Taylor 4.6-4.7 (today) and 4.7-4.8 (Wednesday).
E = T + U
A force is conservative if F is only a function of position, and the ∇×F = 0. Then F = ∇U.
If U is a function of time as well as position, then we can still define a force from F = ∇U, but now the force is not conservative, and dE/dt = ∂U/∂t.
What is meant by a linear one-dimensional system? The author means a system whose position is characterized by a single variable, and that moves in a straight line, as if on a track. For such systems, the requirements for a force to be conservative that (i) the work done must depend only on the position (let's call that x) and (ii) the work done must be independent of path are equivalent.
We can graph U as a function of x. Such a graph has the property that the force, F = -dU/dx, points downhill. There is a similarity with hilly terrain, or a roller coaster where the potential energy is given by mgh. Here the potential changes linearly with height, just like on a graph of U as a function of x.
The force is zero where the derivative dU/dx = 0, at the tops of hills or the bottom of valleys. These can be points of stable or unstable equilibrium: if the potential curves upward, then the object will tend to return to the equilibrium position if it is displaced slightly; if the potential curves downward, then the object will tend to move further away from the equilibrium position if displaced slightly.
Since the total (mechanical) energy is E = T + U, then a line drawn across the graph at U=E leads to a qualitative understanding of the motion. First, the object can be located only where U≤E. Locations where U=E are called turning points, since as these points are approached, the kinetic energy goes to zero, but the force doesn't, so the object then moves away from the point -- the motion turns at these points. If the object is in a region bounded by two turning points, then it moves back and forth between them. If it is in a region bounded by one turning point, then the other direction must be open to infinity, so if an object approaches the turning point, it will turn around and move away to infinity.
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):
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
Many systems that are constrained to move in one dimension, but not necessarily on a straight path, can be treated in the same way.
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
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
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.
This is positive if r>b, and negative if r<b. So the equilibrium is stable if the cube is smaller than the cylinder.