Questions
A circular disc of moment of inertia is rotating in a horizontal plane, about its geometrical axis, with a constant angular speed . Another disc of moment of inertia 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 . The energy lost by the initially rotating disc due to friction is:
detailed solution
Correct option is D
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-KEf= 12Iiωi2 -(12IiIfωi2)/(Ii+If)=12IiIf(Ii+If)2ωi2Talk to our academic expert!
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A disc rotates freely with rotational kinetic energy E about a normal axis through centre. A ring having the same radius but double the mass of disc is now, gently placed on the disc. The new rotational kinetic energy of the system would be
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