Assignment 4, solutions

Problem 1:

Two spheres are of the same radius R and mass M, but one is solid and the other is a hollow shell (of negligible thickness).  Both spheres roll (without sliding) down a ramp of incline θ.
(a)  Which sphere will have the greater acceleration down the ramp?
(b)  Determine the Lagrangian for the motion of the sphere and derive the equation of motion for both cases.

Solution:

bulletConcepts:
Lagrange's Equations
bulletReasoning:
All forces except the forces of constraint are derivable from a potential.  The Lagrangian formalism is well suited for such a system.  We are asked to derive Lagrange's equation of motion.
bulletDetails of the calculation:
(a)  We have a conservative system.  Potential energy is converted into rotational and translational kinetic energy ½Iω2 + ½Mv2 with v = ωR.  The moment of inertia of the solid sphere I = (2/5)MR2 is smaller than the moment of inertia of the thin spherical shell I = (2/3)MR2, so the solid sphere will have the greater linear acceleration down the ramp.
(b)  Set up a coordinate system:

   

L = T – U.  T = ½(I/R2 + M)(dx/dt)2,  U = Mgxsinθ.
L = ½(I/R2 + M)(dx/dt)2 - Mgxsinθ.
Equation of motion: (d/dt)(∂L/∂v) – ∂L/∂x = 0, with v = dx/dt.
(I/R2 + M)d2x/dt2 + Mgsinθ = 0.
a =  d2x/dt2 = (M/(I/R2 + M))gsinθ.
For the solid sphere:  a = (5/7) gsinθ.
For the spherical shell:  a = (3/5) gsinθ.

Problem 2:

Obtain Lagrange's equations of motion for a spherical pendulum (a mass point suspended by a rigid, weightless rod).

Solution:

bulletConcepts:
Lagrange's Equations: d/dt(∂L/∂(dqj/dt)) -  ∂L/∂qj = 0; generalized coordinates
bulletReasoning:
We are asked to obtain Lagrange's equation of motion.  Let g point into the -z direction.

The system has only two degrees of freedom.  Choose the generalized coordinates to be θ and Φ.  The constraints are holonomic, the applied forces are conservative.
bulletDetails of the calculation:
d/dt(∂L/∂(dqj/dt)) -  ∂L/∂qj = 0,  and  q1 = θ,  q= Φ.
L = T – U
T = ½m(r2(dθ/dt)2 + r2sin2θ(dΦ/dt)2),  U = mg r cosθ.
∂L/∂(dθ/dt) = mr2 dθ/dt,  d/dt(∂L/∂(dθ/dt)) = mr2 d2θ/dt2,
∂L/∂θ = mr2 sinθ cosθ(dΦ/dt)2 +  mg r sinθ.
∂L/∂(dΦ/dt) = mr2 sin2θ dΦ/dt,  
d/dt(∂L/∂(dΦ/dt)) = mr2 sin2θ d2Φ/dt2 + 2mr2 sinθ cosθ dθ/dt dΦ/dt,
∂L/∂Φ = 0.
In this problem Φ is cyclic, ∂L/∂(dΦ/dt) is a constant of motion.
sin2θ(dΦ/dt) = constant = C.  dΦ/dt = C/sin2θ.

Equations of motion (second order differential equations):
d2θ/dt2 = sinθ cosθ(dΦ/dt)2 + (g/r)sinθ = (cosθ/sin3θ)C2 + (g/r)sinθ.
d2Φ/dt2 = -2 dΦ/dt dθ/dt cotθ = 2 C dθ/dt (cosθ/sin3θ) = 0.

Note:
Because Φ is cyclic, we already have a first integral. 
We can insert dΦ/dt = C/sin2θ into the expressions for d2θ/dt2 and d2Φ/dt2.

Problem 3:

Consider a bead of mass m sliding freely on a smooth circular wire of radius b which rotates in a horizontal plane about one of its points O, with constant angular velocity Ω.  Let θ be the counterclockwise angle between the diameter that passes through the mass and the diameter that passes through the point O, with θ = 0 the case where the mass is farthest from O.

 

(a)  Find the equation of motion for θ.  Compare this equation with the equation of motion for a simple pendulum (point mass and massless rod).
(b)  For the initial conditions θ = 0, dθ/dt = ω0 at t = 0, describe the θ motion that occurs for |ω0| < 2Ω and for |ω0| > 2Ω.  (Note:  The same equations have the same solutions.)
(c)  Describe the θ motion that occurs for |ω0| << 2Ω.
(d)  Find the force that the wire exerts on the bead as a function of θ and dθ/dt.

Solution:

bulletConcepts:
Lagrangian Mechanics
bulletReasoning:
All forces except the forces of constraint are derivable from a potential.  The Lagrangian formalism is well suited for such a problem.
bulletDetails of the calculation:
(a)  Let Ωt be the angle between the diameter going through the rotation point O and the x-axis.  Then we may write the Cartesian coordinates of the mass as
x = bcos(Ωt) + bcos(θ + Ωt),  y = bsin(Ωt) + bsin(θ + Ωt).
T = ½m((dx/dt)2 + dy/dt)2) = ½mb22 + (Ω + ω)2 + 2Ω(Ω + ω) cosθ),  where ω = dθ/dt.
L = T.
Lagrange’s equation  d/dt(∂L/∂ω) - ∂L/∂θ = 0 yields
d2θ/dt2 + Ω2sinθ = 0.  A simple pendulum has the same equation of motion if we let Ω2 = g/ℓ.

