Two discs of same moment of inertia rotating their regular axis passing through centre and perpendicular to the plane of disc with angular velocities `omega_(1)` and `omega_(2)`. They are brought into contact face to the face coinciding the axis of rotation. The expression for loss of enregy during this process is :

Options

A. `(1)/(2)I( omega_(1)+omega_(2))^(2)`
B. `(1)/(4)I( omega_(1)-omega_(2))^(2)`
C. `2I( omega_(1)-omega_(2))^(2)`
D. `(1)/(8)I( omega_(1)-omega_(2))^(2)`

Answer

B

Explanation

The angular momenta of the two discs can be written as:
`L_(1)=Iomega_(1)`
`L_(2)=Iomega_(2)`
Let the angular speed of the discs when they are brought in contact be `omega`
Applying conservation of angular momentum
`Iomega_(1)+Iomega_(2)=2Iomega`
`[:.` Moment of inertia of the system of the two discs is same as each of them]
`impliesomega=(omega_(1)+omega_(2))/(2)`
Total initial kinetic energy of the two discs:
`(K.E.)_(i)=(1)/(2)Iomega_(1)^(2)+(1)/(2)Iomega_(2)^(2)`
Total final kinetic energy of the two discs:
`(K.E.)_(f)=(1)/(2)xx2Iomega^(2)=I((omega_(1)+omega_(2))/(2))^(2)`
Loss in kinetic energy `=(K.E.)_(i)-(K.E.)_(f)`
`=(1)/(4)I(omega_(1)-omega_(2))^(2)`