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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
![](https://images.topperlearning.com/topper/tinymce/imagemanager/files/c5346f64236151e230dd214a88bd52e56406195038a3a5.58091687f5.png)
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](https://images.topperlearning.com/topper/tinymce/cache/c184b0fd92135fe6df726c73b2a75947.png)
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](https://images.topperlearning.com/topper/tinymce/cache/1dc48da09abca873e12c551c161ee491.png)
F2 and F3 each make angle 45o 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](https://images.topperlearning.com/topper/tinymce/cache/9c07334ee4dc33104470bb04382aee37.png)
![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](https://images.topperlearning.com/topper/tinymce/cache/a6d627d29faad81f7e58ca1c55bbbbb9.png)
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](https://images.topperlearning.com/topper/tinymce/cache/b156edae0a81658b77c04f13f708a154.png)
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](https://images.topperlearning.com/topper/tinymce/cache/c4c51d85f116ed433d4360eed4268361.png)
![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](https://images.topperlearning.com/topper/tinymce/cache/581bbe1c2b2420d3c58cbe3103e87069.png)
Answered by Thiyagarajan K | 06 Mar, 2023, 08:48: AM
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