Question

10. Determine the magnitude of energy and the angular momentum of a particle with mass m moving through the circle path with radius ro because of the attraction of the magnitude inversely proportional to the radius F2

Answer 1

Particle rotate in circular path, centripetal fore = mv^2/r here force of attraction is inversely proportional to radius of path F = -k/r^2 (here negative sign indicate force in attractive force) here force of attraction is balanced by centripetal force, mv^2/r = k/r^2 mv^2 = k/r equation (1) here, we want to find total energy of particle, E_total = kinetic energy + potential energy \r\n first we find kinetic energy k middot F = 1/2 mv^2 k middot E = 1/2 [k/r] mv^2/r = k/r From eqn (1) K middot E = k/2r equation (20 and potential energy = -integral F middot dr = -integral (-k/r^2) dr = integral k/r^2 dr = k integral [1/r^2] dr = k integral r^-2 dr = k[r^-2 + 1/-2 + 1] \r\n P middot E = k[r^-1/(-1)] P middot E = -k/r equation (3) total energy = k middot E + P middot E = k/2r + (-k/r) = k/2r -k/r