A uniform pole of length `L` and mass `M` is pivoted on the ground with a frictionless hinge. The pole makes an angle `theta` with the horizontal. The moment of inertia of the pole about one end is `( 1/3) ML^(2)`. If it starts falling front the position shown in the accompanying figure, the linear acceleration of the free end of the pole immediately after release would be

Options

A. `(2)/(3)g cos theta`
B. `(2)/(3)g`
C. `g`
D. `(3)/(2)g cos theta`

Answer

D

Explanation

A uniform …………..
About point `O`
Torque `tau = I alpha`
`Mg((L)/(2)cos theta) = ((ML^(2))/(3)) alpha implies (3)/(2)(g)/(L)cos theta = alpha`
Initially centripetal accelration of point `P` is zero
`:' a_(c) = (V^(2))/(r ) = (0)/(r ) = 0`
Acceleration of point `P` is `sqrt(a_(c )^(2)+a_(1)^(2))`
`= a_(t) = L alpha = (3)/(2)g cos theta`