# Angular Momentum Conundrum

1. Oct 3, 2009

### e2m2a

I have searched high and low on the internet for the correct equation to determine the angular velocity of a rigid body under the following circumstances, and have not been able to find one. Could someone show me the correct equation?
Suppose a long thin rigid rod lies on the surface of a cart, which initially is at rest. One end of the rod can rotate freely around a vertical axis attached to the cart. A small object prevents the rod from rotating clockwise. The cart can move parallel to the y-axis of an x-y coordinate system. The rod initially lies 45 degrees with respect to the x-axis. At some point in time we push the cart, giving the cart and rod an initial velocity equal to v. At a later point in time the cart makes an inelastic collision with a front bumper. Immediately, the rod rotates in the counter-clock wise direction. I need to know what the angular velocity of the rod is after the collision. We denote the mass of the rod as m, the mass of the cart as M, the initial angle of the rod as theta, the length of the rod as l, and the moment of inertia of the rod as:

I = 1/3 m l sq

I know we CANNOT use the below equation to determine the final angular velocity rod:

L = m v r cos theta

because this equation applies to point particles, but the rod is a solid, rigid body. (Assume for the sake of symmetry, there is an equal small hole at the other end of the rod so that the center of mass of the rod is at its geometric center.)
I have no idea what the correct equation is. Could someone please show me and also a general equation that applies for any shape of a rigid body.

2. Oct 3, 2009

### tiny-tim

Hi e2m2a!

(have a theta: θ and try using the X2 tag just above the Reply box )

Yes, use 1/3 ml2 for the angular momentum after the collision.

For the angular momentum before the collision (about the point where the pivot ends up), use dL = (m/l)dr vrcosθ, and integrate.

Last edited: Oct 3, 2009
3. Oct 3, 2009

### e2m2a

Hello Tiny-Tim,

In your reply you stated: "Yes, use 1/3ml² for the angular momentum after the collision."
Did you mean the moment of inertia, not the angular momentum? Also, I assume the equation for the angular momentum before the collision applies to after the collsion, correct? Also, I am not clear how to use your equation. I integrate the right side of:
dL = (m/l)dr v r cosθ with respect to r, treating everything else as a constant. Correct? Also, is r the distance from the axis or pivot point to the center of mass of the rod? Thanks for your help.

4. Oct 3, 2009

### tiny-tim

Hello e2m2a!

Yes, I meant use it for the moment of inertia to calculate the angular momentum.

r is the distance from the pivot point to the element dr of the rod.

5. Oct 3, 2009

### e2m2a

ok, if I integrate the equation dL = (m/l)vrcos(theta)dr, taking the limits of integration from 0 to l, I get as the final answer:

L = m v r cos(theta).

But this is the same result for a point particle. Is this correct?

6. Oct 4, 2009

### tiny-tim

(what happened to that θ i gave you? )

Noooo … that would be ∫ (m/l) vcosθ dr

get some sleep! :zzz:​

7. Oct 5, 2009

### e2m2a

Tiny-Tim,

Unless I am missing something on integration, when I integrate dL = ∫(m/l)vrcos(θ)dr, taking the limit from 0 to l, where l is the length of the rod, m is the mass of the rod, v is the velocity of the cart right at collision, and θ is the angle of the rod with respect to the x-axis at the time of collision, I get:

L = (m/l) v cos(θ) (r²/2)

Evaluated from the limits 0 to l, I get:

L =(l²/2) m (v/l) cos(θ)

Which reduces to:

L = (l/2) m v cos(θ)

Now, l/2 is simply the distance from the axis to the center of mass of the rod, or:

L = m v rcom cos(θ)

Where, rcom is the distance from the axis to the
center of mass of the rod.

8. Oct 6, 2009

### tiny-tim

oh, your r was rcom … i assumed it was l.