directions). At this point we can write:

Expand our terms for force:

We know the acceleration due to gravity, but we need to determine the centripetal accerleration. The formula for that is:

We know the radius (length of the string), so we need to develop an expression for velocity. We can use the period and circumference of circle:

Here, we use  to denote period.

Substituting this into the expression for centripetal acceleration:

Substituting this back into the equation for tension, we get:

We have all of these values, allowing us to solve:

The equation for centripetal force is:

.

If is doubled and becomes ,  is quadrupled. Centripetal force is proportional to the square of velocity.