A rod of mass M and length l is suspended freely from its endandit canoscillate in the vertical plane about the point of suspension. It is pulled to one side and then released. It passes through the equilibrium position withangular speed `omega` . What is the kinetic energy while passing through the mean position ?

Options

A. `(Ml^(2)omega^(2))/(2)`
B. `(Ml^(2) omega^(2))/(6)`
C. `(Ml^(2)omega^(2))/(4)`
D. `(Ml^(2)omega^(2))/(8)`

Answer

B

Explanation

`KE =(1)/(2)I omega^(2)=(1)/(2)(M l^(2)omega^(2))/(3)=(M l^(2)omega^(2))/(6)`