A 2 , kg stone at the end of a string 1 , m l | vedclass

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(C) The tension $T$ in a string for a stone moving in a vertical circle is given by the formula $T = \frac{mv^2}{r} + mg \cos 	heta$,where $	heta$ i

Vedclass · Accepted Answer

At the bottom of the circle

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(C) The tension $T$ in a string for a stone moving in a vertical circle is given by the formula $T = \frac{mv^2}{r} + mg \cos 	heta$,where $	heta$ is the angle with the downward vertical.Given: mass $m = 2 \, kg$,length $r = 1 \, m$,speed $v = 4 \, ms^{-1}$,and tension $T = 52 \, N$. Taking $g = 10 \, ms^{-2}$:$52 = \frac{2 	imes (4)^2}{1} + 2 	imes 10 	imes \cos 	heta$$52 = 32 + 20 \cos 	heta$$20 = 20 \cos 	heta$$\cos 	heta = 1$,which implies $	heta = 0^{\circ}$.Since $	heta = 0^{\circ}$ represents the position at the bottom of the circle,the correct option is $C$.

$A$ $2 \, kg$ 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 $4 \, ms^{-1}$. The tension in the string will be $52 \, N$ when the stone is

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