Question

A light rod of length $\ell$ is pivoted at the upper end. Two masses (each m), are attached to the rod, one at the middle and the other at the free end. What horizontal velocity must be imparted to the lower end mass, so that the rod may just take up the horizontal ?

• A. $\sqrt{\frac{6\ell g}{5}}$
• B. $\sqrt{\frac{\ell g}{5}}$
• C. $\sqrt{\frac{12\ell g}{5}}$
• D. $\sqrt{\frac{2\ell g}{5}}$

Answer 1

The correct option is C $\sqrt{\frac{12\ell g}{5}}$

By C.O.M.E.,

$\frac{1}{2}m{v}_0^2+\frac{1}{2}m{\left(\frac{v_0}{2}\right)}^2=mg\frac{\ell }{2}+mg\ell =\frac{3mg\ell }{2}$

$\Rightarrow \frac{1}{2}m{v}_0^2(1+\frac{1}{4})=\frac{3mg\ell }{2}\Rightarrow v_0=\sqrt{\frac{12g\ell }{5}}$