Two masses m and `(m)/(2)` are connected at the two ends of a massless rigid rod of length l. The rod is suspended by a thin wire of torsional constant k at the centre of mass of the rod-mass system(see figure). Because of torsional constant k, the restoring torque is `tau=ktheta` for angular displacement `theta`. If the rod is rotated by `theta_(0)` and released, the tension in it when it passes through its mean position will be :

Options

A. ` (3ktheta_(0)^(2))/(l)`
B. `(Ktheta_(0)^(2))/(2l)`
C. `(2ktheta_(0)^(2))/(l)`
D. `Ktheta_(0)^(2))/(l)`

Answer

D

Explanation

Total moment of Inertia of masses
`I = m((l)/(3))^(2) + (m)/(2)((2l)/(3))^(2) = (ml^(2))/(3)`
P.E. at `theta_(0) = (1)/(2)ktheta_(0)^(2)`
P.E. is conveted in to rotational K.E.
`(1)/(2)ktheta_(0)^(2) = (1)/(2)Iomega^(2)`
`omega^(2) = (ktheta_(0)^(2))/(I) = (3ktheta_(0)^(2))/(ml^(2)) " "...(i)`
Tension in rod from equation (i)
`T = momega^(2)(l)/(3) = m((3ktheta_(0)^(2))/(ml^(2)))(l)/(3) = (theta_(0)^(2)k)/(l)`