Circular Motion - Physics
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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?
Ellen is swinging a
What is the velocity of the yo-yo at the bottom of the circle?
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 
.
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
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Two children sit on a merry-go-round. One sits 
from the center, and the other sits 
from the center. If the children are in a straight line from the center, which child has a greater speed?
Two children sit on a merry-go-round. One sits
Velocity is given by distance per unit time.
When moving in a circle, the distance is the circumference, and each rotation takes exactly one period. We can substitute into the velocity formula.
If the children are in a straight line, that means that their periods (how long it takes to make one revolution) will be the same. The only thing that changes is 
, the distance from the center. Since radius is in the numerator, we can conclude that increasing the distance from the center will increase the velocity.
Velocity is given by distance per unit time.
When moving in a circle, the distance is the circumference, and each rotation takes exactly one period. We can substitute into the velocity formula.
If the children are in a straight line, that means that their periods (how long it takes to make one revolution) will be the same. The only thing that changes is
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Which of the following will increase the torque on a wrench, without changing the force applied?
Which of the following will increase the torque on a wrench, without changing the force applied?
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.
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.
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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)?
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)?
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.
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.
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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?
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?
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.
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.
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A penny is placed 12.0cm from the axis of a rotating turntable of variable speed. When the speed of the turntable is slowly increased, the penny remains fixed on the turntable until a rate of 40 revolutions per minute is reached, at which point the penny slides off. Calculate the coefficient of static friction between the penny and the turntable.
A penny is placed 12.0cm from the axis of a rotating turntable of variable speed. When the speed of the turntable is slowly increased, the penny remains fixed on the turntable until a rate of 40 revolutions per minute is reached, at which point the penny slides off. Calculate the coefficient of static friction between the penny and the turntable.
Known
which should be converted to meters (
)
which should 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 force of friction which is holding the penny in place. Therefore we know that the centripetal force is equal to the force of friction.
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 penny and the radius that the penny sits at.
Therefore 
We also know that the frictional force is related to the coefficient of friction and the normal force.
In this case the penny is sitting on the surface and the gravitational force is equal in size to the normal force. Therefore
The force of gravity is equal to the mass times the gravitational acceleration.
We can put all this together to a larger equation that says that
Notice that mass falls out of both sides of the equation
We can now rearrange and solve for the coefficient of friction
Now we can plug in our known values
Known
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 force of friction which is holding the penny in place. Therefore we know that the centripetal force is equal to the force of friction.
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 penny and the radius that the penny sits at.
Therefore
We also know that the frictional force is related to the coefficient of friction and the normal force.
In this case the penny is sitting on the surface and the gravitational force is equal in size to the normal force. Therefore
The force of gravity is equal to the mass times the gravitational acceleration.
We can put all this together to a larger equation that says that
Notice that mass falls out of both sides of the equation
We can now rearrange and solve for the coefficient of friction
Now we can plug in our known values
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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?
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 
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.
Knowns
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.
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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?
A 
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.
Knowns
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.
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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?
Computational
A 
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.
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 
Plug in the given values and solve for the acceleration.
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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?
A 
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.
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 
Plug in the given values and solve for the acceleration.
Plug the acceleration and given mass into the first equation to solve for force.
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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 
.
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 
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
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
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Conceptual
Two children sit on a merry-go-round. One sits 
from the center, and the other sits 
from the center. If the children are in a straight line form the center, which child has a greater speed?
Conceptual
Two children sit on a merry-go-round. One sits 
Velocity is given by distance per unit time.
When moving in a circle, the distance is the circumference, and each rotation takes exactly one period. We can substitute into the velocity formula.
If the children are in a straight line, that means that their periods (how long it takes to make one revolution) will be the same. The only thing that changes is r, the distance from the center. Since radius is in the numerator, we can conclude that increasing the distance from the center will increase the velocity.
Velocity is given by distance per unit time.
When moving in a circle, the distance is the circumference, and each rotation takes exactly one period. We can substitute into the velocity formula.
If the children are in a straight line, that means that their periods (how long it takes to make one revolution) will be the same. The only thing that changes is r, the distance from the center. Since radius is in the numerator, we can conclude that increasing the distance from the center will increase the velocity.
Compare your answer with the correct one above
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)?
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)?
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.
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.
Compare your answer with the correct one above
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?
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?
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.
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.
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A rider on a Ferris wheel moves in a vertical circle of radius r at constant speed v. Is the normal force that the seat exerts on the rider at the top of the wheel?
A rider on a Ferris wheel moves in a vertical circle of radius r at constant speed v. Is the normal force that the seat exerts on the rider at the top of the wheel?
