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A string $0.5 \mathrm{~m}$ long is used to whirl a $1 \mathrm{~kg}$ stone in a vertical circle at a uniform velocity of $5 \mathrm{~m} \mathrm{~s}^{-1}$. What is the tension in the string when the stone is at the top of the circle?
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Verified Answer
The correct answer is:
$40.2 \mathrm{~N}$
The centripetal force needed to keep the stone moving at $5 \mathrm{~m} \mathrm{~s}^{-1}$ is
$$
F_c=\frac{m v^2}{r}=\frac{(1 \mathrm{~kg})\left(5 \mathrm{~ms}^{-1}\right)^2}{0.5 \mathrm{~m}}=50 \mathrm{~N}
$$
The weight of the stone is $W=m g=(1 \mathrm{~kg})\left(9.8 \mathrm{~m} \mathrm{~s}^{-2}\right)=9.8 \mathrm{~N}$.
At the top of the circle,
$$
T=F_c-W=50 \mathrm{~N}-9.8 \mathrm{~N}=40.2 \mathrm{~N}
$$
$$
F_c=\frac{m v^2}{r}=\frac{(1 \mathrm{~kg})\left(5 \mathrm{~ms}^{-1}\right)^2}{0.5 \mathrm{~m}}=50 \mathrm{~N}
$$
The weight of the stone is $W=m g=(1 \mathrm{~kg})\left(9.8 \mathrm{~m} \mathrm{~s}^{-2}\right)=9.8 \mathrm{~N}$.
At the top of the circle,
$$
T=F_c-W=50 \mathrm{~N}-9.8 \mathrm{~N}=40.2 \mathrm{~N}
$$
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