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 $\frac{L}{4}$ from the centre. The force is always perpendicular to the rod. Find the angle rotated by the rod during the time l after the motion starts.

Answer 1

A rod of mass m and length L, lying horizontally if free to rotate about a vertical axis passing through its centre. A force F is acting perpendicular to the road at a distance $\frac{L}{4}$ from the centre.

Therefore Torque about the centre due to this force,

$\tau =F\times r=F\frac{L}{4}$

This torque will produce an angular acceleration a therefore

${\tau }_c=I_c\times \alpha$

$\Rightarrow {\tau }_c=\frac{m{L}^2}{12}\times \alpha$

$(I_c\text{of a rod}=\frac{m{L}^2}{12})$

$\Rightarrow F\frac{L}{4}=\frac{m{L}^2}{12}\times \alpha$

$\Rightarrow \alpha =\frac{3F}{mL}$

Therefore, $\theta =\frac{1}{2}\alpha t^2$ (initially at rest)

$\Rightarrow \theta =\frac{1}{2}\times \left(\frac{3F}{mL}\right)t^2$