I found this solution for the nature frequency
but here it does not include the Ball weight and centrifugal force in the moment balance equation about the pivot (O), it is wrong answer...is not it?
I tried to solve the problem this way, moment balance about pivot(O):
$$(mb^2)θ'' = -1/2*k*(1/100+a sinθ)a conθ + mgbsinθ + mrw^2bcosθ$$
$-1/2*k*(1/100+a sinθ)a conθ$ : is the moment of the spring force
$mgbsinθ$ : is the moment of the weight of the ball.
$mrw^2bcosθ$ : is the moment of the centrifugal force.
$$a = 12 cm, b = 20 cm, k = 10^4 N/m, mg = 25N$$
for small displacment $sinθ = θ$ , $cosθ = 1$. then
$$(mb^2)θ'' = -1/2k(1/100+aθ)a + mgbθ + mrw^2b$$
r is the distance between ball center and the center of rotation.
$$ r = (16/100 + bsinθ) = (16/100 + bθ)$$
the equation becomes
$$(mb^2)θ'' = -1/2*k(1/100+aθ)a + mgbθ + m(16/100 + bθ)w^2b$$
rearrangement of the equation
$$(m*b^2)θ'' + (1/2*ka^2 + - mgb - mb^2w^2)θ + 1/2*k*1/100*a -m*16/100*w^2b = 0$$
from this equation we could say that the natural frequency is
$$ (w_n)^2 = (1/2*ka^2 + - mgb - mb^2w^2)/(mb^2) $$
which shows that the natural frequency changes with rotation speed. Is this a right solution?