Circular and Rotational Motion - AP Physics 1
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In an isolated system, the moment of inertia of a rotating object is halved. What happens to the angular velocity of the object?
In an isolated system, the moment of inertia of a rotating object is halved. What happens to the angular velocity of the object?
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In an isolated system, there is no net torque. If there is no net torque on the system, then the total angular momentum of the system remains the same. The angular momentum of a rotating object is equal to the moment of inertia of the object multiplied by the object's angular velocity.
is the symbol for angular momentum, 
is the moment of inertia, and 
is the angular velocity.
Therefore, if the moment of inertia, 
, is halved, then for the angular momentum, 
, to remain constant, the angular velocity, 
, must be doubled. This is because 
, which is the multiplicative identity. Anything multiplied by one remains the same. So, the final angular velocity would be twice as large as it was originally.
In an isolated system, there is no net torque. If there is no net torque on the system, then the total angular momentum of the system remains the same. The angular momentum of a rotating object is equal to the moment of inertia of the object multiplied by the object's angular velocity.
Therefore, if the moment of inertia,
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A satellite is rotating once per minute. It has an moment of inertia of 
. Erin, an astronaut, extends the satellite's solar panels, increasing its moment of inertia to 
. How quickly is the satellite now rotating?
A satellite is rotating once per minute. It has an moment of inertia of
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The formula for angular momentum is
where
= angular momentum
= moment of inertia
= angular speed
Angular speed is defined as
The initial period of the satellite is 1 minute, so:
Plugging this in, we can solve for the initial angular momentum:
After Erin extends the the solar panels, momentum is the same (conservation of momentum), but the moment of inertia is now 
. Thus,
Therefore, 
Plugging this back into the definition, we get
Therefore,
Thus, the satellite now rotates once every 3 minutes.
The formula for angular momentum is
where
Angular speed is defined as
The initial period of the satellite is 1 minute, so:
Plugging this in, we can solve for the initial angular momentum:
After Erin extends the the solar panels, momentum is the same (conservation of momentum), but the moment of inertia is now
Therefore,
Plugging this back into the definition, we get
Therefore,
Thus, the satellite now rotates once every 3 minutes.
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I start pushing a merry-go-round with a torque of 10 Newton-meters. It has a moment of inertia of 
. What is its rotational speed after 3 seconds assuming it starts at rest?
I start pushing a merry-go-round with a torque of 10 Newton-meters. It has a moment of inertia of
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The angular moment of the merry-go-round after 3 seconds is simply
Angular momentum is also given by
Plugging in 30 for 
gives us
The angular moment of the merry-go-round after 3 seconds is simply
Angular momentum is also given by
Plugging in 30 for
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A blob of clay of mass 
is dropped on top of a rotating disk's edge of mass 
spinning at a speed of 
. What will the resulting rotational speed 
be?
A blob of clay of mass
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Angular momentum is always conserved.
The equation relating angular momentum, moment of inertia and angular velocity is:
Moment of inertia can also be written as follows:
Plug in given values to find 
and 
Set the initial angular momentum equal to the final angular momentum and solve.
Angular momentum is always conserved.
The equation relating angular momentum, moment of inertia and angular velocity is:
Moment of inertia can also be written as follows:
Plug in given values to find
Set the initial angular momentum equal to the final angular momentum and solve.
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A comet is orbiting a star in a far distant galaxy wth only gravity acting upon it. After a period of time, the sun's radiation partially melts the comet, decreasing the mass to a fourth of what it was. Simultaneously, because of an irregular orbit, its distance to the star doubles. What expression best describes the comet's angular velocity, 
, compared to its original conditions (before the melting/distance change)?
A comet is orbiting a star in a far distant galaxy wth only gravity acting upon it. After a period of time, the sun's radiation partially melts the comet, decreasing the mass to a fourth of what it was. Simultaneously, because of an irregular orbit, its distance to the star doubles. What expression best describes the comet's angular velocity,
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Lets examine the equation 
, where 
= angular momentum, 
= mass of the satellite, 
= its angular velocity, and 
= its radius from the center of its orbit. Because of a lack of outside forces (besides the gravity of its star) acting on the satellite, we know that its angular momentum is conserved. Therefore, as its mass is divided by four, and its radius doubled, its angular velocity must also double, making the correct answer 
.
