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A thin rod of length ' $L$ ' is bent in the form of a circle. Its mass is ' $M$ '. What force will act on mass ' $\mathrm{m}$ ' placed at the centre of this circle?
$(\mathrm{G}=\text { constant of gravitation) }$
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$(\mathrm{G}=\text { constant of gravitation) }$
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Verified Answer
The correct answer is:
zero
Consider two diametrically and equal small and equal mass segments $\mathrm{dM}_1$ and $\mathrm{dM}_2$
$\therefore \quad$ Force at the centre due to $\mathrm{dM}_1$ is
$\mathrm{F}_1=\frac{\mathrm{GmdM}_1}{\mathrm{r}^2}$
Similarly,
Force at the centre due to $\mathrm{dM}_2$ is
$\mathrm{F}_2=\frac{\mathrm{GmdM}_2}{\mathrm{r}^2}$
But $\mathrm{F}_1=-\mathrm{F}_2$
$\Rightarrow \mathrm{F}_1+\mathrm{F}_2=0 \quad(\because$ the forces cancel each other out as they are equal and opposite)
If this process is done for all such dM segments, we find the net force at the centre of the circle to be zero.
$\therefore \quad$ Force at the centre due to $\mathrm{dM}_1$ is
$\mathrm{F}_1=\frac{\mathrm{GmdM}_1}{\mathrm{r}^2}$
Similarly,
Force at the centre due to $\mathrm{dM}_2$ is
$\mathrm{F}_2=\frac{\mathrm{GmdM}_2}{\mathrm{r}^2}$
But $\mathrm{F}_1=-\mathrm{F}_2$
$\Rightarrow \mathrm{F}_1+\mathrm{F}_2=0 \quad(\because$ the forces cancel each other out as they are equal and opposite)
If this process is done for all such dM segments, we find the net force at the centre of the circle to be zero.
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