### All AP Physics C: Mechanics Resources

## Example Questions

### Example Question #1 : Understanding Conservation Of Angular Momentum

An ice skater begins to spin, starting with his arms spread out as far as possible, parallel to the ice. He pulls his arms into his body, and then raises them completely vertically towards the ceiling. As the skater pulls his arms inward, his angular velocity will __________, and as he raises his arms vertically his angular velocity will __________.

**Possible Answers:**

increase . . . stay the same

decrease . . . stay the same

increase . . . decrease

decrease . . . increase

**Correct answer:**

increase . . . stay the same

Through conservation of angular momentum, as moment of inertia decreases, the angular velocity increases. Moment of inertia is dependent on the distribution of the spinning body's mass away from the center of mass—as the skater brings his arms in, he lowers his moment of inertia, thus increasing his angular velocity. If he raises his arms completely vertically from this point, it does not change the radius of mass distributions (from the center of his body), thus maintaining his angular velocity.

### Example Question #71 : Ap Physics C

Sarah spins a ball of mass attached to a string of length around her head with a velocity . If the ball splits in half, losing exactly one-half of its mass instantaneously, what is its new velocity, ?

**Possible Answers:**

(no change in velocity)

**Correct answer:**

By the conservation of angular momentum, the angular momentum , is equal to the product of the mass, angular velocity, and radius (or length of the rope in this case). The equation relating these terms is:

Here, is the initial mass, is the initial angular velocity, and is the length of the rope, which remains constant. Angular momentum must be conserved, thus:

Substitute.

We are given that and is the final velocity. Plug in and solve.

### Example Question #71 : Mechanics Exam

A child is standing at the center of a frictionless, rotating platform. Both the child and the platform are rotating with and initial angular velocity, . The child begins to walk slowly toward the edge of the platform. Which quantity will decrease as the child walks?

**Possible Answers:**

Total angular momentum

Rotational inertia of the disk

Total momentum

Total rotational kinetic energy

Total angular inertia

**Correct answer:**

Total rotational kinetic energy

Since the platform is frictionless, as the child walks, angular momentum is conserved. However, since there is an term in the kinetic energy expression, as decreased due to the increase of inertia, it affects the energy more, and it decreases. From the reference frame of the disk, the child feels an outward directed force, and thus is doing negative work. This is the same principle that allows ice skaters to increase their spinning speed.

### Example Question #21 : Circular And Rotational Motion

Two solid cylinderical disks have equal radii. The first disk is spinning clockwise at and the second disk is spinning counterclockwise at . The second disk has a mass three times larger than the first. If both spinning disks are combined to form one disk, they end up rotating at the same angular velocity and same direction. Find this angular velocity after combination.

**Possible Answers:**

**Correct answer:**

For the first disk, we have the information below.

For the second disk, we have the information below.

(The minus sign indicates counterclockwise while the positive indicates clockwise.)

To do this problem, we use conservation of angular momentum. Before the disks are put in contact, the initial total angular momentum is given by the equation below.

It is just the sum of the angular momentums of each disk. When the disks are combined together, then final angular momentum can be found by the following equation.

Set this equal to the initial angular momentum.

Simplify and solve for .

### Example Question #21 : Circular And Rotational Motion

A piece of space debris is travelling in an elliptical orbit around a planet. At its closest point to the planet, the debris is travelling at . When the debris is from the planet, it travels at . How close does the debris get to the planet in its orbit?

**Possible Answers:**

**Correct answer:**

The equation for conservation of angular momentum is:

This means that the angular momentum of the object at the two points in its orbit must be the same. Since the mass of the debris does not change, this gives us an equality of the product of the velocity and distance of the debris at any two points of its orbit:

Plug in known values.

There is disagreement between units; the velocities are given in both and . Glance down at the answer choices and note that they are all lengths in kilometers.

Solve for :

### Example Question #71 : Motion

A mechanic is using a wrench to loosen and tighten screws on an engine block and wants to increase the amount of torque he puts on the screws to adjust them more easily. Which of the following steps will not help him to do so?

**Possible Answers:**

Pulling or pushing at an angle that is more perpendicular to the wrench

Using a wrench of a different material

Increasing the magnitude of the force he puts on the wrench

Pulling or pushing at an angle that is less perpendicular to the wrench

Using a longer wrench

**Correct answer:**

Pulling or pushing at an angle that is less perpendicular to the wrench

Remember the torque equation:

Exerting force on the wrench at an angle less perpendicular to the wrench will reduce and thus reduce the torque.

The other answer options will all increase the torque being applied, making the twisting motion easier.

### Example Question #1 : Circular Motion Concepts And Equations

A 1.6kg ball is attached to a 1.8m string and is swinging in circular motion horizontally at the string's full length. If the string can withstand a tension force of 87N, what is the maximum speed the ball can travel without the string breaking?

**Possible Answers:**

**Correct answer:**

The ball is experiencing centripetal force so that it can travel in a circular path. This centripetal force is written as the equation below.

Remember that centripetal acceleration is given by the following equation.

Since the centripetal force is coming from the tension of the string, set the tension force equal to the centripetal force.

Since we're trying to find the speed of the ball, we solve for v.

We know the following information from the question.

We can use this information in our equation to solve for the speed of the ball.

### Example Question #1 : Circular Motion

In uniform circular motion, the net force is always directed ___________.

**Possible Answers:**

away from the center of the circle

in the same direction as the tangential velocity

toward the center of the circle

nowhere; the net force will be zero

**Correct answer:**

toward the center of the circle

The correct answer is "toward the center of the circle." Newton's second law tells us that the direction of the net force will be the same as the direction of the acceleration of the object.

In uniform circular motion, the object accelerates towards the center of the circle (centripetal acceleration); the net force acts in the same direction.

### Example Question #152 : Mechanics Exam

A car moves around a circular path of radius 100m at a velocity of . What is the coefficient of friction between the car and the road?

**Possible Answers:**

**Correct answer:**

The force of friction is what keeps the car in circular motion, preventing it from flying off the track. In other words, the frictional force will be equal to the centripetal force.

We can cancel mass from either side of the equation and rearrange to solve for the coefficient of friction:

We can use our given values to solve:

### Example Question #1 : Circular Motion Concepts And Equations

An object of mass 10kg undergoes uniform circular motion with a constant velocity of at a radius of 10m. How long does it take for the object to make one full revolution?

**Possible Answers:**

**Correct answer:**

The time for one full revolution can be calculated simply by manipulating the defintion of velocity, where the distance is just the circumference of the circlular path. The time it takes is modeled by the following equation:

Use the given radius and velocity to solve for the time per revolution: