Situation: We suppose a rod is attached by its end to some pivot, and is allowed to fall from a horizontal position. As this occurs, the rod loses gravitational potential energy ($∆U_G$), say 10J, thus gaining the same amount in kinetic energy. The question provides three options, p.t. the form of energy to which this $∆U_G$ is converted:

  1. $∆U_G = K_{cm}$
  2. $∆U_G = K_{rot}$
  3. $∆U_G = K_{cm} + K_{rot}$

My professor selected, (2), giving the following justification:

  • $∆K_i = 1/2∆m_iv_i^2$
  • $∆K_i = 1/2∆m_i(r_iw)^2$
  • $∆K_i = 1/2(∆m_ir_i^2)w^2$
  • $∆K = 1/2(\sum∆m_ir_i^2)w^2$
  • $∆K = 1/2Iw^2$

I do not understand this result, because centre of mass, surely, gains kinetic energy due to loss in gravitational potential, as well?

  • $\begingroup$ your teacher is right only if K_rot is relative to an axis on the pivot $\endgroup$ – Wolphram jonny Jan 31 at 0:07
  • $\begingroup$ Is this because $K_{cm} + K_{rot}$ constitutes a double-count? $\endgroup$ – Julia Kim Jan 31 at 0:29

If you consider an axis around the pivot, the total velocities of each dm is being considered, and thus that is the total kinetic energy. If you consider an axis around the center of mass, then the total velocity of each dm is being decomposed into a component around the CM and another that is the velocity of the CM itself. That is why in the last case you end up with two terms, and in the first case with only one. Notice than in the two cases the I is different, that is why both ways to calculate the total K energy it give the same answer.

  • $\begingroup$ I wonder what makes the CM so special, that we are able to decompose its energy like such ($𝐾=𝐾𝑐𝑚+𝐾𝑟𝑜𝑡-𝑎𝑏𝑡-𝑐𝑚$). Why is it that we cannot do this for any other particle in the system? Moreover, by your response, it seems as if I am able to 'mentally' shift the axis of rotation; even though the rod is rotating about its end, I can nonetheless imagine that the axis passes through centre of mass do to this decomposition. This seems odd... is this correct? $\endgroup$ – Julia Kim Feb 1 at 15:06
  • $\begingroup$ ... or is it (in the second question) that we are simply shifting the reference point (and not the axis)? $\endgroup$ – Julia Kim Feb 1 at 15:12
  • $\begingroup$ the center of mass is special , but actually you can decompose the energy around any axis which is not the axis of pure rotation. You can always put your axis anywehere, the issue is that for any axis' place other than the pivot, you will have rotation plus traslation. $\endgroup$ – Wolphram jonny Feb 1 at 15:58

The kinetic energy of the center of mass along with the kinetic energy of the rest of the rod is included in the final energy your professor computed in the equation

$\text{$\Delta $K}_i=\frac{1}{2} \Delta\ m_i v_i^2$

but if you use the mass of the rod and the velocity of the center of mass to compute the energy change, you will get a wrong answer. If we assume a uniform rod the velocity of the center of mass is $L/2\ \omega$ where L is the rod's length.

$\frac{1}{2}{m\ v^2}=\frac{1}{8}m\ L^2 \omega ^2$

while if we integrate $\frac{1}{2}{m\ v^2}$ over the length of the bar

$\frac{1}{2}{\int v^2 \, dm}=\frac{1}{2} \rho \int_0^L (l\ \omega )^2 \, dl=\frac{1}{6}m\ L^2 \omega ^2$

where $dm = \rho\ dl$ and $\rho=m/L$ and the final answer is $\frac{1}{2}I\ \omega^2$. The center of mass is just one point on the rod with negligible mass, whereas the entire rod is rotating, so I believe rotational kinetic energy is the right way to think of this system.


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