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.

© Robert Harr 2006