A disc of mass m1 is freely rotating with constant angular speed ω . Another disc of mass m2 & same radius is gently kept on the first disc. If the contacting surfaces are rough, the fractional decrease in kinetic energy of the system will be

Q: A disc of mass m1 is freely rotating with constant angular speed ω . Another disc of mass m2 & same radius is gently kept on the first disc. If the contacting surfaces are rough, the fractional decrease in kinetic energy of the system will be
(A)m1/m2
(B)m2/(m1+m2)
(C)m2/m1
(D) m1/(m1+ m2)

Solution: The upper disc speeds up and the lower disc slows down by the accelerating & rotating frictional retarding torques. However, the torque acting on the system is zero. Therefore, the angular momentum of the system remains constant.
 Linitial = Lfinal = L
rotation = 1-Iinitial/Ifinal
Iinitial = ½m1
Ifinal = ½m1r² + ½m2
B putting these values ,
ΔKE/KEinitial = m2/(m1+m2)

Author: Rajesh Jha

QuantumStudy.com for CBSE|IIT-JEE|NEET

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