Lets examine the equation
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A rectangular rod on a horizontal table top is shown in the left side of the diagram. The rod is hinged at one end, as shown. It rotates in such a way that it hits a stationary ball as shown in the diagram's right side. The rod rotates without friction at a rate of 
until it contacts the ball. As a result of the collision, the rod comes to rest and the ball moves to the right. The mass of the rod is 0.2kg. The mass of the ball is 0.067kg. The length of the rod is 0.15m. The moment of inertia of a rod rotated about the end is: 
. What is the ball's speed as a result of the collision?
A rectangular rod on a horizontal table top is shown in the left side of the diagram. The rod is hinged at one end, as shown. It rotates in such a way that it hits a stationary ball as shown in the diagram's right side. The rod rotates without friction at a rate of
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This is a conservation of angular momentum problem, so we set the angular momentum of the rod 
to the equivalent angular momentum of the ball 
Note that the 
here is length, not angular momentum. Be careful not to cancel the 
since it refers to the rod on the left and the ball on the right. Finally, the 
on the right is the effective radius of the ball at the moment of impact, so it's simply the length of the rod.
This is a conservation of angular momentum problem, so we set the angular momentum of the rod
Note that the
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Consider a 
disk 
rotating with an angular speed of 
. At a later time, a 
ring 
is suddenly placed on top of the disk. Calculate the new angular speed of the system.
Consider a
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To solve this problem, we will be using the conservation of angular momentum. Initially we only have a disk rotating. This gives us information about the initial angular momentum of the system. Later, we have both a disk and a ring rotating together at some new angular velocity.
To solve this problem, we will be using the conservation of angular momentum. Initially we only have a disk rotating. This gives us information about the initial angular momentum of the system. Later, we have both a disk and a ring rotating together at some new angular velocity.
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The moment of inertia of two masses separated by distance 
is 
Two balls, each of mass 
are attached by a massless rod of length 
and spinning at 
around the point directly on the midpoint of the rod. Then rod then stretches to a length of 
. Assuming no loss of momentum and that the axis remains constant, determine the new rotational velocity.
The moment of inertia of two masses separated by distance
Two balls, each of mass
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First convert 
into 
:
Conservation of angular momentum:
Where, in this case:
Combine equations:
Plug in values and solve for the final rotational velocity:
First convert
Conservation of angular momentum:
Where, in this case:
Combine equations:
Plug in values and solve for the final rotational velocity:
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Two 
cars are racing side by side on a perfectly circular race track. The inner car is 
from the center of the track. The outer car is 
from center of the track.
If the outer car is moving at 
, determine it's angular momentum.
Two
If the outer car is moving at
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Consider the following system:
Two spherical masses, A and B, are attached to the end of a rigid rod with length l. The rod is attached to a fixed point, p, which is at the midpoint between the masses and is at a height, h, above the ground. The rod spins around the fixed point in a vertical circle that is traced in grey. 
is the angle at which the rod makes with the horizontal at any given time (
in the figure).
The rod is spinning, and as it goes through its horizontal position, the total angular momentum, including both masses, is 
. What is the instantaneous linear velocity of mass A?
Neglect the mass of the rod.
Consider the following system:
Two spherical masses, A and B, are attached to the end of a rigid rod with length l. The rod is attached to a fixed point, p, which is at the midpoint between the masses and is at a height, h, above the ground. The rod spins around the fixed point in a vertical circle that is traced in grey.
The rod is spinning, and as it goes through its horizontal position, the total angular momentum, including both masses, is
Neglect the mass of the rod.
