In Taylor, read sections 7.5 to 7.6 for today, and 7.7 to 7.8 for Monday.
A particle of mass m is constrained to move on a frictionless cylinder of radius R. The particle is subject to a 3-D Hooke's law force F = -kr where r is the vector from the origin located at a point on the axis of the cylinder. Find the Lagrangian, determine the equations of motion, and solve them.
Call the coordinates of the mass on the cylinder are (ρ, φ, z). Note that v² = (ds/dt)² = ds²/dt². In cylindrical coordinates, ds² = dρ² + ρ²dφ² + dz², so v² = ρdot² + ρ²φdot² + zdot². On the cylinder, ρ = R is constant, so the kinetic energy of the mass is T = ½m(R²φdot² + zdot²). The potential energy for a spring-like force is U = ½kr² = ½k(R² + z²). The Lagrangian is
The generalized coordinates are φ and z. The Lagrange equations for φ and z yield
The first equation says φdot = constant = Ω0 and φ(t) = Ω0t + φ0. The second equation is simple harmonic motion in z, z(t) = A cos(ωt) + B sin(ωt), with A and B determined from the initial conditions and ω = √(k/m). We find that the mass moves around the cylinder at a constant rate while executing simple harmonic motion in the direction parallel to the cylinder's axis.
A bead of mass m is constrained to move on a wire formed into a hoop of radius R. The hoop sits in a vertical plane and is rotated about the vertical diameter at a constant angular velocity ω. Specify the bead's position by the angle θ measured from the downward vertical direction. Write down the Lagrangian, determine the equation of motion, and find any equilibrium positions that may exist. Determine the stability of any equilibrium positions.
We can use spherical coordinates, with the origin at the center of the hoop, to describe the bead's position, (r, θ, φ) = (R, θ, ωt). In spherical coordinates, ds² = dr² + r²dθ² + r²sin²θdφ², and v² = ds²/dt² = rdot² + r²θdot² + r²sin²θφdot². For the constrained coordinates of the bead the kinetic energy is T = ½m(R²θdot² + R²sin²θω²). The potential energy is all gravitational, U = -mgR&cosθ. The Lagrangian is
The generalized coordinate is θ. The Lagrange equations for θ yield
An equilibrium position is a value of θ where both θdot and θddot are zero. Set θddot = 0 and solve for any values of θ that satisfy the equation.
This is satisfied for θ = 0, π, or ±arccos(g/Rω²), where the last one(s) exists only if |g/Rω²| ≤ 1. To evaluate the stability of these equilibria, we need to consider what will happen if the bead is moved slightly from the equilibrium location, say by an amount Δθ, with a very small velocity Δθdot. If the equilibrium is stable, the bead will try to return to its starting location, meaning it will accelerate opposite to the motion such that θddot/Δθ < 0.
Near the θ = 0 equilibrium, a small displacement produces an acceleration of
So θddot/Δθ ≈ (ω² - g/R) is less then zero when g/Rω² > 1. The equilibrium at θ = 0 is stable if g/Rω² > 1, and unstable otherwise.
Near the θ = π equilibrium, a small displacment produces an acceleration of
So θddot/Δθ ≈ (ω² + g/R) which is always positive. The equilibrium at θ = π is unstable.
For convenience, call the other equilibrium angles θ0 = arccos(g/Rω²). We will need to use the relations cos(θ0 + Δθ) = cosθ0cosΔθ - sinθ0sinΔθ ≈ (g/Rω²) - sqrt(1 - (g/Rω²)²)Δθ and sin(θ0 + Δθ) = cosθ0sinΔθ + sinθ0cosΔθ ≈ (g/Rω²)Δθ + sqrt(1 - (g/Rω²)²). Near the θ = θ0 equilibrium, a small displacement produces an acceleration of
This gives θddot/Δθ ≈ ((g/Rω²)² - 1)ω² < 0 when g/Rω² < 1 as required for this to be an equilibrium point. Therefore, when this equilibrium point exists, it is stable.
What is the frequency of small oscillations of the bead near the points of stable equilibrium?
We have almost all the information needed now to answer this question. If we expand the expression for θddot to first order in Δtheta;, as done above, then the frequency for small oscillations, Ω, appears as the constant in the relation θddot = -Ω²Δθ.
For the stable equilibrium near θ = 0 the expression is
Therefore the frequency is Ω = sqrt(g/R - ω²).
For the stable equilibria near θ = ±arccos(g/Rω²)
Therefore the frequency is Ω = ωsqrt(1 - (g/Rω²)²) = sqrt(ω² - (g/Rω)²).