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

Explanation

For rotatiobnal equation the tension in spring on both sides must be same .
Let T be the tension in the string in equilibrium position and `l_(0)` the extension of spring . Then for equilibrum of any part of spring on RHS.
`2 T = mg implies2 k l_(0) = mg or l_(0) = (mg)/(2 K)`
For translational equation of pulley `l_(0) = (mg)/(2 K)`
Thus when pulley attains equilibrium , the spring is streched by a distance
`l_(0) = (mg)/(2 K)` (i)
Now suppose that the pulley is down a little and released. The pulley starts up and down oscillation . Let y be instant neous displacement of center of pulley from equilibrium. Then total increase in length of (string + spring) is `2y` (y to the left of pulley and y to the right).
As steing is inextensible, total extension of spring is `2 y`.
As pulley also ratates energy of system is
`E = (1)/(2) l omega^(2) + (1)/(2) m v^(2) - mgy + (1)/(2) K (l_(0) + 2 y)^(2)`
But `omega = (v)/( r)`
`E = (1)/(2) l (v^(2))/( r^(2)) + (1)/(2) K (l_(0) + 2 y)^(2)`
`= (1)/(2) ((l)/r^(2) + m) v^(2) = mgy + (1)/(2) K (l_(0) + 2 y)^(2)`
Asd total energy is constant (system is conservative)
`(dE)/(dt) = 0`
` (1)/(2) ((l)/r^(2) + m) 2v (dv)/(dt) - mg (dv)/(dt) + (1)/(2) K 2 (l_(0) + 2 y) , 2 (dy)/(dt) = 0`
and `(dy)/(dt) = v and (dv)/(dt) = acceleration a`
and `l_(0) = (mg)/(2K)`
we get `((l)/r^(2) + m) a - mg + 2 K ((mg)/(2 k) + 2y) = 0`
`a prop - y` (ii)
As `a prop - y`, motion of pulley is SHM.
Standard equiation of SHM
`a = - omega^(2) y`
Comparing Eqs, (ii) and (iii), `omega^(2) = (4 k)/((l)/r^(2) + m)`
time period.
`T = (2 pi)/(omega) = 2 pi sqrt({{(l)/(r^(2)) + m)/(4K)}} = pi sqrt({{m + (l)/(r^(2)))/(K)}}`