Question

Solve the previous problem if the pulley has a moment of inertia I about its axis and the string does not slip over it.

Answer 1

Let us try to solve the problem using energy method.
If δ is the displacement from the mean position then, the initial extension of the spring from the mean position is given by,
$$\delta =$$ $$\frac{mg}{k}$$

Let x be any position below the equilibrium during oscillation.
Let v be the velocity of mass m and ω be the angular velocity of the pulley. 
If r is the radius of the pulley then
v = rω
As total energy remains constant for simple harmonic motion, we can write:

$$\frac{1}{2}M{v}^2+\frac{1}{2}I{\omega }^2+\frac{1}{2}k[(x+\delta {)}^2-{\delta }^2]-Mgx=\text{ Constant }$$ 

$$\Rightarrow \frac{1}{2}M{v}^2+\frac{1}{2}I{\omega }^2+\frac{1}{2}k{x}^2+kxd-Mgx=\text{ Constant }$$ 

$$\Rightarrow \frac{1}{2}M{v}^2+\frac{1}{2}I\left(\frac{v^2}{r^2}\right)+\frac{1}{2}k{x}^2=\text{ Constant }[\because \delta =\frac{Mg}{k}]$$ 

By taking derivatives with respect to t, on both sides, we have:

$$Mv.\frac{dv}{dt}+\frac{I}{r^2}v.\frac{dv}{dt}+kx\frac{dx}{dt}=0$$ 

$$Mva+\frac{I}{r^2}va+kxv=0(\because v=\frac{dx}{dt}\text{ and }a=\frac{dv}{dt})$$ $$a(M+\frac{I}{r^2})=-kx$$ $$\Rightarrow \frac{a}{x}=\frac{k}{M+\frac{I}{r^2}}={\omega }^2$$  $$T=\frac{2\pi }{\omega }$$ 

$$\Rightarrow T=2\pi \sqrt{\frac{M+\frac{I}{r^2}}{k}}$$