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

About point `O`
Torque `tau =Ialpha`
`Mg((L)/(2)cos theta)=(ML^(2))/(3)alpharArr(3g)/(2L) cos theta =alpha`
Initially centripetal acceleration of point `P` is zero
` ( :'a_(c)=(v^(2))/(r)=(0)/(r)=0)`
Acceleration of point `P` is `sqrt(a_(C)^(2)+a_(t)^(2))`
`=a_(t)=Lalpha=(3)/(2)g cos theta`