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Consider a rigid body suspended about an axis of rotation which, in general does not pass through it's center of mass (COM) and has a moment of inertia (MOI) $0 < I_{axis}$ about that axis. Let $I_C$ denote the moment of inertial of the object about an axis parallel to the one mentioned before and passing through the COM. The parallel axis theorem implies that $I_{axis} = I_C + ml^2$ where $0 \leq l$ is the distance between the two axes. From the rotational dynamics of the object $I_{axis} \ddot{\theta} = -mgl\theta$, where $0 \leq \theta \rightarrow 0$ is the (small) angular displacement of the object, we can fine the time period of oscillation as $T = \frac{2\pi}{\sqrt{\frac{mgl}{I_{axis}}}}$ so that $T \propto \sqrt{\frac{I_{axis}}{ml}}$.

In case of the simple pendulum, $I_C \approx 0$ so that $I_{axis} = ml^2$, and thus $T \propto \sqrt{l}$ leading to the usual conclusion (matching our intuitive physical understanding) that the time-period of small oscillations increases as the length $l$ increases. However, in the case of general rigid bodies, i.e. not point-masses, the algebra results in $T \propto \sqrt{\frac{I_C}{ml} + l}$. From the plot one can see that this expression explains the apparently counter-intuitive observation that for general rigid bodies (i.e. not point-masses), for small $l$, the time-period of small oscillations reduce as the length $l$ increases, in a mathematical sense.

In the case of a simple pendulum, it is physically intuitive that the time-period should increase with the increase in $l$ (distance traveled over an oscillation is increasing linearly with $l$, while the motivating force remains roughly of the same magnitude regardless of $l$). In the same sense what is the intuitive physical explanation of this apparently counter-intuitive behavior?

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Consider a pendulum that consists of 2 points of mass $m$ connected by a massless rigid rod of length $h$. Suspend it at the center of mass. It doesn't oscillate.

Suspend it a small distance, $\delta x$, above the center of mass. Turn it 90 degrees. It oscillates slowly. The moment of intertial, $I$, is almost the same as if it was suspended at the center. The torque is $\tau = mg \delta x$. The angular acceleration is $\alpha = \tau / I$.

Suspend it $2 \delta x$ above the center of mass. $I$ is still almost the same. $\tau$ has doubled, and so has $\alpha$.

Consider other properties of the two cases.

  • The distance the center of mass travels in a half oscillation has doubled from $\pi \delta x$ to $\pi 2\delta x$.
  • The potential energy decrease when rotated upright has doubled from $mg \delta x$ to $mg 2\delta x$. So has the maximum rotational kinetic energy.
  • The maximum angular velocity, $\omega$ has quadrupled.

These ratios hold at each angle during the half oscillations of each case. $\omega$ is quadrupled over a trajectory that is twice as long. The period has halved.

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  • $\begingroup$ The direction of the explanation seems great to me. Please allow me to point out that the potential energy is associated with the factor $(\delta x)^2$ (thus introducing $\theta^2$ associated with the spring-mass analogy of the harmonic-oscillator with spring-constant $mgl$ considered here). Further, the answer does not address the intuition behind this apparent counter-intuitive behavior only in the case of low absolute length $l$ (rather than the change in length $\delta x$ in your qualitative analysis). $\endgroup$
    – kbakshi314
    Mar 7 '21 at 0:48
  • $\begingroup$ $U = mgh$, where $h$ is the change in height of the COM. In this case, $h = \delta x$ or $2 \delta x$. There is no $(\delta x)^2$ dependency. The period only decreases in special cases like when the support is very near the COM. For a usual pendulum, increasing the length a little would cause an increase in period. This is true for a Grandfather clock pendulum with a big block of brass as much as an ideal pendulum with a point mass. $\endgroup$
    – mmesser314
    Mar 7 '21 at 1:01
  • $\begingroup$ Agreed. It seems I misunderstood the variable $\delta x$. However the answer does not address the apparently counter-intuitive of behavior of decreasing time-period with increasing $l$ and why it happens only in the case of low absolute length $l$. $\endgroup$
    – kbakshi314
    Mar 7 '21 at 1:08
  • $\begingroup$ I am not sure how much I can add to this. It is pretty much all the answer I have. It happens for low absolute $l = \delta x$ because $I$ doesn't change much and $\tau$ does. At bigger $l$, both $I$ and $\tau$ change. $\endgroup$
    – mmesser314
    Mar 7 '21 at 1:16
  • $\begingroup$ Ok, thanks for the very subtle and nice answer! $\endgroup$
    – kbakshi314
    Mar 7 '21 at 1:26
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In the case of a simple pendulum, it is physically intuitive that the time-period should increase with the increase in l (distance traveled over an oscillation is increasing linearly with l). In the same sense what is the intuitive physical explanation of this apparently counter-intuitive behavior?

In my opinion, the behaviour of a simple pendulum is not so intuitive. I guess if a common visitor of the Paris Pantheon is asked about the period of the Foucault's pendulum, it would be no surprise if some answers put the mass as a variable for example, and miss the role of the length.

But, we can have an "educated" intuition of the rigid body oscillations: as the moment of inertia is $\alpha mr^2$, where $\alpha$ is some constant, increasing $l$ means decreasing the length parameter ($\frac{r^2}{l}$) inside the square root. The period is proportional to the square root of the "length" so to speak, if l is small compared to the other term. And the mass cancels.

So its behavior is intuitive in this meaning.

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  • $\begingroup$ Indeed, it is non-trivial at-first-glance that the time-period is independent of the mass. Thanks for the answer since it clarifies the basis of the posed question. However it does not address the apparently counter-intuitive of behavior of decreasing time-period with increasing $l$ and why it happens only in the case of low absolute length $l$. I understand why this happens mathematically, but I think there must be an intuitive explanation to it. $\endgroup$
    – kbakshi314
    Mar 7 '21 at 1:06

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