Question

A rod of length $L$ is rotated in horizontal plane with constant velocity $\omega$. A mass $m$ is suspended by a light string of length $L$ from the other end of the rod. If the angle made by vertical with string is $\theta$ then angular speed, $\omega =$

• A. ${\left[\frac{g\sin \theta }{L(1+\tan \theta )}\right]}^{\frac{1}{2}}$
• B. ${\left[\frac{L(1+\tan \theta )}{g\tan \theta }\right]}^{\frac{1}{2}}$
• C. ${\left[\frac{g\tan \theta }{L+\sin \theta }\right]}^{\frac{1}{2}}$
• D. ${\left[\frac{g\tan \theta }{L(1+\sin \theta )}\right]}^{\frac{1}{2}}$

Answer 1

The correct option is D ${\left[\frac{g\tan \theta }{L(1+\sin \theta )}\right]}^{\frac{1}{2}}$

Due to the rotation motion a centrifugal force will act along the horizontal direction.
From FBD of mass $m$:
$m{w}^{2}r=T\sin \theta$
or $m{w}^2(L+L\sin \theta )=T\sin \theta ....\left(1\right)$
also, $mg=T\cos \theta ....\left(2\right)$
$\left(1\right)/\left(2\right),\frac{w^2(L+L\sin \theta )}{g}=\tan \theta$
or $w^2=\frac{g\tan \theta }{L(1+\sin \theta )}$
$\therefore w=[\frac{g\tan \theta }{L(1+\sin \theta )}{]}^{1/2}$