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Q.

A circular disc of moment of inertia Ii is rotating in a horizontal plane, about its geometrical axis, with a constant angular speed ωi. Another disc of moment of inertia If is dropped coaxially onto the rotating disc. Initially the second disc has zero angular speed. Eventually both the discs rotate with a constant angular speed ωf. The energy lost by the initially rotating disc due to friction is:

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a

12IiIf(Ii+If)ωi2

b

12If2(Ii+If)ωi2

c

12Ii2(It+If)ωi2

d

12If-Ii(If+Ii)ωi2

answer is D.

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Detailed Solution

Initial KE of system=  KE1=  12Iiωi2+12If02=12Iiωi2

  Conserving angular momentum, common angular velocity    ωf=Iiωi+If0Ii+If

            Final KE=KE2=12[Ii+If] ωf2 = 12(Ii+If)[Iiωi/(Ii+If)]2=IiIfω2i2(Ii+If)

                     KE lost=KE=KEi-KEf12Iiωi2 -(12IiIfωi2)/(Ii+If)=12IiIf(Ii+If)2ωi2

 

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A circular disc of moment of inertia Ii is rotating in a horizontal plane, about its geometrical axis, with a constant angular speed ωi. Another disc of moment of inertia If is dropped coaxially onto the rotating disc. Initially the second disc has zero angular speed. Eventually both the discs rotate with a constant angular speed ωf. The energy lost by the initially rotating disc due to friction is: