Angular and Linear Momentum Problem

[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?

Homework Equations

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.

 

[TSny]

Welcome to PF!

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!

 

[haruspex]

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?

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?

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.

 

[TSny]

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?

 

[TSny]

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]

Good points.

Yep. Given in a diagram that X is on one end.
Correct. The tip of the rod strikes the ball.

 

[Bobby345]

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.

 

[haruspex]

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]

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

 

[TSny]

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]

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?

 

[TSny]

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.

 

[Bobby345]

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?

 

[TSny]

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]

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]

Even though the question asks why the angular momentum of the system is conserved?

Perhaps this has something to do with net torques?

 

[Bobby345]

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.

 

[TSny]

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?