A small block of mass m is pushed on a smooth track from position A with a velocity `2sqrt5` times the minimum velocity required to reach point D. The block will leave the contact with track at the point where normal force between them becomes zero.

Find where the maximum contact force occurs between the block and the track.

Options

A. (a) At B
B. (b) At C
C. (c) Somewhere between A and B
D. (d) At A

Answer

D

Explanation

`v_1=2/sqrt5sqrt(5gR)=2sqrt(gR)`
Now, `v_2^2=v_1^2-2g(R+R sin theta)=2gR-2gR sin theta`
`N+mg sin theta=(mv_2^2)/(R)`
Put `N=0`
`implies mg sin theta=m2g(1-sin theta)`
`implies sin theta=2/3`
`implies theta=sin^-1(2/3)`

Maximum contact force will occur at the lowest piont A.