Question

A rod of mass m and length L, lying horizontally, is free to rotate about a vertical axis through its centre. A horizontal force of constant magnitude F acts on the rod at a distance of L/4 from the centre. 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 about the centre due to force,
$$\begin{gathered} \overrightarrow{\tau }=\overrightarrow{F}\times \overrightarrow{r}=F\times \frac{L}{4}\times \sin 90° \\ \tau =F\times \frac{L}{4} \end{gathered}$$
Let the torque produces an angular acceleration α.
$$\begin{gathered} \tau =I\alpha \\ \Rightarrow \tau =\frac{m{L}^2}{12}\times \alpha \left(I\mathit{}\mathrm{of}\mathrm{a}\mathrm{rod}=\frac{m{L}^2}{12}\right) \\ \Rightarrow F\frac{L}{4}=\frac{m{L}^2}{12}\times \alpha \\ \Rightarrow \alpha =\frac{3F}{mL} \\ \mathrm{Now},\theta =\frac{1}{2}\alpha t^2(\mathrm{initially}\mathrm{at}\mathrm{rest}) \\ \Rightarrow \mathrm{\theta }=\frac{3F{t}^2}{2mL} \end{gathered}$$