Asked by Md shaheen Mahmud
a thin metal ring of diameter is 0.6m and mass 1 kg starts rest and rolls down inclined plan.Its linear velocity is 5m/s .Calculate the moment of inertia and kinetic energy of rotation.
Answers
Answered by
Damon
omega is angular velocity in radians/second
omega R = 5 m/s
so
omega = 5/.3 = 16.7 radians/second
Ke = (1/2)I omega^2
omega R = 5 m/s
so
omega = 5/.3 = 16.7 radians/second
Ke = (1/2)I omega^2
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