A particle of mass 2m is attached to a rigid support by a spring with a force constant k. At equilibrium, the spring hangs vertically downward. A particle of mass m is attached to this mass-spring combination by and identical spring with a force constant k. Find the eigenfrequencies and describe the normal modes for this system.

Solution:

- Concepts:

Lagrangian mechanics, coupled oscillations - Reasoning:

We are asked to find the normal modes of two coupled harmonic oscillators. - Details of the calculation:

Let x_{1}and x_{2}be the displacements from the equilibrium positions in the gravitational potential near the surface of Earth. Then

T = ½2m(dx_{1}/dt)^{2}+ ½m(dx_{2}/dt)^{2}, U = ½k x_{1}2 +½k(x_{2}- x_{1})^{2}= kx_{1}^{2}+ ½kx_{2}^{2}- ½kx_{1}x_{2}- ½kx_{2}x_{1}, L = T - U.

The Lagrangian is L = T - U. This can be put into the form

L = ½∑_{ij}[T_{ij}(dq_{i}/dt)(dq_{j}/dt) - k_{ij}q_{i}q_{j}] with T_{ij}= T_{ji}, k_{ij}= k_{ji}.

Here T_{11}= 2m, T_{22}= m, T_{ij}(i≠j) = 0,

k_{11}= 2k, k_{22}= k, k_{12}= k_{21}= -k.Solutions of the form x

_{j}= Re(A_{j}e^{iωt}) can be found.

We can find ω^{2}from det(k_{ij}- ω^{2}T_{ij}) = 0.-2mω ^{2}+ 2k-k -k -mω ^{2}+ k= 0.

ω

^{2}= [1 ± 1/√2](k/m), ω_{1}^{2}= [1 + 1/√2](k/m), ω_{2}^{2}= [1 - 1/√2](k/m).

The eigenfrequencies are ω_{1}= 1.3 (k/m)^{½}, ω_{2}= 0.54 (k/m)^{½}.For mode ω = ω

_{1}we have A_{2}= -√2 A_{1}. The displacements are in opposite directions and the magnitude of the displacement of mass m is √2 times the displacement of mass 2m.

For mode ω = ω_{2}we have A_{2}= √2 A_{1}. The displacements are in the same direction and the magnitude of the displacement of mass m is √2 times the displacement of mass 2m.

A block of mass M is rigidly connected to a massless circular track of radius a on a frictionless horizontal table as shown.

A particle of mass m is confined to move without friction on the
circular track which is vertical.

(a) Set up the Lagrangian, using θ as one of the coordinates.

(b) Find the equations of motion.

(c) In the limit of small angles, solve the equation of motion
for θ as function of time

Solution:

- Concepts:

Lagrangian Mechanics - Reasoning:

We are asked to find the Lagrangian and to solve Lagrange's equations. - Details of the calculation:

(a) Choose the x coordinate of the center of the circular track and the angle θ giving the location of m on the circular track as the generalized coordinates.

The coordinates of the mass m are then (x + asinθ, -acosθ). As M is rigidly connected to the circular track its velocity is (dx/dt, 0).

The Lagrangian is

L = T - U

= ½M(dx/dt)^{2}+ ½m[(dx/dt) + a(dθ/dt)cosθ]^{2}+ a^{2}(dθ/dt)^{2}sin^{2}θ + mgacosθ

= ½M(dx/dt)^{2}+ ½m[(dx/dt)^{2}+ a^{2}(dθ/dt)^{2}+ 2a(dx/dt)(dθ/dt)cosθ] + mgacosθ.

(b) ∂L/∂x = 0,

∂L/∂(dx/dt) = M(dx/dt) + m(dx/dt) + ma(dθ/dt)cosθ = constant, since x is cyclic.

(M + m)(dx/dt) + ma(dθ/dt)cosθ = constant is a first-order equation of motion.

(d/dt)∂L/∂(dx/dt) = (M + m)d^{2}x/dt^{2}+ ma d^{2}θ/dt^{2}cosθ - ma(dθ/dt)^{2}sinθ.

