A thin uniform rod. of mass M and length l is rotating about a frictionless axis passing through one of its ends and perpendicular to the rod .The rod is heated uniformly to increase its temperature by "∆x". Calculate the percentage change in rotational kinetic energy of the rod . Explain why the answer is not zero.Take coefficient of linear expansion of the material of the rod to be "k". On calculating the solution says that the kinetic energy decreases by 200k∆x%.

My question is: If the length increased as much as it was supposed to on increasing the temperature, which means that the internal energy increased as much as it was supposed to, then why did the kinetic energy decrease? What about the conservation of energy?


Let us assume that the rod is rotating in a vacuum. We assume that the rod is slender, i.e. its diameter is much smaller than its length $L$. The moment of inertia of such a rod rotating about one of its ends is $I(L)=mL^2/3$. Due to temperature increase $\Delta T$ the length of the rod increases by $\delta x$. The new moment of inertia is $I(L+\delta x)=m(L+\delta x)^2/3$. Let $\omega$ and $\omega'$ be the initial and final angular velocity of the rod. Then since no torque is applied to the rod due to heating, angular momentum is conserved: \begin{align} I(L+\delta x)\omega' &=I(L)\omega\\ (L+\delta x)^2\omega' &=L^2\omega\\ \omega' &=\frac{1}{(1+\delta x/L)^2}\omega\\ &\approx\left(1-2\frac{\delta x}{L}\right)\omega\quad\textrm{for }\frac{\delta x}{L}\ll 1 \end{align}

Fractional change in kinetic energy of the rod is: \begin{align} \eta&=\frac{I(L+\delta x)\omega'^2}{I(L)\omega^2}-1\\ &=\frac{\omega'}{\omega}-1\\ &\approx -2\frac{\delta x}{L}=-2k\Delta T\quad\textrm{for }\frac{\delta x}{L}\ll 1 \end{align}

See that we have not used energy conservation equation so far. This becomes: Change in K.E. of the rod+Change in internal energy of the rod+Change in some other energy of the rod = Heat input. Energy conservation only tells you that there should be at least one other type of energy involved, but not what it is. That is up to you to decide depending on specifics of the problem. In a real rod one possible type would be strain energy stored inside the rod due to its elongation. Of course there could be other sources of energy storage too: for example, if there are attracting magnets attached to the ends of the rod, then the rod must do work in elongating against this attraction.

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  • $\begingroup$ But the internal energy increased only due to the increase in temperature. It must not be due to the conversion of kinetic energy into internal energy. $\endgroup$ – Amit Hegde Nov 6 '17 at 16:09
  • $\begingroup$ @AmitHegde Sorry a heat input term must be included. Edited. Kinetic energy can get converted into other forms such as strain energy (which is a kind of potential energy). $\endgroup$ – Deep Nov 7 '17 at 4:40

Anyway, I think I can pull enough from your question to give you a reasonable answer.

No external force is applied, it is just spinning on a friction less axis. So torque with respect to the object spinning and its axis is 0, and angular momentum will remain the same. Angular momentum, $L$, is given by $\mathrm{L = mvr}$. Since $m$ and $L$ are constants, $v$ and $r$ would be inversely proportional due to conservation. In order to maintain $\frac{L}{m}$ as a constant value, as $r$ increases, $v$ must decrease.

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  • $\begingroup$ Sorry, it was supposed to be a rod $\endgroup$ – Amit Hegde Oct 25 '17 at 15:16
  • $\begingroup$ I am deleting the comment part of your answer, since there is a Comment's section for it. And, at present, you don't have enough rep to comment. Also, try to use MathJaX for writing scientific formulae in future answers. It sure helps in understanding. $\endgroup$ – SchrodingersCat Oct 25 '17 at 15:26
  • $\begingroup$ Thanks SchrodingerCat. Yes, I need to build rep before commenting, sorry about that. And thank you, I will have to learn how to implement that into my posts. $\endgroup$ – CuriousOne Oct 25 '17 at 15:33
  • $\begingroup$ No problem Amit, I thought so. Did my answer help at all? $\endgroup$ – CuriousOne Oct 25 '17 at 15:36
  • $\begingroup$ I just don't understand why the KE loss occurs and where does it go? $\endgroup$ – Amit Hegde Oct 25 '17 at 16:26

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