ONLINE: MATHEMATICS EXTENSION 2
Topic 6
MECHANICS
EXERCISE p6503
Consider the vertical motion an object of mass m near the Earth’s surface.The object
was released with an initially velocity v0. The resistive force due acting on the object is
of the form FR   v 2 and directed in the opposite direction to the motion. Derive the
following result and comment on the acceleration a, velocity v and displacement x as
Down is the positive direction.
t  ?
ag
Object falling v > 0 only
vT 
mg


m
v2
2g
 t 

v
  v  v    v0  vT  e T 
v  vT  0 T
2g
 t 
  v  v    v  v  e vT 
0
T
 0 T

Object rising v < 0 only

v
v  vT tan atan  0
 vT

  g

  vT
ag
 
 t
 

m
v 2 v 2 
V 2 
x   T  loge  T 2 02 
 2g 
 vT  v 
v2
 v 2  v2 
v 2 
x   T  loge  T2
2 
2g 
 vT  v0 
physics.usyd.edu.au/teach_res/hsp/math/math.htm
p6503
1
Solution
The forces acting on the object are the gravitational force FG (weight) and the resistive
force FR. In our frame of reference, we will take down as the positive direction.
physics.usyd.edu.au/teach_res/hsp/math/math.htm
p6503
2
The equation of motion of the object is determined from Newton’s Second Law.
ma  m
dv
 FG  FR  mg   v 2  v / v 
dt
ag

m
v 2 v / v 
where a is the acceleration of the object at any instance.
The initial conditions are
t  0 v  v0
v 
x  0 a  g    / m  v0 2  0 
 v0 
When a = 0, the velocity is constant v = vT where vT is the terminal velocity
0  m g   vT 2
vT 
vT 2 
mg

mg

We start with the equation of motion then integrate this equation where the limits of
the integration are determined by the initial conditions (t = 0 and v = v0) and final
conditions (t and v)
Since the acceleration depends upon v2 it a more difficult problem then for the linear
resistive force example. We have to do separate analytical calculations for the motion
when the object is falling or rising.
physics.usyd.edu.au/teach_res/hsp/math/math.htm
p6503
3
Velocity of the object is always positive (falling object) v0  0 v  0
Equation of motion
ag

m
v2
dv
 
 g    v2
dt
m
dv
dv
dt 

  mg 2
 
g    v 2    
v 
m
 m   

a
vT 2 
mg

 1  1
dv
1 
 
   dt  2



 dv
2
v  vT
m
 2vT   v  vT v  vT 
 mg
2


  
 1
1 

   dt  
 dv
 v  vT v  vT 
 m 

 1
4 g
1
dt  

m
 v  vT v  vT

v
4 g t
1
1
dt   


v0 v  v
m 0
v  vT
T


v
4 g
t    log e  v  vT   log e  v  vT   v
0
m

 dv


 dv

physics.usyd.edu.au/teach_res/hsp/math/math.htm
p6503
4
v
4 g
t  log e  v  vT   log e  v  vT   v
0
m

 v  vT
4 g
t  log e 
m
 v0  vT



 v  vT  
  log e 


 v0  vT  
4 g
 v  vT
t  log e 
m
 v0  vT
 v  vT   v0  vT


 v0  vT   v  vT


e

  v0  vT

  v  vT
4 g
t
m
 

 
4 g
4 g 2
4 g2 2 g



m
mg
vT 2
vT
v v   t
v  vT   v  vT   0 T  e vT   v  vT  K
 v0  vT 
1 K 
v 1  K   vT 1  K 
v  vT 

1 K 
2g
  v0  vT
1 
v v
v  vT   0 T
 v v
 1   0 T
  v0  vT
v v   t
K   0 T  e vT
 v0  vT 
2g
  vT t 

e


2g 

t
 vT

e


2g
2g
 t 

v

v

v

v
e
  0 T  vT 

v  vT  0 T
2g
 t 
  v  v    v  v  e vT 
0
T
 0 T

valid only if v0  0 v  0
physics.usyd.edu.au/teach_res/hsp/math/math.htm
p6503
5
We can now calculate the displacement x as a function of velocity v
a
dv v dv

 g   / m  v 2
dt
dx
v dv
  / m   m g /   v 2 
vT 2  m g / 
dx
 VT 2   2v  dv
 m  v dv
dx    2


 2
2
2
    vT  v   2 g   vT  v 

x
0
  v 2  v  2 v 
dx   T  
dv
2
2
 2 g  v0  vT  v 
v
 v 2 
x   T  log e  vT 2  v 2 
v0
 2g 
v 2 v 2 
v 2 
x   T  log e  T 2 02 
2g 
 vT  v 
valid only if v0  0 v  0
physics.usyd.edu.au/teach_res/hsp/math/math.htm
p6503
6
We can now investigate what happens as time t  
 v  v   0 
v  t     vT  0 T

  v0  vT   0 
v  t     vT

e
2g
t
vT
0
In falling, the object will finally reach a constant velocity vT (a = 0 ) which is known as
the terminal velocity.
t   v  vT
vT  v  0
1

