This question came up in a discussion section today. Some fellow students and I broke our heads on it for a while but couldn't come to a satisfactory conclusion.

Imagine you have a thin disc of mass $M$, radius $R$ and uniform density that's rotating about its symmetry axis with a constant angular velocity $\omega$. Ignore friction. You immerse the disc into a heat bath that's $\Delta T$ degrees warmer than the disc, which causes the size of the disc to expand with some coefficient $\alpha > 0$:

$\Delta R = \alpha\,\Delta T\,R$ and hence to first order in $\alpha \Delta T$

$\Delta R^2 \approx 2\alpha\, \Delta T\,R^2$

This clearly increases the moment of inertia $I = \frac{1}{2}MR^2$ of the disc.

The question is, what now happens to the angular momentum $L$ and the energy $E$ of the disc? If you assume $L$ is conserved since (as far as I can see) there is no external torque working on the system, then $\omega$ will decrease by the same factor $I$ increases by; but this decreases the overall rotational energy $E = I\omega^2/2$. This seems counterintuitive - energy somehow needs to escape the disc, but it cannot flow into the heat bath since the latter is at equal or higher temperature than the disc. If you assume that $L$ is not conserved, then does it increase or decrease and where does the corresponding torque come from or go to?

  • $\begingroup$ Would you like me to provide a detailed analysis of the stresses, strains, displacements, and deformation of the disk so that the problem can be resolved definitively? $\endgroup$ Aug 30, 2016 at 22:52
  • $\begingroup$ Well if you're offering then sure, I wouldn't say no to that :) $\endgroup$
    – user28400
    Aug 31, 2016 at 0:38
  • $\begingroup$ Question to the community: I've upvoted all the answers because they've all shed light on one or more aspects of this problem. I don't feel like there's a single best answer that fully resolves and clarifies the problem, so I'm refraining from accepting one. Is this acceptable (no pun intended) or should I go ahead and choose one to accept anyway? $\endgroup$
    – user28400
    Aug 31, 2016 at 0:43
  • $\begingroup$ I will edit my analysis into my existing answer. But, I won't do it all in one installment because it will take to long. I'm going to edit it in piecemeal. I hope you can be patient. $\endgroup$ Sep 1, 2016 at 0:22
  • $\begingroup$ I finally got around to completing my analysis. It is documented beneath the word EDIT in my answer. The analysis confirms that the decrease in kinetic energy is precisely compensated for by the increase in stored elastic energy, so that total mechanical energy is conserved (and there is no need to invoke heat as a means for making good on energy conservation). $\endgroup$ Sep 8, 2016 at 0:38

5 Answers 5


This is a really interesting problem. To expand on what @alephzero said, before the disk temperature even changes, there will already be a radial displacement distribution u(r), and accompanying strains and stresses in the radial and hoop directions. So the disk will already be deformed to begin with, and this will play a role in determining both the initial radial mass distribution and the stored elastic energy.

When the temperature changes, the radial displacement distribution will change again, and the stresses will change as a result of both the temperature change and the radial and hoop strains, over and above those that would exist from unconstrained thermal expansion. This will result in a change in the stored elastic energy of the disk. So, while the angular momentum of the disk will remain constant, its kinetic energy will change such that the sum of the kinetic energy plus stored elastic energy before the temperature rise is the same as after the temperature rise.

This can all be modeled precisely. The basis for such a model would be primarily the determination of the radial displacement distribution u(r) before and after. The displacements in the tangential direction would be zero, and, in the thickness direction, one could assume plane stress.


Below is a simplified problem that captures the essence of what you are asking. The reason I have introduced this simplified problem is that, if we can't solve this problem, we certainly won't be able to solve the much more complicated disk problem. Plus, the fundamental physical mechanisms present in the simplified problem are precisely those that characterize the disk problem.

Consider a mass M at the end of a massless elastic wire that is traveling in a horizontal circle at an angular velocity $\omega_i$. The cross sectional area of the wire is A, its elastic modulus is E, its coefficient of linear expansion is $\alpha$, and its unextended length is $R_0$. Initially, the length of the wire attached to the rotating mass is $R_i$. After this initial rotation state is established, the temperature of the wire is increased by $\Delta T$ and, as a result, its length increases to $R_f$ and its angular velocity decreases to $\omega_f$. Using linear stress-strain analysis (i.e., for small strains), find (in terms of $R_0$, M, E, A, $\alpha$, and $\omega_i$) the initial extended length $R_i$, the final extended length $R_f$, and the final angular velocity $\omega_f$. Also show that the decrease in kinetic energy of the mass as a result of heating is equal to the increase in stored elastic energy of the wire.

