Download Lecture 17 - De Anza College

Dynamics
Applying Newton’s Laws
Air Resistance
Lana Sheridan
De Anza College
April 30, 2015
Overview
• resistive forces
• two models for resistive forces
• terminal velocities
Resistive Forces
Galileo predicted (correctly) that all objects at the Earth’s surface
accelerate at the same rate, g .
This was a revolutionary idea because it seems obvious that less
massive objects should fall more slowly: consider a feather and a
bowling ball.
What is happening there?
Resistive Forces
Galileo predicted (correctly) that all objects at the Earth’s surface
accelerate at the same rate, g .
This was a revolutionary idea because it seems obvious that less
massive objects should fall more slowly: consider a feather and a
bowling ball.
What is happening there?
Air resistance can play a big role in determining an object’s motion.
e
Resistive Forces
Resistive forces act on an object when it moves through a fluid
medium, like a liquid or gas.
Is this object accelerating?
S
R
S
v
S
mg
Resistive Forces
where D is a dimensionle
density of air, and A is the
plane perpendicular to its
spherical objects but can
Let us analyze the mo
force of magnitude R 5 12
As Figure 6.14 shows,
the
S
gravitational force Fg 5 m
tude of the net force is
Air resistance increases with speed.
Will the object continue to accelerate?
S
R
S
R
S
S
vT
v
S
S
mg
a
where we have taken dow
object as a particle under
find that the object has a
mg
b
Figure 6.14 (a) An object
falling through air experiences
S
a resistive force R and a graviS
tational force Fg 5 mS
g . (b) The
We can calculate the te
force is balanced by the re
fore its acceleration is zero
Resistive Forces
where D is a dimensionle
density of air, and A is the
plane perpendicular to its
spherical objects but can
Let us analyze the mo
force of magnitude R 5 12
As Figure 6.14 shows,
the
S
gravitational force Fg 5 m
tude of the net force is
Air resistance increases with speed.
Will the object continue to accelerate? No.
S
R
S
R
S
S
vT
v
S
where we have taken dow
object as a particle under
find that the object has a
S
mg
mg
We can calculate the te
force
is balanced by the re
The velocity will not Figure
exceed
some
6.14
(a)terminal
An object value.
fore
its
acceleration is zero
falling through air experiences
a
b
S
a resistive force R and a graviS
tational force Fg 5 mS
g . (b) The
object reaches terminal speed
ming the only forces
acting on the sphere are the resistive force
S
Resistive force
ForcesFg , let us describe its motion.1 We model the
ravitational
What is happening to the acceleration vector?
v!0 a!g
The sphere
maximum
speed vT .
v
vT
S
R
S
v
v ! vT
S
mg
a!0
0.632vT
t
Resistive Forces
Resistive Forces resist motion.
They can play a big role in determining a an object’s motion.
Resistive Forces
Resistive Forces resist motion.
They can play a big role in determining a an object’s motion.
There are two main models for how this happens. Either the
resistive force R
• is proportional to v, or
• is proportional to v 2
Either way, this leads to an upward force on a falling object that
increases with the object’s speed through the fluid.
Model 1: Stokes Drag
The resistive force behaves like:
R = −bv
where b is a constant that depends on the falling object and the
medium.
Model 1: Stokes Drag
The resistive force behaves like:
R = −bv
where b is a constant that depends on the falling object and the
medium.
This model applies when the Reynolds Number of the fluid is very
low and/or the object is moving very slowly.
Reynolds Number, Re ∼
inertial forces
viscous forces .
More viscous fluids have lower Reynolds Numbers.
Model 1: Stokes Drag
Low Reynolds Number and/or low speed means there will be no
turbulence. Flow will be laminar, or close to laminar. (ie. smooth
flow lines)
Model 1: Stokes Drag
To summarize, cases where the resistive force will be proportional
to v :
• slow moving objects
• very viscous fluids
• laminar (smooth) flow in the fluid
Model 1: Stokes Drag
If R = −bv, what is the terminal velocity?
where D is a dimensionle
density of air, and A is the
Model 1: Stokes Drag
plane perpendicular to its
spherical objects but can
If R = −bv, what is the terminal velocity?
Let us analyze the mo
force of magnitude R 5 12
Remember, at terminal velocity, the object is in equilibrium.
As Figure 6.14 shows,
the
S
gravitational force Fg 5 m
S
R
tude of the net force is
S
R
S
S
vT
v
S
S
mg
a
where we have taken dow
object as a particle under
find that the object has a
mg
b
Figure 6.14 (a) An object
falling through air experiences
We can calculate the t
force is balanced by the re
fore its acceleration is zer
Model 1: Stokes Drag
Equilibrium ⇒ Fnet = 0.
Fnet = bvT − mg
vT =
mg
b
= 0
is (b) 50.0 m/s and (c) 30.0 m/s?
Example
on in
aises
first
d ridve of
k fores on
vator
29. Calculate the force required to pull a copper ball of
radius 2.00 cm upward through a fluid at the constant speed 9.00 cm/s. Take the drag force to be proto the speed, with proportionality constant
Page portional
171, #30
0.950 kg/s. Ignore the buoyant force.
30. A small piece of Styrofoam packing material is dropped
W from a height of 2.00 m above the ground. Until it
reaches terminal speed, the magnitude of its acceleration is given by a 5 g 2 Bv. After falling 0.500 m, the
Styrofoam effectively reaches terminal speed and then
takes 5.00 s more to reach the ground. (a) What is the
value of the constant B? (b) What is the acceleration at
t 5 0? (c) What is the acceleration when the speed is
0.150 m/s?
