Taylor 2.1 (today), 2.2 - 2.3 (Monday)
Polar coordinates, r and φ, with unit vectors rhat and φhat.
First, notice that since rhat and φhat are unit vectors, their lengths are fixed; any change to a unit vector must be in a direction perpendicular to the vector. Make a sketch of some arbitrary motion in two dimensions and pick two nearby points with slightly different rhat vectors, r1hat and r2hat. Call the angle between the two vectors Δφ, and notice that Δrhat = r2hat - r1hat ≈ Δ&phi φhat. Let the time it takes for the object to move from point 1 to point 2 be Δt. Then
Similarly we can show that
Now
The second derivative of r is
Notice that this is the familiar expression for the radial and tangential components of acceleration.
Now we're ready to solve problems in polar coordinates.
Consider the simple pendulum problem in polar coordinates, that is a mass m at the end of a string of length l, and solve for the motion for small oscillations.
Put the center of the coordinate system at the fixed end of the string and let φ=0 at the equilibrium position where the pendulum hangs vertically. There are two forces acting on the pendulum, the tension in the string, and the weight of the mass. The components of the weight are Fr = mg cosφ and Fφ = -mg sinφ. The tension in the string acts in the radial direction, T = Trhat.
Since the length of the string is fixed, there can be no motion in the r direction. Begin by examining the φ component of the equation of motion,
Since r = l is constant, the second term on the right is zero, leaving
The mass term can be divided out. For small amplitudes, sinφ ≈ φ and the equation of motion becomes
This is a differential equation, a linear, homogeneous second order differential equation to be precise. This differential equation can't be solved by simple integration, but we can "guess" the form of the solution. It says that φ is a function of time whose second derivative with respect to time returns the same dependence on time. For instance, if φ(t) is a polynomial of order n, then the second derivative is a polynomial of order n-2, so this can't be the functional form for φ. What functions do you know of with second derivatives that are the same as the original function? The sin and cos functions have this property. So we try a solution of the form φ(t) = A sin(ωt) + B cos(ωt). The second derivative of this function is
and the equation of motion becomes
This equation is satisfied if either φ(t) = 0 [the trivial solution, not considered further], or ω² = g/l.
Use the initial conditions to find the constants A and B (the specific solution). Since this is a second order equation, two initial conditions are needed to completely specify the solution, for instance, the position and velocity at some time.
What about the r equation? This equation isn't needed for a solution (I'm telling you this, knowing the answer). This equation does give other information, but nothing that isn't derivable from the solution obtained with the φ equation.