∂L/∂θ = -ma(dx/dt)(dθ/dt)sinθ - mgasinθ,

∂L/∂(dθ/dt) = ma^{2}(dθ/dt) + ma(dx/dt)cosθ,

(d/dt)∂L/∂(dθ/dt) = ma^{2}d^{2}θ/dt^{2}+ madx^{2}/dt^{2}cosθ - ma(dx/dt)(dθ/dt)sinθ.

The second-order equations of motion therefore are

(M + m) d^{2}x/dt^{2}+ ma d^{2}θ/dt^{2}cosθ - ma(dθ/dt)^{2}sinθ = 0,

ad^{2}θ/dt^{2}+ d^{2}x/dt^{2}cosθ + gsinθ = 0.

(c) For small oscillations we set sinθ = θ, cosθ = 1.

In the equations of motion we only keep terms up to first order in the small quantities.

(M + m) d^{2}x/dt^{2}+ ma d^{2}θ/dt^{2}= 0,

ad^{2}θ/dt^{2}+ d^{2}x/dt^{2}+ gθ = 0.

Eliminating d^{2}x/dt^{2}we have

d^{2}θ/dt^{2}+ (M + m)gθ/(Ma) = 0.

The most general solution is θ = A cos(ωt + Φ), with ω = ((M + m)g/(Ma))^{½}.

The constants A and Φ are determined by the initial conditions.

We can also view the problem as a coupled small oscillation problem.

Keeping only terms to second order in the Lagrangian we have

L = ½M(dx/dt)^{2}+ ½m[(dx/dt)^{2}+ a^{2}(dθ/dt)^{2}+ 2a(dx/dt)(dθ/dt)] - ½mgaθ^{2}.

= ½(M + m)(dx/dt)^{2}+ ½ma^{2}(dθ/dt)^{2 }+ ½ma(dx/dt)(dθ/dt) + ½ma(dθ/dt)(dx/dt) - ½mgaθ^{2}.

L = ½∑_{ij}[T_{ij}(dq_{i}/dt)(dq_{j}/dt) - k_{ij}q_{i}q_{j}] with T_{ij}= T_{ji}, k_{ij}= k_{ji}.

T_{11}= (M + m), T_{22}= ma^{2}, T_{12}= T_{21}= ma, k_{22}= mga, k_{11}= k_{12}= k_{21}= 0.

We can find ω^{2}from det(k_{ij}- ω^{2}T_{ij}) = 0.

-ω ^{2}(M + m)-ω ^{2}ma-ω ^{2}mamga - ω ^{2}ma^{2}= 0.

-ω

^{2}(M + m)mga + ω^{4}Mma^{2}= 0.

We have two solutions, ω = 0 and ω = ((M + m)g/(Ma))^{½}.

There is only one normal mode with non-zero ω.

For this mode x = B cos(ωt + Φ), B = -maA/(M + m), the CM stays fixed.

The other mode has ω = 0 and corresponds to uniform translation of the CM.

A star of mass M and radius R is moving with
constant velocity **v** through a cloud of particles of density ρ.

If all the particles which collide with the star are trapped by it, show that the mass of
the star will increase at a rate

dM/dt = πρv(R^{2} + 2GMR/v^{2}).

Solution:

- Concepts:

Motion in a central potential conservation of energy and angular momentum - Reasoning:

The particles in the cloud approach the star. At a large distance from the star they had speed v and varying impact parameters. We find the impact parameter b for the particles that will just graze the surface of the star. The particles that are trapped by the star were, at a large distance, contained in a cylinder of radius b with its axis pointing towards the star and moving towards the star with speed v.

The volume dV = πb^{2}vdt of this cylinder sweeps over the star in time dt. Therefore dM/dt = ρπb^{2}v. We find b using energy and angular momentum conservation. - Details of the calculation:

For a particle that just grazes the surface of the star we have

L = mbv = mRv_{R}, where v_{R}is the speed at R and v = v_{∞}.

