A uniform rod of mass M and length L is moving on a smooth horizontal plane, such that its one end is moving with a velocity `V_(0)` and other end is moving with velocity `2V_(0)` in the same direction as shown. Then which of the following is kinetic energy of the rod?

Options

A. `(13)/(2L)MV_(0)^(2)`
B. `(24)/(31)MV_(0)^(2)`
C. `(7)/(6)MV_(0)^(2)`
D. `(31)/(24)MV_(0)^(2)`

Answer

A

Explanation

Normal reaction N = `sqrt(m^(2)g^(2)+(m^(2)V_(0)^(4))/(r^(2)))`

`therefore` Friction = `mu_(k)N`
So tangential acceleration = `(mu_(k)N)/(m)`