Consider the situation of previous example. If the platform is rotating intially with angular velocity `omega_(0)` and then person jumps off tangentially. Determine the new angular velocity of the platform.

Explanation

Let the angular velocity of platfrom after jumps off the mass is `omega`. Then velocity Of man.
`v_(m) = v_(mp) + v_(p)`
`v_(m) = u - omega R`
From Angular momentum conservation
`(I + mR^(2)) omega_(0) = I omega - m (u - omega R) R `
`Iomega_(0) + mR^(2) omega_(0) = I omega - m u R + m omega R^(2)`
` rArr omega = ((1 + mR^(2)) omega _(0) + m u R)/((l + mR^(2)))`