Two stones of masses m and 2,m are whirled in | vedclass

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Question

$\begin{array}{l}
\,\,\,\,\,\,\,\,\,\,Let\,v\,be\,{m{tangential}}\,{m{speed}}\,{m{of}}\,{m{heavier}}\
{m{stone}}{m{.}}\,{m{Then}}\,{\

Vedclass · Accepted Answer

$2$

Vedclass · Answer

$\begin{array}{l}
\,\,\,\,\,\,\,\,\,\,Let\,v\,be\,{m{tangential}}\,{m{speed}}\,{m{of}}\,{m{heavier}}\
{m{stone}}{m{.}}\,{m{Then}}\,{m{,}}\,{m{centripetal}}\,{m{force}}\,{m{experienced}}\
{m{by}}\,{m{lighter}}\,{m{stone}}\,{m{is}}\,\
{\left( {{F_c}} ight)_{lighter}} = \frac{{m{{\left( {nv} ight)}^2}}}{r}\,and\,that\,of\,heavier\,stone\,is\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\left( {{F_C}} ight)_{heavier}} = \frac{{2m{v^2}}}{{\left( {r/2} ight)}}
\end{array}$

$\begin{array}{l}
But\,{\left( {{F_c}} ight)_{lighter}} = {\left( {{F_c}} ight)_{heavier}}\,(given)\
	herefore \,\,\,\frac{{m{{\left( {nv} ight)}^2}}}{r} = \frac{{2m{v^2}}}{{\left( {r/2} ight)}}\
\,\,\,\,\,\,\,\,\,\,{n^2}\left( {\frac{{m{v^2}}}{r}} ight) = 4\left( {\frac{{m{v^2}}}{r}} ight)\
\,\,\,\,\,\,\,\,\,\,{n^2} = 4\,\,\,or\,\,\,n = 2
\end{array}$

Two stones of masses $m$ and $2\,m$ are whirled in horizontal circles, the heavier one in a radius $\frac{r}{2}$ and the lighter one in radius $r.$ The tangential speed of lighter stone is $n$ times that of the value of heavier stone when they experience same centripetal forces. The value of $n$ is

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