Physics - Circular Motion Question with Solution | TestHub

Let the tension in the string be $T$. Since the bead is rotating about B with angular speed $\omega$, its acceleration is ${\omega }^{2}r$ towards B.

From the figure:

$${\mathrm{a}}^2+{\mathrm{r}}^2=(\ell -\mathrm{r}{)}^2$$

$$\Rightarrow {\mathrm{a}}^2+{\mathrm{r}}^2={\ell }^2+{\mathrm{r}}^2-2\ell \mathrm{r}$$

$$\Rightarrow \mathrm{r}=\frac{{\ell }^2-{\mathrm{a}}^2}{2\ell }\dots (\mathrm{i})$$

Now, from the Free Body Diagram (F.B.D.) of the bead:

$$T\sin \theta +T=m{\omega }^{2}r..\dots \dots \left(2\right)$$

$$T\cos \theta =mg..\dots \dots \left(3\right)$$

Dividing the two equations above, we have:

$$\frac{1+\sin \theta }{\cos \theta }=\frac{{\omega }^{2}r}{g}$$

From the figure:

$$\sin \theta =\frac{r}{\ell -r}and\cos \theta =\frac{a}{\ell -r}$$

Substituting these into the equation:

$$\Rightarrow \frac{1+\frac{r}{\ell -r}}{\frac{a}{\ell -r}}=\frac{{\omega }^{2}r}{g}$$