# High School Physics : Circular Motion

## Example Questions

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### Example Question #1 : Circular Motion

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.

Explanation:

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.

### Example Question #2 : Circular Motion

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.

Explanation:

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.

### Example Question #3 : Circular Motion

The torque applied to a wrench is . If the force applied to the wrench is , how long is the wrench?

Explanation:

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.

### Example Question #4 : Circular Motion

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?

Explanation:

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.

### Example Question #1 : Understand Torque

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?

Both exert equal non-zero torques

The second applied perpendicular to the door

The first applied at a angle

Both exert zero torques

The second applied perpendicular to the door

Explanation:

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.

### Example Question #6 : Circular Motion

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?

The side the girl is setting on will tilt downward

The side the boy is sitting on will tilt downward

It is impossible to say without knowing the masses and the distances

Nothing, the seesaw will still be balanced

Nothing, the seesaw will still be balanced

Explanation:

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.

### Example Question #7 : Circular Motion

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?

Both exert equal non-zero torques

The first at the midpoint

The second at the doorknob

Both exert zero torques

The second at the doorknob

Explanation:

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.

### Example Question #8 : Circular Motion

A child spins a top with a radius of  with a force of . How much torque is generated at the edge of the top?

Explanation:

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.

### Example Question #9 : Circular Motion

A shop sign weighing  hangs from the end of a uniform  beam as shown.  Find the tension in the support wire at .

Explanation:

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.

### Example Question #10 : Circular Motion

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 .

Explanation:

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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