# Angular Momentum

1. Oct 14, 2008

### Kites

1. The problem statement, all variables and given/known data

Consider a particle of mass m and a particle of mass 2m. They are connected by a horizontal, massless, rigid rod of length 2a. The rod is fixed to a vertical stick that connects to the rod's midpoint. The stick is spinning with constant angular speed.

Consider a point P, located on the stick a distance d below the rod. Show that angular momentum of the two particle system, when taken about point P, is not conserved.

2. Relevant equations

L = r X p

theta = 90 degrees

3. The attempt at a solution

So what I've done is find angular momentum for either mass, separately.
Doing the following:

L_1 = R x P
= |r||p| sin(theta)
= (a^2+d^2)^(1/2)* ma(omega)

L_2 = R x P
= |r||p| sin(theta)
= (a^2+d^2)^(1/2)*2ma(omega)

What I want to do... by intuition... is just add these two together. but they're magnitudes not vectors... so i am a bit confused if I can.

If not, what the heck do I do?

Last edited: Oct 14, 2008
2. Oct 15, 2008

### tiny-tim

angular momentum is a vector

Hi Kites!

Angular momentum is a vector (strictly, a pseudovector), not a scalar …

it's the cross-product of two vectors!!

The moment form of good ol' Newton's second law is Net Torque = rate of change of angular momentum … and both sides of the equation are (pseudo-)vectors.

Most exam questions on angular momentum are two-dimensional, so the angular momentums are all perpendicular to the plane of the exam paper, and you can just add them!

But this question is three-dimensional … so you must treat them like the vectors they really are.

(btw, applying Newton's second law, can you see why it's not conserved? )