A cord is wound around the circumference of wheel of radius r. The axis of the wheel is horizontal and `MI` 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. `sqrt((2gh)/(I+mr^(2)))`
B. `((2mgh)/(I+mr^(2)))^(1//2)`
C. `((2mgh)/(I+2mr^(2)))^(1//2)`
D. `sqrt(2gh)`

Answer

B

Explanation

Velocity, `v = r omega`
Decrease in gravitational potential energy = Increase in kinetic energy

`:. mgh=(1)/(2)Iomega^(2)+(1)/(2)mv^(2)=(1)/(2)Iomega^(2)+(1)/(2)m(r omega)^(2)`
`:.` Angular velocity, `omega=sqrt((2mgh)/(I+mr^(2)))`.