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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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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. Find the speed of the particle.

$\begin{array}{1 1} v=rg\;\tan\;\theta \\ v= \sqrt{(rg\;\tan \theta)} \\v=mg \;\cos \theta\\ v=\sqrt{(mg \cos \theta )}\end{array} $

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Answer : $ v= \sqrt{(rg\;\tan \theta)}$
Consider the string to make an angle $\theta$ with vertical as shown in the figure and the particle moves in a horizontal circular path of radius $r$.
Let the observer be standing on the ground.
According to the observer, only two forces act on the particle namely weight (mg downward) and tension (T).
On resolving tension along X and Y axes, T sin $\theta$ will contribute to centripetal force as it is the net force towards the centre of circle.
Along X-axis : $F_{net},x=T \sin \theta =\large\frac{mv^2}{r}$ ------(1)
Along Y-axis : $F_{net},y=0$ (as the particle has no motion along Y-axis)
$T \cos \theta =mg$                                             ---------(2)
Therefore tension in the string $T= \large\frac{mg}{\cos \theta}$
(Where $\cos \theta =\large\frac{\sqrt {l^2-r^2}}{r}$)
Dividing equation (1) by equation (2)
Speed of the particle $v= \sqrt {rg \tan \theta}$
(Where $\tan \theta=\large\frac{r}{\sqrt {l^2-r^2}} $)


answered Aug 26, 2014 by meena.p
edited Sep 2, 2014 by balaji.thirumalai

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