Question

AB is a uniform rod of mass $1kg$ and length $1m$. Rod is placed on a smooth horizontal table in XY plane along X-axis. Weight is acting toward negative z axis. Initially centre of mass C of the rod is at origin as shown. A constant force $\overrightarrow{F}=1\hat{j}N$ starts acting on the rod at end A. Choose the correct options.

• A. C will move along Y-axis only
• B. C will move along line $y=x$
• C. Angular velocity of rod will be proportional to $\sqrt{sin\theta }$ where $\theta$ is angle made by rod with X-axis, $(0\le \theta <\frac{\pi }{2})$
• D. Angular velocity of rod will be proportional to $\sqrt{cos\theta }$ where $\theta$ is angle made by rod with X-axis, $(0\le \theta <\frac{\pi }{2})$

Answer 1

Answer: (A), (C)

The correct options are
A C will move along Y-axis only
C Angular velocity of rod will be proportional to $\sqrt{sin\theta }$ where $\theta$ is angle made by rod with X-axis, $(0\le \theta <\frac{\pi }{2})$


Since, surfaces are smooth, net forces are acting along Y axis and as centre of mass was initially at rest in X-direction, so, point C will remain at rest in X-direction and only move in Y- direction (direction in which net force is acting).

Torque about point C
$T_c=I_c\alpha =\frac{m{l}^2}{12}\alpha$
$T_c=F\left(\frac{1}{2}cos\theta \right)=\frac{1}{2}cos\theta$

$\Rightarrow \left(\frac{1}{2}cos\theta \right)1=\frac{1(1{)}^2}{12}\frac{d^2\theta }{d{t}^2}$

But, using chain rule,
$\frac{d^2\theta }{d{t}^2}=\frac{d\omega }{dt}=\frac{d\omega }{d\theta }\times \frac{d\theta }{dt}=\frac{d\omega }{d\theta }\omega$

Substituting in previous equation,

$\left(\frac{1}{2}cos\theta \right)d\theta =\frac{1}{12}\omega d\omega$

Integrating both sides,

${\int }_0^{\omega }\omega d\omega =6{\int }_0^{\theta }cos\theta d\theta$

$\Rightarrow \frac{{\omega }^2}{2}=6sin\theta$

$\Rightarrow \omega \propto \sqrt{sin\theta }$