A cubical block of side a is moving with velocity V on a horizontal smooth plane as shown in Figure. It hits a ridge at point O. The angular speed of the block after it hits O is

Options

A. `3V//(4a)`
B. `3V//(2a)`
C. `sqrt(3V)//(sqrt(2)a)`
D. `zero`

Answer

A

Explanation

`r=sqrt(2a)/2` or `r^(2)=sqrt(2)(a^(2))/2`
Net torque about `O` is zero.
Therefore, angular momentum `L` about `O` will be conserved
or `L_(i)=L_(f)`
`MV(a/2)=I_(0)omega=(I_(CM)+Mr^(2))omega`
`{(Ma^(2))/6+M(a^(2))/2}omega`
`omega=(3v)/(4a)`