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Is it possible to derive the time-period of a physical pendulum using forces instead of torque ? Normally when we encounter a physical pendulum like a oscillating ruler for example, we use concepts of moment of Inertia and torque to derive it's time period.

Is it at all possible, to calculate this time period by using forces, mass and distance from pivot, instead of torque and moment of inertia ?

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  • $\begingroup$ since this is rotational rather than translational motion, you’re bound to use torques either directly or in disguise. $\endgroup$ Commented Oct 31, 2021 at 23:11
  • $\begingroup$ calculate this time period by using forces, mass and distance from pivot” In other words, you’re still going to have to use torque. $\endgroup$
    – joseph h
    Commented Oct 31, 2021 at 23:24

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Yes, at least in the case of a simple pendulum: all the mass concentrated in a very small 'bob' at the end of a light string of length $l$.

If the maximum angular displacement from the vertical is small (say < 10°), we can make these approximations...

• the centripetal acceleration of the bob is always negligible,

$\cos \theta \approx 1$ so the tension in the string = the weight of the bob, that is $T=mg$,

Suppose that the pendulum is passing through an angular displacement $\theta$ from the vertical. Then the horizontal force on the bob is $$F_x=-T\sin\theta \approx -mg\sin\theta=-mg\frac xl$$ in which $x$ is the horizontal displacement of the bob from the equilibrium position.

So, using Newton's second law $$m\ddot x\approx-mg\frac xl\ \ \ \ \text{that is}\ \ \ \ \ddot x\approx -g\frac xl$$ So we have SHM of period $T=2\pi \sqrt \frac lg$.

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  • $\begingroup$ what about the compound pendulum ? Is it still possible ? $\endgroup$
    – RayPalmer
    Commented Nov 1, 2021 at 5:30
  • $\begingroup$ Yes, in principle. You treat each small part of the pendulum as a simple pendulum and then integrate, but this procedure forces you to 're-invent' the concepts of moment of inertia and torque. You should try it for yourself. $\endgroup$ Commented Nov 1, 2021 at 10:51
  • $\begingroup$ Ah, so that is why we use torque and moment of inertia - because it already encodes information about the shape of the body, and hence we can again treat the body by considering only the center of mass. Using force, we would have had to consider each point mass in the body. Is that correct ? $\endgroup$
    – RayPalmer
    Commented Nov 1, 2021 at 11:01
  • $\begingroup$ Yes, except that I'm a little uneasy about "we can again treat the body by considering only the center of mass". I suppose it's alright if you acknowledge that you've already "treated the body" by finding its moment of inertia about the relevant axis! $\endgroup$ Commented Nov 1, 2021 at 11:08
  • $\begingroup$ What I meant was, when we use the force approach, we usually care about the forces acting on the center of mass. We consider the entire body to be concentrated about the center of mass. Using moments, we are doing something analogous. Instead of caring about the point center of mass, here we care about the pivot point which is the axis of rotation. This is easier, as it already encodes all information about the shape of the body. Hence we use this approach over the force approach. Is this a better way of putting it ? $\endgroup$
    – RayPalmer
    Commented Nov 1, 2021 at 11:13

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