The pulley shown in figure has a moment of inertias I about its axis and mass m. find the time period of vertical oscillation of its centre of mass. The spring has spring constant k and the string does not slip over the pulley.

Explanation

Let us first find the equilibrium position. For rotatioN/Al equilibrium of the pulley, the tensions in the two strings should be equal. Only then the torque on the pulley wil be zero. Let this tension be T. The extension of the spring wil be `y=T/k`, as teh tension in the spring will be the same as the tension in the string. For translatioN/Al equilibrium of the pulley.
`2T=mgor 2ky=mg, or ly=(mg)/(2k)`
The spring is extended by a distance `(mg)/(2k)` when the pulley is in equilibrium.
Now suppose the centre of the pulley goes down further by a distance x. The total inrease in tehlengthof the stringplus the springnis `2x(x` on the left of the pulley and x on the right). As the string has a constant length, the extension of the spring is 2x. The energy of the system is
`U=1/2iomega^2+1/2mv^2-mgx+1/2k((mg)/(2k)+2x)^2 `
`=1/2(1/r^2+m)v^2+(m^2g^2)/(8k)+2kx^2`
As the system is conservative `(dU)/(dt)=0, ` ltbr. giving `0=(I/r^2+m)v(dv)/(dt)+4kxv`
`or (dv)/(dt)=-(4kx)/((I/r^2+m))`
or `a=-omega^2x, where omega^2=(4k)/((I/r^2+m))`
Thus the centre of mass of the pulley executes a simple harmonic motion with time period
`T=2pisqrt((I/r^2+m)/(4k))`.