vT  v
v 2 v 2 
V 2 
x   T  log e  T 2 02   
 2g 
 vT  v 
In falling, as time t increases the objects displacement x just gets larger and larger.
physics.usyd.edu.au/teach_res/hsp/math/math.htm
p6503
7
Example Small rock dropped from rest:
m = 0.010 kg  = 1.0010-4 kg.m-1 v0 = 0 m.s-1  vT = 31.3 m.s-1
physics.usyd.edu.au/teach_res/hsp/math/math.htm
p6503
8
Example Small rock thrown vertically downward (v < vT)
m = 0.010 kg  = 1.0010-4 kg.m-1 v0 = +5.00 m.s-1  vT = 31.3 m.s-1
physics.usyd.edu.au/teach_res/hsp/math/math.htm
p6503
9
Example Small rock thrown vertically downward (v > vT)
m = 0.010 kg  = 1.0010-4 kg.m-1 v0 = +40.0 m.s-1  vT = 31.3 m.s-1
physics.usyd.edu.au/teach_res/hsp/math/math.htm
p6503
10
For problems in which the object is projected vertically upward, you have to divide the
problem into two parts. (1) Calculate the time to reach its maximum height and
calculate the maximum height reached for the upward motion. (2) Reset the initial
conditions to the position at maximum height where the initial velocity becomes v0 = 0
and do the calculations for the downward movement of the object.
Velocity of the object negative and moving up v0  0 and
v0
Equation of motion
ag

m
v2
valid only if
v0  0 and
v0
Note: up is the positive direction
physics.usyd.edu.au/teach_res/hsp/math/math.htm
p6503
11
dv
 
 g    v2
dt
m
dv
dv
dt 

  mg 2
 
g    v 2    
v 
m
 
 m   

a
vT 2 
mg

dv
 
  dt  2
v  vT 2
m
v
dv
  t
  0 dt  v0 2
v  vT 2
m
a
Standard Integral
2
v
dx
1
 x
   atan    C
2
x
a
a
 v 
   1 
  t    atan   
 m   vT  
 vT   v0
v
t T
 g
 v

 atan  v

 T
v
 m

  vT
  m g   vT 

 

v
g
  T   g

 v
 vT  
     atan 
  v0  g  
 vT

 v0
  atan 

 vT



The time tup to reach maximum height occurs when v = 0
 0
 v 
tup   T  atan 
 g 
 vT

 v0  
 v0 
 vT 
  atan        atan  
 g

 vT  
 vT 
physics.usyd.edu.au/teach_res/hsp/math/math.htm
p6503
v0  0
12
The velocity v as a function of time t is
 v
atan 
 vT

 v0
  atan 

 vT
  g

  vT

t


v   g  
v  vT tan atan  0     t 
 vT   vT  

v0  0 and
v0
atan  tan 
physics.usyd.edu.au/teach_res/hsp/math/math.htm
-1
p6503
13
The displacement x as a function of velocity v is
a
dv v dv

 g   / m  v 2
dt
dx
v dv
  / m   m g /   v 2 
vT 2  m g / 
dx
 vT 2   2v  dv
 m  v dv
dx    2


 2
2
2
    vT  v   2 g   vT  v 

x
0
 v 2  v  2v 
dx   T  
dv
2
2
 2 g  v0  vT  v 
v0  0 and
v0
v
v 2 
x   T  log e  vT 2  v 2 
v0
2g 
 v 2  v2 
v 2 
x   T  log e  T2
2 
2g 
 vT  v0 
The maximum height xup reached by the object occurs when v = 0
 v 2 
v 2 
xup   T  loge  2 T 2 
2g 
 vT  v0 
physics.usyd.edu.au/teach_res/hsp/math/math.htm
p6503
14
Example Small rock thrown vertically upward (v0 < 0 v0 = - u u > 0)
m = 0.010 kg  = 1.0010-4 kg.m-1 v0 = - 12.0 m.s-1  vT = 31.3 m.s-1
physics.usyd.edu.au/teach_res/hsp/math/math.htm
p6503
15
The terminal velocity vT is
vT 2  m g / 
vT  m g /  
10   9.8 / 10  m.s
2
4
-1
vT  31.31 m.s-1
When v = 0 the object reaches its maximum height xup (up is negative)
 v 2 
v 2 
xup   T  log e  2 T 2 
2g 
 vT  v0 
xup  6.855 m
The time tup to reach the maximum height
v 
v 
tup    T  atan  0 
 g
 vT 
tup  1.169 s
The calculations agree with the values for tup and xup determined from the graphs.
physics.usyd.edu.au/teach_res/hsp/math/math.htm
p6503
16
From the graphs:
x=0
time t = 2.275 s
velocity v = 11.21 m.s-1
Time to fall from max height to origin x = 0
tdown = (2.375 – 1.169) s = 1.206 s
takes slight longer to fall then rise to and from origin to max height
Launch speed = 12.00 m.s-1 slightly greater than return speed = 11.21 m.s-1
In the absence of any resistive forces a  g v  v0  a t v 2  v0 2  2 a s
At maximum height v  0 tup  1.2245 s xup  7.3469 m
physics.usyd.edu.au/teach_res/hsp/math/math.htm
p6503
17