Physics - Rotation Question with Solution | TestHub

The moment of inertia of a disc about an axis passing through its centre and perpendicular to its plane is given by

$I=\frac{1}{2}M{R}^2$

The torque acting on the disc is related to the angular acceleration by

$\tau =I\alpha$

Since the torque is provided by a tangential force $F$ applied at the rim, we have

$\tau =F\times R$

Equating the two expressions for torque:

$F\times R=\frac{1}{2}M{R}^2\alpha$

$F=\frac{1}{2}MR\alpha$

Substituting the given values ($M=1$ kg, $R=0.4$ m, $\alpha =10$ rad/s${}^2$):

$F=\frac{1}{2}\times 1\times 0.4\times 10$

$F=2$ N

Answer: $2$ N