PROBLEM SET: Terminal Velocity

Topics and skills: Integration, graphing

When an object falls in Earth's gravitational field (think of a skydiver jumping from an airplane or a marble falling in a tank of oil), it accelerates due to the force of gravity. If gravity were the only force acting on the object, then all objects-elephants and feathers alike-would fall at the same rate. But gravity is not the only force present. Moving objects also experience resistance or friction from the surrounding medium; it would be air resistance for a skydiver and fluid resistance for a marble falling in oil. The strength of the resistance depends on several factors, among them the shape of the object and the thickness (or viscosity) of the surrounding medium. The effect of resistance is that a falling object does not accelerate forever, as it would without resistance. Eventually the gravitational force acting downward and the resistance force acting upward balance each other. As this balance is reached, the object approaches a constant terminal velocity.

The motion of moving objects is described by Newton's second law of motion, which says that

Mass x acceleration = sum of external forces.

For a falling object there are two significant external forces: gravity and resistance. We let x(t) be the position of the falling object where x = 0 is the point at which the object is released and the positive direction is downward (Figure 1).

The velocity of the object is V(t) = dx/dt and its acceleration is a(t) = dx/dt. The force due to gravity is mg, where m is the mass of the object and g 9.8 m/s² is the acceleration due to gravity; it acts in the positive (downward) direction. We denote the resistance force R; it acts in the negative (upward) direction.

The equation of motion now takes the form ma = m(dv/dt) = mg - R.

The goal is to solve this equation for the velocity of the object. All of the terms have been specified except the resistance force. Air or fluid resistance is usually modeled in one of two ways. For small velocities, often in a heavy medium such as water or oil, it is common to assume that R = kv, where k > 0 is a coefficient of resistance. This says that the resistance force increases linearly with the velocity. For large velocities, often in air, a better assumption is R = Kv², where K > 0 is a (different) coefficient of resistance. With this model, the resistance force increases with the square of the velocity. Our immediate task is to solve the equation of motion for the velocity using both types of resistance.

1. Let's begin with the assumption that R = kv. What are the units (dimensions) of the coefficient k in terms of kilograms, meters, and/or seconds?

2. The equation of motion becomes m(dv/dt) = mg-kv.

This equation is said to be separable because of the terms involving the unknown v can be collected on one side of the equation: dv/dt(1/g-(k/m)v) = 1.

Now it is a matter of integrating both sides of the equation with respect to t:

∫(1/g-(k/m)v)(dv/dt)dt = ∫dt.

A change of variables on the left side results in an integral with respect to v: ∫(1/g - (k/m)v) =∫dt.

Evaluate the integrals on both sides of this equation. Then, use the initial condition that v(0) = 0 to determine the arbitrary constant of integration. Show that the velocity function is given by

v(t) = mg/k(1-e^-kt/m).

3. Assume m = 0.1 kg and graph the velocity function for k = 0.1, 0.5, 1.0. Describe the graph and check that the initial condition v(0) = 0 is satisfied.

4. The terminal velocity is lim_t→∞v(t). What is the terminal velocity in each case in Step 3? Use Step 2 to find an expression for the terminal velocity in terms of m, g, and k. How does the terminal velocity vary with k?

5. We now turn to the assumption that R = Kv². What are the units (dimensions) of the coefficient K in terms of kilograms, meters, and/or seconds? Note that k and K have different units, which means we cannot compare numerical values of the two parameters.

6. The equation of motion now becomes m(dv/dt) = mg-kv².

This equation is also separable because all of the terms involving v can be brought to the left side of the equation: dv/dt(1/g-(k/m)v²) = 1.

As before, we integrate both sides of the equation with respect to t and use a change of variables.

The resulting equation is ∫dv/g-(k/m)v² = ∫dt.

It's easiest to write the equation as ∫dv/a²-v² = k/m∫dt, where a² = mg/k.

Evaluate the integrals on both sides of this equation and use the initial condition v(0) = 0 to determine the arbitrary constant. Show that the velocity function is given in either of the two forms

v(t) = √(mg/k)(e^2√(kg/mt)-1/ e^2√(kg/mt)+1) = √(mg/k)(1- e-^2√(kg/mt)/1+ e-^2√(kg/mt)).

7. Assume m = 0.1 kg and graph the velocity function for K = 0.1, 0.5, 1.0. Describe the graph and check that the initial condition v(0) = 0 is satisfied.

8. What is the terminal velocity in each case in Step 7? Find an expression for the terminal velocity in terms of m, g, and K. How does the terminal velocity vary with K?

9. Consider the following velocity measurements of a marble falling in light oil.

Graph the data and graph the velocity functions in Steps 2 and 6 with m = 0.1 for various values of k and K. Which model and which value of k or K give the best fit to the data?

10. Integrate the velocity function found in Step 2 to find the position function x(t) where x(0) = 0. Graph the position function for m = 0.1 kg and k = 0.1, 0.5, and 1.0. Discuss how the graphs vary with k. How is the terminal velocity reflected in these graphs?

11. Integrate the velocity function found in Exercise 6 to find the position functions x(t) where x(0) = 0. Graph the position for m = 0.1 kg and K = 0.1, 0.5, and 1.0. Discuss how the graphs vary with K. How is the terminal velocity reflected in these graphs?