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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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