Question

A weightless rod of length $2\ell$ carries two equal masses ${}^{\prime }{m}^{\prime }$ one tied at lower end $A$ and the other at the middle of the rod at $B$. The rod can rotate vertical plane about a fixed horizontal axis passing through $C$. The rod is released from rest in horizontal axis passing through $C$. The rod is released from rest in horizontal position. The speed of the mass $B$ at the instant rod, becomes vertical is:

• A. $\sqrt{\frac{3g\ell }{5}}$
• B. $\sqrt{\frac{4g\ell }{5}}$
• C. $\sqrt{\frac{6g\ell }{5}}$
• D. $\sqrt{\frac{7g\ell }{5}}$

Answer 1

The correct option is A $\sqrt{\frac{3g\ell }{5}}$

Potential Energy of the mass A in horizontal position, $U_a=mgl$

Potential Energy of the mass B in horizontal position, $U_b=mgl$

Total energy of the system in horizontal position $=2mgl$

In vertical position:

Potential energy of the masses $A\&B$ will convert into kinetic energy and the center of mass of the system will still have a potential energy of $\frac{mgl}{2}$.

Thus;

By conservation of energy:

$2mgl=mg\frac{l}{2}+\frac{1}{2}m{v}_a^2+\frac{1}{2}m{v}_b^2\frac{3}{2}mgl=\frac{1}{2}m(v_a^2+v_b^2)v_a^2+v_b^2=3gl$

The angular velocity of the masses $A\&B$ will be same, say $\omega$, then:

$(2\omega l{)}^2+(\omega l{)}^2=3gl\omega =\sqrt{\frac{3g}{5l}}$

Therefore,

Velocity of mass $B=l\omega v_b=\sqrt{\frac{3gl}{5}}$

Therefore, the velocity of mass B will be $\sqrt{\frac{3gl}{5}}.$