Rotational MotionHard
Question
A ring of mass m and radius R has three particles attached to the ring as shown in the figure. The centre of the ring has a speed v0. The kinetic energy of the system is : (slipping is absent)
Options
A.6 mv02
B.12 mv02
C.4 mv02
D.8 mv02
Solution
KEsystem = 
mv2(1 + 1) + 
(m + 2m) 2v2 +
m(2v)2 = 6mv2
mv2(1 + 1) + (m + 2m) 2v2 +
m(2v)2 = 6mv2Create a free account to view solution
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