Question

A particle of mass $m$ is suspended from a ceilling through a string of length $L$ the particle moves in a horizontal circle of radius r. what is the speed of the particle and the tension in the string for conical pendulum ?

• A. $\upsilon =\frac{r\sqrt{grl}}{(L^2-r^2{)}^{1/4}},T=\frac{mgL}{\sqrt{L^2-r^2}}$
• B. $\upsilon =\frac{r\sqrt{g}}{(L^2-r^2{)}^{1/4}},T=\frac{mgL}{\sqrt{L^2-r^2}}$
• C. $\upsilon =\frac{r\sqrt{g}}{(L^2-r^2{)}^{1/4}},T=\frac{mgL}{r}$
• D. $\upsilon =\frac{\sqrt{grL}}{(L^2-r^2{)}^{1/4}},T=\frac{mgr}{\sqrt{L^2-r^2}}$

Answer 1

The correct option is B $\upsilon =\frac{r\sqrt{g}}{(L^2-r^2{)}^{1/4}},T=\frac{mgL}{\sqrt{L^2-r^2}}$

$sin\theta =\frac{r}{L}$(a) Tension T along the string
(b) The weight mg vertically downward In radial direction,
$Tsin\theta =\frac{m{v}^2}{r}$
In Vertical direction, $Tcos\theta =mg$
Equation (i) and (ii)
$tan\theta =\frac{{\upsilon }^2}{rg}$
$\upsilon =\sqrt{\left(rgtan\theta \right)}=\sqrt{\left(rg\right)\times \left(\frac{r}{\sqrt{L^2-r^2}}\right)}$
$\upsilon =\frac{r\sqrt{g}}{(L^2-r^2{)}^{1/4}}$
$\text{By equation (ii)}T=\frac{mg}{cos\theta }=\frac{mgL}{(L^2-r^2)}$
$T=\frac{mgL}{(L^2-r^2{)}^{1/2}}$