# Angular Momentum Collision Problem

## Homework Statement

A uniform bar of Length L and Mass M is hit by a blob of putty moving with velocity v. The blob hits the bar at a distance d from the center of the bar and sticks to the bar at the point of contact.

Obtain expressions for the velocity of the systems center of mass and for the angular speed of the following collision.

p = mv
I = cMr^2
L = Iw

## The Attempt at a Solution

So to solve for the velocity of the systems center of mass I just used conservation of linear momentum.

mv1 = (M+m)v2
(mv1) / (M+m) = v2

I'm having problems understanding the solution for finding the angular velocity about the COM.
So first it finds the COM of the system.

y|COM| = (md) / (M+m)

Solving for the angular momentum about the COM gives
L = mv(d - y|COM|) = (Mmvd) / (M+m)

Okay now here is where I don't understand the equations...
It uses the parallel axis theorem to find the moment of Inertia of the system about it's center.

I = (1/12)MR^2 + My|COM|^2 + m(d - y|COM|)^2

So there are three terms in this...
(1/12)MR^2 represents the moment of Inertia of JUST the pole... cool

m(d-y|COM|)^2 represents the moment of Interia the putty adds from being attached to the stick?

My|COM|^2 represents the moment of Inertia of the stick adjusted for the fact the putty is disrupting it's COM? This is term that confuses me. Why do you have to add this? Don't you just take the moment of inertia of the pole and then add the putties moment of inertia? You have to add in a term for the poles moment of Inertia because the putty throws off it's center of mass? I looked at it like the pole retains the same center of mass so you don't need this term you just need to add a term for the putty.. Idk anyone clarify? Sry for **** notation

Also: The problem gives the Angular Momentum as
L = mv(d-y|COM|), how come they only use the mass of the putty in this equation to determine the angular momentum? Wouldn't they also have to account for the mass of the pole that needs to be taken in account due to the fact that the center of mass of the system has shifted away from the center of the mass of just the pole by itself?

I'm having problems understanding what these formulas mean I guess X_x. If anyone wants to be a complete boss and break them down for me that would be superb and I would be eternally in your debt.

Last edited:

gneill
Mentor

## Homework Statement

A uniform bar of Length L and Mass M is hit by a blob of putty moving with velocity v. The blob hits the bar at a distance d from the center of the bar and sticks to the bar at the point of contact.

Obtain expressions for the velocity of the systems center of mass and for the angular speed of the following collision.

p = mv
I = cMr^2
L = Iw

## The Attempt at a Solution

So to solve for the velocity of the systems center of mass I just used conservation of linear momentum.

mv1 = (M+m)v2
(mv1) / (M+m) = v2

I'm having problems understanding the solution for finding the angular velocity about the COM.
So first it finds the COM of the system.

y|COM| = (md) / (M+m)

Solving for the angular momentum about the COM gives
L = mv(d - y|COM|) = (Mmvd) / (M+m)

Okay now here is where I don't understand the equations...
It uses the parallel axis theorem to find the moment of Inertia of the system about it's center.

I = (1/12)MR^2 + My|COM|^2 + m(d - y|COM|)^2

So there are three terms in this...
(1/12)MR^2 represents the moment of Inertia of JUST the pole... cool
Actually it represents the moment of inertia of the pole about its own center. But when the putty will be added there will be a new center of rotation that is shifted away from the pole's center of mass. That's why the parallel axis theorem is applied, to find the moment of inertia of the pole about that new center of rotation.

m(d-y|COM|)^2 represents the moment of Interia the putty adds from being attached to the stick?
Yes. It's the moment of inertia of the putty about the new center of rotation.

My|COM|^2 represents the moment of Inertia of the stick adjusted for the fact the putty is disrupting it's COM? This is term that confuses me. Why do you have to add this? Don't you just take the moment of inertia of the pole and then add the putties moment of inertia? You have to add in a term for the poles moment of Inertia because the putty throws off it's center of mass? I looked at it like the pole retains the same center of mass so you don't need this term you just need to add a term for the putty.. Idk anyone clarify? Sry for **** notation
My|COM|^2 represents the adjustment to the moment of inertia of the rod to account for it being rotated a point other than its center of mass. That's the point of the Parallel Axis Theorem, it allows you to find the moment of inertia for a body about a point other than the one specified in the basic formula for the object.

Also: The problem gives the Angular Momentum as
L = mv(d-y|COM|), how come they only use the mass of the putty in this equation to determine the angular momentum? Wouldn't they also have to account for the mass of the pole that needs to be taken in account due to the fact that the center of mass of the system has shifted away from the center of the mass of just the pole by itself?
Initially the pole is stationary. It's not rotating, so it contributes zero angular momentum. Only the moving putty has angular momentum with respect to the (future) center of rotation of the rod+putty. So taken as a whole, the initial angular momentum for the system is due to the putty alone.

• 1 person
so if the moment of inertia of just the pole is 1/12*MR^2, and the Moment of Inertia of the pole about a point y away from the COM is 1/2*MR^2 + My^2.. an object has a higher moment of Inertia when it's not rotated about it's COM?

gneill
Mentor
so if the moment of inertia of just the pole is 1/12*MR^2, and the Moment of Inertia of the pole about a point y away from the COM is 1/2*MR^2 + My^2.. an object has a higher moment of Inertia when it's not rotated about it's COM?

Sure. Think about the effort required to spin a rod around its middle versus rotating it around its end; in the latter case more of the rod's mass is located further from the center of rotation.