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A system consists of N identical particles of mass m placed rigidly on the vertices of a regular polygon with each side of length 'a'. If k1 be the kinetic energy imparted to one of the particles so that it just escapes the gravitational pull of the system and thereafter kinetic energy k2 is given to the adjacent particle to escape, then the difference k1-k2 is

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one year ago

```							A regular polygon with n sides can always have a circle made which touches all the vertices of thepolygon, so all the particles are the same distance from the center of the circle.Unless you are working this problem for a very high-level college physics course, you must make a simplifying assumption:The mass of all n particles acts as though it is concentrated at the center of the circle and does not change its location when either of the particles leaves and goes to "infinity".The first assumption is approximately valid for large enough values of n: that is for many particlesand many sides to the polygon. With this assumption, the gravitational potential energies p1 and p2 are both directly proportional to the product of the smaller mass "m" times the larger mass "M" in each case. In each case, all other factors in calculating a p-value do not change between p1 and p2. The gravitational energy in each case must be equal in magnitude to the kinetic energy lost by the particle of mass going from k1 or k2 to zero at infinity, as per the definition of "escape velocity" for that kinetic energy. This means thatk1 = -p1 and k2 = -p2 , and thusk1/k2 = -p1/(-p2) =p1/p2The mass left behind for the first particle is just (n-1)*m , andthe mass left behind for the second particle is (n-2)*m.Thusp1/p2 = A*m[(m)(n-1)/{A*[m(m)(n-2) = (n-1)/(n-2) , where the "A" is a factor which does not change between the two cases.So nowk1/k2 = p1/p2 = (n-1)/(n-2)k1 = [(n-1)/(n-2)]*k2Thus k1-k2 = [(n-1)/(n-2)]*k2 - k2 = [(n-1)/(n-2)-1]*k2 ,Or k1-k2 = k1/[(n-1)/(n-2)-1] = {1/[(n-1)/(n-2)-1]}*k1This equation gives the energy difference as a fraction of the energy of the firstparticle to leave. If you know calculus you can integrate and find an expression for the gravitational potential energy (or if you have been given an equation for it)p = -2G(m)(M)/R , where G is Newton's gravitational constant, R is the radius of the circle, and M is the total mass left behind in each case.Thusp1 = -2G(m)[m(n-1)]/RThe chord of the circle between any two adjacent vertices is "a", so the central angle for the chord is 2*pi/n (pi = 3.1416). The chord length is related to the radius bya =2R*sin[(2*pi/n)/2] , soR = (a/2)/sin(pi/n) and 1/R = 2*sin(pi/n)/aThenk1 = -p1 givesk1 = 2G(m)[m(n-1)]*2*sin(pi/n)/a simplify this by combining factors and then substitute it into the expression found above
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one year ago
```							Dear student CM of the two adjacent masses situated at a distance L/2, now CM of the whole system will be found at the circumcentre O of the polygon.The apothem a of a regular polygon is given bya = ½Ltan(π/n)  Mass of the CM = NmSo total PE of the system = - Gm(Nm)/a  = - GNm^2/aTo escape the system KE required for the first particle will beK1 =  GNm^2/aSince the masses are rigidly supported, loss of the first particle does not displace the system so the CM maintains its original position. Now KE = K2 will be required to overcome the PE of N-1 particle. K2 = G(N-1)m^2/a=> K2 – K1 = Gm^2/a=> K2 – K1 = Gm^2/{½Ltan(π/n)}
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one year ago
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