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Since we are given the angular momentum, let's being with that and its formula:
Where:
L = angular momentum
I = moment of inertia
w = angular velocity
Since there are two masses on the rod (and we are neglecting the mass of the rod itself), we can expand this expression to:
Since the masses are attached to a rigid rod and going in a uniform circle, we can say that:
Since we're asked to determine the velocity of mass A, we'll substitute the angular velocity of mass A for mass B:
Rearranging for the angular velocity of A:
Now we need to determine what the moment of inertia is for each mass. Since they are spheres, we can treat them as point masses and use the following formula:
Thus for each mass, we get:
Where r is half the length of the rod and equal for both masses. Plugging these into the equation, we get:
Substituting half the length of the rod in for radius, we get:
We know the value of each variable in this equation, so we can solve it. But first, let's check to make sure our units are correct. The units for angular velocity are 
:
It checks out, so time to plug and chug:
We can then use the expression for angular velocity to get linear velocity:
Since we are given the angular momentum, let's being with that and its formula:
Where:
L = angular momentum
I = moment of inertia
w = angular velocity
Since there are two masses on the rod (and we are neglecting the mass of the rod itself), we can expand this expression to:
Since the masses are attached to a rigid rod and going in a uniform circle, we can say that:
Since we're asked to determine the velocity of mass A, we'll substitute the angular velocity of mass A for mass B:
Rearranging for the angular velocity of A:
Now we need to determine what the moment of inertia is for each mass. Since they are spheres, we can treat them as point masses and use the following formula:
Thus for each mass, we get:
Where r is half the length of the rod and equal for both masses. Plugging these into the equation, we get:
Substituting half the length of the rod in for radius, we get:
We know the value of each variable in this equation, so we can solve it. But first, let's check to make sure our units are correct. The units for angular velocity are
It checks out, so time to plug and chug:
We can then use the expression for angular velocity to get linear velocity:
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Consider the following system:
Two spherical masses, A and B, are attached to the end of a rigid rod with length l. The rod is attached to a fixed point, p, which is at a height, h, above the ground. The rod spins around the fixed point in a vertical circle that is traced in grey. is the angle at which the L side of the rod makes with the horizontal at any given time ( in the figure and can be negative if mass A is above the horizontal).
The system is currently at rest and being held so that mass A is below the horizontal and 
. The rod is then released and allowed to rotate freely. What is the maximum angular momentum achieved by mass B? Neglect air resistance and internal frictional forces
Consider the following system:
Two spherical masses, A and B, are attached to the end of a rigid rod with length l. The rod is attached to a fixed point, p, which is at a height, h, above the ground. The rod spins around the fixed point in a vertical circle that is traced in grey.
The system is currently at rest and being held so that mass A is below the horizontal and
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We can begin with the expression for conservation of energy to solve this problem:
Since we are told that the system is initially at rest, we can eliminate initial kinetic energy to get expression (1):
Expanding each term from left to right to:
We are told that mass A is held at an angle of 15 degrees below the horizontal, so we can use the sine function to determine the height of each mass. We will assume that the lowest point of rotation has a height of 0.
Where d is the vertical distance below horizontal, and r is the radius from the center of the rod, which is a distance equal to half the length of the rod:
Rearranging for d:
We know that this is the distance below horizontal that mass A begins at and the distance above the horizontal that mass B begins at. Also, horizontal is at a height of half a length rod above our reference height, so we can say:
Plugging these into our expanded expression for initial potential energy, we get:
Moving on to final potential energy:
We are asked to find the maximum angular momentum achieved by mass B. This occurs when mass B is traveling at it's highest velocity, which is when the larger of 2 masses, mass A, is traveling through the low point of rotation. This means that mass A is at a height of 0, and thus has no final potential energy. We can then say:
Also, we know that mass B will be at the high point of rotation, which means that its height will be the length of the rod. Therefore, we can say:
Moving on to final kinetic energy:
We don't have to expand the velocity term since the two masses are always traveling at the same speed.
Now, substituting our expanded terms back into expression (1), we get:
In order to determine angular momentum, we are going to need to solve this equation for final velocity, so let's begin rearranging:
We have all of these values, so time to plug and chug:
This is the maximum velocity that mass B will experience. We can now use the expression for angular momentum:
where:
Where r is half the length of the rod:
We have all of these values, so time to plug and chug:
We can begin with the expression for conservation of energy to solve this problem:
Since we are told that the system is initially at rest, we can eliminate initial kinetic energy to get expression (1):
Expanding each term from left to right to:
We are told that mass A is held at an angle of 15 degrees below the horizontal, so we can use the sine function to determine the height of each mass. We will assume that the lowest point of rotation has a height of 0.
