Vector Addition
Vector Addition: Place the vectors tip to tail. A vector may be
moved any way you please provided that you do not change
its length nor rotate it. The resultant points from the tail of
the first vector to the tip of the second (A+B).
2
To add vectors together they must first be resolved into
components. The x (y) component of a vector is found by
projecting the vector onto the x (y) axis.
y
A
Ay


Ax
x
3
Example: Vector A has a length of 5.00 meters and points
along the x-axis. Vector B has a length of 3.00 meters and
points 120 from the +x-axis. Compute A+B (=C).
y
B
C
120
A
x
4
opp
sin  
hyp
adj
cos 
hyp
sin  opp
tan  

cos adj
Example continued
y
B
By
60
Bx
sin 60 
By
120
A
x
 B y  B sin 60  3.00m sin 60  2.60 m
B
 Bx
cos60 
 Bx   Bcos60  3.00m cos60  1.50 m
B
And Ax = 5.00 m and Ay = 0.00 m
5
Example continued
The components of C:
C x  Ax  Bx  5.00 m  - 1.50 m  3.50 m
C y  Ay  By  0.00 m  2.60 m  2.60 m
y
The length of C is:
C
2
Cy = 2.60 m


Cx = 3.50 m
C  C  Cx  C y
x
2
3.50 m 2  2.60 m 2
 4.36 m
Cy
2.60 m

 0.7429
The direction of C is: tan  
C x 3.50 m
  tan 1 0.7429  36.6 From the +x-axis
6
Example: At the instant a traffic light turns green, an automobile starts with a
constant acceleration of 2.2 m/s2. At the same instant a truck, traveling with a
constant speed of 9.5 m/s, overtakes and passes the automobile.
(a) How much time will elapse before the automobile overtakes the truck?
7
Example continued
(b) How fast will the car be traveling at that instant?
(c) Where do they meet?
8
Example: A penny is dropped from the observation deck of the Empire State
Building 369 m above the ground. With what velocity does it strike the ground?
Ignore air resistance.
y
Given: v0y = 0 m/s; ay= –9.8 m/s2; y0 = 0 m;
and yf= –369 m
x
ay
Unknown: vyf
369 m
9
Example: You throw a ball into the air with speed 15.0 m/s, how high does the
ball rise?
y
Given: v0y = +15.0 m/s; ay = –9.8 m/s2
v0y
x
ay
10
Example: An arrow is shot into the air with  = 60° (from the horizontal) and v0 =
20.0 m/s. The arrow is released from a height of 1.80 m above the ground.
(a) What are vx and vy of the arrow when t=3 sec?
y
The components of the initial velocity
are:
v0
v0 x  v0 cos   10.0 m/s
60°
x
At t = 3 sec:
v0 y  v0 sin   17.3 m/s
vfx  v0 x  10.0 m/s
vfy  v0 y  gt  12.1 m/s
11
(b) What are the x and y components of the displacement of the arrow during the
3.0 sec interval?
y
r0
rf
x
r  rt  t   rt 
12
Example continued
The initial position of the arrow is
r0  0 xˆ  1.8 m yˆ
The final position of the arrow is
xf  x0  v0 x t  30.0 m
1 2
yf  y0  v0 yt  gt  9.60 m
2
The displacement is
r  30 m xˆ  7.8 m yˆ
13
Example: How far does the arrow in the previous example land from where it is
released?
The arrow lands when y = 0
Determine the roots with quadratic
formula
The distance traveled is:
1 2
y  y0  v0 yt  gt  0
2
t  0.10 sec or  3.64 sec
x  v0 x t  36.4 m
14
Example: How high does the arrow go?
The arrow rises until
dy
0
dx
The max height is reached when
dy
with
dy
 dt
dx dx
dt
dy
 0  v0 y  gt
dt
v0 y
t
 1.77 sec
g
The y-coordinate of the arrow is
1 2
y t  1.77 s   y0  v0 yt  gt  17.1 m
2
15
Relative motion
Example: You are traveling in a car (A) at 60 miles/hour east on a long straight
road. The car (B) next to you is traveling at 65 miles/hour east. What is the
speed of car B relative to car A?
16
Example continued:
+x
t=0
A
B
From the picture:
t>0
rAG
rBA
A
rBG
B
rBG  rAG  rBA
rBA  rBG  rAG
Divide by t:
v BA  v BG  v AG
v BA  65 miles/hr east  60 miles/hr east
 5 miles/hour east
17
Example: You are traveling in a car (A) at 60 miles/hour east on a long straight
road. The car (B) next to you is traveling at 65 miles/hour west. What is the
speed of car B relative to car A?
18
Example continued:
+x
t=0
t>0
A
B
From the picture:
Divide by t:
rBG
B
t>0
rAG
A
rBA
rBA  rBG  rAG
v BA  v BG  v AG
 65 miles/hr w est  60 miles/hr east
 125 miles/hr w est
19
Example: The current in a river has a steady speed of 0.5 m/s. A student swims
upstream a distance of 1 km and then swims back to the starting point. If the
student can swim 1.2 m/s in still water, how long does the round trip in the river
take? How long would the same trip take in still water?
In still water:
d 2000 m
t 
 1670 s
v 1.2 m/s
Or 835 seconds per leg
20
Example continued
vs
v up  v s  v r  0.7 m/s
Upstream
vr
vs
d 1000 m
t 
 1430 s
v 0.7 m/s
v up  v s  v r  1.7 m/s
Downstream
vr
d 1000 m
t 
 590 s
v 1.7 m/s
The total time is 2020 seconds!
21
Example: A jet moving initially with v = 300 mph due east enters a region
where the wind is blowing at 100 mph at 30 north of east. What is the new
velocity of the jet?
y (north)
vwind
vjet
x (east)
Place the vectors tip-to-tail:
vnew
vwind
vjet
22
Example continued
v new  v jet  v wind
 300 mph  xˆ  100 mph  cos 30 xˆ  100 mph sin 30 yˆ
 387 xˆ  50 yˆ  mph
The magnitude and direction of the velocity are
v new  vx2  v y2  390 mph
tan  
vy
vx
 0.129
The plane travels 390 mph 7.4 north of east
23
Free Body Diagrams
Use idealized models to account for all forces acting on each mass (body)
involved in the system being analyzed.
24
Example: Find the tension in each cord of the system shown in the figure.
25
Example: A box slides across a rough surface. If the coefficient of kinetic friction is
0.3, what is the acceleration of the box?
y
FBD for
box:
Apply Newton’s 2nd Law:
Nsb
F  N
 f k;sb  web  ma
sb
Fk;sb
x
web
F
F
x
  f k;sb  ma
y
 N sb  web  0
26
Example continued
(1)
 f k;sb  ma
(2)
N sb  web  0  N sb  web  mg
From (1):
Solving for a:
 f k;sb    k N sb    k mg  ma
a  k g


