Four particles of equal mass are moving round a circle of radius r due to their mutual gravitational attraction. Find the angular velocity of each particle
Answers
The angular velocity of each particle is ω = GMm / r^3 ( 1/√2 × 1/4 )
Explanation:
Let us consider particle is moving from AA to BB.
Force on A due to B
F 1 = GMm / AB^2
Where AB is the distance between A and B
AB = r √2
F = GMm / 2r^2 in right hand direction.
Force on A due to C
F(2) = GMm / 2r^2 as AC = r √2 in downward direction
Force on A due to D
F(3) = GMm / AD^2
AD=2r
So, F3 = GMm / 4r^2 passes through center.
F1 and F 2 will have component along the center and cancel out each other.
Hence F1 (cos45∘ )+F2 (cos45 ∘ )+F3
= GMm /2 × 2r^ 2 + 1/√2 GMm / 4r^2
Net force = GMm / r^2 ( 1/√2 × 1/4 ) = mω^2 r
Solving angular velocity
ω = GMm / r^3 ( 1/√2 × 1/4 )
Hence the angular velocity of each particle is ω = GMm / r^3 ( 1/√2 × 1/4 )
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