A $2\, kg$ stone at the end of a string $1 \,m$ long is whirled in a vertical circle at a constant speed. The speed of the stone is $4 \,m/sec$. The tension in the string will be $52\, N$, when the stone is
A $2\, kg$ stone at the end of a string $1 \,m$ long is whirled in a vertical circle at a constant speed. The speed of the stone is $4 \,m/sec$. The tension in the string will be $52\, N$, when the stone is
At the top of the circle
At the bottom of the circle
Halfway down
None of the above
A $2\, kg$ stone at the end of a string $1 \,m$ long is whirled in a vertical circle at a constant speed. The speed of the stone is $4 \,m/sec$. The tension in the string will be $52\, N$, when the stone is
$mg = 20N$ and $\frac{{m{v^2}}}{r} = \frac{{2 \times {{(4)}^2}}}{1} = 32N$
It is clear that $52 \,N$ tension will be at the bottom of the circle. Because we know that
${T_{{\rm{Bottom}}}} = mg + \frac{{m{v^2}}}{r}$
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