First midterm exam will be next Monday, Feb. 9. Friday will be a review. The exam will cover chapters 9 and 6, and 7.1.
In Taylor, read sections 7.4 to 7.5 for today, and 7.5 to 7.6 for next Wednesday.
Please read this section. I will not cover this in class. This is a proof that Lagrange's equations for the generalized coordinates of a constrained system are consistent with Newton's laws.
An Atwood machine consists of two masses, m1 and m2, connected by a string of fixed length l strung over a frictionless pulley of radius R and moment of inertia I. Write down the Lagrangian for the system in terms of position of m1, x, find the Lagrange equation for x, and solve for the acceleration. Compare with the Newtonian solution.
If m1 is at x, then m2 is at y = l - x - πR. The angular velocity of the wheel is related to xdot by ω = xdot/R Therefore the kinetic energy is
The potential energy is
where we can ignore the terms independent of x, xdot and t. The Lagrangian is
The Lagrange equation for x is
which becomes
This yields an acceleration of
To solve this problem with Newton's laws we need three equations:
Adding the first two equations and using the third to substitute for F1 - F2, yields the same equation as above. The fact that you must realize that the tension on the two sides of the pulley are different occurs automatically in the Lagrangian approach.
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
and
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 + r² dθ² + r² sin²θ dφ², and v² = ds²/dt² = rdot² + r² θdot² + r² sin²θ φdot². For the constrained coordinates of the bead the kinetic energy is T = ½m(R² θdot² + R²Ω² sin² θ ). The potential energy is all gravitational, U = -m g R &cosθ. The Lagrangian is
The generalized coordinate is θ. The Lagrange equation for θ yields
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/R Ω²), where the last one(s) exists only if g/R Ω² ≤ 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/R Ω² > 1. The equilibrium at θ = 0 is stable if g/R Ω² > 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/R Ω²). We will need to use the relations cos(θ0 + Δθ) = cosθ0cosΔθ - sinθ0sinΔθ ≈ (g/R Ω²) - sqrt(1 - (g/R Ω²)²) Δθ and sin(θ0 + Δθ) = cosθ0 sinΔθ + sinθ0 cosΔθ ≈ (g/R Ω²) Δθ + sqrt(1 - (g/R Ω²)²). Near the θ = θ0 equilibrium, a small displacement produces an acceleration of
This gives θddot/Δθ ≈ ((g/R Ω²)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 = -Ω² Δθ.
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/R Ω²)
Therefore the frequency is ω = Ω sqrt(1 - (g/R Ω²)² ) = sqrt(Ω² - (g/R Ω)² ).