All High School Physics Resources
Example Questions
Example Question #4 : Circular Motion
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 first applied at a angle
Both exert zero torques
The second applied perpendicular to the door
The second applied perpendicular to the door
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 #5 : 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?
Nothing, the seesaw will still be balanced
The side the boy is sitting on will tilt downward
The side the girl is setting on will tilt downward
It is impossible to say without knowing the masses and the distances
Nothing, the seesaw 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.
Example Question #1 : 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?
The first at the midpoint
Both exert zero torques
Both exert equal non-zero torques
The second at the doorknob
The second at 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.
Example Question #1 : 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?
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 #6 : Circular Motion
A shop sign weighing hangs from the end of a uniform beam as shown. Find the tension in the support wire at .
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 #1 : 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 .
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
Example Question #11 : Circular Motion
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.
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.
Example Question #12 : Circular Motion
An automobile engine slows down from to in . Calculate its angular acceleration.
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
Example Question #13 : Circular Motion
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 ?
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.
Example Question #911 : High School Physics
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?
It does not change
It increases
It changes but it is impossible to tell which way
It decreases
It does not change
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