Question

AB is a uniform rod of mass $1kg$ and length $1m$. The rod is placed on a smooth horizontal table on the x- y plane along x-axis. Initially, centre of mass C of the rod is at origin as shown. A constant force $\overrightarrow{F}=1N$ starts acting on the rod at end A. Choose the correct option(s).

• A. C will move on Y-axis.
• B. After a finite time, rod will be parallel to Y-axis momentarily.
• C. Angular velocity of rod will be proportional to $\sqrt{\cos \theta }$ where $\theta$ is the angle made by the rod with x-axis. $(0\le \theta <\frac{\pi }{2})$
• D. Angular velocity of rod will be proportional to $\sqrt{\sin \theta }$ where $\theta$ is the angle made by the rod with x-axis. $(0\le \theta <\frac{\pi }{2})$

Answer 1

Answer: (A), (B), (D)

The correct options are
A C will move on Y-axis.
B After a finite time, rod will be parallel to Y-axis momentarily.
D Angular velocity of rod will be proportional to $\sqrt{\sin \theta }$ where $\theta$ is the angle made by the rod with x-axis. $(0\le \theta <\frac{\pi }{2})$

When the rod makes angle $\theta$ with the x-axis,

$T_c=I_c\alpha$
$\left(\frac{l}{2}\cos \theta \right)1=\frac{m{l}^2}{12}\alpha$
$\left(\frac{1}{2}\cos \theta \right)1=\frac{1(1{)}^2}{12}\omega \frac{d\omega }{d\theta }$
$(\alpha =\frac{d\omega }{dt}=\frac{d\omega }{d\theta }\times \frac{d\theta }{dt}=\omega \frac{d\omega }{d\theta })$

Integrating,
${\int }_0^{\omega }\omega d\omega =6{\int }_0^{\omega }\cos \theta d\theta$
$\frac{{\omega }^2}{2}=6\sin \theta$
$\Rightarrow \omega \propto \sqrt{\sin \theta }$ (D)

C.O.M. will move on Y-axis as the direction of net force is along Y-axis. (A)

$T_C=\frac{1}{2}\cos \theta$ is positive so $\theta$ increases as $t$ increases. So rod will be parallel to y-axis in finite time. (B)