Circular Motion - High School Physics
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A potter’s wheel is rotating around a vertical axis through its center a frequency of 
. The wheel can be considered a uniform disk of mass 
and diameter 
. The potter then throws a 
chunk of clay, approximately shaped as a flat disk of radius 
, onto the center of the wheel. What is the angular velocity of the wheel after the clay sticks to it?
A potter’s wheel is rotating around a vertical axis through its center a frequency of 
We can use the conservation of angular momentum in order to solve this problem. The law of conservation of angular momentum states that the momentum before the collision must equal to the momentum after the collision. Angular momentum is calculated with the equation
Before the collision we only have the potter’s wheel rotating.
We know that the moment of inertia of the wheel can be considered as a uniform disk.
We can convert the velocity of the wheel to rad/s
We can now calculate the momentum before the collision.
Now it is time to analyze the momentum after the collision. At this point we have added a piece of clay which is now moving at the same angular velocity as the pottery.
We know that the moment of inertia of the clay can be considered as a uniform disk.
We know the angular momentum at the beginning equals the angular momentum at the end.
We can now solve for the angular velocity
We can use the conservation of angular momentum in order to solve this problem. The law of conservation of angular momentum states that the momentum before the collision must equal to the momentum after the collision. Angular momentum is calculated with the equation
Before the collision we only have the potter’s wheel rotating.
We know that the moment of inertia of the wheel can be considered as a uniform disk.
We can convert the velocity of the wheel to rad/s
We can now calculate the momentum before the collision.
Now it is time to analyze the momentum after the collision. At this point we have added a piece of clay which is now moving at the same angular velocity as the pottery.
We know that the moment of inertia of the clay can be considered as a uniform disk.
We know the angular momentum at the beginning equals the angular momentum at the end.
We can now solve for the angular velocity
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To get a flat, uniform cylindrical spacecraft spinning at the correct speed, astronauts fire four tangential rockets equidistance around the edge of the cylindrical spacecraft. Suppose the spacecraft has a mass of 
and a radius of 
, and the rockets each add a mass of 
. What is the steady force required of each rocket if the satellite is to reach 
in 
.
To get a flat, uniform cylindrical spacecraft spinning at the correct speed, astronauts fire four tangential rockets equidistance around the edge of the cylindrical spacecraft. Suppose the spacecraft has a mass of 
In order to get the spacecraft spinning, the rockets must supply a torque to the edge of the spacecraft.
We can calculate the moment of inertia of the spacecraft and the 4 rockets along the edge.
The spacecraft can be considered a uniform disk.
The rocket can be calculated
So the total moment of inertia
We can also calculate the angular acceleration of the rocket
Since the spacecraft starts from rest the initial angular velocity is 
.
The final angular velocity needs to be converted to radians per second.
We also need to convert the 4 minutes to seconds
Therefore the total torque applied by the rockets is
Each rocket contributes to the torque. So to determine the torque contributed by one rocket we would divide this by 4
We can now determine the force applied by one rocket through the equation
We can approximate that to about 
In order to get the spacecraft spinning, the rockets must supply a torque to the edge of the spacecraft.
We can calculate the moment of inertia of the spacecraft and the 4 rockets along the edge.
The spacecraft can be considered a uniform disk.
The rocket can be calculated
So the total moment of inertia
We can also calculate the angular acceleration of the rocket
Since the spacecraft starts from rest the initial angular velocity is 
The final angular velocity needs to be converted to radians per second.
We also need to convert the 4 minutes to seconds
Therefore the total torque applied by the rockets is
Each rocket contributes to the torque. So to determine the torque contributed by one rocket we would divide this by 4
We can now determine the force applied by one rocket through the equation
We can approximate that to about
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A merry-go-round has a mass of 
and radius of 
. How much net work is required to accelerate it from rest to a ration rate of 
revolution per 
seconds? Assume it is a solid cylinder.
A merry-go-round has a mass of 
We know that the work-kinetic energy theorem states that the work done is equal to the change of kinetic energy. In rotational terms this means that
In this case the initial angular velocity is 
.
We can convert our final angular velocity to radians per second.
We also can calculate the moment of inertia of the merry-go-round assuming that it is a uniform solid disk.
We can put this into our work equation now.
We know that the work-kinetic energy theorem states that the work done is equal to the change of kinetic energy. In rotational terms this means that
In this case the initial angular velocity is 
We can convert our final angular velocity to radians per second.
We also can calculate the moment of inertia of the merry-go-round assuming that it is a uniform solid disk.
We can put this into our work equation now.