The centripetal force is what is acting on the rider. At the top of the Ferris wheel, the normal force is pointing up, and the gravitational force is pointing down. The sum of these two forces must equal the centripetal force pointing downward toward the center of the circle. Therefore the normal force must be smaller than the gravitational force. At the bottom of the Ferris wheel, the same forces are present. However, the sum of these forces must equal the centripetal force point upward toward the center of the circle. Therefore the normal force must be greater than the gravitational force. Since the normal force must be greater than the gravitational force at the bottom and less than the gravitational force at the top, the force at the bottom must be greater than the force on the top.
The centripetal force is what is acting on the rider. At the top of the Ferris wheel, the normal force is pointing up, and the gravitational force is pointing down. The sum of these two forces must equal the centripetal force pointing downward toward the center of the circle. Therefore the normal force must be smaller than the gravitational force. At the bottom of the Ferris wheel, the same forces are present. However, the sum of these forces must equal the centripetal force point upward toward the center of the circle. Therefore the normal force must be greater than the gravitational force. Since the normal force must be greater than the gravitational force at the bottom and less than the gravitational force at the top, the force at the bottom must be greater than the force on the top.
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A child spins a top with a radius of 
with a force of 
. How much torque is generated at the edge of the top?
A child spins a top with a radius of
Torque is a force times the radius of the circle, given by the formula:
In this case, we are given the radius in centimeters, so be sure to convert to meters:
Use this radius and the given force to solve for the torque.
Torque is a force times the radius of the circle, given by the formula:
In this case, we are given the radius in centimeters, so be sure to convert to meters:
Use this radius and the given force to solve for the torque.
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Two children are trying to balance on a 
see-saw. One child has a mass of 
the other has a mass of 
. If the see-saw is balanced perfectly in the middle and the 
child is sitting at one end of the see-saw, how far from the center should the 
child sit so that the system is perfectly balanced?
Two children are trying to balance on a
If the see-saw is 
in total, then it has 
on either side of the fulcrum.
The question is asking us to find the equilibrium point; that means we want the net torque to equal zero.
Now, find the torque for the first child.
We are going to use the force of gravity for the force of the child.
When thinking of torque, treat the positive/negative as being clockwise vs. counter-clockwise instead of up vs. down. In this case, child one is generating counter-clockwise torque. That means that since 
, child two will be generating clockwise torque.
Solve for the radius (distance) of the second child.
If the see-saw is
The question is asking us to find the equilibrium point; that means we want the net torque to equal zero.
Now, find the torque for the first child.
We are going to use the force of gravity for the force of the child.
When thinking of torque, treat the positive/negative as being clockwise vs. counter-clockwise instead of up vs. down. In this case, child one is generating counter-clockwise torque. That means that since
Solve for the radius (distance) of the second child.
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A 
force is applied perpendicularly to a 
wrench. How much torque is generated?
A
The formula for torque is:
In this formula, 
is the angle the force makes with the lever arm. Since our force is applied perpendicularly, this angle will be 
. Use this angle, the force applied, and the length of the lever arm to calculate the torque.
The formula for torque is:
In this formula,
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In ancient Greece, construction workers would use long levers to lift up boulders. Assume a boulder has a mass of 
and a certain construction worker has a mass of 
. The pivot point of the lever is placed 
from the boulder and the construction worker stands on the opposite end. How long does the lever need to be between the pivot point and the construction worker in order for the system to be in equilibrium?
In ancient Greece, construction workers would use long levers to lift up boulders. Assume a boulder has a mass of
We know that the system will be in equilibrium at the final point, meaning that the final net torque must be zero. That means that the torque generated by the boulder and the torque generated by the construction worker should be equal.
The negative sign is due to the placement of the two masses on opposite sides of the fulcrum. Essentially, one mass will have a positive radius and the other will have a negative radius. Use the definition of torque to expand this equation.
In this case, the forces will be the weights of the two objects due to gravity.
We can cancel the acceleration due to gravity from the equation to simplify.
Using the given information from the question, we can solve for the distance between the worker and the fulcrum. We know the mass of the worker, the mass of the boulder, and the distance between the boulder and the fulcrum.
We know that the system will be in equilibrium at the final point, meaning that the final net torque must be zero. That means that the torque generated by the boulder and the torque generated by the construction worker should be equal.
The negative sign is due to the placement of the two masses on opposite sides of the fulcrum. Essentially, one mass will have a positive radius and the other will have a negative radius. Use the definition of torque to expand this equation.
In this case, the forces will be the weights of the two objects due to gravity.
We can cancel the acceleration due to gravity from the equation to simplify.
Using the given information from the question, we can solve for the distance between the worker and the fulcrum. We know the mass of the worker, the mass of the boulder, and the distance between the boulder and the fulcrum.
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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?
A
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.
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
Plug in the given values and solve for the acceleration.
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