Question

Consider a 20 kg uniform circular disk of radius 0.2 m. It is pin supported at its center and is at rest initially. The disk is acted upon by a constant force F = 20 N through a massless string wrapped around is periphery as shown in the figure.

Suppose the disk makes n number of revolutions to attain an angular speed of 50 rad s^-1.
The value of n, to the nearest integer, is __________.
[Given : In one complete revolution, the disk rotates by 6.28 rad]

Hint:

Torque \(= FR\) gives the angular acceleration; then use \(\omega^2 = 2\alpha\theta\) and divide the angle by 6.28 rad per revolution.

Answer 1

Correct Answer: 20

Torque about the centre: \(\tau = FR = 20\times0.2 = 4\ \mathrm{N\,m}\).
Moment of inertia: \(I = \frac{1}{2}MR^2 = \frac{1}{2}\times20\times(0.2)^2 = 0.4\ \mathrm{kg\,m^2}\).
\(\alpha = \frac{\tau}{I} = \frac{4}{0.4} = 10\ \mathrm{rad\,s^{-2}}\).
From rest, \(\omega^2 = 2\alpha\theta\): \(\theta = \frac{50^2}{2\times10} = 125\ \mathrm{rad}\).
\(n = \frac{125}{6.28} = 19.9 \approx 20\).
Answer: 20