Question

A thin uniform rod of length $L$ is bent at its midpoint as shown in the figure. The distance of the center of mass from the point ${}^{\prime }{O}^{\prime }$ is

• A. $\frac{L}{2}\cos \frac{\theta }{2}$
• B. $\frac{L}{4}\sin \frac{\theta }{2}$
• C. $\frac{L}{4}\cos \frac{\theta }{2}$
• D. $\frac{L}{2}\sin \frac{\theta }{2}$

Answer 1

The correct option is C $\frac{L}{4}\cos \frac{\theta }{2}$

Co-ordinate of point $P=(\frac{l}{4}\sin (\pi -\theta ),\frac{l}{4}\sin (\pi -\theta ))$

$=(\frac{l}{4}\cos \theta ,\frac{l}{4}\sin \theta )$

$Q=[-\frac{l}{4},0)$

Let center of mass $c(x,y)$

$x=\frac{\frac{m}{2}(-\frac{l}{4})+\frac{m}{2}(-\frac{l}{4}\cos \theta )}{m}=-\frac{l}{8}(1+\cos \theta )$

$y=\frac{0+\left(\frac{l}{4}\sin \theta \right)\frac{m}{2}}{m}=\frac{l}{8}\sin \theta$

Now, distance $d$ from $o(0,0)$ is

$d=\sqrt{x^2+y^2}$

$=\frac{l}{8}\sqrt{{\sin }^2\theta }+(1+\cos \theta {)}^2$

$=\frac{l}{8}\sqrt{2(1+\cos \theta )}$

$=\frac{l}{8}\sqrt{2.2{\cos }^2\frac{\theta }{2}}$

$=\frac{l}{4}\cos \frac{\theta }{2}$