# Angular and Linear Momentum Problem

• Bobby345
In summary, the rod and ball collide and the ball moves away with a certain final velocity. The linear momentum of the system is conserved during the collision, but I am not sure why.
Bobby345

## Homework Statement

A system has a ball and a uniform rod. The rod is rotating about point X on a frictionless table until it strikes the ball. The rod stops and the ball moves away.

Variables:
Rod's mass: m1
Ball's mass: m2
Rod's original angular velocity: ω
Ball's final velocity: v
Rod's moment of inertia about point X: (1/3)(m1L^2)
Length of the rod: L

1. What is the linear momentum of the system before the collision in terms of L, m1, and ω?
2. What is the linear momentum of the system after the collision in terms of L, m1, and ω?
3. Why is the angular momentum about point X conserved during the collision?
4. What makes the linear momentum of the system change during the collision?

L=Iω
FΔt=mΔv

## The Attempt at a Solution

I figured out that the speed of the ball after the collision (v) is equal to (m1Lω)/(3m2).
I cannot figure out anything about the linear momentum and conservation of either angular or linear momentum. I suspect that the reason angular momentum is conserved has something to do with torque but I am not positive.

Welcome to PF!
Bobby345 said:
I figured out that the speed of the ball after the collision (v) is equal to (m1Lω)/(3m2).
Looks good. I guess you assumed conservation of angular momentum to get this.
I cannot figure out anything about the linear momentum
For any system of particles, the total linear momentum is equal to the total mass of the system multiplied by the velocity of the center of mass of the system. The rod can be thought of as a system of particles. So, try to use this idea to get the linear momentum of the rod just before the collision.

and conservation of either angular or linear momentum. I suspect that the reason angular momentum is conserved has something to do with torque but I am not positive.
Can you state the condition that is necessary for linear momentum to be conserved? Likewise for angular momentum?

When the rod hits the ball, there will be a strong external force that acts on the rod at the pivot point.

Bobby345

So the linear momentum of the system is just m1Lω, considering v = Lω?The velocity of the rod is now 0, but the ball has a velocity of v now. I am not exactly sure how to express this momentum in terms of m1, L, and ω, considering that m1 doesn't have any momentum anymore, but m2 does. Any ideas?Linear: There must be an external force applied that makes it so the linear momentum is not conserved. Not exactly sure what it is, but it has to exist, otherwise it wouldn't ask that.
Angular: Same goes for angular. If it is conserved, there cannot be an external torque applied.

Excuse my lack of LaTeX and thanks for your help!

Bobby345 said:
Rod's moment of inertia about point X: (1/3)(m1L^2)
Yes, if X is one end of the rod (or anywhere on the circle of which the rod is a diameter). Is that given, in a diagram maybe?
Bobby345 said:
the speed of the ball after the collision (v) is equal to (m1Lω)/(3m2).
I presume from that that the tip of the rod strikes the ball, correct?
Bobby345 said:
So the linear momentum of the system is just m1Lω, considering v = Lω?
Where is the centre of mass of the rod? How fast was that moving?
Also, try not to use the same name for two variables. v is the final speed of the ball.

Bobby345
Bobby345 said:
So the linear momentum of the system is just m1Lω, considering v = Lω?
This is not quite right. The speed of the center of mass is not Lω.

Re: the final linear momentum of the system
The velocity of the rod is now 0, but the ball has a velocity of v now. I am not exactly sure how to express this momentum in terms of m1, L, and ω, considering that m1 doesn't have any momentum anymore, but m2 does. Any ideas?
After the collision, the linear momentum is just the momentum of the ball. You have already found an expression for this velocity in terms of m1, m2, L, and ω as stated in your first post.

Re: conservation of angular and linear momentum
Linear: There must be an external force applied that makes it so the linear momentum is not conserved. Not exactly sure what it is, but it has to exist, otherwise it wouldn't ask that.
Angular: Same goes for angular. If it is conserved, there cannot be an external torque applied.
The end of the rod that is rotating on an axle is held in place by the axle. When the rod hits the ball, the axle exerts a force on the rod to prevent this end of the rod from moving. That's the external force acting on the system during the collision. Does this force produce any torque about the axle?

Bobby345
haruspex said:
Yes, if X is one end of the rod (or anywhere on the circle of which the rod is a diameter). Is that given, in a diagram maybe?

I presume from that that the tip of the rod strikes the ball, correct?.
Good points.

Bobby345
TSny said:
Good points.
Yep. Given in a diagram that X is on one end.
Correct. The tip of the rod strikes the ball.

TSny said:
This is not quite right. The speed of the center of mass is not Lω.

After the collision, the linear momentum is just the momentum of the ball. You have already found an expression for this velocity in terms of m1, m2, L, and ω as stated in your first post.

The end of the rod that is rotating on an axle is held in place by the axle. When the rod hits the ball, the axle exerts a force on the rod to prevent this end of the rod from moving. That's the external force acting on the system during the collision. Does this force produce any torque about the axle?

