# Find density of a ball on a pendulum

1. Oct 18, 2010

### brahle

First of all, sorry for my bad English as it isn't my native language. Also, sorry for poor formating.

1. The problem statement, all variables and given/known data
A pendulum consists of a rod of length $$L=1 m$$ and mass $$m_1=0.5 kg$$ and a ball (radius $$r=5 cm$$) attached to one end of the rod. The axis of oscillation is perpendicular to the rod, and goes through the other end of the rod. The period of oscillation is $$T=2s$$. Find the density $$\rho$$of the ball.

2. Relevant equations
$$T=2 \pi \sqrt{\frac{I}{mgx}}$$
$$m= \rho V= \rho \frac{4}{3}r^3 \pi$$

3. The attempt at a solution
For this pendulum, the period is:
$$T=2 \pi \sqrt{\frac{I}{mgx}}$$
where $$x$$ is the distance between the axis of oscillation and the center of mass and $$I$$ is the moment of inertia.

So, according to parallel axis theorem, the moment of inertia for the rod $$I_{rod}$$ is equal to:
$$I_{rod}=\frac{1}{3}m_1 L^2 + m_1 (\frac{1}{2}L - x)^2$$
Similarly, the moment of inertia of the ball $$I_{ball}$$ is equal to:
$$I_{ball}=\frac{2}{5}m_2 r^2 + m_2 (L - x + r)^2$$
Considering that $$I=I_{rod} + I_{ball}$$, we are left with two unknowns - $$x$$ and $$\rho$$. That means we need another equation, and that's the part that isn't clear to me.

I suspect it has to do something with the fact that this is pretty much a one-dimensional problem and certain properties of center of mass. Unfortunately, I do not know what to do next.

Thanks a lot,
Bruno