### All High School Physics Resources

## Example Questions

### Example Question #210 : High School Physics

Ellen is swinging a yo-yo in a circular path perpendicular to the ground. The yo-yo moves in a clockwise direction with a constant speed of .

What is the velocity of the yo-yo at the bottom of the circle?

**Possible Answers:**

**Correct answer:**

Remember, when working with circular motion, the velocity is ALWAYS tangential to the circle. This means that, even though the speed is constant, the direction is always tangent to the edge of the circle. If the circle below represents the path of the yo-yo, and it moves in a clockwise direction, then the velocity at the bottom of the path will be to the left.

The magnitude of the velocity is constant, so the final answer will be .

### Example Question #1 : Circular Motion

Which of the following will increase the torque on a wrench, without changing the force applied?

**Possible Answers:**

Use a longer wrench

None of these will change the torque generated

Change the angle of rotation

Use a wrench of a stronger material

Use a shorter wrench

**Correct answer:**

Use a longer wrench

The formula for torque is:

If the force applied remains the same, then that means that the torque and the lever arm are directly proportional: when one increases the other increases as well. The length of the lever arm, in this problem, is the length of the wrench; thus, increasing the length of the wrench will increase the amount of torque generated, without requiring a change in the force applied.

### Example Question #2 : Circular Motion

A car driving on the highway is moving at 60 miles per hour. As the car nears an exit ramp, the car slows to 35 miles per hour, a speed that is maintained throughout the circular path of the exit ramp. What force is keeping the car on its path (i.e. in circular motion)?

**Possible Answers:**

Momentum

Weight of the car (gravity)

Normal force

Centripetal force

**Correct answer:**

Centripetal force

The correct answer is centripetal force. In a free body diagram, this is the force that would be directed towards the center of the circular path; in circular motion, acceleration and net force are always in this direction.

Normal force and weight act perpendicular to the line that is directed towards the center of the circle, and will not prevent the car from traveling off the circular path. Momentum is not a measure of force and is not particularly relevant to this question.

### Example Question #3 : Circular Motion

A ball attached to a string is moving counterclockwise in a vertical circle. If the string is cut exactly at the point where the ball is at the top of its motion (the top of the circle), what direction will the ball move in initially?

**Possible Answers:**

Left

Downward

Upward

Right

**Correct answer:**

Left

In circular motion, velocity is tangential to the circular path. Since the object is moving counterclockwise, at the top of the circle this tangent line points to the left. It may help to draw a diagram to better visualize this motion.

### Example Question #4 : Circular Motion

In a rotating carnival ride, people rotate in a vertical cylindrical walled room. During the ride, the floor drops out. If the room radius is , and the rotation frequency is when the floor drops out, what minimum coefficient of static friction keeps the people from slipping down?

**Possible Answers:**

**Correct answer:**

Knowns

revolutions per second which needs to be converted to

We are dealing with circular motion so the first thing to do is to identify the force that is causing the centripetal motion. In this case it is the normal force of the wall that is turning the person toward the center. The force of gravity is pulling the person down and the friction force is what is counteracting the force of gravity keeping the person from falling down.

The centripetal force is equal to the mass times the centripetal acceleration.

To calculate the centripetal acceleration you will need to know the velocity of the person and the radius that the person is at.

Therefore

We will now need to use the frictional force to get a relationship that will help us solve for the normal force.

Since the force of gravity is equal to the frictional force

The force of gravity is equal to the object’s mass times the acceleration due to gravity

We can rearrange this equation for normal force by itself

We can now put this all together

Notice that mass falls out of both sides of the equation

We can now solve for the coefficient of friction

We can now plug in our known variables.

### Example Question #5 : Circular Motion

A car rounds a curve of radius banked at an angle of . If the car is traveling at , will friction force be required? If so, in what direction?

**Possible Answers:**

A friction force will be required down the slope of the curve

A friction force will not be required

A friction force will be required up the slope of the curve

**Correct answer:**

A friction force will be required down the slope of the curve

Knowns

(need to convert to )

The first thing to do is determine the magnitude of the centripetal force acting on the car.

Next we need to determine if the component of the normal force that contributes to the centripetal acceleration is equal to, greater than or less than this value.

This number is less than the magnitude of the centripetal force acting on the car meaning that there is friction involved pointing down the slope to help keep the car moving around the circle.

### Example Question #6 : Circular Motion

Computational

A ball rolls around the edge of a circle with a radius of . If it is rolling at a speed of , what is the centripetal acceleration?

**Possible Answers:**

**Correct answer:**

Centripetal acceleration is the acceleration towards the center when an object is moving in a circle. Though the speed may be constant, the change in direction results in a non-zero acceleration.

The formula for this is , where is the perceived tangential velocity and is the radius of the circle.

Plug in the given values and solve for the acceleration.

### Example Question #7 : Circular Motion

A ball rolls around the edge of a circle with a radius of . If it is rolling at a speed of , what is the centripetal force?

**Possible Answers:**

**Correct answer:**

Centripetal force is the force that constantly moves the object towards the center; it is what keeps the object moving in a circle rather than flying off tangentially to the circle.

The formula for force is

To find the centripetal force, we need to find the centripetal acceleration. We do this with the formula , where v is the perceived tangential velocity and r is the radius of the circle.

Plug in the given values and solve for the acceleration.

Plug the acceleration and given mass into the first equation to solve for force.

### Example Question #8 : Circular Motion

At what minimum speed must a roller coaster be traveling so that the passengers upside down at the top of the circle do not fall out? The radius of curvature of the roller coaster is .

**Possible Answers:**

It is not possible to solve this problem without additional information

**Correct answer:**

At the top of the roller coaster the minimum force that is applied is the gravitational force pulling the passengers down. Therefore the centripetal force at the top of the coaster is caused by the force of gravity.

We know that the centripetal force is equal to the centripetal acceleration times the mass.

To calculate the centripetal acceleration you will need to know the velocity of the roller coaster and the radius that the coaster is at.

Therefore

We also know that the force of gravity is equal to mass times the acceleration due to gravity.

Putting this together we get

Notice that mass falls out of both sides of the equation

We can now solve for the velocity by itself

Plug in our values

### Example Question #9 : Circular Motion

A car driving on the highway is moving at 60 miles per hour. As the car nears an exit ramp, the car slows to 35 miles per hour, a speed that is maintained throughout the circular path of the exit ramp. What force is keeping the car on its path (i.e. in circular motion)?

**Possible Answers:**

Weight of the car (gravity)

Normal force

Centripetal force

Momentum

**Correct answer:**

Centripetal force

The correct answer is centripetal force. In a free body diagram, this is the force that would be directed towards the center of the circular path; in circular motion, acceleration and net force are always in this direction.

Normal force and weight act perpendicular to the line that is directed towards the center of the circle, and will not prevent the car from traveling off the circular path. Momentum is not a measure of force and is not particularly relevant to this question.

Certified Tutor