3-2.Motion in Plane
hard

A particle of mass ${m}$ is suspended from a ceiling through a string of length $L$. The particle moves in a horizontal circle of radius $r$ such that ${r}=\frac{{L}}{\sqrt{2}}$. The speed of particle will be:

A$\sqrt{{rg}}$
B$\sqrt{2 {rg}}$
C$2 \sqrt{{rg}}$
D$\sqrt{\frac{r g}{2}}$
(JEE MAIN-2021)

Solution

Conical pendulum
${r}=\frac{\ell}{\sqrt{2}}$
$\sin \theta=\frac{{r}}{\ell}=\frac{1}{\sqrt{2}}$
$\theta=45^{\circ}$
${T} \sin \theta=\frac{{mv}^{2}}{{r}}$
${T} \cos \theta={mg}$
$\tan \theta=\frac{{v}^{2}}{{rg}} \Rightarrow {v}=\sqrt{{rg}}$
Standard 11
Physics

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