Question 326: Gravitation

Question

A thin rod of length '$$L$$' is bent in the form of a circle. Its mass is '$$M$$'. What force will act on mass '$$m$$' placed at the centre of this circle?
( $$\mathrm{G}=$$ constant of gravitation)

Accepted Answer

Consider two diametrically and equal small and equal mass segments $$\mathrm{dM}_1$$ and $$\mathrm{dM}_2$$
$$	herefore \quad$$ 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
$$F_2=\frac{G m d M_2}{r^2}$$
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 $$\mathrm{dM}$$ segments, we find the net force at the centre of the circle to be zero.

Answer

zero

Answer

$$\frac{\mathrm{GMm}}{4 \mathrm{~L}^2 \pi^2}$$

Answer

$$\frac{4 \pi^2 \mathrm{GMm}}{\mathrm{L}}$$

Answer

$$\frac{2 \mathrm{GMm}}{\mathrm{L}^2}$$

Gravitation

A thin rod of length '$L' is bent in the form of a circle. Its mass is 'M'. What force will act on mass 'm' placed at the centre of this circle? ( \mathrm{G}=$ constant of gravitation)

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