A uniform ring is rotating about vertical axis with angular velocity `omega` initally. A point insect (S) having the same mass as that of the ring starts walking from the lowest point `P_(1)` and finally reaches the point `P_(2)` (as shown in figure). What is the final angular velocity of the ring ?
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Answer

`omega//3`

Explanation

from conservation of angular momentum
`((1)/(2) MR^(2)) omega = ((1)/(2) MR^(2) + MR^(2)) omega'`
`((1)/(2) MR^(2)) omega = ((3)/(2) MR^(2)) omega' rArr omega' = (omega)/(3)`.