Solve the following :


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.


Torque $=\bar{\tau} \times \bar{r}$

$\left.\mathrm{T}=\mathrm{F}^{\left(\frac{L}{4}\right.}\right) \sin 90^{\circ}$

$1 \propto \frac{F L}{4}$

$\frac{m L^{2}}{12} \cdot 0 C=\frac{F L}{4}$

$\alpha=\frac{3 F}{m L}$

$\theta=\omega_{0} t+\frac{1}{2} \alpha c t^{2}$

$\theta=\frac{3 F t^{3}}{2 m L}$

Now $\omega_{0}=0 ;$ time $=t$

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now