CBSE Class 11-science Answered
pls help
Asked by tanay15 | 12 Jan, 2010, 06:22: PM
Expert Answer
MI = MR2/2 = 2.5 kg m2
Torque = (10 N)(0.25 m) = = 2.5 N m
Torque = MI x Angular acceleration
(10 N)(0.25 m) = ((80 kg)(0.25 m)2/2) x Angular Acceleration,
Angular Acceleration = 1 rad/s2
Final Angular Velocity = Initial Angular Velocity + (Angular Acceleration)(Time)
= 5 + (1)(20) = 25 rad/s
Work done by the torque = Change in KE = I(wf2-wi2)/2 = 2.5(252-52)/2 = 750 J
Regards,
Team,
TopperLearning.
Answered by | 12 Jan, 2010, 10:08: PM
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