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Determine the moment of inertia of a 
sphere of radius 
when the axis of rotation is through its center.
Determine the moment of inertia of a 
A wheel can be looked at as a uniform disk. We can then look up the equation for the moment of inertia of a solid cylinder.The equation is
We can now solve for the moment of inertia.
A wheel can be looked at as a uniform disk. We can then look up the equation for the moment of inertia of a solid cylinder.The equation is
We can now solve for the moment of inertia.
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An automobile engine slows down from 
to 
in 
. Calculate its angular acceleration.
An automobile engine slows down from 
The first thing we need to do is convert our velocities to radians to per second.
We can now find the angular acceleration through the equation
The first thing we need to do is convert our velocities to radians to per second.
We can now find the angular acceleration through the equation
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What is the angular momentum of a 
ball revolving on the end of a thin string in a circle of radius 
at an angular speed of 
?
What is the angular momentum of a 
The equation for angular momentum is equal to the moment of inertia multiplied by the angular speed.
The moment of inertia of an object is equal to the mass times the radius squared of the object.
We can substitute this into our angular momentum equation.
Now we can substitute in our values.
The equation for angular momentum is equal to the moment of inertia multiplied by the angular speed.
The moment of inertia of an object is equal to the mass times the radius squared of the object.
We can substitute this into our angular momentum equation.
Now we can substitute in our values.
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Two spheres have the same radius and equal mass. One sphere is solid, and the other is hollow and made of a denser material. Which one has the bigger moment of inertia about an axis through the center?
Two spheres have the same radius and equal mass. One sphere is solid, and the other is hollow and made of a denser material. Which one has the bigger moment of inertia about an axis through the center?
Since both spheres have the same radius and the same mass, we need to look at the equations for the moment of inertia of a solid sphere and a hollow sphere.
A solid sphere 
A hollow sphere 
If both of these have the same mass and radius, the only difference is the constant that is being multiplied by 
In this case the hollow sphere has a larger constant and therefore would have the larger moment of inertia.
This also conceptually makes sense since all the mass is distributed along the outside of the sphere meaning it all has a larger radius. A solid sphere has mass that is both close to the center and farther away, meaning that it would have a reduced moment of inertia.
Since both spheres have the same radius and the same mass, we need to look at the equations for the moment of inertia of a solid sphere and a hollow sphere.
A solid sphere
A hollow sphere
If both of these have the same mass and radius, the only difference is the constant that is being multiplied by
In this case the hollow sphere has a larger constant and therefore would have the larger moment of inertia.
This also conceptually makes sense since all the mass is distributed along the outside of the sphere meaning it all has a larger radius. A solid sphere has mass that is both close to the center and farther away, meaning that it would have a reduced moment of inertia.
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An ice skater performs a fast spin by pulling in her outstretched arms close to her body. What happens to her angular momentum about the axis of rotation?
An ice skater performs a fast spin by pulling in her outstretched arms close to her body. What happens to her angular momentum about the axis of rotation?
According to the law of conservation of momentum, the momentum of a system does not change. Therefore in the example, the angular momentum of the ice skater is constant. When she pulls her arms in, she is reducing her moment of inertia which causes her angular velocity to increase
According to the law of conservation of momentum, the momentum of a system does not change. Therefore in the example, the angular momentum of the ice skater is constant. When she pulls her arms in, she is reducing her moment of inertia which causes her angular velocity to increase
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Several objects roll without slipping down an income of vertical height H, all starting from rest. The objects are a battery (solid cylinder), a frictionless box, a wedding band (hoop), an empty soup can, and a marble (solid sphere). In what order do they reach the bottom of the incline?
Several objects roll without slipping down an income of vertical height H, all starting from rest. The objects are a battery (solid cylinder), a frictionless box, a wedding band (hoop), an empty soup can, and a marble (solid sphere). In what order do they reach the bottom of the incline?
We can use conservation of energy to compare the gravitational potential energy at the time of the hill to the rotational and kinetic energy at the bottom of the hill.
The box would be the fastest as all of the gravitational potential energy would convert to translational energy.
The round objects would share the gravitational potential energy between translational and rotational kinetic energies.
The moment of inertia is equal to a numerical factor (
) times the mass and radius squared. Since the mass is the same in each term, the speed does not depend on 
.
Additionally we can substitute angular speed for translational velocity using the equation
The radius cancels out and we are left with
Therefore the velocity is purely dependent on the numerical factor (
) in the moment of inertia and the height from which it was released. Since all of these objects were released from the same height, we can examine the moment of inertia for each to determine which will be the fasters.
