Small oscillation frequency of rod and disk pendulum

[Jenny Physics]

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.

 

[TSny]

Your result looks OK to me. In what way does your answer differ from the solution that you were given?

It looks like you have a typographical error where you left out a factor of $\theta$ on the right side of

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

But this doesn't affect your result.

 

[Jenny Physics]

Your result looks OK to me. In what way does your answer differ from the solution that you were given?

The solution given is

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

 

[TSny]

To see which answer cannot be correct, try a special case where you already know the answer. For example, suppose you transform the pendulum into a simple pendulum by letting the mass of the rod equal zero and also letting the radius of the disk equal zero. So, you would have a point mass m at the end of a massless rod of length L.

 

[Jenny Physics]

To see which answer cannot be correct, try a special case where you already know the answer. For example, suppose you transform the pendulum into a simple pendulum by letting the mass of the rod equal zero and also letting the radius of the disk equal zero. So, you would have a point mass m at the end of a massless rod of length L.

In that case my solution would give $\omega=\sqrt{g/L}$ while the official solution would give $\omega=0$ so my solution would seem to win out.

 

[TSny]

In that case my solution would give $\omega=\sqrt{g/L}$ while the official solution would give $\omega=0$ so my solution would seem to win out.

Absolutely.