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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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A merry-go-round is turning at the rate of $10.0$ revolutions per minute. A $50.0\;kg$ youth is located $5.00 \;m$ from the axis of rotation . What is the centripetal force acting on him ?

$\begin{array}{1 1} 21.7 \times 10^4 \;N \\ 6.85 \times 10^4 \;N \\ 4.7 \times 10^4 \;N \\5 \times 10^4 \;N \end{array} $

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Answer : $ 6.85 \times 10^4 \;N $
By Newton's second law the force F producing this acceleration on a mass m towards the center of rotation is $ F= ma= \large\frac{mv^2}{r}$
The speed of an object traveling in a circle of radius $r$ is equal to $2\pi r n$, where $n$ is the number of revolutions per sec. Therefore $F= 4 \pi ^2n^2mr$
$F= mv^2/r= 4\pi n^2 mr$
Since n must be in revolutions per second, $n= 10.0 rpm/60\;sec/min=1/6.00 rps$
Then , $F=\large\frac{4 \pi^2} {(6.00)^2 .50.0 .5.00}$
$\qquad= \large\frac{1}{9.00} .\pi^2 $$.250 $
$\qquad=6.85 \times 10^4 \;N$
answered Aug 25, 2014 by meena.p

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