Hoop (wedding ring) = 
Hollow cylinder (empty can) = 
Solid cylinder (Battery) = 
Solid sphere (Marble) = 
From this we can see that the marble will reach the bottom at the fastest velocity as it has the smallest numerical factor. This will be followed by the battery, the empty can and the wedding ring.
We can use conservation of energy to compare the gravitational potential energy at the time of the hill to the rotational and kinetic energy at the bottom of the hill.
The box would be the fastest as all of the gravitational potential energy would convert to translational energy.
The round objects would share the gravitational potential energy between translational and rotational kinetic energies.
The moment of inertia is equal to a numerical factor (
Additionally we can substitute angular speed for translational velocity using the equation
The radius cancels out and we are left with
Therefore the velocity is purely dependent on the numerical factor (
Hoop (wedding ring) =
Hollow cylinder (empty can) =
Solid cylinder (Battery) =
Solid sphere (Marble) =
From this we can see that the marble will reach the bottom at the fastest velocity as it has the smallest numerical factor. This will be followed by the battery, the empty can and the wedding ring.
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A shop sign weighing 
hangs from the end of a uniform 
beam as shown. Find the tension in the support wire at 
.
A shop sign weighing 
This is a static equilibrium problem. In order for static equilibrium to be achieved, there are three things that must be true. First, the sum of the forces in the horizontal direction must all equal 0. Second, the sum of the forces in the vertical direction must all be equal to zero. Third, the torque around a fixed axis must equal 
.
Let us begin by summing up the forces in the vertical direction.
Then let us sum up the forces in the horizontal direction
Lastly let us analyze the torque, using the hinge as the axis point where 
is the length of the beam.
Looking at these three equations, the easiest to work with would be our torque equation as it is only missing one variable. If we are able to find the tension in the y direction we would then be able to use trigonometry to determine the tension in the cable overall.
We can now use trigonometry to determine the tension force in the wire.
Rearrange to get the tension force by itself.
This is a static equilibrium problem. In order for static equilibrium to be achieved, there are three things that must be true. First, the sum of the forces in the horizontal direction must all equal 0. Second, the sum of the forces in the vertical direction must all be equal to zero. Third, the torque around a fixed axis must equal 
Let us begin by summing up the forces in the vertical direction.
Then let us sum up the forces in the horizontal direction
Lastly let us analyze the torque, using the hinge as the axis point where 
Looking at these three equations, the easiest to work with would be our torque equation as it is only missing one variable. If we are able to find the tension in the y direction we would then be able to use trigonometry to determine the tension in the cable overall.
We can now use trigonometry to determine the tension force in the wire.
Rearrange to get the tension force by itself.
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A shop sign weighing 
hangs from the end of a uniform 
beam as shown. Find the horizontal and vertical forces exerted by the hinge on the beam at the wall.
A shop sign weighing 
This is a static equilibrium problem. In order for static equilibrium to be achieved, there are three things that must be true. First, the sum of the forces in the horizontal direction must all equal 
. Second, the sum of the forces in the vertical direction must all be equal to zero. Third, the torque around a fixed axis must equal 
.
Let us begin by summing up the forces in the vertical direction.
Then let us sum up the forces in the horizontal direction
Lastly let us analyze the torque, using the hinge as the axis point where 
is the length of the beam.
Looking at these three equations, the easiest to work with would be our torque equation as it is only missing one variable. If we are able to find the tension in the 
direction we would then be able to use trigonometry to determine the tension in the cable overall.
We can now substitute this value back into our vertical direction equation to determine the force on the hinge in the 
direction.
We can go back to our tension in our 
direction and use trigonometry to determine the force of the tension in 
direction.
Rearrange and solve for the tension in the 
direction.
We can now go back to our second equation and substitute this value in.
This is a static equilibrium problem. In order for static equilibrium to be achieved, there are three things that must be true. First, the sum of the forces in the horizontal direction must all equal 
Let us begin by summing up the forces in the vertical direction.
Then let us sum up the forces in the horizontal direction
Lastly let us analyze the torque, using the hinge as the axis point where 
Looking at these three equations, the easiest to work with would be our torque equation as it is only missing one variable. If we are able to find the tension in the 
We can now substitute this value back into our vertical direction equation to determine the force on the hinge in the 
We can go back to our tension in our 
Rearrange and solve for the tension in the 
We can now go back to our second equation and substitute this value in.
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A pulley has a bucket of weight 
hanging from the cord over a well. The pulley has a mass of 
and radius of 
. There is a frictional torque of 
at the axle. Assume the cord has negligible mass and does not stretch or slip on the pulley. Calculate the linear acceleration of the bucket.
