question_answer

A metre stick of length 1m is held vertically with one end in contact of the floor and is then allowed to fall. If the end touching the floor is now allowed to slip, the other end will hit the ground with a velocity of \[\left( g = 9.8 m/{{s}^{2}} \right)\]

A) 3.2 m/s

B)        5.4 m/s

C) 7.6 m/s

D)        9.2 m/s

Correct Answer: B

Solution :

In this process potential energy of the metre stick will be converted into rotational kinetic energy. \[\operatorname{P}.E. of meter stick=mg\left( \frac{l}{2} \right)\] Because its centre of gravity lies at the middle point of the rod. Rotational kinetic energy \[E=\frac{1}{2}\,I\,{{\omega }^{2}}\] \[\operatorname{I} = M.I.\] of metre stick about point \[A=\frac{m{{l}^{2}}}{3}\] \[\omega =\] Angular speed of the rod while striking the ground \[{{\nu }_{B}}=\] Velocity of end B of metre stick while striking the ground By the law of conservation of energy, \[mg\left( \frac{l}{2} \right)=\frac{1}{2}\,I\,{{\omega }^{2}}=\frac{1}{2}\,\frac{m{{l}^{2}}}{3}\,{{\left( \frac{{{\nu }_{B}}}{l} \right)}^{2}}\] By solving we get, \[{{\nu }_{B}}=\,\,\sqrt{3gl}=\,\,\sqrt{3\times 10\times 1}=\,\,5.4\,\,m/s\]