Question

A particle of mass M is placed on the axis of a ring of radius R and mass M at a point P as shown in figure gravitational force on the particle due to ring will be​

Answer 1

Answer:

Explanation:

Mass M is placed at point P consider a mass element of mass dx of mass dm at a point T.

Answer 2

Explanation:

Refer the attachement for diagram !!

Consider an element of length dx and mass dm on the ring.

The distance between the element and point P is:

$=>a=\sqrt{3R^2+R^2}\\\\=> a=2R$

If you observe closely, diametrically opposite Fsin(θ) components cancel each other, hence Force contribution is due to the Fcos(θ) components only

Force experienced by particle at P due to the element is

$2dFcos(\theta)=\frac{Gmdm}{(2R)^2}----(1)$

Since $dm=\frac{M}{2\pi R}dx----(2)$

∴ $\boxed{dFcos(\theta)=\frac{GmM}{16\pi R^3}dx}$

Now integrating the expression w.r.t x from 0 to πR

$Fcos(\theta)=\int\limits^{\pi R} _0{\frac{GmM}{16\pi R^3}dx}$

$Fcos(\theta)={\frac{GmM}{16 R^2}$

Since $cos(\theta)=\frac{\sqrt{3}}{2}$;

We get $\boxed{\boxed{F={\frac{\sqrt{3}GmM}{8 R^2}}}}$

Hope this answer helped you :)