A pulley has a bucket of weight 
First let us analyze the torque that is happening on the bucket. There is a torque from the friction on the pulley and there is a torque from the bucket pulling on the pulley.
Next let us analyze the forces involved. There is a tension force pulling up on the bucket and there is the force of gravity pulling down on the bucket.
We can rearrange this to find 
so that we can substitute it into our torque equation.
Now substitute this into our torque equation
We know that there is a relationship between acceleration and angular acceleration.
So we can substitute this into our equation so that only linear acceleration is present.
We also know that the moment of inertia of the system is equal to the moment of inertia of the pulley plus the bucket.
We can now substitute this into our equation.
We can now starting putting in our known variables to solve for the missing acceleration.
First let us analyze the torque that is happening on the bucket. There is a torque from the friction on the pulley and there is a torque from the bucket pulling on the pulley.
Next let us analyze the forces involved. There is a tension force pulling up on the bucket and there is the force of gravity pulling down on the bucket.
We can rearrange this to find 
Now substitute this into our torque equation
We know that there is a relationship between acceleration and angular acceleration.
So we can substitute this into our equation so that only linear acceleration is present.
We also know that the moment of inertia of the system is equal to the moment of inertia of the pulley plus the bucket.
We can now substitute this into our equation.
We can now starting putting in our known variables to solve for the missing acceleration.
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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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The torque applied to a wrench is 
. If the force applied to the wrench is 
, how long is the wrench?
The torque applied to a wrench is 
The formula for torque is:
We are given the total torque and the force applied. Using these values, we can solve for the length of the wrench.
The formula for torque is:
We are given the total torque and the force applied. Using these values, we can solve for the length of the wrench.
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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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Two equal forces are applied to a door. The first force is applied at the midpoint of the door, the second force is applied at the doorknob. Both forces are applied perpendicular to the door. Which force exerts the greater torque?
Two equal forces are applied to a door. The first force is applied at the midpoint of the door, the second force is applied at the doorknob. Both forces are applied perpendicular to the door. Which force exerts the greater torque?
Torque is equal to the force applied perpendicular to a surface, multiplied by the radius or distance to the pivot point.
In this case, both forces are equal to one another. Therefore the force that is applied at the point furthest from the axis of rotation (the hinge) will have the greater torque. In this case, the furthest distance is the doorknob.
Torque is equal to the force applied perpendicular to a surface, multiplied by the radius or distance to the pivot point.
In this case, both forces are equal to one another. Therefore the force that is applied at the point furthest from the axis of rotation (the hinge) will have the greater torque. In this case, the furthest distance is the doorknob.
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Two equal forces are applied to a door at the doorknob. The first force is applied 
to the plane of the door. The second is applied perpendicular to the door. Which force exerts a greater torque?
Two equal forces are applied to a door at the doorknob. The first force is applied 
Torque is equal to the force applied perpendicular to a surface, multiplied by the radius or distance to the pivot point.
In this case, one force is applied perpendicular and the other at an angle. The one that is applied at an angle, only has a small component of the total force acting in the perpendicular direction. This component will be smaller than the overall force. Therefore the force that is already acting perpendicular to door will provide the greatest torque.
Torque is equal to the force applied perpendicular to a surface, multiplied by the radius or distance to the pivot point.
In this case, one force is applied perpendicular and the other at an angle. The one that is applied at an angle, only has a small component of the total force acting in the perpendicular direction. This component will be smaller than the overall force. Therefore the force that is already acting perpendicular to door will provide the greatest torque.
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A heavy boy and a light girl are balanced on a massless seesaw. If they both move forward so that they are one-half their original distance from the pivot point, what will happen to the seesaw?
A heavy boy and a light girl are balanced on a massless seesaw. If they both move forward so that they are one-half their original distance from the pivot point, what will happen to the seesaw?
Torque is equal to the force applied perpendicular to a surface, multiplied by the radius or distance to the pivot point.
In this example the boy of mass M is a distance R away and is balancing a girl of mass m at a distance r away.
If both of these kids move to a distance that is one half their original distance.
The half cancels out of the equation and therefore the boy and girl will still be balanced.
Torque is equal to the force applied perpendicular to a surface, multiplied by the radius or distance to the pivot point.
In this example the boy of mass M is a distance R away and is balancing a girl of mass m at a distance r away.
If both of these kids move to a distance that is one half their original distance.
The half cancels out of the equation and therefore the boy and girl will still be balanced.
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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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