Question

A uniform circular disc of mass $1.5kg$ and radius $0.5m$ is initially at rest on a horizontal frictionless surface. Three forces of equal magnitude $0.5N$ are applied simultaneously along the three sides of an equilateral triangle $XYZ$ with its vertices on the perimeter of the disc (see figure). One second after applying the forces, the angular speed of the disc in $rad{s}^{-1}$ is:

Answer 1

The torque acting on each of the points X, Y and Z are in anticlockwise direction and equal in magnitude.
${\tau }_{total}=3FR\sin {30}^o=\frac{3FR}{2}$
Moment of inertia $I=\frac{M{R}^2}{2}$
$\therefore {\tau }_{total}=I\alpha =\frac{M{R}^2}{2}\alpha =\frac{3FR}{2}$
$\therefore \frac{1.5\times \left({0.5}^2\right)}{2}\alpha =\frac{3\times 0.5\times 0.5}{2}$
$\Rightarrow \alpha =2$
From rotational kinematics equation
$\omega ={\omega }_0+\alpha t$
At $t=1s$
$\therefore \omega =2t=2\times 1=2rad/s$