E = ½mv^{2}= ½mv_{R}^{2}- GMm/R. v_{R}^{2}= v^{2}+ 2GM/R.

b = Rv_{R}/v_{∞}= R(1 + 2GM/(Rv^{2}))^{½}.

dM/dt = πρv(R^{2}+ 2GMR/v^{2}).

Note: This is an approximate result unless ρ(**r**,t) is constant in space and time.

A satellite is in a circular orbit of radius r around an airless spherical planet of radius R. An asteroid of equal mass falls radially towards the planet, starting at zero velocity from a very large distance. The satellite and the asteroid collide inelastically and stick together, moving in a new orbit that just misses the planet's surface. What was the radius r of the satellite's original circular orbit in terms of R?

Solution:

- Concepts:

Kepler orbits - Reasoning:

Since M >> m, assume the CM is the center of the planet. In a central potential the energy E and the angular momentum L are conserved. The motion is in a plane. For an object with mass m' in a circular orbit about a fixed center with mass M we have

T = ½m'v^{2}= ½GMm'/r, U = -GMm'/r, E = -½GMm'/r,

p = m'v = (GMm'^{2}/r)^{½}, L = rp = (GMm'^{2}r)^{½}. - Details of the calculation:

Just before the collision:

For the orbiting mass m, E_{1}= -½GMm/r, T_{1}= ½GMm/r, p_{1}= (GMm^{2}/r)^{½}, L_{1}= (GMm^{2}r)^{½}.

For the falling mass m, E_{2}= 0, T_{2}= GMm/r, p_{2}= (2GMm^{2}/r)^{½}, L_{2}= 0.Just after the collision:

Momentum and angular momentum are conserved in the totally inelastic collision. The momenta of the two colliding objects are perpendicular to each other. Therefore the magnitude of the momentum of the combined mass is p = (p_{1}^{2}+ p_{2}^{2})^{½}, p^{2}= 3GMm^{2}/r.

The magnitude of the angular momentum of the combined object is L = (GMm^{2}r)^{½}.

Its energy is E = T + U = p^{2}/4m - 2GMm/r = -(5/4)GMm/r.

From now on we have motion of a single object in a central potential.

Energy and angular momentum are conserved.

At the distance of closest approach L = p_{max}r_{min}= p_{max}R = (GMm^{2}r)^{½}, p_{max}^{2}R^{2}= GMm^{2}r.

E = p_{max}^{2}/(4m) - 2GMm/R = -(5/4)GMm/r.

GMm^{2}r/(4mR^{2}) - 2GMm/R = -(5/4)GMm/r.

r^{2}- 8Rr + 5R^{2}= 0, r = 4R ± √(11)R.

The perihelion of the orbit is at R, so r > R.

r = 4R + √(11)R.

A two electron system has (in its center of mass frame) energy E and angular momentum L. What is the closest distance the electrons can approach each other? (Ignore radiation.)

Solution:

- Concepts:

The Coulomb potential, motion in a central potential, energy conservation - Reasoning:

The problem of two interacting particles in their CM frame is equivalent to the problem of a fictitious particle of reduced mass μ moving in a central potential U(r).

Here U(r) = -e^{2}/r,^{ }r = distance between their centers.

E = ½μ(dr/dt)^{2}+ L^{2}/(2μr^{2}) + e^{2}/r, μ = m_{e}/2, e^{2}= q_{e}^{2}/(4πε_{0}).

At the distance of closest approach dr/dt = 0. - Details of the calculation:

E = L^{2}/(2μr_{min}^{2}) + e^{2}/r_{min}.

r_{min}^{2}- e^{2}r_{min}/E - L^{2}/(2μE) = 0.

r_{min}^{2}- e^{2}r_{min}/E - L^{2}/(m_{e}E) = 0.

r_{min}= e^{2}/(2E) + (e^{4}/(4E^{2}) + L^{2}/(m_{e}E))^{½}.