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Let’s say we have a pendulum that consist of a light string hanging a disk-like object. It is allowed to undergo simple harmonic motion with small oscillations.

My question: Is the energy of the disk pendulum at anytime written as

  • (a) $$E_\text{total}= \frac12mv^2 + mgh + \frac12Iω^2,$$ where $v$ is tangential velocity of the center of mass of the pendulum and $I$ is the moment of inertia of disk $+ m(\text{length of string})^2,$ or

  • (b) $$E_\text{total}= mgh + \frac12Iω^2$$ and $I$ is the moment of inertia of disk $+ m(\text{length of string})^2$?

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  • $\begingroup$ Is the system in the question rotating or it is only oscillating like a normal pendulum? $\endgroup$ Aug 20 '21 at 6:20
  • $\begingroup$ Just a disk shape pendulum $\endgroup$
    – Eugene
    Aug 20 '21 at 6:58
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Using the parallel axis theorem, a disk pendulum that is allowed to rotate around a fixed point with a distance $l$ from it's center, has total rotational kinetic energy $$E= \frac{1}{2}\omega^2(I_{cm}+ml^2)=\frac{1}{2}I_{cm}\omega^2 +\frac{1}{2}m\omega^2l^2$$ and since we can write the angular velocity in terms of the tangential velocity, i.e., $$\omega=\frac{v}{l}$$ where $v$ is the linear (tangential) velocity of the center of mass of the disk, then the above equation for the kinetic energy can simply be written $$E= \frac{1}{2}I_{cm}\omega^2 +\frac{1}{2}mv^2$$ and the disk has a moment of inertia $$I_{cm}=\frac{mR^2}{2}$$ which is the moment of inertia of thin disk (similar to a cylinder of small height) of radius $R$.

This means we can write its total energy as $$T=\frac{1}{2}I_{cm}\omega^2 +\frac{1}{2}mv^2+mgh$$

So in your above question, option (a) is the correct answer since option (b) ignores this second kinetic energy term altogether. It's also important to note that the second term in this expression has the velocity of the center of mass, which again can be expressed in terms of the angular velocity that the disc has around the fixed point i.e., $$KE_{p}=\frac{1}{2}m\omega^2l^2$$ and to note that the first term is the kinetic energy the disc has about its own axis. i.e., $$KE_{ax}=\frac{1}{2}I_{cm}\omega^2$$

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    $\begingroup$ Your first expression can be written as $E=\frac12(I+ml^2)\omega^2$ which is clearly the rotational kinetic energy about fixed point. Am I correct? $\endgroup$
    – ACB
    Aug 20 '21 at 7:15
  • $\begingroup$ Then why we talk about translational kinetic energy according to the last paragarph? $\endgroup$
    – ACB
    Aug 20 '21 at 7:22
  • $\begingroup$ The angular velocity and linear velocity are related by v=ωL We need to talk about both rotational and translational kinetic energy, since this disc has both. $\endgroup$
    – joseph h
    Aug 20 '21 at 7:31
  • $\begingroup$ Does that mean my expression in the comment above contains both rotational and translational kinetic energy? $\endgroup$
    – ACB
    Aug 20 '21 at 7:33
  • $\begingroup$ Yes it does. The "part (a) answer" of your question has a term for linear kinetic energy in terms of $v$, but this can also be expressed in terms of rotational kinetic energy (in terms of $\omega$) since the disk center of mass is translating and the disc is also rotating about cm. Cheers. $\endgroup$
    – joseph h
    Aug 20 '21 at 7:40
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The kinetic energy of a rigid body is invariant (does not change) with the location where it is measured if the parallel axis theorem is used to transfer mass moment of inertia from point to point.

fig1

For your example, consider the following two locations

  1. About the center of mass the object has mass moment of inertia of $I$ and kinetic energy $$ K = \tfrac{1}{2} m v^2 + \tfrac{1}{2} I \omega^2 $$
  2. About the pivot point the system has mass moment of inertia of $I+m \ell^2$ and kinetic energy $$\require{cancel} K = \cancel{ \tfrac{1}{2}m 0^2} + \tfrac{1}{2} (I + m \ell^2) \omega^2$$

Both of the above calculations yield the same result, as you can prove yourself by using the kinematic relationship $v = \ell \omega$.

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  • $\begingroup$ What does $\require{cancel} \cancel{ \tfrac{1}{2}m 0^2}$ represent? Is that translational KE of pivot point? $\endgroup$
    – ACB
    Aug 20 '21 at 15:26
  • $\begingroup$ @ACB - yes correct. It reflects the fact that speed is 0 for the pivot. $\endgroup$ Aug 20 '21 at 23:34
  • $\begingroup$ But translational KE depends on the motion of centre of mass. So this seems odd to me. $\endgroup$
    – ACB
    Aug 21 '21 at 4:12
  • $\begingroup$ @ACB - I am demonstrating that you can pick other points besides the CM and you will get the same answer as long as the MMOI is adjusted. $\endgroup$ Aug 21 '21 at 14:25
  • $\begingroup$ @ACB - As far as a rigid body is concerned there is one kinetic energy. The artificial separation of rotating and translating components isn't really meaningful in terms of thinking of a body consisting of many particles that move together. It is all translational KE in that sense. $\endgroup$ Aug 21 '21 at 14:53

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