A uniform disc of mass `m` and radius `R` rotates about a fixed vertical axis passing through its centre with angular velocity `omega`. A particle of same angular velocity mass m and moving horizontally with velocity `2omegaR` towards centre of the disc collides and sticks to its rim. The impulse on the particle due to the disc is

Options

A. `(sqrt(17))/3momegaR`
B. `(sqrt(35))/3momegaR`
C. `(sqrt(37))/3momegaR`
D. `(sqrt(29))/3momegaR`

Answer

C

Explanation

Apply conservation of angular momentum,
`(mR^(2))/2omega=((mR^(2))/2+mR^(2))omega'`
`omega'=omega/3`

`|P_(l)|=2mRomega`
Impulse on particle` =` change in linear momentum

`vecJ=/_\vecp=vecp_(j)-vecp_(i)`
`|/_\vecp|=sqrt(((mRomega)/3)^(2)+(2mromega)^(2))=sqrt(37/3)mRomega`
Impulse on hinge is negative of impulse on disc.