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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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