31. A small, spherical bead of mass 3.00 g is released from
M rest at t 5 0 from a point under the surface of a viscous liquid. The terminal speed is observed to be vT 5
2.00 cm/s. Find (a) the value of the constant b that
appears in Equation 6.2, (b) the time t at which the
Example
Page 171, #30
(a) Terminal velocity vT =
∆x
∆t
=
2.00−0.50 m
5.00 s
At terminal velocity a = 0 ⇒ g = Bv , B =
= 0.30 m/s.
g
v
= 32.7 s−1 .
Example
Page 171, #30
(a) Terminal velocity vT =
∆x
∆t
=
2.00−0.50 m
5.00 s
At terminal velocity a = 0 ⇒ g = Bv , B =
(b) at t = 0, a = g , directed downward.
= 0.30 m/s.
g
v
= 32.7 s−1 .
Example
Page 171, #30
(a) Terminal velocity vT =
∆x
∆t
=
2.00−0.50 m
5.00 s
At terminal velocity a = 0 ⇒ g = Bv , B =
= 0.30 m/s.
g
v
(b) at t = 0, a = g , directed downward.
(c) at v = 0.150 m/s, a = g − Bv = 4.9 ms−2 .
= 32.7 s−1 .
Model 1: Stokes Drag
Can we find an expression for how the velocity changes with time
before reaching vT ?
Model 1: Stokes Drag
Can we find an expression for how the velocity changes with time
before reaching vT ?
Yes! In general, (calling down positive)
Fnet = mg − bv
ma = mg − bv
dv b
+ v
dt m
A differential equation for v .
= g
Model 1: Stokes Drag
Solving
dv
dt
b
+m
v = g:
Model 1: Stokes Drag
Solving
dv
dt
b
+m
v = g:
Use an integrating factor µ(t). This is just some unknown function
of t that we multiply through the equation.
µ(t)
b
dv
+µ(t) v
dt
m
= µ(t)g
Model 1: Stokes Drag
Solving
dv
dt
b
+m
v = g:
Use an integrating factor µ(t). This is just some unknown function
of t that we multiply through the equation.
µ(t)
b
dv
+µ(t) v
dt
m
= µ(t)g
b
But µ(t) was arbitrary, so let it have the property µ 0 (t) = µ(t) m
.
Model 1: Stokes Drag
Solving
dv
dt
b
+m
v = g:
Use an integrating factor µ(t). This is just some unknown function
of t that we multiply through the equation.
µ(t)
b
dv
+µ(t) v
dt
m
= µ(t)g
b
But µ(t) was arbitrary, so let it have the property µ 0 (t) = µ(t) m
.
bt/m
⇒ µ(t) = e
Model 1: Stokes Drag
Solving
dv
dt
b
+m
v = g:
Use an integrating factor µ(t). This is just some unknown function
of t that we multiply through the equation.
µ(t)
b
dv
+µ(t) v
dt
m
= µ(t)g
b
But µ(t) was arbitrary, so let it have the property µ 0 (t) = µ(t) m
.
bt/m
⇒ µ(t) = e
Our equation becomes
µ(t)
dv
+µ 0 (t)v
dt
= µ(t)g
Now we can integrate both sides! (Product rule...)
Model 1: Stokes Drag
µ(t)
dv
+µ 0 (t)v
dt
= µ(t)g
d
(µ(t)v ) = µ(t)g
dt
Model 1: Stokes Drag
µ(t)
dv
+µ 0 (t)v
dt
= µ(t)g
d
(µ(t)v ) = µ(t)g
dt
Integrating with respect to t:
Z
µ(t)v
µ(t)g dt + C
=
R
v
=
µg dt + C
µ
We have an expression for v . All we need to do is substitute back
for µ(t).
Model 1: Stokes Drag
µ(t) = e bt/m :
R
v
e bt/m g dt + C
e bt/m
mg
= e −bt/m
e bt/m + C
b
mg
=
+ Ce −bt/m
b
=
Model 1: Stokes Drag
µ(t) = e bt/m :
R
v
e bt/m g dt + C
e bt/m
mg
= e −bt/m
e bt/m + C
b
mg
=
+ Ce −bt/m
b
=
C is an unknown constant, but we can find a value for it using an
initial condition. We can set v = 0 at t = 0, then C = − mg
b :
v (t) =
mg
(1 − e −bt/m )
b
Model 1: Stokes Drag
Exercise: Check that the solution v (t) =
the equation
dv b
+ v =g
dt m
mg
b (1
− e −bt/m ) satisfies
ns of the object and v is
the velocity of the object relative to
S
ve sign indicates that R is in the opposite direction to S
v.
Model
1: Stokes
re of
mass m released
from restDrag
in a liquid as in
Figure
6.13a.
S
s acting on the sphere are the resistive force R 5 2bS
v and
mg
(1 − e −bt/m ) satisfies
Exercise:
Check
that
the
solution
v (t) =
1
,
let
us
describe
its
motion.
We
model
the
sphere
as
abparg
the equation
dv
v!0 a!g
+
b
v =g
Thedt
spherem
approaches a
maximum (or terminal)
speed vT .
v
vT
v ! vT
a!0
0.632vT
t
b
c
The time constant t is the
t
time at which the sphere
Time constant, τ = m
b . The solution can also be written
−t/τ ). reaches a speed of 0.632vT .
v (t) = mg
(1
−
e
b
Summary
• Stokes drag
• terminal velocity
Collected Homework!
I will post it this afternoon.
(Uncollected) Homework Serway & Jewett,
• Read Chapter 6 if you haven’t already.
• Ch 6, onward from page 171. Questions: Section Qs 31, 33,
35
• Read ahead into the first 2 sections of Chapter 7.