A disc of mass m and radius r is gently placed on another disc of mass 2m and radius r. The disc of mass 2m is rotating with angular velocity `omega_(0)` initially.
The disc is placed such that axis of both are concident. The coefficient of friction is `mu` for surfaces of contact Assume that pressure on disc is uniformely distributed.

Loss in kinetic energy of system

Options

A. `(1)/(6)mr^(2)omega_(0)^(2)`
B. `(1)/(3)mr^(2)omega_(0)^(2)`
C. `(1)/(2)mr^(2)omega_(0)^(2)`
D. None of these

Answer

A

Explanation

By COE
`Mg.(L)/(2)=(1)/(2)((ML^(2))/(3))omega^(2)implies omega^(2)=(3g)/(L)`
To find tension at mid point
`T-(M)/(2)g=(M)/(2)(a_(cm))_("lower half")`
`T-(M)/(2)g=(M)/(2)(omega^(2)(3L)/(4))`
`T-(M)/(2)g=(M)/(2)((3g)/(L)(3L)/(4))`
`T=(Mg)/(2)+(9Mg)/(8)=(13Mg)/(8)`
Stress at mid point = `(T)/(A) = (13Mg)/(8A)`
`S = (13dALg)/(8A)=(13dLg)/(8)`