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?

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

⇒ V = −GM/r, where

  • M is the mass
  • r, 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

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