An object moving through a medium such as a gas or a liquid is acted on by a resistive force. This force always points in a direction opposite to the direction of the velocity of the object, and its magnitude depends on the speed of the object.

(a) For objects moving through a liquid or for small objects moving
**slowly**
through a gas, the magnitude of the resistive force, R, is often
proportional to the speed. We write

**R **= -b**v**.

Here b is a constant that depends on properties of the fluid and the dimensions of the object.

The equation of motion for or a stone falling through water to the bottom of a lake (motion in one dimension, along a line) is

**F **= m**a**

If we neglect the buoyant force (which we will study later), then this leads to

mg - bv = mdv/dt

or

dv/dt = g - (b/m)v.

The solution to this **differential equation**
is

v = (g/k) - [(kv_{0}+g)/k]e^{-kt},

where k = b/m.

For a stone starting from rest v_{0 }= 0 and
therefore v = (g/k)(1 - e^{-kt}).

At t = 0, e^{-kt }= 1 and the speed of the stone is v_{0}. As t becomes very large, e^{-kt} approaches zero, and v approaches v_{t
}= g/k = mg/b, the **terminal speed**. A stone falling
through water to the bottom of a lake (motion in one dimension, along a line)
reaches terminal speed when the drag force pointing upward exactly balances the
force of gravity pointing downward, i.e. when the two forces have exactly
the same magnitude. Then the net force is zero, and the acceleration
is zero. We have for the magnitudes

mg = bv_{terminal},
v_{terminal} = mg/b.

Particles in liquids achieve terminal velocity very quickly.
One can measure the time it takes for a particle to fall a certain distance and
then calculate the constant b. Suppose a
steel ball bearing (density 7.8*10^{3} kg/m^{3},
diameter 3.0 mm) is dropped in a
container of motor oil. It takes 12 s to fall a distance of 0.60 m. Find
the terminal speed v_{terminal} and calculate b.
(Neglect the buoyant force.)

Solution:

v_{terminal}*(12 s) = 0.6 m, v_{terminal} = 0.05 m/s.

v_{terminal} = mg/b, b = mg/v_{terminal}.

The mass of the sphere equals its density times its volume.

m = (7.8*10^{3} kg/m^{3})*4πr^{3}/3
= (7.8*10^{3} kg/m^{3})*4π(1.5*10^{-3}
m)^{3}/3 = 1.1*10^{-4} kg.

b = (1.1*10^{-4} kg)*(9.8 m/s^{2})/0.05 m/s = 2.16*10^{-2}
kg/s.

(b) For objects moving at **high speed** through air the magnitude of the
resistive force is often proportional to the square of the speed, and can be
written as

R = ½CρAv^{2}.

Here A is the cross sectional area of the falling object in a plane
perpendicular to its velocity, ρ is the density of the air, and
C is the
drag coefficient, which depends on the shape of the object. For a spherical
object C has a value of approximately 0.5.

If we neglect the buoyant force, then a high-speed
falling object reaches a **terminal speed **when

mg = ½CρA(v_{terminal})^{2}.

he equation of motion for or a stone falling through water to the bottom of a lake (motion in one dimension, along a line) is

**F **= m**a**

which leads for 1-dimensional motion in the vertical direction near the surface of the earth to

mg - ½DρAv^{2 }= mdv/dt

or

dv/dt = g - DρAv^{2}/(2m).

The object again reaches a terminal speed v_{t}. We find v_{t}
without solving the differential equation by simply setting dv/dt = 0.

v_{t} = (2mg/(DρA))^{½}

Link: Typical drag coefficient values

On long journeys, jet aircraft usually fly at high altitudes of about
30000 ft. What is the main advantage of flying at
these altitudes from an economic viewpoint?

Solution:

The magnitude of the resistive force for an object moving with
high speed through air increases as the density of the air
increases. At high altitudes, the air density is lower and
therefore the magnitude of the resistive force is smaller than at
low altitudes.

A 70 kg skydiver descending through air head first will have an area of
approximately 0.18 m^{2} and a drag coefficient of approximately 0.7.
What is his terminal velocity? The density of air is 1.21 kg/m^{3}.

Solution:

mg = ½CρA(v_{terminal})^{2}, (v_{terminal})^{2}
= mg /(0.5 C ρ A).

(v_{terminal})^{2} = (70 kg)*(9.8 m/s^{2})/(0.5*0.7*(1.21
kg/m^{3})*(0.18 m^{2})) = 9*10^{3} (m/s)^{2}.

v_{terminal} = 94.9 m/s =
212 mph.

If the sky diver opens a parachute, the drag coefficient does not
change much, but the area increases by more than two orders of magnitude.

A diver of mass 80 kg jumps from a slow-moving aircraft and reaches a
terminal speed of 50 m/s.

(a) What is the acceleration of the skydiver, when her speed is 30 m/s?

(b) What is the drag force on the diver when her speed is 50 m/s?

(c) What is the drag force on the diver when her speed is 30 m/s?

Solution:

Given: m = 80 kg, v_{t }= 50 m/s.

(a) Using v_{t} = (2mg/(DρA))^{½} we can find DρA.
DρA = 2mg/v_{t}^{2 }= 0.6272 kg/m.

Therefore a = dv/dt = 9.8 m/s^{2
}- (0.6272/(160 m))v^{2}.

When v = 30 m/s, a = 6.272 m/s^{2}.

(b) When v = v_{t
}= 50 m/s, the drag force is equal in magnitude
to mg = 784 N.

(c) When v = 30 m/s, the total force is
equal in magnitude to

ma = (80 kg)(6.272 m/s^{2}) = 502 N. The magnitude of the drag
force is mg - ma = 282 N.

Links: