# Angular Momentum Term Equals Zero?

1. Sep 6, 2014

### KleZMeR

Hi All,

This is from a classical mechanics problem, and I already 'solved' the problem, but I'm interested in why a certain term is set to zero. I think I understand the concept but just want to clarify.

The problem is a table with a hole in it and two masses on a string, one mass is hanging through the hole with only a Z component, and the other is on the table with an X and Y component (Z plane).

When I take the cross product of R x mV, I get the angular momentum vector K which has only a 'vertical' component:
R x mV = [m*(r^2)*dθ + m*r*dr*sin(2θ)] K

But I am told that:
R x mV = m*(r^2)*dθ K

The sin(2θ) came from some trig identity work. So I am wondering is this because there is no effect on the K vector from a sin(2θ) factor which is only in the Z plane? Why is this term 0? Is dr = 0 ? I think r is fixed but the problem does say that gravity affects the hanging mass, so perhaps dr in the Z plane is not zero? Any help understanding this is appreciated.

2. Sep 6, 2014

### voko

There is an error in your derivation of equation for angular momentum. Without seeing the derivation, I cannot say what this error is. A simple, if somewhat cumbersome/ way to obtain the angular momentum in polar coordinates is by writing $x = r \cos \theta, \ y = r \sin \theta$, then writing $\dot x = ..., \ \dot y = ...$ and taking their cross product.

3. Sep 6, 2014

### KleZMeR

Here is my attempt, I uploaded it

File size:
67.2 KB
Views:
94
4. Sep 6, 2014

### voko

So you have $mr [\dot r \cos \theta \sin \theta + ... - \dot r \sin \theta \cos \theta + ... ]$ yet you write $= mr [ ... + 2\dot r \cos \theta \sin \theta]$.