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:

Question

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:

Infinity Learn · Accepted Answer

Initial $$KE$$ of $$system=$$  $$KE1=$$  $$12Ii$$ω$$i2+12If02=12Ii$$ω$$i2$$  Conserving angular momentum, common angular velocity    ω$$f=Ii$$ω$$i+If0Ii+If$$            Final $$KE$$$$=KE2=12$$[$$Ii+If$$] ω$$f2$$ $$=$$ $$12(Ii+If)$$[Iiω$$i/(Ii+If)$$]$$2=IiIf$$ω$$2i2(Ii+If)$$                     $$KE$$ $$lost=$$∆$$KE$$$$=KEi-KEf=$$ $$12Ii$$ω$$i2$$ $$-(12IiIf$$ω$$i2)/(Ii+If)=12IiIf(Ii+If)2$$ω$$i2$$

Detailed Solution

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