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## Homework Statement

Consider a rod of length ##L## and mass ##M## attached on one end to the ceiling and on the other end to the edge of a disk of radius ##r## and mass ##m##. This system is slightly moved away from the vertical and let go. Let ##\theta## be the angle the pendulum makes with the vertical. What is the frequency ##\omega## of small oscillations of this pendulum?

## Homework Equations

Newton's rotation law gives

$$\vec{\tau}=I\ddot{\theta}$$

where ##I## is the moment of inertia of the rod and disk relative to the pivot in the ceiling and ##\vec{\tau}## is the torque due to the gravity force.

## The Attempt at a Solution

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The torque due to gravity around the pivot ##P## divides into the torque of gravity at the center of mass of the rod (located at ##L/2##) plus the torque of the gravity force at the center of mass of the disk (located at ##L+r##). This is

$$\vec{\tau}=\vec{r}_{P,M}\times M\vec{g}+\vec{R}_{P,cm}\times m\vec{g}$$

This ends up giving

$$\vec{\tau}=-\left[M\frac{L}{2} +m(L+r)\right]g\sin\theta\hat{z}$$

So the equation of motion $$\vec{\tau}=I\ddot{\theta}$$ becomes

$$-\left[M\frac{L}{2} +m(L+r)\right]g\sin\theta=I\ddot{\theta}$$

If we make this an harmonic type equation where ##\ddot{\theta}=-\omega^{2}\theta,\sin\theta\approx \theta ## it becomes ##-\left[M\frac{L}{2} +m(L+r)\right]g\theta=-I\omega^{2}\theta##

and so

$$\omega=\sqrt{\frac{\left[M\frac{L}{2} +m(L+r)\right]g}{I}}$$

This is not the solution for the frequency of oscillation, which means I am missing something perhaps in the center of mass calculation.

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