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3-2.Motion in Plane
hard
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 $\omega$ about the vertical diameter $AB$, as shown in figure, the bead is at rest with respect to the circular ring at position $P$ as shown. Then the value of $\omega^2$ is equal to
A$\frac{{\sqrt 3 g}}{{2r}}$
B$\left( {g\sqrt 3 } \right)/r$
C$2g / r$
D$2g/\left( {r\sqrt 3 } \right)$
(JEE MAIN-2019)
Solution
$\begin{array}{l}N\,\sin \,\theta = m\frac{r}{2}{\omega ^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,…\left( i \right)\\
N\,\cos \,\,\theta = mg\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,…\left( {ii} \right)\\
\tan \,\,\theta \, – \frac{{r{\omega ^2}}}{{2g}}\\
\frac{r}{{2\frac{{\sqrt 3 \,r}}{2}}} = \frac{{r{\omega ^2}}}{{2g}}\,\,\,;\,\,{\omega ^2} = \frac{{2g}}{{\sqrt 3 \,r}}
\end{array}$
Standard 11
Physics
