A uniform rod of mass `m` and length `l` is applied pivoted at point `O`. The rod is initially in vertical position and touching a block of mass `M` which is at rest on a horizontal surface. The rod is given a slight jerk and it starts rotating about point `O` this causes the block to move forward as shown The rod loses contact with the block at `theta=30^(@)` all surfaces are smooth now answer the following questions.
Q. The acceleration of centre of mass of rod, when it loses contact with the block is

Options

A. `5g//4`
B. `5g//2`
C. `3g//2`
D. `3g//4`

Explanation

`a_(n)=(l)/(2)omega^(2)=(l)/(2)((4v^(2))/(l^(2)))`
`=((2)/(l))((3gl)/(16))=(3)/(8)g`
`a_(t)=(l)/(2)alpha=(l)/(2)((tau)/(I))=(l)/(2)[(mg((l)/(2)cos30^(@)))/(ml^(2)//3)]`
`=(3sqrt(3)g)/(8)`
`thereforea=sqrt(a_(n)^(2)+a_(t)^(2))`
`=(3g)/(4)`