Question

Consider a $20\text{kg}$ uniform circular disk of radius $0.2\text{m}$. It is pin supported at its center and is at rest initially. The disk is acted upon by a constant force $F=20\text{N}$ 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 $50{\text{rad s}}^{-1}$. The value of n, to the nearest integer is . [Given: In one complete revolution, the disk rotates by 6.28 rad]

Answer 1

Using $\alpha =\frac{\tau }{I}=\frac{F.R.}{m{R}^2/2}=\frac{2F}{mR}$
$\alpha =\frac{2\times 20}{20\times \left(0.2\right)}=10{\text{rad/s}}^2$
Using Kinematic equation,
${\omega }^2={\omega }_0^2+2\alpha \Delta \theta$
${\left(50\right)}^2=0^2+2\left(10\right)\Delta \theta$
$\Rightarrow \Delta \theta =\frac{2500}{20}$
$\Delta \theta =125\text{rad}$
No. of revolution $=\frac{128}{2\pi }\approx 20$ revolutions.
$\therefore 20$ is the required value.