There is quite a lot going on in this problem. However, we will take it one step at a time so that you can understand the logical progression of thought it takes to solve this problem.

The first major step will be converting the frequency into a centripetal acceleration.

Convert the frequency into a velocity:

This is how fast the outer walls of the ride are traveling. We can convert this into centripetal acceleraton using the expression:

Then, we can create an expression for centripetal force for a student with mass :

Now that we have a term for the centirpetal force, we can work toward finding the force of friction. The centripetal force is also the normal force used in calculating frictional force. If the student is pinned to the wall, then the frictional force is exactly equal to the student's weight.

Plugging in our expression for the normal force:

Cancel out mass and rearrange for the coefficient of friction:

In this case, the kinetic energy of the spring is at a minimum. This is because, as the question indicates, the spring is not moving. At the same time, because the spring is compressed as far is it can be compressed, we know that its potential energy is at a maximum. The total energy of the spring therefore cannot be zero.

There is quite a lot going on in this problem. However, we will take it one step at a time so that you can understand the logical progression of thought it takes to solve this problem.

The first major step will be converting the frequency into a centripetal acceleration.

Convert the frequency into a velocity:

This is how fast the outer walls of the ride are traveling. We can convert this into centripetal acceleraton using the expression:

Then, we can create an expression for centripetal force for a student with mass :

Now that we have a term for the centirpetal force, we can work toward finding the force of friction. The centripetal force is also the normal force used in calculating frictional force. If the student is pinned to the wall, then the frictional force is exactly equal to the student's weight.

Plugging in our expression for the normal force:

Cancel out mass and rearrange for the coefficient of friction:

In this case, the kinetic energy of the spring is at a minimum. This is because, as the question indicates, the spring is not moving. At the same time, because the spring is compressed as far is it can be compressed, we know that its potential energy is at a maximum. The total energy of the spring therefore cannot be zero.

Choice 2 is correct; this is how a classical slingshot operates. A body in motion tends to remain in motion unless a force disturbs that motion. At any point in time, the insect is moving in a straight line along a line tangent to the circumference of the wheel. When the little creature leaves the wheel, there is no longer any force acting on it, so it will move in a straight line, not a curvilinear path. Newtonian kinetics of course includes these concepts, because they relate closely to the motion of celestial bodies.

Choice 2 is correct; this is how a classical slingshot operates. A body in motion tends to remain in motion unless a force disturbs that motion. At any point in time, the insect is moving in a straight line along a line tangent to the circumference of the wheel. When the little creature leaves the wheel, there is no longer any force acting on it, so it will move in a straight line, not a curvilinear path. Newtonian kinetics of course includes these concepts, because they relate closely to the motion of celestial bodies.

Choice 2 is correct. Think about trying to change a tire…you apply the force on the tire iron as close to perpendicular as possible in order to generate the most force.

Since the sine of 90 degrees is 1, then you are applying the maximum torque you can generate according to the equation τ = F sin θ.

The center of mass is the point where we can consider all of the substance's mass to be concentrated. Finding the center of mass greatly facilitates mechanics problems, because we can perceive the force to be acting on only this one point.

Since the lead side of the cylinder is heavier than the aluminum side, we can conclude that the center of mass is located inside the lead side of the cylinder. Since the lead side has a mass of 60g, each inch of the lead side will have a mass of 15g. By going one inch into the lead side, there are 45g on each side of the point. As a result, we say that this is the location for the center of mass for the cylinder, as either side would weigh the same individually.

The only factor that affects the period of a pendulum is the length of the pendulum. Therefore we can ignore any of the other answers which include other factors. The equation for the period of a pendulum is:

The center of mass is the point where we can consider all of the substance's mass to be concentrated. Finding the center of mass greatly facilitates mechanics problems, because we can perceive the force to be acting on only this one point.

Since the lead side of the cylinder is heavier than the aluminum side, we can conclude that the center of mass is located inside the lead side of the cylinder. Since the lead side has a mass of 60g, each inch of the lead side will have a mass of 15g. By going one inch into the lead side, there are 45g on each side of the point. As a result, we say that this is the location for the center of mass for the cylinder, as either side would weigh the same individually.