All High School Physics Resources
Example Questions
Example Question #911 : High School Physics
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 #911 : High School Physics
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
Example Question #14 : Circular Motion
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?
Empty Soup Can, Wedding Band, Marble, Battery, Box
Box, Marble, Battery, Empty Soup Can, Wedding Band
Marble, Empty Soup Can, Battery, Box, Wedding Band
Wedding Band, Empty Soup Can, Battery, Marble, Box
Wedding Band, Box, Empty Soup Can, Marble, Battery
Box, Marble, Battery, Empty Soup Can, Wedding Band
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.
Example Question #15 : Circular Motion
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?
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
Example Question #1 : Rotational Angular Momentum
Determine the moment of inertia of a sphere of radius when the axis of rotation is through its center.
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.
Example Question #17 : Circular Motion
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?
Correct Answer
Both the same
The hollow one
The solid one
The hollow one
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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