(b)  Using the simple pendulum (point mass on a massless rod) as a guide we know that the θ motion is oscillatory if ½mℓ2ω02 < 2mgℓ, or ω02 < 4g/ℓ and the θ motion is circular if ω02 < 4g/ℓ.  So for our problem the θ motion is oscillatory if ω02 < 4Ω2 and the θ motion is circular if ω02 < 4Ω2.

(c)  If |ω0| << 2Ω then sin(θ) ≈ θ and we have simple harmonic motion with a period T = 2π/Ω that does not depend on the amplitude.
As |ω0| increases, and approaches 2Ω the period increases and becomes very long near |ω0| ≈ 2Ω.
For |ω0| > 2Ω the period (of the now circular motion) increases again and approaches T = 2π/|ω0| for |ω0| >> 2Ω.

(d)  Use Lagrange multipliers.  Let r become a generalized coordinate.
x = bcos(Ωt) + rcos(θ + Ωt),  y = bsin(Ωt) + rsin(θ + Ωt).
T = ½m(b2Ω2 + (dr/dt)2 + r2(Ω + ω)2 + 2bΩ(dr/dt)sinθ + 2brΩ(Ω + ω) cosθ) = L.
The equation of constraint is r = b, dr = 0, 
λ1(a1r dr + a dθ) = 0, a1r  = 1, a = 0.
Lagrange’s equation for the r degree of freedom is d/dt(∂L/∂vr) -  ∂L/∂r = λ1a1r = λ1.
λ1 = Fr, the force that the wire exerts on the bead.
Fr = -mb[Ω2cosθ + (Ω + ω)2].


Problem 4:

The Lagrangian of a system of N degrees of freedom is

What is the Hamiltonian for a symmetric mass matrix Mij = Mji?

Solution:

bulletConcepts:
The Lagrangian and the Hamiltonian
bulletReasoning:
Given the Lagrangian, we are asked to find the Hamiltonian of a system.
bulletDetails of the calculation:
 

Therefore:

We then have:

In matrix notation:

Problem 5:

A bead, of mass m, slides without friction on a wire that is in the shape of a cycloid with equations
x = a(2θ + sin2θ),
y = a(1 - cos2θ),
- π/2 ≤ θ ≤ π/2.


A uniform gravitational field g points in the negative y-direction.
(a)  Find the Lagrangian and the second order differential equation of motion for the coordinate θ.
(b)  The bead moves on a trajectory s with elements of arc length ds.
Integrate ds = (dx2 + dy2)½ = ((dx/dθ)2 + (dy/dθ)2)½dθ with the condition s = 0 at θ = 0 to find s as a function of θ.
(c)  Rewrite the equation of motion, switching from the coordinate θ to the coordinate s and solve it.  Describe the motion.

Solution:

bulletConcepts:
Lagrangian Mechanics
bulletReasoning:
We are asked to obtain Lagrange's equation of motion, using two different generalized coordinates.
bulletDetails of the calculation:
(a)  L = T – U.   T = ½m(vx2 + vy2),  vx = dx/dt = 2a(dθ/dt)(1 + cos2θ), 
vy = dy/dt = 2a(dθ/dt)sin2θ.
T = 2ma2(dθ/dt)2[(1 + cos2θ)2 + sin22θ] = 4ma2(dθ/dt)2(1 + cos2θ)
= 8ma2(dθ/dt)2cos2θ.
U = mgy = mga(1 – cos2θ) = 2mga sin2θ.
L = 8ma2(dθ/dt)2cos2θ - 2mga sin2θ.

Lagrange’s equations:  d/dt(∂L/∂(dqi/dt)) -  ∂L/∂qi = 0.
We have only one generalized coordinate.
∂L/∂(dθ/dt)) = 16ma2(dθ/dt)cos2θ.
d/dt(∂L/∂(dθ/dt)) = 16ma2d2θ/dt2cos2θ - 32ma2(dθ/dt)2cosθ sinθ.
∂L/∂θ = -16ma2(dθ/dt)2cosθ sinθ - 4mga cosθ sinθ.
Equation of motion: 
16ma2d2θ/dt2cos2θ - 16ma2(dθ/dt)2cosθ sinθ + 4mga cosθ sinθ = 0.
d2θ/dt2cosθ - (dθ/dt)2sinθ + (g/(4a)) sinθ = 0.

(b)  ds = 4a cosθ dθ.  s(θ) = ∫0θds = ∫0θ4a cosθ’dθ’ = 4a sinθ.

(c)  ds/dt = 4a cosθ (dθ/dt).  d2s/dt2 = 4a cosθ d2θ/dt2 - 4a sinθ (dθ/dt)2.
Therefore d2θ/dt2 cosθ - (dθ/dt)2 sinθ = (4a)-1d2s/dt2,
and  (g/(4a))sinθ = (g/(4a)2)s.

The equation of motion in terms of the coordinate s is d2s/dt2 = -(g/(4a))s. 
This is Hooke’s law with the solution s(t) = s0 cos(ωt + φ).

ω = (g/(4a))½, s0 and φ depend on the initial conditions.
The bead executes simple harmonic motion in s.