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:

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

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

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

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

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

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

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