Issue about rotational and translational kinetic energy of a pendulum

Question

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 $+ m(\text{length of string})^2$?

Answer 1

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., $$KE_{ax}=\frac{1}{2}I_{cm}\omega^2$$

Answer 2

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.

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 $v = \ell \omega$.