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  4. A string 0.5 m long is used to whirl a 1 kg stone in a vertical circle at a uniform velocity of 5 m s-1. What is the tension in the string when the stone is at the top of the circle?

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Q. A string $0.5\, m$ long is used to whirl a $1\,kg$ stone in a vertical circle at a uniform velocity of $5 \,m \,s^{-1}$. What is the tension in the string when the stone is at the top of the circle?

AIIMSAIIMS 2015Work, Energy and Power

Solution:

The centripetal force needed to keep the stone moving at 5 m s-1 is
$F_{c}=\frac{mv^{2}}{r}=\frac{\left(1 kg\right)\left(5 ms^{-1}\right)^{2}}{0.5 m}=50 N$
The weight of the stone is
W = mg = (1 kg) (9.8 m s-2) = 9.8 N.
At the top of the circle,
T = Fc - W = 50 N - 9.8 N = 40.2 N