A smooth horizontal disc rotates about the vertical axis `O` (figure) with a constant angular velocity `omega`. A thin uniform rod `AB` of length `l` performs small oscillation about the vertical axis `A` fixed to the disc at a distance `a` from the axis of the disc. Find the frequency `omega_(0)` of these oscillations.

Explanation

Let us go to the rotating frame, in which the disc is stationary. In this frame the rod is subjected to coriolis and centrigugal forces, `F_(co r)` and `F_(c f )`, where `F_(co r)=int2 dm (v^(')xxvec(omega_(0)))` and `F_(c f )=int d m omega_(0)^(2)r`, where `r` is the position of an elemental mass of the rod. with respect to point `O` ( disc's centre `0` and
`v^(')=(dr^('))/(dt)`
As, `r=OP=OA+AP`
So, `(dr)/(dt)=(d(AP))/(dt)=v^(')` `(` as `OA` is constant`)`
As the rod is vibrating transversely , so `v^(')` is directeld perpendicular to the length of the rod. Hence `2 dm (v^(')xxvec(omega))` for each elemental mass of the rod is directed along `PA`. Therefore the net torque about `A` becomes zero. The not torque of centrifulgal force about point `A:`
Now, `vec(tau)_(cf(A))=intAPxxdm omega_(0)^(2)r = int AP xx ((m)/(l))ds omega_(0)^(2)(OA+AP)`
`=intAPxx((m)/(l)ds)omega_(0)^(2)OA=int (m)/(l)d s omega_(0)^(2)s a sin theta(-k)`
`=(m)/(l) omega_(0)^(2)a sin theta (-k)int _(0)^(t)s ds =m omega_(0)^(2)a (l)/(2) sin theta (-k)`
So `tau _(cf (Z))=vec (tau)_(c f (A)). k=-m omega_(0)^(2) a (l)/(2) sin theta`
According to the equation of rotational dynamics `: tau _(A(Z))=l_(A)alpha_(Z)`
or ` - m omega_(0)^(2)a (l)/(2) sin theta=(ml^(2))/(3)ddot(theta)`
or ` .ddot(theta)=-(3)/(2)(omega_(0)^(2)a)/(l)sin theta`
Thus, for small `theta`, ` ddot(theta)=-(3)/(2)(omega_(0)^(2)a)/(l)sin theta`
This implies that the frequency `omega_(0)` of oscillation is `omega_(0)=sqrt((3 omega^(2)a)/(2l))`