Question

The initial speed given to the ball is
Note: Attempt the above question after studying chapter of rotational motion.

• A. $\sqrt{\frac{Mgl+2mgl}{m}}$
• B. $\sqrt{2gl}$
• C. $\sqrt{\frac{2Mgl+mgl}{m}}$
• D. None the these

Answer 1

The correct option is D None the these

Centre of mass of the rod only lie at point C and C' in horizontal and vertical state.Thus when rod becomes vertical its COM rises by $\frac{l}{2}$

Also, it is given that velocity at the vertical upward position is Zero i.e $v_1=0$ and $w_1=0$

Let initial speed given be $v$ and the rod rotates with $w$ when it is horizontal.

Work-energy theorem: $-mgl-Mg\frac{l}{2}=0-\frac{1}{2}{I}_O{w}^2-\frac{1}{2}m{v}^2$

$mgl+Mg\frac{l}{2}=\frac{1}{2}\times \frac{1}{3}M{l}^2{w}^2+\frac{1}{2}m{v}^2$

As $v=lw$

$⟹mgl+Mg\frac{l}{2}=\frac{1}{2}\times \frac{1}{3}M{v}^2+\frac{1}{2}m{v}^2$

$⟹v=\sqrt{\frac{(3M+6m)gl}{(M+3m)}}$