A smooth wire of length 2 π r is bent into a circle and kept in a vertical plane. A bead can slide…

A smooth wire of length $2\pi r$ is bent into a circle and kept in a vertical plane. A bead can slide smoothly on the wire. When the circle is rotating with angular speed $\mathrm{\omega }$ about the vertical diameter $\mathrm{A}\mathrm{B},$ as shown in figure, the bead is at rest with respect to the circular ring at position $\mathrm{P}$ as shown. Then the value of ${\mathrm{\omega }}^2$ is equal to:

1. $2\mathrm{g}/\mathrm{r}$
2. $\frac{\sqrt{3}\mathrm{g}}{2\mathrm{r}}$
3. $2\mathrm{g}/\left(\mathrm{r}\sqrt{3}\right)$
4. $\left(\mathrm{g}\sqrt{3}\right)/\mathrm{r}$

Solution

From free body diagram of bead,
$\mathrm{N}\mathrm{c}\mathrm{o}\mathrm{s}30°=\mathrm{m}\mathrm{g}$ .... (1)
$\mathrm{N}\mathrm{s}\mathrm{i}\mathrm{n}30°=\mathrm{m}\frac{\mathrm{r}}{2}{\mathrm{\omega }}^2$ ..... (2)
By dividing equation (2) by equation (1)
$\mathrm{t}\mathrm{a}\mathrm{n}30°\hspace{0.17em}=\hspace{0.17em}\frac{\mathrm{r}{\mathrm{\omega }}^2}{2\mathrm{g}}$
${\mathrm{\omega }}^2\hspace{0.17em}=\hspace{0.17em}\frac{2\mathrm{g}}{\mathrm{r}\sqrt{3}}$

Asked in: JEE Main 2019 (12 Apr Shift 2)

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