PHY5200 F06

Chapter 4: Energy

Reading

Taylor 4.5-4.7 (today) and 4.7-4.9 (Friday).

Recall

It is possible to treat spherically symmetric potentials as linear one-dimensional motion.

For one-dimensional motion, it is possible to write down an integral for xdot = v and x. Note that T = ½mv²(x) = E - U(x). This expression can be solved for v(x):

v(x) = ±(2/m)½(E - U(x))½
The sign of the velocity is ambiguous and must be determined from some information about the motion, for instance, the initial conditions.

The right hand side of the expression for v depends only on x. Therefore, we can separate the x and t dependencies, and integrate to find x as a function of t

tf - ti = ∫xixf dx/v(x)

Curvilinear one-dimensional systems

Many systems that are constrained to move in one dimension, but not necessarily on a straight path, can be treated in the same way.

Example: Stability of a cube balanced on a cylinder

As an example, let's consider the stability of a cube balanced on top of a cylinder. Let the side of the cube be of length 2b, and the radius of the cylinder be r. Position the cube centered directly atop the cylinder, parallel to the axis of the cylinder, and let the cube rock on the cylinder without slipping.

Line up the center of the cube over the axis of the cylinder. When the cube rotates, its point of contact moves and can be conveniently quantified by the angle θ with respect to the center of the cylinder. The potential energy of the cube is U = mgh where h is the height of the center of the cube (its CM) above some reference. When the cube is rocked to an angle θ its height above the axis of the cylinder is h = (r+b)cosθ + rθsinθ so

U(θ) = mg[(r+b)cosθ + rθsinθ].
When the force goes to zero we have an equilibrium position; the force is given by minus the derivative of U with respect to θ
∂U/∂θ = mg[rθcosθ - bsinθ] = 0
This is satisfied for θ=0, as we would guess. To see if this is a point of stable equilibrium, we must determine the sign of the second derivative, positive is stable while negative is unstable.
∂²U/∂θ² = mg(r-b)
This is positive if r>b, and negative if r<b. So the equilibrium is stable if the cube is smaller than the cylinder.

Further Generalizations

Central Froces

Central forces are forces that act along a straight line from source to object; for example gravity and the Coulomb force. It is particularly convenient to express central forces in spherical coordinates where F = F(r)rhat.

Spherical (Polar) Coordinates

Spherical coordinates are extremely useful in central force problems: planets orbiting the sun, 3-D harmonic oscillator, or electron bound to a proton. Spherical coordinates define a location by its

We can relate (r, θ, φ) to (x, y, z) using geometry. First, it is easy to see that z = r cosθ. Then the x and y components are projections of the length projected onto the x-y plane, r sinθ. We find that x = r sinθ cosφ and y = r sinθ sinφ.

The surface of the earth is mapped with spherical coordinates with r approximately constant, so usually neglected. The latitude is a measure of the angle from the equator (positive for north and negative for south). This is an azimuthal angle, but measured from the equator rather than the north pole; θ = 90o - latitude. The longitude is just like our azimuthal angle φ, where for earth we arbitrarily choose a particular direction for the x axis (φ=0).

A spherically symmetrical function f(r) means that f depends only on the distance from the origin, r, not the directions θ or φ, meaning that f(r) = f(r). Since r, θ, and φ are independent coordinates (we can change one without changing the others), then ∂f(r)/∂θ = ∂f(r)/∂φ = 0. Only the partial derivative with respect to r is non-zero.

The unit vectors are defined as the direction in which the corresponding coordinate increases while the other two don't change. The unit vectors are mutually orthogonal, so the dot product of two vectors in spherical coordinates is still the sum of the products of the corresponding coordinates, that is, if a = arrhat + aθθhat + aφφhat and b = brrhat + bθθhat + bφφhat then

a⋅b = arbr + aθbθ + aφbφ.
© 2007 Robert Harr