PHY5200 F06

Chapter 1: Newton's Laws of Motion

Reading

Taylor 2.1 (today), 2.2 - 2.3 (Monday)

Recall

drhat / dt = dφ/dt φhat
dφhat / dt = -dφ/dt rhat
r[ddot] = (r[ddot] - r(φ[dot])²) rhat + (rφ[ddot] + r[dot]φ[dot]) φhat.

Example: A pendulum

Consider the simple pendulum problem in polar coordinates, that is a mass at the end of a string, 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 force are Fr = mg cosφ and Fφ = -mg sinφ. The tension in the string will be equal and opposite to the radial component of the force, T = -Fr yielding no motion in the radial direction.

In the φ direction we have

Fφ = -mg sinφ = mrφ[ddot] + 2mr[dot]φ[dot]
Since there is no motion in the r direction, the second term on the right is zero, leaving
-mg sinφ = mrφ[ddot]
For small amplitudes, sinφ ≈ φ and the equation of motion becomes
φ[ddot] + (r/g)φ = 0

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)

Use the initial conditions to find the specific solution.



Chapter 2: Projectiles and Particles

Here we revisit the motion of projectiles near the Earth's surface, this time adding air resistance to the problem, and we'll discuss the motion of charged particles in magnetic fields.

Air Resistance

Frictional forces are complex, involving microscopic and mesoscopic processes. Yet, we can treat many problems with the simple approximation for sliding friction f=μN, possibly with a different coefficient of friction for static, μs, and kinetic friction, μk.

In general, air resistance is complex. But in a similar way, a simple approximation can be used to treat many problems. On what quantity do we expect air friction to depend, in analogy to the normal force for sliding? Obviously, the normal force plays no role for a projectile, so what should take its place? Velocity is the appropriate quantity, and we can approximate air friction with two terms, one proportional to velocity (the linear term) and one proportional to the square of velocity (the quadratic term)

f(v) = bv + cv²
The frictional force is directed opposite the motion of the projectile
f = -f(v) vhat
Note that we are clearly neglecting certain cases, for instance the lift on an airplane wing. Nevertheless, this approximation is useful in many situations, and allows us to move from the simple problems of introductory physics to more realistic situations.

The two constants, b and c, must be determined, but there is a useful approximation for spherical objects. For spherical objects, the constants can be parameterized in terms of the diameter, D,

b = βD   and   c = γD²
where, for air at STP
β = 1.6×10-4 N⋅s/m²   and   γ = 0.25 N⋅s² / m4

Now we need to solve the equation of motion when air resistance is included. The difficulty varies depending on whether one drag force dominates so that the other can be neglected, or if both must be included. The easiest case to solve is when the linear drag force dominates; the case where the quadratic force dominates is more difficult; and when both must be included, numerical calculations are required.

To determine if one force dominates, it is convenient to look at their ratio

fquad / flin = (cv²) / (bv) = (γD / β) v = (1.6×103 s/m²) Dv

Some examples: baseball and liquid drops

Calculate the ratio of quadratic to linear drag force for:

For the baseball

fquad / flin = (1.6×103 s/m²)(0.07m)(5m/s) = 560

For the raindrop

fquad / flin = (1.6×103 s/m²)(0.001m)(0.6m/s) = 1

For the oil drop

fquad / flin = (1.6×103 s/m²)(1.5×10-6m)(5×10-5m/s) = 1.2×10-7

The quadratic force dominates for the baseball; the linear force dominates for the oil drop; and both forces are comparable for the raindrop.


© Robert Harr 2006