Question

A thin uniform rod of length $L$ is bent at its midpoint as shown in the figure. The distance of the centre of mass from the point $O$ is

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

Answer 1

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

${\overrightarrow{X}}_{cm}=\frac{\frac{m}{2}\left(\frac{L}{4}\right)+\frac{m}{2}\frac{L}{4}\cos \theta }{m}\hat{i}$

$=\frac{L}{8}(1-\cos \theta )\hat{i}$

$\overrightarrow{Y_{cm}}=\frac{\frac{m}{2}\left(\frac{L}{4}\right)\sin \theta \times m}{m}$

$=\frac{L}{8}\sin \theta \hat{j}$

Distance of center of mass fromO=$\sqrt{{X_{com}}^2+{Y_{com}}^2}$

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

$=\frac{L}{8}\sqrt{1-2\cos \theta +{\sin }^2\theta +{\cos }^2\theta }$

$=\frac{L}{4}\sin \frac{\theta }{2}$