# Homework Help: Angular Momentum of a beam problem

1. Dec 3, 2011

### vicTURBO

1. The problem statement, all variables and given/known data
Hey everyone I'm new here and I'm completely stuck on the logic behind this question.

A 230 kg beam 2.7m in length slides broadside down the ice with a speed of 18 m/s. A 65 kg man at rest grabs one end as it goes past and hangs on as both he and the beam go spinning down the ice. Assume friction less motion. With what angular velocity does the system rotate about its Center of Mass?

2. Relevant equations

L= Iw
L_initial = L_final
(I_rod + I_man )w = mvr

3. The attempt at a solution

So first I found the new center of mass when the man grabs the beam.
New CM = [m_beam(0) + m_man(2.7/2)]/(m_beam + m_man)

L_initial = L_final

The answer key uses this as their primary equation, with r_beam = the new center of mass of the man with the beam

(m_beam)(v_initial)(r_beam) = (I_rod + I_man) w

My question is why does the initial angular momentum include the new center of mass (with the man and the beam)? Shouldn't it only include variables of only the beam since the beam is moving by itself initially?

Thanks for your help

2. Dec 3, 2011

### BruceW

It is because angular momentum around the same axis is conserved. They have chosen the axis to be the centre of mass of the man and rod at the instant the man grabs hold of it. So therefore, the angular momentum of the system just before the man grabs hold must be with respect to the same axis.

Incidentally, you can choose any axis you want (as long as you use the same axis before and after). It just happens that the axis through the centre of mass makes the calculation simpler.

In your answer, the next thing you need to do is calculate the moment of inertia of the rod and man. So this will be with respect to the axis through the centre of mass.