Drag

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 = -bv.

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 = ma

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) - [(kv0+g)/k]e-kt

where k = b/m.
For a stone starting from rest v0 = 0 and therefore v = (g/k)(1 - e-kt).
At t = 0, e-kt = 1 and the speed of the stone is v0.   As t becomes very large, e-kt approaches zero, and v approaches vt = 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 = bvterminal,    vterminal = mg/b.

Problem:

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*103 kg/m3, 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 vterminal and calculate b.    (Neglect the buoyant force.)

Solution:
vterminal*(12 s) = 0.6 m, vterminal = 0.05 m/s.
vterminal = mg/b,  b = mg/vterminal.
The mass of the sphere equals its density times its volume.
m = (7.8*103 kg/m3)*4πr3/3 = (7.8*103 kg/m3)*4π(1.5*10-3 m)3/3 = 1.1*10-4 kg.
b = (1.1*10-4 kg)*(9.8 m/s2)/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 = ½CρAv2.

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(vterminal)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 = ma

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

mg - ½DρAv2 = mdv/dt

or

dv/dt = g - DρAv2/(2m).

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

vt = (2mg/(DρA))½

Link:  Typical drag coefficient values

Problem:

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.

Problem:

A 70 kg skydiver descending through air head first will have an area of approximately 0.18 m2 and a drag coefficient of approximately 0.7.  What is his terminal velocity?  The density of air is 1.21 kg/m3.

Solution:
mg = ½CρA(vterminal)2, (vterminal)2 = mg /(0.5 C ρ A).
(vterminal)2 = (70 kg)*(9.8 m/s2)/(0.5*0.7*(1.21 kg/m3)*(0.18 m2)) = 9*103 (m/s)2.
vterminal = 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.

Problem:

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, vt = 50 m/s.
(a)  Using vt = (2mg/(DρA))½ we can find DρA.  DρA = 2mg/vt2 = 0.6272 kg/m.
Therefore a = dv/dt = 9.8 m/s2 - (0.6272/(160 m))v2.
When v = 30 m/s, a = 6.272 m/s2.
(b)  When v = vt = 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/s2) = 502 N.  The magnitude of the drag force is mg - ma = 282 N.

Links: