Question

a point P lies on the axis of the fixed ring of mass M and radius R and distance are root 3 from the centre of the particle starts from p and reaches at O and the gravitational attraction between the ring and the particle the speed of the particle at distance r from the centre is

I know its ans is option 4 but how ..
can anyone help plz.
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Answer 1

Given: ring of mass M and radius R,  distance are √3 from the centre of the particle.

To find: The speed of the particle at distance r from the centre?

Solution:

• As we know that total energy conservation states that the total energy of the system should always be a constant.

• So, by using theorem of the conservation of total energy, we get:

              (KE + PE) at point P = (KE + PE) at point O

• First we need to find the potential energy at point P, so

              u = -GMm/√r²+x²

• where M is mass of ring, m is mass of the body and x is the distance.

• putting this malue in conservation theorem, we get:

               0 + (-GMm/(√R²+(R√3)²) = 1/2 x mv² - GMm/√R²+0

               -GMm/(2R) + GMm/R = 1/2 x mv²

               GMm/2R = 1/2 x mv²

• cancelling m and 2 from both sides, we get:

               GM/R = v²

                v = √GM/R

Answer:

            Speed of the particle at distance r from the centre is √GM/R.