# Period of a pendulum with two masses.

1. Mar 29, 2015

### 1bigman

1. The problem statement, all variables and given/known data
We have a light rigid pendulum with length $l$. A mass $M$ is placed at the end and a mass $m$ is placed a distance $x$ from the pivot. What is the period of the pendulum?

2. Relevant equations

3. The attempt at a solution

Reduce the problem to a single mass situation using the center of mass. From the pivot, I get the center of mass $r = \frac{mx+Ml}{m+M}$

Using the small angle approximation we end up with $\frac{d^2\theta}{dt^2} + \frac{g(M+m)}{mx + Ml} \theta = 0$ and hence $T = 2\pi \sqrt{\frac{mx + Ml}{g(M+m)}}$

I suspect this is incorrect as the second part of the questions asks for the value of $x$ which has the greatest impact on $T$ which cannot be obtained as $\frac{dT}{dx} = 0$ has no solutions in $0<x<l$. Help is much appreciated.

2. Mar 29, 2015

### TSny

Problems like this are usually given after a discussion of the "physical pendulum" as opposed to the "simple pendulum". So, you might check your notes or text to see if a general formula for the period of a physical pendulum is given. Or see http://hyperphysics.phy-astr.gsu.edu/hbase/pendp.html

3. Mar 29, 2015

### Matternot

The 2 masses cannot be reduced to a single mass at the centre of mass. This is because the inertia of the system is dependant of the moment of inertia of the 2 masses (Which is different to that of 1 mass at the centre of mass).

If you haven't seen this before, take a look at this: http://hyperphysics.phy-astr.gsu.edu/hbase/pendp.html

If you are yet to visit moment of inertia, consider the potential and kinetic energy of the system (V+T)=E. By differentiating with respect to time and setting dE/dt to 0, the equation of motion can be obtained.

4. Mar 29, 2015

### 1bigman

Ah yes of course! Many thanks for the help guys!