A uniform dis of mass `m` and radius `R` rotates about a fixed vertical axis passing through its centre with angular velocity `omega`. A particle of same mass `m` and having `2 omega R` towards centre of the disc collides with the disc moving horizontally and sticks to its rim. Find
(A) the angular velocity of the disc.
(B) the impulse on the particle due to disc
( C) the impulse on the disc due to hinge.

Answer

(a) `omega//3`,
(b) `(sqrt(37))/(3) m omega R`
( c) `(sqrt(37))/(3) m omega R`.

Explanation

(a) From conservation of angular momentum

:. `I_(1) omega_(1) = I_(2) omega_(2)`
`((1)/(2) mR^(2)) omega =((1)/(2) mR^(2) + mR^(2)) omega_(f)`
`(1)/(2) mR^(2) omega = ((3)/(2) mR^(2)) omega_(f)`
`omega_(f) = (omega)/(3)`
(b) Impulse = change in linear momentum
`vec J = mv_(f) hat j - m(2 omega R) (- hat i)`
`vec J = m((R omega)/(3)) hat j + 2 m omega R hat i`
`|vec J| = sqrt((m(R omega)/(3))^(2) + 2 m omega R^(2))`
`|vec J| = m R omega sqrt((1)/(9) + 4) rArr |vec J| = (sqrt(37))/(3) mR omega`
( c) Impulse on disc is equal and opposite to impulse on particle due to disc.
:. `J = (sqrt(37))/(3) mR omega`.