Question

A particle of mass m is whirled in horizontal circle with the help of two threads of length l each as shown in figure.Angular velocity equals $\omega$, then

Answer 1

$\sin {30}^{\circ }=\frac{d}{l}$

$d=l\sin {30}^{\circ }$

$=\frac{l}{2}$

$T_1\cos {30}^{\circ }=T_2\cos {30}^{\circ }+mg$

$\Rightarrow T_1\frac{\sqrt{3}}{2}=\frac{T_2\sqrt{3}}{2}+mg$ __(i)

$\Rightarrow (T_1-T_2)\frac{\sqrt{3}}{2}=mg$

$\underset{\Leftarrow }{a_r=w^{2}d}$ $(T_1+T_2)\sin {30}^{\circ }=m{w}^{2}d$

$(T_1+T_2)\sin {30}^{\circ }\Rightarrow (T_1+T_2)\frac{1}{2}=m{w}^2\frac{l}{2}$ __(ii)

$T_1+\frac{T}{2}=m{w}^{2}l$

from (i) $\frac{T_1-\frac{T}{2}=\frac{2}{\sqrt{3}}mg}{2T_1=m{w}^{2}l+\frac{2}{\sqrt{3}}mg}$

$T_1=\frac{m{w}^{2}l}{2}+\frac{mg}{\sqrt{3}}$

$T_2=m{w}^{2}l-T_1$

$=m{w}^{2}l-\frac{m{w}^{2}l}{2}-\frac{mg}{\sqrt{3}}$

$T_2=\frac{m{w}^{2}l}{2}-\frac{mg}{\sqrt{3}}$