Question

A weightless rod of length $2l$ carries two equal masses ${}^{\prime }{m}^{\prime }$, one secured at lower end $A$ and the other at the middle of the rod at $B$. The rod can rotate in vertical plane about a fixed horizontal axis passing through $C$. What horizontal velocity must be imparted to the mass at $A$ so that it just completes the vertical circle :

• A. $\sqrt{\frac{48gl}{5}}$
• B. $\sqrt{\frac{64gl}{5}}$
• C. $\sqrt{\frac{72gl}{5}}$
• D. $\sqrt{\frac{120gl}{5}}$

Answer 1

The correct option is A $\sqrt{\frac{48gl}{5}}$

Energy conservation

$\frac{1}{2}m{v}_2^a+\frac{1}{2}m{v}_b^2=2Lmg+4Lmg$

$\frac{v_a}{2L}=\frac{v_b}{L}\Rightarrow v_a=2v_b$

and substituting

$\frac{1}{2}m{v}_a^2+\frac{1}{2}m(\frac{v_a}{2}{)}^2=6Lmg$

and finally

$v_a=4\sqrt{\frac{3}{5}Lg}$ or $\sqrt{\frac{48}{5}Lg}$