A uniform cylindrical pulley of mass `M` and the radius `R` can freely rotate about the horizontal axis `O` (figure). The free end of a thread tightly wound on the pulley carries a deadweight `A`. At a certain angle `alpha` it counterbalanes a point mass `m` fixed at the rim of the pulley. Find the frequency of small osicllations of the arrangement.

Explanation

At equillibrium position, `N_(o z)=0(` Net torque about 0)
So, `m_(A)gR-mg R sin alpha=0 `or ` m_(A)= m sin alpha ...(1)`
From the equation of rotational dynamics of a solid body about the stationary axis ( say `z-` axis) rotation i.e. from `N_(z)=I beta_(z)`
when the pulley is rotated by the small angular displacemetn `theta` in clockwise sense relative to the equilibrium position (figure) , we get `:`
`m_(A)gR-mgRsin (alpha+theta)`
`=[(MR^(2))/(2)+mR^(2)+m_(A)R^(2)]ddot(theta)` ltbr. Using Eqn. `(1)`
`mg sin alpha -mg( sin alpha cos theta+ cos alpha sin theta)`
`={(MR+2m (1+ sin alpha)R)/(2)}ddot(theta)`
But for small `theta` , we amy write `cos theta ~= 1 ` and `sin theta~=theta`
Thus we have
`mg sin alpha- mg ( sin alpha + cos alpha theta)=({MR+2m (1+ sin alpha )R})/(2)ddot(theta)`
Hence the sought angular frequency `omega_(0)sqrt((2mg cos alpha)/(MR+2mR(1+sin alpha)))`