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:

Concepts: Lagrange's Equations | |

Reasoning: 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. | |

Details of the calculation: (a) We have a conservative system. Potential energy is converted into rotational and translational kinetic energy ½Iω ^{2} + ½Mv^{2} with v = ωR. The moment of inertia
of the solid sphere I = (2/5)MR^{2} is smaller than the moment of
inertia of the thin spherical shell I = (2/3)MR^{2}, 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/R |

**Problem 2:**

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

Solution:

Concepts: Lagrange's Equations: d/dt(∂L/∂(dq _{j}/dt)) - ∂L/∂q_{j} = 0;
generalized coordinates | |

Reasoning: 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. | |

Details of the calculation: d/dt(∂L/∂(dq _{j}/dt)) - ∂L/∂q_{j} = 0,
and q_{1 }= θ, q_{2 }= Φ.L = T – U T = ½m(r ^{2}(dθ/dt)^{2} + r^{2}sin^{2}θ(dΦ/dt)^{2}),
U = mg r cosθ.∂L/∂(dθ/dt) = mr ^{2} dθ/dt,^{ } d/dt(∂L/∂(dθ/dt)) = mr^{2} d^{2}θ/dt^{2},^{
}∂L/∂θ = mr^{2 }sinθ cosθ(dΦ/dt)^{2} + mg r sinθ.∂L/∂(dΦ/dt) = mr ^{2} sin^{2}θ dΦ/dt,^{ } d/dt(∂L/∂(dΦ/dt)) = mr ^{2} sin^{2}θ d^{2}Φ/dt^{2}
+ 2mr^{2} sinθ cosθ dθ/dt dΦ/dt,^{
}∂L/∂Φ = 0.^{
}In this problem Φ is cyclic, ∂L/∂(dΦ/dt) is a constant of motion.sin ^{2}θ(dΦ/dt) = constant = C. dΦ/dt = C/sin^{2}θ.Equations of motion
(second order differential equations):
We can insert dΦ/dt = C/sin ^{2}θ into the expressions for d^{2}θ/dt^{2}
and
d^{2}Φ/dt^{2}. |

**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 = ω_{0} at t = 0,
describe the θ motion that occurs for
|ω_{0}| < 2Ω and for |ω_{0}| > 2Ω. (Note: The same equations
have the same solutions.)

(c) Describe the θ motion that occurs for |ω_{0}| << 2Ω.

(d) Find the force that the wire exerts on the bead as a function of θ and dθ/dt.

Solution:

Concepts: Lagrangian Mechanics | |

Reasoning: All forces except the forces of constraint are derivable from a potential. The Lagrangian formalism is well suited for such a problem. | |

Details 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}) = ½mb^{2}(Ω^{2}
+ (Ω + ω)^{2} + 2Ω(Ω + ω) cosθ), where ω = dθ/dt.L = T. Lagrange’s equation d/dt(∂L/∂ω) - ∂L/∂θ = 0 yields d ^{2}θ/dt^{2} + Ω^{2}sinθ = 0. A simple pendulum has
the same equation of motion if we let Ω^{2} = 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 ½mℓ ^{2}ω_{0}^{2}
< 2mgℓ, or ω_{0}^{2} < 4g/ℓ and the θ motion is circular if
ω_{0}^{2} < 4g/ℓ. So for our problem the θ motion is
oscillatory if ω_{0}^{2} < 4Ω^{2} and the θ motion is
circular if ω_{0}^{2} < 4Ω^{2}.(c) If |ω _{0}| << 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 |ω _{0}| increases, and approaches 2Ω the period increases and
becomes very long near |ω_{0}| ≈ 2Ω.For |ω _{0}| > 2Ω the period (of the now circular motion) increases again
and approaches T = 2π/|ω_{0}| for |ω_{0}| >> 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(b ^{2}Ω^{2} + (dr/dt)^{2} + r^{2}(Ω + ω)^{2}
+ 2bΩ(dr/dt)sinθ + 2brΩ(Ω + ω) cosθ) = L.The equation of constraint is r = b, dr = 0, λ _{1}(a_{1r }dr + a_{1θ} dθ)_{ }= 0, a_{1r
} = 1, a_{1θ} = 0.Lagrange’s equation for the r degree of freedom is d/dt(∂L/∂v _{r}) -
∂L/∂r = λ_{1}a_{1r} = λ_{1}.λ _{1} = F_{r}, the force that the wire exerts on the bead.F _{r} = -mb[Ω^{2}cosθ + (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 M_{ij} = M_{ji}?

Solution:

Concepts: The Lagrangian and the Hamiltonian | |

Reasoning: Given the Lagrangian, we are asked to find the Hamiltonian of a system. | |

Details 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 = (dx

(c) Rewrite the equation of motion, switching from the coordinate θ to the coordinate s and solve it. Describe the motion.

Solution:

Concepts: Lagrangian Mechanics | |

Reasoning: We are asked to obtain Lagrange's equation of motion, using two different generalized coordinates. | |

Details of the calculation: (a) L = T – U. T = ½m(v _{x}^{2} + v_{y}^{2}),
v_{x} = dx/dt = 2a(dθ/dt)(1 + cos2θ), v _{y} = dy/dt = 2a(dθ/dt)sin2θ. T = 2ma ^{2}(dθ/dt)^{2}[(1 + cos2θ)^{2} + sin^{2}2θ]
= 4ma^{2}(dθ/dt)^{2}(1 + cos2θ) = 8ma ^{2}(dθ/dt)^{2}cos^{2}θ.U = mgy = mga(1 – cos2θ) = 2mga sin ^{2}θ.L = 8ma ^{2}(dθ/dt)^{2}cos^{2}θ - 2mga sin^{2}θ.Lagrange’s equations: d/dt(∂L/∂(dq _{i}/dt)) - ∂L/∂q_{i} = 0.We have only one generalized coordinate. ∂L/∂(dθ/dt)) = 16ma ^{2}(dθ/dt)cos^{2}θ.d/dt(∂L/∂(dθ/dt)) = 16ma ^{2}d^{2}θ/dt^{2}cos^{2}θ
- 32ma^{2}(dθ/dt)^{2}cosθ sinθ.∂L/∂θ = -16ma ^{2}(dθ/dt)^{2}cosθ sinθ - 4mga cosθ sinθ.Equation of motion: 16ma ^{2}d^{2}θ/dt^{2}cos^{2}θ
- 16ma^{2}(dθ/dt)^{2}cosθ sinθ + 4mga cosθ sinθ = 0.d ^{2}θ/dt^{2}cosθ - (dθ/dt)^{2}sinθ + (g/(4a)) sinθ = 0.(b) ds = 4a (c) ds/dt = 4a The equation of motion in terms of the coordinate s is d ω = (g/(4a)) |