JEE Class main Answered

Four particles of equal masses are moving round a circle of radius r due to their mutual gravitation attraction. Find the angular velocity.

Asked by aadityakumar0603 | 05 Mar, 2023, 22:17: PM

Expert Answer

Figure shows four particles moving in a circular path due to gravitational force of attraction .

F₁ is the force acting on each particle due to attraction by other particle that is diametrically opposite .

magnitude of F₁ is

begin mathsize 14px style F subscript 1 space equals space G space m squared over left parenthesis 2 r right parenthesis squared space equals space 1 fourth G space m squared over r squared end style

where G is universal gravitationla constant , m is mass of each particle and r is radius of circular path.

F₂ and F₃ are the forces acting on each particl due to left side and right side neighbour.

Magnitude of F₂ and F₃ are equal .

begin mathsize 14px style F subscript 2 space equals space F subscript 3 space equals space G space m squared over left parenthesis square root of 2 space r right parenthesis squared space equals space 1 half G m squared over r squared end style

F₂ and F₃ each make angle 45^o with F₁ .

Hence resultant of F₂ and F₃ is in the same direction of F₁ .

begin mathsize 14px style F subscript 2 plus F subscript 3 equals space 2 space cross times 1 half G space m squared over r squared cross times cos 45 end style

begin mathsize 14px style F subscript 2 plus F subscript 3 equals space G space m squared over r squared cross times fraction numerator 1 over denominator square root of 2 end fraction end style

Magnitude of net force F acting on each particle is

begin mathsize 14px style F space equals space F subscript 1 plus F subscript 2 plus F subscript 3 space equals space G space m squared over r squared open parentheses 1 fourth plus fraction numerator 1 over denominator square root of 2 end fraction close parentheses space equals 0.9571 cross times G m squared over r squared end style

Above force F provides the centripetal force required for circular motion .

Centripetal force of circular motion equals gravitational force of attraction . Let ω be the angular speed of rotation of masses.

begin mathsize 14px style m space omega squared space r space equals space 0.9571 space cross times G space m squared over r squared end style

begin mathsize 14px style omega space equals space space square root of 0.9571 space cross times space G space cross times m over r cubed end root space space space end style

Answered by Thiyagarajan K | 06 Mar, 2023, 08:48: AM

Please Login or Sign up to continue!