Where d is the vertical distance below horizontal, and r is the radius from the center of the rod, which is a distance equal to half the length of the rod:
Rearranging for d:
We know that this is the distance below horizontal that mass A begins at and the distance above the horizontal that mass B begins at. Also, horizontal is at a height of half a length rod above our reference height, so we can say:
Plugging these into our expanded expression for initial potential energy, we get:
Moving on to final potential energy:
We are asked to find the maximum angular momentum achieved by mass B. This occurs when mass B is traveling at it's highest velocity, which is when the larger of 2 masses, mass A, is traveling through the low point of rotation. This means that mass A is at a height of 0, and thus has no final potential energy. We can then say:
Also, we know that mass B will be at the high point of rotation, which means that its height will be the length of the rod. Therefore, we can say:
Moving on to final kinetic energy:
We don't have to expand the velocity term since the two masses are always traveling at the same speed.
Now, substituting our expanded terms back into expression (1), we get:
In order to determine angular momentum, we are going to need to solve this equation for final velocity, so let's begin rearranging:
We have all of these values, so time to plug and chug:
This is the maximum velocity that mass B will experience. We can now use the expression for angular momentum:
where:
Where r is half the length of the rod:
We have all of these values, so time to plug and chug:
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Moment of inertia of disk: 
A car with wheels of mass 
and wheels of radius 
is traveling at 
. Treating the wheels as disks of uniform mass density, calculate the angular momentum of one wheel.
Moment of inertia of disk:
A car with wheels of mass
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Finding angular velocity:
Using
and
Where 
is angular momentum
Combining equations and plugging in values:
Finding angular velocity:
Using
and
Where
Combining equations and plugging in values:
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A solid bowling ball of mass 
and radius 
is traveling at a linear velocity of 
. What is the angular momentum of the ball?
A solid bowling ball of mass
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The expression for angular momentum is:
Where I is the moment of inertia for a solid sphere:
And w is the angular velocity:
Plugging these in, we get:
Plugging in our values, we get:
The expression for angular momentum is:
Where I is the moment of inertia for a solid sphere:
And w is the angular velocity:
Plugging these in, we get:
Plugging in our values, we get:
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A bowling ball of mass 
and radius 
is traveling down a slope with a vertical height change of 
. If the ball starts from rest, what is the final linear velocity of the ball as it reaches the bottom of the slope? Neglect air resistance.
A bowling ball of mass
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Let's begin with the expression for conservation of energy:
Since the ball begins from rest, we can eliminate initial kinetic energy. Also, if we assume that the bottom of the slope has a height of 0, we can eliminate final potential energy to get:
Plugging in expressions, we get:
Where I is the moment of inertia for a solid sphere:
And w is the angular velocity:
Plugging these in, we get:
Rearranging for final velocity, we get:
Plugging in our values, we get:
Let's begin with the expression for conservation of energy:
Since the ball begins from rest, we can eliminate initial kinetic energy. Also, if we assume that the bottom of the slope has a height of 0, we can eliminate final potential energy to get:
Plugging in expressions, we get:
Where I is the moment of inertia for a solid sphere:
And w is the angular velocity:
Plugging these in, we get:
Rearranging for final velocity, we get:
Plugging in our values, we get:
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Two identical cars are racing side by side on a circular race track. Which has the greater angular momentum?
Two identical cars are racing side by side on a circular race track. Which has the greater angular momentum?
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Where 
is the radius of the circle
is the mass of the object
is the linear velocity of the object
The car on the outside has a larger 
and a larger 
, and the same 
, thus it has a higher angular momentum.
Where
The car on the outside has a larger
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A 
weight on a 
rope is swung around at 
until it comes into contact with a resting 
brick on flat ground. The brick travels 
. What is the coefficient of kinetic friction between the the brick and ground?
