Request a call back

Join NOW to get access to exclusive study material for best results

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, 10:17: PM
answered-by-expert 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 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
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
JEE main - Physics
Asked by krajasekharnaidu13 | 04 Feb, 2024, 03:37: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Physics
t
question image
Asked by jpmeena2222 | 31 Dec, 2023, 10:27: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Physics
.
question image
Asked by swayamagarwal2114 | 18 Aug, 2022, 09:04: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Physics
Asked by swayamagarwal2114 | 21 Jul, 2022, 05:18: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Physics
Asked by bathularohith19 | 27 Mar, 2022, 03:05: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Physics
Asked by Amit | 21 Mar, 2019, 09:48: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
Get Latest Study Material for Academic year 24-25 Click here
×