High School Physics : High School Physics

Study concepts, example questions & explanations for High School Physics

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Example Questions

Example Question #6 : Circular Motion

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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 .

Possible Answers:

Correct answer:

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 #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 .

 

Possible Answers:

Correct answer:

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 

 

 

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.

 

Possible Answers:

Correct answer:

Explanation:

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.

 

Possible Answers:

Correct answer:

Explanation:

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 ?

 

Possible Answers:

Correct answer:

Explanation:

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?

 

Possible Answers:

It does not change

It increases

It changes but it is impossible to tell which way

It decreases

Correct answer:

It does not change

Explanation:

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?

 

Possible Answers:

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

Correct answer:

Box, Marble, Battery, Empty Soup Can, Wedding Band

Explanation:

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?

Possible Answers:

Correct answer:

Explanation:

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. 

Possible Answers:

Correct answer:

Explanation:

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? 

Possible Answers:

Correct Answer

Both the same

The hollow one

The solid one

Correct answer:

The hollow one

Explanation:

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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