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)/(1+2mr^(2))]^(1//2)`
C. `[(2mgh)/(1+mr^(2))]^(1//2)`
D. `[(mgh)/(1+mr^(2))]^(1//2)`

Answer

C

Explanation

Applying energy conservation, we have
`U_(i)+K_(i)=U_(f)+K_(f)`
where,
`U_(i)`=initial potential energy of the (block + pulley) system
`U_(f)`= final potential energy of the (block + pully) system
`K_(i)`= initial kinetic energy of the system
`k_(f)`= final kinetic energy of the system

Here, initial situation corresponds to rest position of the system and fianl situation corresponds to position after falling through height h.
Eq. (i) gives
`0+0=-mgh +(1)/(2)mv^(2)+(1)/(2)l omega^(2)`
`rArr" "mgh=(1)/(2)m(omegar)^(2)+(1)/(2)l omega^(2)`
`=(1)/(2)m omega^(2)r^(2)+(1)/(2)l omega^(2)`
`rArr" "2mgh = omega^(2)[mr^(2)+l]`
`rArr" "omega^(2)=(2mgh)/(1+mr^(2)) or omega=[(2mgh)/(l+mr^(2))]^(1//2)`