Question

A thin rod of length L is bent to form a semicircle. The mass of the rod is M. What will be the gravitational potential at the center of the circle?

Answer 1

The gravitational potential due to any point mass is given by the formula,

⇒ V = −GM/r, where

M is the massr, is the distance from the point mass where gravitational potential is being calculated.

 

Now, in the instance of the half-circle, we must compute its gravitational potential at the circle’s centre. All of the mass ejected from the rod is now concentrated solely along the circle’s circumference. As a result, the rod’s whole mass is positioned at a constant distance from the circle’s centre, which is determined by the radius.

 

Therefore, we can write the gravitational potential due to the semicircle at the centre of the circle as,

⇒ VC = −GM/r

 

Now, the rod is of the length L. So as the rod is bent in the form of a semicircle, the length of the rod will be half of the perimeter of the circle of the radius r.

Hence we can write,

⇒ L = 2πr/2 where 2πr is the perimeter of the circle of the radius r.

Hence, the length is given as,

⇒ L = πr

So we can write the length in the form of the radius as,

⇒ r = L/π

So substituting this value of the radius in the formula for the gravitational potential, we get,

⇒ VC = −GM/L/π

 

We can simplify this as,

⇒ VC = −πGM/L

 

So the gravitational potential due to the mass of a rod bent in the shape of a semicircle at the centre is given as, −πGM/L

 

 

Gravitational potential at the center of the circle Is −πGM/L