This is my attempt at deriving the exact formula for the rotational energy of a disk.

Let the disk have radius $R > 0$, mass $m > 0$, angular velocity of rotation $\omega > 0$, and the constant density $\rho = m / (\pi R^2)$. Now, consider the area element $\mathrm{d}A = r \, \mathrm{d}r \, \mathrm{d}\alpha$. The mass of that area element is given by $$\mathrm{d}m = \rho \, \mathrm{d}A = \frac{mr \, \mathrm{d}r \, \mathrm{d}\alpha}{\pi R^2}.$$ Next, the energy of that area is given by $$\mathrm{d}E = \frac{1}{2} \mathrm{d}mv^2 = \frac{1}{2} \mathrm{d}m(r\omega)^2 = \frac{mr^3 \omega^2 \, \mathrm{d}r \, \mathrm{d}\alpha}{2\pi R^2}.$$

Finally, the rotational energy is \begin{aligned} E_{rotational} &= \int_0^{2\pi} \int_0^R \mathrm{d}E \\ &= \int_0^{2\pi} \int_0^R \frac{mr^3 \omega^2 }{2\pi R^2} \, \mathrm{d}r \, \mathrm{d}\alpha\\ &= \frac{m\omega^2}{2\pi R^2} \int_0^{2\pi} \int_0^R r^3 \, \mathrm{d}r \, \mathrm{d}\alpha \\ &= \frac{m\omega^2}{2\pi R^2} \int_0^{2\pi} \Bigg[ \frac{1}{4} r^4 \Bigg]_{r = 0}^{r = R} \mathrm{d}\alpha \\ &= \frac{m\omega^2R^2}{8\pi} \int_0^{2\pi}\mathrm{d}\alpha \\ &= \frac{m\omega^2R^2}{8\pi} \Bigg[\alpha\Bigg]_{\alpha = 0}^{\alpha = 2\pi} \\ &= \frac{1}{4}m\omega^2R^2. \end{aligned}

As you can see, I have an error by the factor of two. Where my calculation goes wrong?

Thank you in advance.


1 Answer 1


Your answer is correct. $I = \frac{1}{2}mR^2$ and rotational energy is $E = \frac{1}{2}I\omega^2 = \frac{1}{4}mR^2\omega^2$.


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