A long uniform rod of length `L` and mass `M` is free to rotate in a horizontal plane about a vertical axis through its one end 'O'. A spring of force constant `k` is connected between one end of the rod and `PQ`. When the rod is in equilibrium it is parallel to `PQ`.

(a)What is the period of small oscillation that result when the rod is rotated slightly and released ?
(b) What will be the maximum speed of the displacement end of the rod, if the amplitude of motion is `theta_(0)`?

Answer

A, B, C, D

Explanation

(a) Restoring torque about 'O' due to elastic of the spring
`tau = - FL = - kyL" "(F = ky)`
`tau = - kL^(2)theta` (as `y = L theta`)
`tau = I alpha = (1)/(3) ML^(2)(d^(2) theta)/(dt^(2))`
`(1)/(3) ML^(2)(d^(2)theta)/(dt^(2)) = - kL(2)theta`
`(d^(2))/(dt^(2)) = - (3k)/(M)theta`
`omega = sqrt((3k)/(M))`
`rArr T = 2pisqrt((M)/(3k))`
(b) In angular SHM, maximum angular velocity
`((d theta)/(dt))_(max) = theta_(0)omega = sqrt((3k)/(M))`
`v = r ((d theta)/(dt))`
So , `v_(max) = L((d theta)/(dt))_(max) = L theta_(0)sqrt((3k)/(M))`.