InterviewSolution
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InterviewSolution
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A cord is wound round the circumference of wheel of radius `r`. The axis of the wheel is horizontal and fixed and moment of inertia about it is `I`. A weight `mg` is attached to the end of the cord and falls from rest. After falling through a distance `h`, the angular velocity of the wheel will be.A. `[mgh]^(1//2)`B. `[(2mgh)/(I+2mr^(2))]^(1//2)`C. `[(2mgh)/(I+mr^(2))]^(1//2)`D. `[(mgh)/(I+mr^(2))]^(1//2)` |
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Answer» Correct Answer - C PE = Rolling KE `mgh = (1)/(2) mv^(2) [1+k^(2)//r^(2)]` `v^(2)=(2mgh)/(m[1+k^(2)//r^(2)])` `therefore v = sqrt((2mgh)/(m[1+k^(2)//r^(2)]))` `omega = sqrt((2mgh)/(mr^(2)+mk^(2)))=sqrt((2mgh)/(I+mr^(2)))` |
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