# Where does the kinetic energy go?

A uniform cylinder was placed on a frictionless bearing and set to rotate about its vertical axis. After a cylinder has reached a specific state of rotation it is heated without any mechanical support from temperature $T$ to $T+\Delta T$. Angular momentum will be conserved but rotational kinetic energy will decrease as angular velocity decreases. What I fail to understand is where does this energy go?

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Does the object expand as it is heated? –  BMS Jan 13 '14 at 13:21
@BMS I guess it does. That is how the moment of inertia would rise and the angular velocity decrease. –  Satwik Pasani Jan 13 '14 at 13:23
Interesting... what if the cylinder is made of a rubber-like substance which shrinks as it heats? Looks like you can reach certain conclusions about the material type from the statements given :-) –  Carl Witthoft Jan 13 '14 at 13:30

We know from conservation of angular momentum that $I$$\omega$ = constant. So when the object is heated, the body expands leading to a change in moment of inertia. If the new M.O.I. is $I$2; then as $I$2 > $I$1 , automatically $\omega$1 < $\omega$2. Therefore the problem of decreasing rotational kinetic energy is solved, as work is done to expand the material.

Note: Expansion takes place partly from the heat supplied too. As values are not given we cannot calculate for sure.

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You cant say that, the expansion may as well have been done due to the provided heat which is increasing the temperature –  Rijul Gupta Jan 13 '14 at 14:19
Then again the problem appears doesnt it? If the energy doesnt go into doing some work, where does it go ultimately? I am making an edit, see if that is okay. –  Sagnik Jan 13 '14 at 14:22
same thing @rijulgupta. –  Sagnik Jan 13 '14 at 14:26
How would you explain change in KE when a person folds his arms while spinning/rotating ? Nothing expands or compresses there ! –  Rijul Gupta Jan 13 '14 at 14:38
See here an external heat source is applied isnt it? in a ballet dancer's case, no external source is involved. –  Sagnik Jan 13 '14 at 14:39

This seems to be applicable to the systems in which $I$ is changed through other means such as a dancing balle dancer, I.e.

$I_1 \omega_1 = I_2 \omega_2$
$KE_2 = (1/2) I_2{\omega_2}^2$
$KE_2 = (1/2) I_1{\omega_1}^2 (I_1/I_2)$
$KE_2 = KE_1 (I_1/I_2)$

There is a clear dependence with change in moment of inertia, what i noticed is that when we reduce or increase the moment of inertia, paritcles move close or away from the axis or rotation i.e. movement in the direction of centripetal force According to me there is work done by/against centripetal force when change in $I$ takes place this work is where the $KE$ is going or coming from.

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Imagine you are pushing on the side of an enormous aluminum cube, with some constant force F=20N. At first, the wall of the cube doesn't budge and of course no work is done. But then someone turns on the air conditioning and the cube cools to a lower temperature. The width of the cube contracts by, say, 10cm as you continue to apply a constant force. Did you do any work on the cube?

Now imagine the cylinder... and let's suppose it's also metal. There is a radial centripetal force "pushing" on each molecule of the cylinder. As the heat increases, the average distance between molecules is increased, and the cylinder expands. Is work done on the molecules as they expand outward?

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