I don't believe it produces a torque because the radius at point X is 0, correct? That must be the reason why the angular momentum is conserved, but not the linear momentum. Now I just need to establish the initial and final linear momentum of the system.

Bobby345 said:
That must be the reason why the angular momentum is conserved
To be pedantic, it is the reason the angular momentum about point X is conserved.
For angular motion, it is usually important to be clear about the axis being used.

Bobby345
TSny said:
This is not quite right. The speed of the center of mass is not Lω.

After the collision, the linear momentum is just the momentum of the ball. You have already found an expression for this velocity in terms of m1, m2, L, and ω as stated in your first post.

Ohhh. So you can just negate m2 by means of multiplying it by v. I was just confused because I hadn't tried the multiplication and automatically assumed that m2 would be in the answer. Thus, final linear momentum = (m1Lω)/3

Bobby345 said:
Ohhh. So you can just negate m2 by means of multiplying it by v. I was just confused because I hadn't tried the multiplication and automatically assumed that m2 would be in the answer. Thus, final linear momentum = (m1Lω)/3
I think that's right.

Bobby345
haruspex said:
Where is the centre of mass of the rod? How fast was that moving?
Also, try not to use the same name for two variables. v is the final speed of the ball.

Center of mass should be at L/2, right? Thus the rod's initial linear velocity is (Lω)/2 at the center of mass, making it's momentum (m1Lω)/2?

Bobby345 said:
Center of mass should be at L/2, right? Thus the rod's initial linear velocity is (Lω)/2 at the center of mass, making it's momentum (m1Lω)/2?
Right.

So back to the original questions.
1. What is the linear momentum of the system before the collision in terms of L, m1, and ω?
(m1Lω)/2
2. What is the linear momentum of the system after the collision in terms of L, m1, and ω?
(m1Lω)/3
3. Why is the angular momentum about point X conserved during the collision?
The angular momentum about point X is conserved during the collision because no external torque is applied to the system. Despite the force applied to the rod at point X by the axle upon the collision of the rod and the ball, no torque is applied due to a radius of 0.
4. What makes the linear momentum of the system change during the collision?
The linear momentum of the system is not conserved during the collision due to the force applied to the rod by the axle at point X.

How does this look?

Bobby345 said:
How does this look?
Looks good to me. As haruspex pointed out, for part 3 it is good to emphasize that there is no external torque about point X. If you chose some other point as your origin for torques, there would be an external torque.

Bobby345
TSny said:
Looks good to me. As haruspex pointed out, for part 3 it is good to emphasize that there is no external torque about point X. If you chose some other point as your origin for torques, there would be an external torque.
Even though the question asks why the angular momentum of the system is conserved?

Bobby345 said:
Even though the question asks why the angular momentum of the system is conserved?
Perhaps this has something to do with net torques?

I think all my answers are sorted except part 3 could be buffed up a little bit. I was thinking to maybe discuss how there is 0 torque from the force applied by the axle at point 0 and therefore assert that the net torque on the system remains unchanged so angular momentum is conserved.

The only external force acting on the rod-ball system is the force at the axle (gravity force is canceled by normal force). This force does not produce any torque about point X. So angular momentum of the rod-ball system about point X is conserved. If you picked some other point as the origin for calculating torques, then the torque about that point might not be zero. Then, angular momentum of the rod-ball system would not be conserved about that point.

For example, you might think about this for the case where you choose the origin for torques to be a fixed point on the table at the location of the initial position of the ball. Would the rod-ball system have angular momentum about this point before the collision? How about after the collision?

Bobby345

## 1. What is the difference between angular and linear momentum?

Angular momentum is a measure of the rotational motion of an object, while linear momentum is a measure of the translational motion of an object. In other words, angular momentum involves the spinning or turning of an object, while linear momentum involves the movement of an object in a straight line.

## 2. How are angular and linear momentum related?

Angular momentum and linear momentum are related through the concept of torque. Torque is the rotational equivalent of force, and it is responsible for changing an object's angular momentum. When a torque is applied to an object, it causes a change in both the object's angular and linear momentum.

## 3. What is the conservation of angular and linear momentum?

The conservation of angular and linear momentum is a fundamental law of physics that states that the total angular and linear momentum of a closed system remains constant over time, unless acted upon by an external force. This means that in a closed system, the total amount of angular and linear momentum before an event is equal to the total amount of angular and linear momentum after the event.

## 4. How do you solve problems involving angular and linear momentum?

To solve problems involving angular and linear momentum, you can use the equations for each type of momentum and apply the principle of conservation of momentum. It is important to carefully define your system and consider all forces acting on the system, including external forces and internal forces between objects within the system.

## 5. What are some real-life examples of angular and linear momentum?

Some examples of angular momentum include the spinning of a top, the rotation of a planet around the sun, and the rotation of a figure skater during a spin. Examples of linear momentum include the movement of a car on a highway, the flight of a baseball after being hit by a bat, and the motion of a person on a skateboard.

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