A disc is freely rotating with an angular speed `omega` on a smooth horizontal plane. If it is hooked at a rigid pace `P` and rotates without bouncing about a point on its circumference. Its angular speed after the impact will be equal to.

Options

A. `(2omega)/3`
B. `omega/3`
C. `omega/2`
D. none of the above

Answer

B

Explanation

During the impact the impact forces pass through point `P`. Therefore, the torque produced by it about `P` is equal to zero. Cosequently the angular momentum of the disc about `P`, just before and after the impact, remains the same
`impliesL_(2)=L_(1)`

where `L_(1) =` angular momentum of the disc about `P` just before the impact
`I_(0)omega=(1/2mr^(2)+mr^(2))omega'=3/2mr^(2)omega'`
Just before the impact the disc rotates about `O`. But just after the impact the disc rotates about `P`.
`implies 1/2mr^(2)omega=3/2mr^(2)omega'impliesomega'=1/3omega`