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Two concentric rings of uniform density at every point Are placed on a flat surface. Now since the centre of mass of a uniform Ring is its geometrical centre, the distance between the centers of Mass of the two rings are zero. Thus according to newtons law of Gravitation the attraction between themes (or at least tends to be) infinite, Making it impossible to pull one ring radially out from the other. However, as We know, this is not true. Why?

  
Total Answers and Comments: 4 Last Update: June 19, 2008   
  
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April 06, 2005 13:02:44   #1  
Sam        

RE: Two concentric rings of uniform density at every point Are placed on a flat surface. Now since the c...
inverse square law not valid when distance tends to zero.
 
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July 22, 2005 10:45:10   #2  
shweta singh        

RE: Two concentric rings of uniform density at every point Are placed on a flat surface. Now since the c...
centre of mass is a hypothetical point of concentration of mass with no actual mass there...or the concept of unstable equilibrium may be applied..
 
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September 09, 2005 02:27:52   #3  
indu        

RE: Two concentric rings of uniform density at every point Are placed on a flat surface. Now since the c...
how come the inverse sguare law is not valid when distance tends to zero? Suppose if the distance is not zero,if it is of the order of 10^(-30) or above, even then the force is much greater. And comming to the nxt, the effective distance is considered b/w the 2 bodies.
 
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June 19, 2008 02:14:38   #4  
kiranaar10728 Member Since: June 2008   Contribution: 10    

RE: Two concentric rings of uniform density at every point Are placed on a flat surface. Now since the centre of mass of a uniform Ring is its geometrical centre, the distance between the centers of Mass of the two rings are zero. Thus according to newton

if u intend to apply inverse square law to rings, Force between them is not inversely proportional to the distance between their centre of mass as one may comprehend. some log of distance terms will be there and u know log0 is not defined, so in effect cant apply inverse square law if their centres coincide.

GOT IT!!!!!!!


 
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