Skip to main content
Physics Guru is an online education and competitive examinations preparation portal founded by Manish Verma, an IIT Madras alumnus, providing learning resources for competitive examinations.

The emphasis is on conceptual understanding, analytical thinking and problem solving rather than simply memorising formulas and standard methods.

The objective is to help students understand what they are learning and develop the ability to solve problems they have not seen before.

Consider a 20 kg uniform circular disk of radius 0.2 m. It is pin supported at its centre and is at rest initially. The disk is acted upon by a constant force F = 20 N through a massless string wrapped around its periphery as shown in the figure. Suppose the disk makes n number of revolutions to attain an angular speed of 50 rad/s. The value of n, to the nearest integer, is ---------. [Given : In one complete revolution, the disk rotates by 6.28 rad]

Solution

We have, $\omega ^2 = 2\alpha \theta $

Or, $\omega ^2 = 2\alpha .(n.2\pi)$

$\Rightarrow n = \frac {\omega ^2}{4\pi \alpha }$

Torque $\tau = rF = I\alpha $

$\Rightarrow \alpha = \frac {rF}{I}$

So, $n = \frac {I \omega ^2}{4\pi rF}=\frac {mr^2 \omega ^2}{8\pi rF}=\frac {mr \omega ^2}{8\pi F}=\frac {20\times 0.2 \times 50^2}{4\times 6.28 \times 20}$

$\Rightarrow n=19.9 \approx 20$