Question

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.

• A. $V=\sqrt{rg\tan \theta }$
• B. $\frac{r\sqrt{g}}{(L^2+r^2{)}^{1/4}}$
• C. $\frac{(L^2+r^2{)}^{1/4}}{e\sqrt{g}}$
• D. None of these

Answer 1

The correct option is A $V=\sqrt{rg\tan \theta }$

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) from (T).

On resolving tension along $X$ and $Y$ axis, $Tsin\theta$ will contribute to centripetal force as it is the towards the center of the circle.

Along $X-axis$:$F_{net}=x=T\sin \theta =\frac{m{v}^2}{r}.............\left(1\right)$

Along $Y$-axis:$F_{net}=y=0$(as the particle has no motion along Y-axis)

$Tcos\theta =mg............\left(2\right)$

Therefore tension in the string $T=\frac{mg}{\cos \theta }$

(where $\cos \theta =\frac{\sqrt{l^2-r^2}}{r}$)

Dividing equation (1) by equation (2)

Speed of the particle $v=\sqrt{rg\tan \theta }$

(where $\tan \theta =\frac{r}{\sqrt{l^2-r^2}}$)