A stone of mass 900 mathrm{~g} is tied to a s | vedclass

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Question

Given thaat

$ \mathrm{m}=900 \mathrm{gm}=\frac{900}{1000} \mathrm{~kg}=\frac{9}{10} \mathrm{~kg} $

$ \mathrm{r}=1 \mathrm{~m} $

$ \omega=\fra

Vedclass · Accepted Answer

$9.8 \mathrm{~N}$

Vedclass · Answer

Given thaat

$ \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} \omega^2 $

$ \mathrm{~T}=\mathrm{mg}+\mathrm{mr} \omega^2 $

$ =\frac{9}{10} 	imes 9.8+\frac{9}{10} 	imes 1\left(\frac{\pi}{3}ight)^2 $

$ =8.82+\frac{9}{10} 	imes \frac{\pi^2}{9} $

$ =8.82+0.98 $

$ =9.80 \mathrm{~N}$

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 $g=9.8 \mathrm{~m} / \mathrm{s}^2$ )

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