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. `5.0`
B. `6.7`
C. `9.4`
D. `13.3`

Answer

A

Explanation

There is no external torque acting, so angular momentum is constant.
`(dvecL)/(dt)=vectau_(ext)=0tovecL=Iomega`= constant
It means magnitude and direction both are fixed for that rotation of particle, therefore,
`mvr=Iomega`
Using conservation of angular momentum
`mv(a)/(2)=(2)/(3)ma^(2)omega`
`omega=(3v)/(4a)=5` rad/s