# Does a spinning part affect the moment of inertia of a composite object?

I have been going back through some Kleppner problems and have a doubt concerning problem 6.18. It states:

Find the period of a pendulum consisting of a disk of mass $M$ and radius $R$ fixed to the end of a rod of length $l$ and mass $m$. How does the period change if the disk is mounted to the rod by a frictionless bearing so that it is perfectly free to spin?

The first part (with the disk not free to spin) was reasonably straightforward. My only doubt was that I assumed the moments of inertia of the disk and rod could be added; this seems reasonable, but I don't quite know how to justify it rigorously (side-note: if anybody could give a hint on this, it would be highly appreciated). My result was:

$$T=2\pi\sqrt{\frac{MR^2/2+Ml^2+ml^2/3}{gl(M+m/2)}}$$

For the second part, the issue I had was mainly conceptual... So when the disk is free to spin, it's no longer part of the rigid body; so it won't contribute to the moment of inertia, right? The problem comes here: earlier, to calculate the torque on the rigid body about the pivot, I said that:

$$\tau=R_{CM}\times W$$

Where $R_{CM}$ is the center of mass of the rigid body. This formula comes from a summation over the torques on every small mass in the rigid body, so I figured that the torque on the rigid body, once the disk was no longer a part of it, would depend only on the center of mass of the rod (the disk would no longer affect the 'effective' center of mass that the torque acts on). This gives

$$T=2\pi\sqrt{\frac{2l}{3g}}$$

I checked my answer with this website (pages 5-7) afterwards, and the first part agreed but the second part was in disagreement; the problem was that in that site, $R_{CM}$ was still 'affected' by the spinning disk. Why is this so? (I explained above why I think that $R_{CM}$ should not contribute.)

• For adding the moments try the Parallel Axis Theorem. And you can't neglect eh intertia of a disc merely because it's frictionless, it still takes energy to spin.
– user21433
Commented Jul 13, 2014 at 3:45
• Will the disk actually begin to spin though? Intuitively, it seems like it will spin with respect to the plank but not with respect to an inertial observer. What force would be applying the torque on the disk necessary for it to spin about its CM? I'm neglecting the inertia of the disk not because it is frictionless, but essentially because it is no longer part of the rigid body; thus perhaps it no longer contributes to the body's angular momentum? Or maybe that whole approach is wrong and it will contribute just as much translational ang. momentum as before, but with added spin ang.momentum Commented Jul 13, 2014 at 3:58

You are very close.

Just to review what is going on, the period is given by

$$T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{I}{k_{eff}}}$$ where $I$ is the moment of inertia of the system and the torque is proportional to the angle by which the pendulum has been displaced with a coefficient that I'm calling $k_{eff}$ in analogy to Hooke's law: $$\tau = -k_{eff} \theta.$$ You have correctly identified that, when the disk is free to rotate, $I$ is simply the moment of interia of the rod $I=\frac{1}{3}ml^2$.

Let's compute the torque on the rod about the pivot. There are two contributions: one is the gravity acting directly on the rod, which (as you have correctly identified) gives a contribution $k_{eff,rod}=\frac{1}{2}mgl$.

However the disk also applies a torque to the rod. The disk is now essentially a weight hanging off the end of the rod, and so it contributes to the torque by adding to the effective Hooke's law constant: $k_{eff,disk}=Mgl$.

Thus $$k_{eff} = k_{eff,rod} + k_{eff,disk} = \frac{1}{2}mgl + Mgl$$

This resolves your issue.