The force balance on the mass in the initially-rotating and in the heated states is given by:

$$k(R_i-R_0)=M\omega_i^2R_i\tag{1}$$ $$k(R_f-R_0-\alpha \Delta T R_0)=M\omega_f^2R_f\tag{2}$$where $k=EA/R_0$. In addition, conservation of angular momentum requires that:$$\omega_fR_f^2=\omega_iR_i^2\tag{3}$$ The linearized (small strain) solution to these equations for $R_i$, $R_f$, and $\omega_f$ is given by:$$R_i=R_0\left(1+\frac{m\omega_i^2}{k}\right)\tag{4}$$ $$R_f=R_0\left(1+\frac{m\omega_i^2}{k}+\alpha \Delta T\right)\tag{5}$$ $$\omega_f=\omega_i(1-2\alpha \Delta T)\tag{6}$$

The change in kinetic energy of the mass $\Delta (KE)$ is given by: $$\Delta (KE)=\frac{M}{2}(\omega_f R_f)^2-\frac{M}{2}(\omega_i R_i)^2\tag{7}$$ The change in stored elastic energy of the wire $\Delta (SE)$ is given by $$\Delta (SE)=k\left[\frac{(R_f-R_0)^2}{2}-\frac{(R_i-R_0)^2}{2}-\frac{\alpha \Delta TR_0}{2}(R_f-R_i)\right]\tag{8}$$ If we substitute Eqns. 4-6 into Eqns. 7 and 8, and disregard higher order non-linear terms, we obtain $$\Delta (KE) = M(\omega_iR_0)^2(\alpha \Delta T)\tag{9}$$ $$\Delta (SE) = -M(\omega_iR_0)^2(\alpha \Delta T)\tag{10}$$ This verifies that the decrease in kinetic energy of the system is precisely compensated for by the increase in stored elastic energy of the system. This is the way that energy is conserved in the system.

These same qualitative results will apply to the disk problem.


I would simplify this to a mass M at the end of a wire of length R, swinging at tangential velocity $v$ around a central pivot.

Now by some process the wire is lengthened by a factor of 2. In so doing, the angular momentum is conserved, so the tangential velocity of mass M decreases by a factor of 2, and the kinetic energy decreases by a factor of 4. As the wire increased in length, the mass's centripetal force acted on it, doing work, most likely by putting heat into the wire.

That is where the energy went.

If you want to see how a change in radius can affect kinetic energy, my favorite example is the incense burner at the Cathedral of Santiago Compostela. If you can tolerate the music, you can see that shortening the radius in the middle of the swing, when centripetal force is greatest, puts energy into the system, and lengthening the radius at the ends of the swing, when force is least, takes less energy out. So the net is a gain in energy.

To slow it down, they can just do the opposite. Lower it in the center, and raise it at the ends. Lowering it in the center is analogous to what is happening in your problem.

  • $\begingroup$ If the disk (or wire) is elastic, the deformation work will not go into heat. It will go into stored elastic energy. $\endgroup$ Aug 30, 2016 at 0:43

The amount of energy required to raise the temperature of the disc is much larger than the change in kinetic energy resulting from the expansion. So even though the KE does go into heat, it does not follow that the net flow from the disc to the bath would be outward. It would just reduce the total heat transfer (by a small amount).


Even if you simplify the problem by assuming the disk is rigid, there is an internal stress field in it when it is rotating.

If the disk changes shape, the stress field will do mechanical work.

I suggest you start by considering a thin ring, not a solid disk - you then only have to consider the work done by the hoop stress when the circumference of the ring changes, which is analogous to the change in internal strain energy when you stretch a spring.

This problem is similar to the change in KE of a spinning ice-skater, when he/she extends his/her arms to change angular velocity. Moving the skater's arms radially in or out does mechanical work.


There is a similar discussion here in the past.

In this case, the angular momentum is conserved as there is no apparent torque. However, the rotational kinetic energy increase because heat expand the disk doing the work. When the environment cools down, the disk contract, the work done by disk produces heat.


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