Two small balls, each of mass m are connected by a light rigid rod of length L. The system is suspended from its centre by a thin wire of torsional constant k. The rod is rotated about the wire through an angle `theta_0` and released. Find the tension in the rod as the system passes through the mean position.

Options

A. `sqrt(k^(2)theta_(0)^(4)+4m^(2)g^(2)L^(2))/(2L)`
B. `sqrt(k^(2)theta_(0)^(4)+m^(2)g6(2)L^(2))/(L)`
C. `(ktheta_(0)^(2))/(2L)`
D. `(ktheta_(0)^(2)+mgL)/(L)`

Answer

B

Explanation

at mean position, tension is
`T = sqrt((mg)^(2)+((m omega^(2)L)/(2))^(2))rarr(1)`
from law of conservation of enegry `(1)/(2)I omega^(2) =(1)/(2) k theta_(0)^(2)`
`rArr omega^(2) = (k theta_(0)^(2))/(I) =(ktheta_(0)^(2))/(2((mL^(2))/(4)))=(2ktheta_(0)^(2))/(mL^(2))rarr(2)`
from (1) & (2) `T` can be solved.