A small block of mass m is pushed on a smooth track from position A with a velocity `(2)/(sqrt(5))` 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.
Q. When the block reaches the point B, what is the direction (in terms of angle with horizontal) of acceleration of the block?

Options

A. `tan^(-1)((1)/(2))`
B. `sin^(-1)((3)/(4))`
C. `sin^(-1)((2)/(3))`
D. The block never reackes B

Answer

A

Explanation

`a_(1)=(mu mg)/(m)=mu g, a_(2)=(mu mg)/(m)=mu g`
`V_(1)=a_(1)t=mu g t, " "V_(2)=a_(2)t =mu g t`
` tau=mu m g r`
`omega = omega_(0)+ alpha t`
Where `alpha=-(tau)/(I)=-(mu mgr)/((2)/(5)mr^(2))=-(5)/(2)(mug)/(r )implies omega=omega_(0)-(5)/(2)(mu g)/(r ) t`
When the pure rolling starts
`omega=(V_(1)+V_(2))/(r )impliesomega_(0)-(5)/(2)(mu g t)/(r )=(2mu g t)/(r ) implies t= (2omega_(0)r)/(9 mu g)`
Velocity of sphere `V_(2)= mu g t=(2omega_(0)r)/(9)`
`s_(p)=(1)/(2)a_(1)t^(2)=(1)/(2)mu g((2omega_(0)r)/(9 mu g)) = (2)/(81)(omega_(0)^(2)r^(2))/(mu g)`