The path of a particle moving under the action of force is given by r=ae^bθ. Find the law of force.
Answers
mass m moves in a central force field. The force is F = f(r)(r/r), where f(r) = -kr and k > 0. Assume the mass moves at a constant speed in a circular path of radius R. Calculate the angular velocity of the mass, and show that its energy is E = kR2.
Solution:
Concepts:
Motion in a central potential, F = f(r)(r/r), f(r) = (-∂U/∂r)
Reasoning:
f(r) = (-∂U/∂r) = -kr, U(r) = ½kr2. We have a central potential.
The energy of the mass moving at a constant speed v in a circular path of radius R is
E = T + U = ½mv2 + ½kR2.
Details of the calculation:
For circular motion at constant speed we have |F| = mv2/R = kR. v = R(k/m)½.
The energy is E = T + U = ½mv2 + ½kR2 = ½mR2(k/m) + ½kR2 = kR2.
Problem:
A particle of mass m moves in a central force field such that its potential energy is given by V = krn, where r is the distance from the center of force and k and n are constants.
(a) Write down the Lagrangian for this system and determine the equations of motion in polar coordinates.
(b) Show that angular momentum is conserved for the system.
(c) Find an expression for the total energy of the system that depends only on the radial variable.
(d) Find the conditions (sign and magnitude of n and k) for a stable circular orbit by investigating the particle at stable equilibrium.
Solution:
Concept:
Motion in a central potential, Lagrange's equations, stable orbits
Reasoning:
We are asked to write down the Lagrangian, find an expression for the total energy, and investigate the stability of orbits.
Details of the calculation:
(a) L = T - U = ½m[(dr/dt)2 + r2(dΦ/dt)2] - krn.
The generalized coordinates are r and Φ. The coordinate Φ is cyclic.
Lagrange's equations yield the equations of motion.
(d/dt)[mr2(dΦ/dt)] = 0.
md2r/dt2 = M2/mr3 - nkrn-1.
(b) (d/dt)[mr2(dΦ/dt)] = 0. M = mr2(dΦ/dt) = constant, angular momentum is conserved.
(c) E = T + U = ½m(dr/dt)2 + M2/(2mr2) + krn = ½m(dr/dt)2 + Ueff(r).
This expression looks like the total energy of a particle moving in one dimension in a potential Ueff(r).
(d) For a circular orbit at with radius a we need ∂Ueff(r)/∂r|a = 0.
∂Ueff(r)/∂r = nkrn-1 - M2/(mr3).
nkan - 1 = M2/(ma3), an+2 = M2/(mnk), (n ≠ 0).
For a stable orbit of radius a we need a restoring force. Let r = a + ρ, ρ << a.
We need md2ρ/dt2 = -kρ = -∂Ueff/∂ρ = -(∂2Ueff/∂ρ2)|ρ=0ρ,
or ∂2Ueff/∂ρ2 > 0 near ρ = 0.
∂Ueff(r)/∂r = nkrn-1 - M2/(mr3).
∂Ueff(ρ)/∂ρ = nk(a + ρ)n-1 - M2/[m(a + ρ)3]
= nkan-1(1 + ρ/a)n-1 - [M2/(ma3)](1 + ρ/a)-3
= nkan-1(1 + (n-1)ρ/a) - [M2/(ma3)](1 - 3ρ/a).
∂2Ueff/∂ρ2|ρ=0 = n(n-1)kan-2 + 3M2/(ma4) > 0.
mn(n - 1)kan+2 + 3M2 > 0.
mn(n - 1)k[M2/(mnk)] + 3M2 > 0.
(n - 1)M2 + 3M2 > 0.
(n - 1) + 3 > 0.
n + 2 > 0.
Requirements for stable circular orbit with radius a for potential V(r) = krn are:
n > -2 , n ≠ 0,
k = M2/(nman+2).
When n < 0 then k < 0, and when n > 0 then k > 0.
Problem:
Two particles with reduced mass μ orbiting each have a potential energy function U = ½kr2, where k > 0 and r is the distance between them.
(a) Find the equilibrium distance r0 at which the particles can circle each other at a constant distance as a function of the angular momentum M.
(b) Determine if this is a stable equilibrium distance.
(c) Assume the particles orbiting at the equilibrium distance r0 are slightly disturbed. Determine if the disturbed orbits are closed.
Solution:
Concept:
Motion in a central potential
Reasoning:
For motion in a central potential energy and angular momentum are conserved. These two equations are enough to answer the questions.
Details of the calculation:
(a) M = μr2(dΦ/dt) = constant, angular momentum is conserved.
E = T + U = ½μ(dr/dt)2 + M2/(2μr2) + ½kr2 = ½μ(dr/dt)2 + Ueff(r).
This expression looks like the total energy of a particle moving in one dimension in a potential Ueff(r).
(b) For a circular orbit at with radius r0 we need ∂Ueff(r)/∂r|r0 = 0.
∂Ueff(r)/∂r = kr - M2/(μr3).
kr0 = M2/(μr03), r04 = M2/(μk).
(b) For a stable orbit of radius a we need a restoring force. Let r = a + ρ, ρ << a.
We need μd2ρ/dt2 = -½kρ = -∂Ueff/∂ρ = -(∂2Ueff/∂ρ2)|ρ=0ρ.
We need ∂2Ueff/∂ρ2 > 0 near ρ = 0, or ∂2Ueff/∂r2 > 0 near r = r0.
∂2Ueff/∂r2|r0 = k + 3M2/(μr04) = 4k > 0.
Since k > 0, this is a position of stable equilibrium.
(c) μd2ρ/dt2 = -4kρ for small displacements from equilibrium.
The angular frequency of small oscillations about r0 is ωosc = 2(k/μ)½.
The angular frequency of the orbit with radius r0 is ωorbit = M/(μr02) = r02(μk)½/(μr02) = (k/μ)½.
ωosc = 2ωorbit.
The relative distance between the particles goes through exactly two cycles for each full orbit. The orbits are closed.