Possible Duplicate:
A pendulum clock problem

There are two pendulums.

First pendulum consists of a rod of length L and flat heavy disk of radius R (R < L), disk is connected rigidly to the rod such that the plane of the disk is vertical. Mass of the rod is negligible compared to mass of the disk. Center of the disk is at distance L from the upper point of the rod.

Second pendulum has same rod of length L and same disk, but the disk is not connected rigidly to the rod. Disk can rotate freely around its center at which it is attached to the rod at distance L from the upper end of the rod. Disk remains in vertical plane, as if there is ball bearing atached to the rod.

Which pendulum has smaller period at small deviations, if their periods are different.


marked as duplicate by Mark Eichenlaub, Qmechanic, David Z Apr 14 '12 at 21:48

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Indeed, but the answer in the duplicate question is in my opinion too complex. No offense. $\endgroup$ – Pygmalion Apr 14 '12 at 20:02
  • 1
    $\begingroup$ I think that the link posted by Mark is an exact duplicate of this, and the answer there is quite clean. $\endgroup$ – Vijay Murthy Apr 14 '12 at 20:12
  • $\begingroup$ Well, never mind, I guess I don't fit into the club. It was still fun answering the questions, and very instructive for myself too. Even seemingly stupid questions force you to reassess your own knowledge or improve lectures. Greets! $\endgroup$ – Pygmalion Apr 14 '12 at 20:32
  • $\begingroup$ @Pygmalion Don't confuse simple and short. $\endgroup$ – Mark Eichenlaub Apr 14 '12 at 21:09
  • $\begingroup$ @Pygmalion that's a perfect opportunity to post your own, less complex answer to the other question. If others agree with your opinion, there should be plenty of upvotes to be had. $\endgroup$ – David Z Apr 14 '12 at 21:49

In the case of the first pendulum, disk has to rotate and this is classical physical pendulum. In the case of the second pendulum, you don't expect disk to rotate (the only force is in the center of mass, so there is no torque), so therefore you simply have only transitional movement - mathematical pendulum.

Using equation for periods of physical pendulum

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

and noting, that

$$I = I_\textrm{cm} + m l^2,$$

where $I_\textrm{cm} > 0$ is center-of-mass inertia of the disk, you get

$$T = 2 \pi \sqrt{\frac{l}{g} + \frac{I_\text{cm}}{m g l}}.$$

On the other hand equation for periods of mathematical pendulum

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

Obviously, period is smaller for the second pendulum.


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