PHY5200 F06

Chapter 2: Projectiles and Particles

Reading

Taylor 2.2-2.4 (today) and 2.4-2.5 (Friday)

Comments on homework problems

In problem 1, the tangential component of the acceleration means the component parallel to the velocity.

In problem 5, I want to see the calculation of the first four terms of the series. Don't simply write down the answer, the answers are available in many places.

In problem 6 part b, there is a subtraction of two infinite terms. Be sure that you see that.

In problem 7, the cylindrical coordinates are (ρ, φ, z) where z is the z coordinate, ρ is the radius from the z axis, and φ is the angle in the x-y plane. You may find the result of problem 1.39 in Taylor (problem 4 in homework set 2) to be useful.

Recall

The velocity of an object moving horizontally subject to a linear drag force, f=-bv, is

v_x(t) = A e^-kt

where k=m/b. The velocity tends to zero as time goes to infinity.

The horizontal position as a function is found by integrating the velocity

x(t) = x(0) + ∫₀^t v_x(t')dt'

x(t) = x_∞(1 - e^-t/τ)

where I've set x=0 at t=0 and used x_∞ = v_x0τ

Vertical motion with linear drag

The motion in the vertical direction is the same as that for an object dropped from a height and subject to a linear drag force,

mv_y[dot] = mg - bv_y .

The solution of this equation is not much different from the case for horizontal motion and is

v_y = Ae^-kt + mg/b

as can be checked by substitution, where k = b/m. This motion differs from the previous case in that, instead of the velocity tending to zero as t tends to infinity, v_y tends to mg/b = g/k. We call this the terminal velocity, since an object dropped from rest will achieve this velocity,

v_ter = mg/b = g/k

but won't exceeed it.

If the object is dropped from rest, then the velocity as a function of time can be written as

v_y = v_ter(1 - e^-t/τ)

where τ = 1/k = v_terg is the time constant. Expressions like this arise for the problem of charging a capacitor through a resistor. In this case, it says that the velocity will increase from zero and reach about 63% of its final value, v_ter, after one time constant, τ. After two time constants the velocity reaches 86% of its final value, and 95% after 3 time constants.

If the object doesn't start from rest but with initial velocity v_y0, then the velocity as a function of time is

v_y(t) = v_ter + (v_y0 - v_ter)e^-t/τ .

This is basically the result applied to the Millikan oil drop experiment. To determine the charge on a drop, one needs the mass. The above result shows that no matter how the drop forms it will quickly equalibrate and fall through the air with the terminal velocity. By measuring the terminal velocity in air, the mass can be found.

The position as a function of time is found by integrating the velocity

y(t) = ∫₀^t v_y(t')dt'

y(t) = v_tert + (v_y0 - v_ter)τ(1 - e^-t/τ)

Trajectory and Range in a Linear Medium

Now we have the solutions for the motion in both x and y, in particular

x(t) = v_x0τ(1 - e^-t/τ)

y(t) =(v_y0 + v_ter)τ(1 - e^-t/τ) - v_tert

where the sign of v_ter is reversed so that positive y is vertically upward. We'd like to plot the trajectory, and this is easier if we eliminate t in the above equations to give one equation of y as a function of x. This gives

y = [(v_y0+v_ter)/v_x0]x + v_terτ ln[1 - x/(v_x0τ)]

One of your homework problems is to show that as the air drag is reduced to zero, the expression reduces to what we expect from introductory physics.