Question

A rod of length $l$is held vertically stationary with its lower end located at a point P, on the horizontal plane. When the rod is released to topple about P, the velocity of the upper end of the rod with which it hits the ground is

• A. $\frac{g}{l}$
• B. $gl$
• C. $3\frac{g}{l}$
• D. $\frac{5g}{l}$

Answer 1

The correct option is C $3\frac{g}{l}$

I = moment of inertia of metre stick about point $A=\frac{ml{}^2}{3}.$
By the law of conservation of energy
$mg\frac{1}{2}=\frac{1}{2}I{w}^2=\frac{1}{2}\frac{m{l}^2}{3}{\frac{v_B}{l}}^2$
By solving, we get $v_B=3g/l$