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.

Options

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)`

Answer

C

Explanation

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)))`