Question

A ring of mass ‘M’ and radius R is rotating about its axis with angular velocity . Two identical bodies each of mass ‘m’ are now gently attached at the two ends of a diameter of ring. The kinetic energy loss will be 1) m M 2m   2 2 R M  2) Mm 2 2 R M m   3) Mm 2 2 R M 2m   4) M M m   2 2 R

Answer 1

Answer:

Explanation:

Iw = const

MR² X W = (M +2m)R² x w'

w' = M W/(M +2m)

k ini = 1/2 MR² W²

K final = 1/2  (M +2m)R²  [M W/(M +2m)]²

Kf - Ki = Mm/M+2m  W²R²