Question

A smooth horizontal disc rotates about the vertical axis O (Fig. 4.15) with a constant angular velocity $\omega$. A thin uniform rod AB of length l performs small oscillations about the vertical axis A fixed to the disc at a distance a from the axis of the disc. Find the frequency ${\omega }_o$ of these oscillations.

Answer 1

Let us go to the rotating frame, in which the disc is stationary. In this frame the rod is subjected to coriolis and centrifugal forces, $F_{coe}$ and $F_{cf}$ where
$F_{cor}\int 2dm(v^{{}^{\prime }}\times {\overrightarrow{\omega }}_0)$ and $F_{cf}\int dm{\omega }_0^{2}r$
where $r$ is the position of an element mass of the rod (Fig) with respect to point $O$ (disc's centre) and
$v^{{}^{\prime }}=\frac{d{r}^{{}^{\prime }}}{dt}$
A s $r=OP=OA+AP$
So, $\frac{dr}{dt}=\frac{d\left(AP\right)}{dt}=v^{{}^{\prime }}$ (As $OA$ is constant)
As the rod is vibrating transversely, so $v^{{}^{\prime }}$ is directed perpendicular to the length of the rod. Hence $2dm(v^{{}^{\prime }}\times \overrightarrow{\omega })$ for each element mass of the rod is directed along $PA$. Therefore the net torque of coriolis about $A$ becomes zero. The not torque of centrifugal force about point $A:$
Now, ${\overrightarrow{\tau }}_{f\left(A\right)}=AP\times dm{\omega }_0^{2}r=\int AP\times \left(\frac{m}{l}\right)ds{\omega }_0^2(OA+AP)$
$AP\times \left(\frac{m}{l}ds\right){\omega }_0^{2}OA=\int \frac{m}{l}ds{\omega }_0^{2}sa\sin \theta (-k)$
$\frac{m}{l}{\omega }_0^{2}a\sin \theta (-k){\int }_0^{1}sds=m{\omega }_0^{2}a\frac{1}{2}\sin \theta (-k)$
So, ${\tau }_{cf\left(x\right)}={\overrightarrow{\tau }}_{cf\left(A\right)}.K=-m{\omega }_0^{2}a\frac{1}{2}\sin \theta$
According to the equation of rotational dynamic :${\tau }_{A\left(Z\right)}=l_A{\alpha }_Z$
or, $-m{\omega }_0^{2}a\frac{1}{2}\sin \theta =\frac{m{l}^2}{3}\ddot{\theta }$
or, $\ddot{\theta }=-\frac{3}{2}\frac{{\omega }_0^{2}a}{l}\sin theta$
Thus, for small $\theta ,\ddot{\theta }=-\frac{3}{2}\frac{{\omega }_0^{2}a}{2l}\theta$
This implies that the frequency ${\omega }_0$ of oscillation is ${\omega }_0=\sqrt{\frac{3{\omega }^2}{2l}}$