# Find density of a ball on a pendulum

by brahle
Tags: oscillation, pendulum
 P: 1 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

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