Question

A thin solid disk of 1 kg is rotating along its diameter axis at the speed of 1800 rpm. By applying an external torque of 25$\pi$ Nm for 40s, the speed increases to 2100 rpm. The diameter of the disk is ________ m.

Answer 1

40

Answer 2

1. Convert rpm to rad/s: $\omega_i = 1800 \times \frac{2\pi}{60} = 60\pi$ rad/s $\omega_f = 2100 \times \frac{2\pi}{60} = 70\pi$ rad/s

2. Calculate angular acceleration: $\alpha = \frac{\Delta \omega}{\Delta t} = \frac{70\pi - 60\pi}{40} = \frac{10\pi}{40} = \frac{\pi}{4}$ rad/s$^2$

3. Calculate moment of inertia using $\tau = I\alpha$: $25\pi = I \times \frac{\pi}{4} \implies I = 100$ kg m$^2$

4. Calculate radius using $I = \frac{1}{4}MR^2$ for a disk about its diameter: $100 = \frac{1}{4} \times 1 \times R^2 \implies R^2 = 400 \implies R = 20$ m

5. Diameter $D = 2R = 40$ m.