A
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The first part of this problem is a transfer between angular and linear momentum. Conservation of momentum tells us that we can equate the angular momentum of the pendulum to the linear momentum of the brick, or 
, where 
and 
are the masses of the pendulum and brick respectively, 
is the angular velocity of the pendulum in 
, 
is the length of the pendulum rope, and 
is the resultant velocity of the brick. We find that the resulting velocity of the brick can be written as 
The next part of the problem is identifying that the kinematic equations can be used here to find the acceleration the brick undergoes as it comes to a stop. Here we can use 
, where 
is the end velocity of the brick (zero), 
is the deceleration it undergoes, and 
is the distance the brick travels before coming to a stop. Solving for 
we have that 
.
Lastly, we know that the kinetic friction force can be written as 
. Newton's second law says that this can be written as 
. Thus we find that 
.
The first part of this problem is a transfer between angular and linear momentum. Conservation of momentum tells us that we can equate the angular momentum of the pendulum to the linear momentum of the brick, or
The next part of the problem is identifying that the kinematic equations can be used here to find the acceleration the brick undergoes as it comes to a stop. Here we can use
Lastly, we know that the kinetic friction force can be written as
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If it takes a bike wheel 3 seconds to complete one revolution, what is the wheel's angular velocity?
If it takes a bike wheel 3 seconds to complete one revolution, what is the wheel's angular velocity?
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The definition of angular velocity is 
.
By identifying the given information to be 
and 
, we can plug this into the equation to calculate the angular velocity:
The definition of angular velocity is
By identifying the given information to be
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A horizontally mounted wheel of radius 
is initially at rest, and then begins to accelerate constantly until it has reached an angular velocity 
after 5 complete revolutions. What was the angular acceleration of the wheel?
A horizontally mounted wheel of radius
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You may recall the kinematic equation that relates final velocity, initial velocity, acceleration, and distance, respectively:
Well, for rotational motion (such as in this problem), there is a similar equation, except it relates final angular velocity, intial angular velocity, angular acceleration, and angular distance, respectively:
The wheel starts at rest, so the initial angular velocity, 
, is zero. The total number of revolutions of the wheel is given to be 5 revolutions. Each revolution is equivalent to an angular distance of 
radians. So, we can convert the total revolutions to an angular distance to get:
The final angular velocity was given as 
in the text of the question. So, we should use the above equation to solve for the angular acceleration, 
.
You may recall the kinematic equation that relates final velocity, initial velocity, acceleration, and distance, respectively:
Well, for rotational motion (such as in this problem), there is a similar equation, except it relates final angular velocity, intial angular velocity, angular acceleration, and angular distance, respectively:
The wheel starts at rest, so the initial angular velocity,
The final angular velocity was given as
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An object moves at a constant speed of 
in a circular path of radius of 1.5 m. What is the angular acceleration of the object?
An object moves at a constant speed of
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For a rotating object, or an object moving in a circular path, the relationship between angular acceleration and linear acceleration is
Linear acceleration is given by 
, angular acceleration is 
, and the radius of the circular path is 
.
For circular/centripetal motion, the linear acceleration is related to the object's linear velocity by
We know the linear velocity is 
, and the radius is 1.5 m, so we can find the linear acceleration...
Now that we have the linear acceleration, we can use this in the equation at the top to find the angular acceleration...
For a rotating object, or an object moving in a circular path, the relationship between angular acceleration and linear acceleration is
Linear acceleration is given by
For circular/centripetal motion, the linear acceleration is related to the object's linear velocity by
We know the linear velocity is
Now that we have the linear acceleration, we can use this in the equation at the top to find the angular acceleration...
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What is the angular velocity of the second hand of a clock?
What is the angular velocity of the second hand of a clock?
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The angular velocity of the second hand of a clock can be found by dividing the number of radians the second hand will travel over a known period of time. Thankfully for a clock, we know that the second hand will make one revolution, i.e. covering 
in one minute, or 60s. The formula for angular velocity is:
So the angular velocity, 
is 
, which simplifies to our answer, 
The angular velocity of the second hand of a clock can be found by dividing the number of radians the second hand will travel over a known period of time. Thankfully for a clock, we know that the second hand will make one revolution, i.e. covering
So the angular velocity,
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