Question

Q3 (5pt) A particle of mass m is attracted to a force center with the force of magnitude k/r2. Use the plane polar coordinates (r, ?) and the Hamiltonian method to find the equation of motion for r and ?.

Answer 1

The expression for the kinetic energy of the particle in polar coordinates is, T = 2m(32 +r??) The expression for the attraction force is, F=-- Here, the force is attractive, so, consider the negative sign for the force. The expression for the potential energy of the particle is, U=-| Fdr Calculate the potential energy of the particle as follows: U=-| Fdr The expression for the Lagrange equation for the particle is, I=T-U

= = m(j? +r?0?) + The expression for the Hamiltonian of the particle is, H = ip, +ope- The expression for r polar coordinate of the particle's momentum is, Then write the expression for derivative of position of the particle as follows: P = 0 Cm(+r+0%) + 5) (24)= (10)*:03) = -52(27 +0+0)

i-P; m The expression for a polar coordinate of the particle's momentum is, al PA = Po ao Then write the expression for derivative of O as follows: 00 po novom -20 (*(**+r+0%) +) = "( 20 (*)+co(+*0) +:)) --- (0+rº (20)+0) Then calculate the Hamiltonian of the particle as follows: H = rp, +ope -L

+ =ip. +0p{m{x+r20)) =()+(?).-10-17)2) 2mr2 PIE 2m 2mr r The equation of motion of the particle in position vector is, H = ops Calculate the equation of motion of the particle in position vector as follows: aH -

2m = 2P: +0-0 i= (or) pe = mi The equation of motion of the particle in momentum vector is, . -a Para Or Calculate the equation of motion of the particle in momentum vector as follows: P = H + kl

The equation of motion of the particle in direction vector is, 0-OH op Calculate the equation of motion of the particle in direction vector as follows: . ОН =- = Op på k) = 0 +2PO2 - 0 Jė= Po, (or) Po = mr20 The equation of motion of the particle in momentum vector is, Po = -0H 2o

Calculate the equation of motion of the particle in momentum vector as follows: -OH ao 2mr2 Do = 0