Laws Of Motion Question 8
Question 8 - 2024 (29 Jan Shift 2)
A stone of mass $900 \mathrm{~g}$ is tied to a string and moved in a vertical circle of radius $1 \mathrm{~m}$ making $10 \mathrm{rpm}$. The tension in the string, when the stone is at the lowest point is (if $\pi^{2}=9.8$ and $\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2}$ )
(1) $97 \mathrm{~N}$
(2) $9.8 \mathrm{~N}$
(3) $8.82 \mathrm{~N}$
(4) $17.8 \mathrm{~N}$
Show Answer
Answer: (2)
Solution:
Given that
$\mathrm{m}=900 \mathrm{gm}=\frac{900}{1000} \mathrm{~kg}=\frac{9}{10} \mathrm{~kg}$
$\mathrm{r}=1 \mathrm{~m}$
$\omega=\frac{2 \pi \mathrm{N}}{60}=\frac{2 \pi(10)}{60}=\frac{\pi}{3} \mathrm{rad} / \mathrm{sec}$
$\mathrm{T}-\mathrm{mg}=\mathrm{mr}^{2}$
$\mathrm{T}=\mathrm{mg}+\mathrm{mr} \omega^{2}$
$=\frac{9}{10} \times 9.8+\frac{9}{10} \times 1\left(\frac{\pi}{3}\right)^{2}$
$=8.82+\frac{9}{10} \times \frac{\pi^{2}}{9}$
$=8.82+0.98$
$=9.80 \mathrm{~N}$
