Rotational, Circular, and Harmonic Motion
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MCAT Physical › Rotational, Circular, and Harmonic Motion
An amusement ride is used to teach students about centripetal force. The ride is a circlular wall that you place your back on. The wall and floor then begin to spin. Once it reaches a certain rotational velocity, the floor drops, and the students are pinned to the wall as a result of centripital force.
The diameter of the ride is 10m. The minimum frequency of rotation that results in students being pinned to the wall after the floor drops is 0.5Hz. What is the coefficient of static friction between the students and the wall?
Explanation
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:
A spring is compressed as far possible and is not permitted to expand. What can be said about its potential energy and its kinetic energy?
Its potential energy is at a maximum and its kinetic energy is at a minimum
Its potential energy is at a minimum and its kinetic energy is at a maximum
Its potential energy and kinetic energy are both at a minimum
Its potential and kinetic energy are both at a maximum
The total energy of the spring is zero
Explanation
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.
An amusement ride is used to teach students about centripetal force. The ride is a circlular wall that you place your back on. The wall and floor then begin to spin. Once it reaches a certain rotational velocity, the floor drops, and the students are pinned to the wall as a result of centripital force.
The diameter of the ride is 10m. The minimum frequency of rotation that results in students being pinned to the wall after the floor drops is 0.5Hz. What is the coefficient of static friction between the students and the wall?
Explanation
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:
A spring is compressed as far possible and is not permitted to expand. What can be said about its potential energy and its kinetic energy?
Its potential energy is at a maximum and its kinetic energy is at a minimum
Its potential energy is at a minimum and its kinetic energy is at a maximum
Its potential energy and kinetic energy are both at a minimum
Its potential and kinetic energy are both at a maximum
The total energy of the spring is zero
Explanation
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.
An insect sits at the edge of slowly turning wheel. The wheel is accelerated gradually until the insect can no longer hold on. The insect’s path of travel will be:
- The insect will move away from the wheel in a straight line perpendicular to the wheel edge.
- The insect will move away from the wheel in a straight line tangent to the wheel edge.
- The insect will move away from the wheel edge in a curvilinear path.
- The path is not predictable by Newtonian physics.
- None of these is true.
2
1
3
4
5
Explanation
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.
An insect sits at the edge of slowly turning wheel. The wheel is accelerated gradually until the insect can no longer hold on. The insect’s path of travel will be:
- The insect will move away from the wheel in a straight line perpendicular to the wheel edge.
- The insect will move away from the wheel in a straight line tangent to the wheel edge.
- The insect will move away from the wheel edge in a curvilinear path.
- The path is not predictable by Newtonian physics.
- None of these is true.
2
1
3
4
5
Explanation
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.
Which of these is the correct expression for torque, τ, when θ is the angle at which the force, F, acts around the axis of rotation?
- τ = F cos θ
- τ = F sin θ
- τ = F tan θ
- τ = F cotan θ
- τ = F sec θ
2
1
3
4
5
Explanation
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.
An eight inch long cylinder is composed of two substances: the first four inches are composed of aluminum, and the second four inches are composed of lead. The aluminum region has a mass of 30g and the lead region has a mass of 60g.
Where is the center of mass in the cylinder?
In the center of the cylinder, one inch into the lead side
In the center of the cylinder, one inch into the aluminum side
In the center of the cylinder, right in the middle of the length
On the outside of the cylinder, right in the middle of the length
Explanation
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.
Which of the following changes to a pendulum would decrease its period?
Increasing the length of the pendulum
Starting the pendulum from a greater height
Increasing the mass of the weight at the end of the pendulum
Decreasing the gravitational attraction involved
Decreasing the density of the pendulum
Explanation
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:
An eight inch long cylinder is composed of two substances: the first four inches are composed of aluminum, and the second four inches are composed of lead. The aluminum region has a mass of 30g and the lead region has a mass of 60g.
Where is the center of mass in the cylinder?
In the center of the cylinder, one inch into the lead side
In the center of the cylinder, one inch into the aluminum side
In the center of the cylinder, right in the middle of the length
On the outside of the cylinder, right in the middle of the length
Explanation
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.