question_answer

A rod of length; 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)  \[\sqrt{\frac{g}{I}}\]

B)  \[\sqrt{3gI}\]

C)  \[3\sqrt{\frac{g}{I}}\]               

D)  \[\sqrt{\frac{3g}{I}}\]  

Correct Answer: B

Solution :

In this process potential energy of the meter stick will be converted into rotational kinetic energy.   PE of metre stick \[=\frac{mgl}{2}\] Because its centre of gravity lies at the middle                  of the rod. Rotational kinetic energy \[E=\frac{1}{2}I{{\omega }^{2}}\] \[I\]= moment of inertia of metre stick about point \[A=\frac{m{{l}^{2}}}{3}\]. By the law of conservation of energy \[mg\left( \frac{l}{2} \right)=\frac{1}{2}I{{\omega }^{2}}=\frac{1}{2}\frac{m{{l}^{2}}}{3}{{\left( \frac{{{v}_{B}}}{l} \right)}^{2}}\] By solving, we get \[{{v}_{B}}=\sqrt{3gl}\]