# Gravitational torque on a uniform beam: Cavendish experiment (LAB)

1. Mar 5, 2009

### v8volvo

This is my first post here. I'm a senior in Applied Physics at Tufts Univ. Trying to work through a lab report and can't get this part figured out. The lab is a replication of the Cavendish Experiment.

The situation is as follows: there is an aluminum beam of known mass, width, and length, mounted on a pivot that can be assumed to be located at the center of mass. So the beam is free to rotate about its midpoint. For calculation purposes we're allowed to consider the beam as a uniform rod of mass M and length L.

So this uniform rod is, say, hanging, and two spheres of comparatively large mass are situated near its end, one on each side. Looks something like this (crude representation):

OOOO <---- one "sphere" (large point mass)
OOOO
OOOO

the rod:
------------------------------------------------------|-----------------------------------------------------
approximate pivot point (at midpoint of beam) ^
(axis of rotation perpendicular to, or coming out of, your screen)

The question is, what is the gravitational torque these masses exert on the beam? Since the rod can't be treated as two point masses at either end, as it's a uniform piece, some kind of integration will clearly be required. Formula for gravitational attraction is F = GMm/R^2. My brain just can't come up with the formula for this torque.

Thanks in advance for any help!

2. Mar 5, 2009

### lanedance

to set up the intgration ask yourself what is the gravitational force on an element dx of the rod?

The infinitesimal torque dT will be the the perpindicular component times the distance from the centre of rotation

so write dT in terms of dx & integrate over the beam

Last edited: Mar 5, 2009
3. Mar 6, 2009

### v8volvo

Thanks for the help. Figured it out. Turned out to be a pretty nasty equation to integrate, but got it done.

Thanks again!!