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A ring mass `m` and radius `R` has three particle attached to the ring as shown in the figure. The centre of the centre `v_(0)`. Find the kinetic energy of the system. (Slipping is absent). . |
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Answer» Correct Answer - `6mv_(0)^(2)` About instantaneous axis of rotation rolling system can be considered as pure rotation `KE = (1)/(2) I omega^(2)` …(1) here `I` = moment of inertia about instantaneous axis of rotation `I = 2m (sqrt(2) R)^(2) + m(2R)^(2) + m(sqrt(2) R)^(2) + I_(ring)` `I = 2m (sqrt(2) R)^(2) + m(2R)^(2) + m(sqrt(2) R)^(2) + mR^(2)` `I = 12 mR^(2)` putting the value of `I` in equation (1) `KE = 6mR^(2) omega^(2)` `KE = 6m(R omega)^(2)` `KE = 6mv_(o)^(2)`. |
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