A disc of the moment of inertia `'l_(1)'` is rotating in horizontal plane about an axis passing through a centre and perpendicular to its plane with constant angular speed `'omega_(1)'` . Another disc of moment of inertia `'I_(2)'`. having zero angular speed is placed discs are rotating disc. Now, both the discs are rotating with constant angular speed `'omega_(2)'`. The energy lost by the initial rotating disc is

Options

A. `1/2[(l_(1) + l_(2))/(l_(1)l_(2))]omega_(1)^(2)`
B. `1/2[(l_(1) l_(2))/(l_(1)-l_(2))]omega_(1)^(2)`
C. `1/2[(l_(1)-l_(2))/(l_(1)l_(2))]omega_1^(2)`
D. `1/2[(l_(2)l_(2))/(l_(1) + l_(2))]omega_(1)^(2)`

Answer

D

Explanation

Net external torque on the system is zero.
Therefore, angular momentum of the system will remain same.
`rArr l_(1)omega_(1) = (l_(1) + l_(2)) omega_(2)`.
`(omega_(2))/(omega_(1)) = (l_(1))/(l_(1) + l_(2))`
The energy lost, `E_(1) - E_(2)`
`= 1/2 l_(1)omega_(1)^(2) - 1/2 (l_(1) +l_(2)) omega_(2)^(2)`
`= 1/2 omega_(1)^(2) [l_(1) - (l_(1) + l_(2))(omega_(2)^(2))/(omega_(1)^(2))]`
`= 1/2 omega_(1)^(2) [l_1 - (l_(1) + l_(2)) (l_(1)^(2))/((l_(1) + l_(2))^(2))]`
[ `:'` using Eq. (i)]
`= 1/2 omega_(1)^(2) [(l_(1)^(2) + l_(1)l_(2) - l_(1)^(2))/(l_(1) + l_(2))]`
`= 1/2 [(l_(1)l_(2))/(l_(1) + l_(2))] omega_(1)^(2)`