A thin solid disk of $1\ \text{kg}$ is rotating about its diameter axis at a speed of $1800\ \text{rpm}$. By applying an external torque of $25\pi\ \text{N m}$ for $40\ \text{s}$, the speed increases to $2100\ \text{rpm}$. The diameter of the disk is ______ m.
Numerical value type. Enter your answer.
Answer: 40
$\omega_i = 1800\times\dfrac{2\pi}{60} = 60\pi\ \text{rad/s}$, $\omega_f = 70\pi\ \text{rad/s}$, so $\alpha = \dfrac{10\pi}{40} = \dfrac\pi4\ \text{rad/s}^2$.
About a diameter, $I = \dfrac{mR^2}{4}$:
$$\tau = I\alpha \Rightarrow 25\pi = \frac{R^2}{4}\cdot\frac\pi4 \Rightarrow R^2 = 400 \Rightarrow R = 20\ \text{m}$$
Diameter $= 40\ \text{m}$. (An unrealistically large disk, but this is what the data give.)
Solution by Sreeraj P, M.Sc Physics