Question 85 of 210

A stone of mass $900g$ is tied to a string and moved in a vertical circle of radius $1m$ making $10rpm$. The tension in the string, when the stone is at the lowest point is (if ${\pi }^{2}=9.8$ and $g=9.8m{s}^{-2}$)

A$97N$B$9.8N$C$8.82N$D$17.8N$

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Answer: (B) $9.8N$

Given that

Given the mass $m=900g=\dfrac{900}{1000}kg=\dfrac{9}{10}kg$,
The radius of the circular path $r=1m$
And, the angular velocity
$\omega =\dfrac{2\pi N}{60}=\dfrac{2\pi (10)}{60}=\dfrac{\pi }{3}rad{s}^{-1}$
With reference to the above diagram, the formula to calculate the tension at the lowermost point can be written as
$T-mg=mr{\omega }^{2} \Rightarrow T=mg+mr{\omega }^{2}...\left(1\right)$
From equation (1), it follows that
$\begin{matrix}T & = & \dfrac{9}{10}\times 9.8+\dfrac{9}{10}\times 1{\left(\dfrac{\pi }{3}\right)}^{2} \\ & = & 8.82+\dfrac{9}{10}\times \dfrac{{\pi }^{2}}{9} \\ & = & 8.82+0.98 \\ & = & 9.8N\end{matrix}$