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  • A particle of mass is kept on the axis of a fixed circular ring of mass <img alt="\mathrm{M}" src="ht

A particle of mass \mathrm{m} is kept on the axis of a fixed circular ring of mass \mathrm{M} and radius \mathrm{R} at a distance \mathrm{x} from the centre of the ring. If \mathrm{x<<R}, At what value of \mathrm{x} the force between the ring and the particle is maximum:

Option: 1

\mathrm{\frac{R}{\sqrt2}}


Option: 2

\mathrm{\frac{R}{2\sqrt2}}


Option: 3

\mathrm{{R\sqrt2}}


Option: 4

\mathrm{\frac{R}{2}}


Answers (1)

best_answer

We see from the figure that, each portion of the ring having a mass \mathrm{\delta \mathrm{m}} attracts the particle towards itself with a force \mathrm{\delta \mathrm{F}} given as \mathrm{\delta F=\frac{G(\delta m) m}{r^2}}

\mathrm{\Rightarrow} The axial component of this force

\mathrm{=\delta F_x=\delta F \cos \theta=\frac{G(\delta m) m}{r^2} \cdot \frac{x}{r}=\frac{G m x}{r^3}(\delta m)}
\mathrm{\Rightarrow} The total axial force experienced by the particle due to the whole ring

\mathrm{ F_x=\sum \delta F_x=\int d F_x \quad \Rightarrow \quad F_x=\frac{G m x}{r^3} \int_0^M \delta m }

\mathrm{ \Rightarrow F_x=\frac{G M m x}{\left(R^2+x^2\right)^{3 / 2}} }            .........(1)

\mathrm{ \text { (since } \left.r=\sqrt{R^2+x^2}\right) . }

For \mathrm{ F_x} to be maximum

\mathrm{ \frac{d F_x}{d x}=0 \text { and } \frac{d^2 F_x}{d x^2}<0 . }

Using (1),

\mathrm{ \frac{d F_x}{d x}= G M \frac{d}{d x}\left\{\frac{x}{\left(R^2+x^2\right)^{3 / 2}}\right\} }

\mathrm{ =G M\left\{\frac{\left(R^2+x^2\right)^{3 / 2}-x \frac{3}{2}\left(R^2+x^2\right)^{1 / 2}(2 x)}{\left(R^2+x^2\right)^3}\right\} . }

Setting \mathrm{\frac{\mathrm{dF}_{\mathrm{x}}}{\mathrm{dx}}=0} and simplifying the factors, we obtain, \mathrm{x=\frac{\mathrm{R}}{\sqrt{2}}}

Hence option 2 is correct.



 

Posted by

Ajit Kumar Dubey

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