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

Options

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

Answer

A

Explanation

L=Angular Momentum after brought in contact
`L_(1)=` Angular Moment of I disc
`L_(2)=`Angular Momentum of II disc
`L=L_(1)+L_(2)`
`L=Iomega, L_(1)=I omega_(1),L_(2)=Iomega_(2)`
`:.2Iomega=Iomega_(1)+Iomega_(2)`
`omega=(omega_(1)+omega_(2))/2`
`DeltaK.E. =K.E_("initial")-K.E_("initial")`
`=[1//2Iomega_(1)^(2)+1//2Iomega_(2)^(2)]-1//2(2I)omega^(2)`
solving `=1//4I(omega_(1)-omega_(2))^(2)`