Question

A rod of mass m and length L, hinged horizontally, is free to rotate about a vertical axis through its center. A horizontal force of constant magnitude Facts on the rod at a distance of L/4 from the center. The force is always perpendicular to the rod. Find the angle rotated by the rod during the time t after the motion starts.

Answer 1

Torque acting on the rod $=\frac{F\times L}{4},$ (as force is acting perpendicular to the rod at a distance of $L/4$ from center)

We know that $TORQUE=I\left(\alpha \right),$ where $I$ is the moment of inertia and $\alpha$ is the angular acceleration

$I$ for rod moving about its vertical axis about centre$=\frac{M{L}^2}{12}$

$⟹\frac{FL}{4}=\frac{M{L}^2}{12}\alpha$ which gives

$\alpha =\frac{3F}{ML}$

we know that $\theta ={\omega }_o+\frac{1}{2}\alpha t^2,$ where $\theta =$ is angle rotated and ${\omega }_o=$ initial angular velocity

${\omega }_o=0$ (it is given that rod is at rest initially)

$⟹\theta =\frac{3F{t}^2}{2ML}$