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In the last few minutes of Walter Lewin's lecture no.13, Lewin measures the oscillation period of a cart sliding on a bent low-friction track and gets impressive agreement with his prediction. The track seems to be bent into a circular arc with some radius $R$, so its equation of motion is the same as the motion of a simple pendulum.

He then repeats the experiment for a ball bearing rolling on a similarly bent track, and the prediction fails:

$$\begin{align}T\textrm{(prediction)} &= 2\pi \sqrt{\frac{R}{g}} = 1.85 ~\mathrm s\\ T\textrm{(observed)} & = 2.27~\mathrm s\end{align}$$

Why is the observed time period of ball bearing not matching with the predicted time period?

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The difference is that the ball bearing is rolling on a circular track instead of sliding. When analyzing the motion we must take into account the moment of inertia of the ball bearing.

In a simple pendulum the bob rotates about the pivot point. This is equivalent to sliding (without friction) on the circular track. But the ball-bearing pendulum is rotating about its own centre as it also rotates about the imaginary pivot point at the centre of the circular track.

For a certain amplitude of swing, there is a fixed amount of energy. At the lowest point the kinetic energy of the ball bearing is divided between linear motion of the CM and rotation about the CM, whereas for the simple pendulum bob it all goes into linear motion of the CM. So the linear speed of the CM of the ball bearing is slower than that of the simple pendulum bob, resulting in a longer period for the ball bearing. When we come to measure the period, it is only the back-and-forth motion of the CM which we measure; we ignore the rotation of the ball bearing.

According to Oscillation of a rolling sphere in a bowl, the period of small oscillations is
$$T=2\pi \sqrt{\frac{7R}{5g}}=1.1832T_0$$ where $T_0$ is the period of a simple pendulum of the same length $R$. In the lecture $T_0=1.85s$ so we should expect $T=2.19s$. We can only assume the remaining difference with the measured value of $2.27s$ is due to other errors - eg the measurement of $R$.

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    $\begingroup$ When you roll a bearing on a track the ratio $v:\omega$ for the bearing is not $r$ (the radius of the bearing) but a smaller value that depends on the relative separation of the rails and size of the bearing. That means the coefficient applied to the rotational energy is larger than 2/5 and the expected time is still longer. See physics.stackexchange.com/a/119585/520 $\endgroup$ – dmckee --- ex-moderator kitten Nov 4 '16 at 0:01
  • $\begingroup$ @dmckee : Thank you. I didn't think of that. $\endgroup$ – sammy gerbil Nov 4 '16 at 1:54

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