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

When the cord is pulled by the weight (mg), the tension in the cord, provides the necessary torque to the cylinder to rotate it with angular acceleration `(alpha)`.
`tau = Tr and tau = Ialpha" "therefore Ialpha=Tr`
The linear acceleration of the mass is a `= ralpha`
`therefore" "alpha=(a)/(r)" "therefore (Ia)/(r)=Tr" "therefore T=(Ia)/(r^(2))`
To calculate the acceleration, we use
`ma=mg -T=mg-(Ia)/(r^(2))`
`therefore" "mg=am+(Ia)/(r^(2))=a(m+(I)/(r^(2)))`
`therefore" "a=(g)/((1+(1)/(mr^(2))))=(mg^(2))/((I+mr^(2)))`
`because v^(2)=u^(2)+2ah=2ah" "(because u=0)`
`therefore" "v^(2)=2((mg^(2))/(I+mr^(2)))xxh=(r^(2)(2mgh))/(I+mr^(2))`
`because" "v^(2)=u^(2)+2ah=2ah" "(because u=0)`
Second Method:
The initial P.E. of the system `("i.e. wheel "+"weight"+"cord")`
`=U_(i)=0`
The intial K.E. of the system `=K_(i)=0`
Final P.E. of the system, when the weight falls through a distance `h(U_(f))=-mgh`
and Final K.E. of the system `(K_(f))`
`=(1)/(2)mv^(2)+(1)/(2)Iomega^(2)=(1)/(2)m.r^(2)omega^(2)+(1)/(2)Iomega^(2)`
`=(1)/(2)omega^(2)(mr^(2)+I)`
By the principle of conservation of energy
`U_(i)+K_(i)=U_(f)+K_(f)`
`therefore 0+0=-mgh+(1)/(2)omega^(2)(mr^(2)+I)`
By the principle of cnservation of energy
`U_(i)+K_(i)=U_(f)+K_(f)`
`therefore 0+0=-mgh+(1)/(2)omega^(2)(mr^(2)+1)`
`therefore 2mgh=omega^(2)(mr^(2)+I)`
`therefore" "omega^(2)=(2mgh)/(mr^(2)+I)`
`therefore" "omega=[(2mgh)/(I+mr^(2))]^(1//2)`