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:

$$\begin{gathered} \frac{1}{2}M{v}^2+\frac{1}{2}I{\omega }^2+\frac{1}{2}k\left[(x+\delta {)}^2-{\delta }^2\right]-Mgx=\mathrm{Constant} \\ \Rightarrow \frac{1}{2}M{v}^2+\frac{1}{2}I{\omega }^2+\frac{1}{2}k{x}^{\mathit{2}}+kxd\mathit{-}Mgx=\mathrm{Constant} \\ \Rightarrow \frac{1}{2}M{v}^2+\frac{1}{2}I\left(\frac{v^2}{r^2}\right)+\frac{1}{2}k{x}^{\mathit{2}}=\mathrm{Constant}\left[\because \delta =\frac{Mg}{k}\right] \end{gathered}$$
By taking derivatives with respect to t, on both sides, we have:

$$\begin{gathered} Mv.\frac{\mathrm{d}v}{\mathrm{d}t}+\frac{I}{r^2}v.\frac{\mathrm{d}v}{\mathrm{d}t}+kx\frac{\mathrm{d}x}{\mathrm{d}t}=0 \\ Mva+\frac{I}{r^2}va+kxv=0\left(\because v=\frac{\mathrm{d}x}{\mathrm{d}t}\mathrm{and}a=\frac{\mathrm{d}v}{\mathrm{d}t}\right) \\ a\left(M+\frac{I}{r^2}\right)=-kx \\ \Rightarrow \frac{a}{x}=\frac{k}{M+{\displaystyle \frac{I}{r^2}}}={\omega }^2 \\ T=\frac{2\pi }{\omega } \\ \Rightarrow T=2\pi \sqrt{\frac{M+{\displaystyle \frac{I}{r^2}}}{k}} \end{gathered}$$