 0.3 9.8 m/s 2  2.94 m/s 2
27
Example: In the previous example, a box sliding across a rough surface was found
to have an acceleration of -2.94 m/s2. If the initial speed of the box is 10.0 m/s,
how long does it take for the box to come to rest?
Know: ax = –2.94 m/s2, v0x=10.0 m/s, vx= 0.0 m/s
Want: t.
v x  v0 x  a x t  0
v0 x
 10.0 m/s
t  

 3.40 sec
2
ax
 2.94 m/s
28
Example: A 1.00 kg mass is at rest an on ramp that makes an angle of 20 with
respect to the horizontal. For this situation s = 0.400. What is the magnitude
of the static friction force?
FBD for
box
y
Nrb
fs;rb


web
x
Apply Newton’s 2nd law
F  N
rb
 fs;rb  web  0
29
Example continued
(1)
(2)
F
F
x
  f s;rb  mg sin   0
y
 N rb  mg cos   0
The magnitude of the static friction force can be found from (1)
f s;rb  mg sin   3.35 N
What is the magnitude of the maximum static friction force?
f s;rb  s N rb  s mg cos   3.68 N
30
Example continued
If the angle of the ramp is changed to 40, what is the acceleration of the
mass? Take k = 0.35.
The FBD is unchanged, except fs;rb is now
the kinetic friction force of the ramp on the
box.
Apply Newton’s 2nd law
(1)
(2)
F
F
F  N
rb
 f k;rb  w eb  ma
x
  f k;rb  mg sin   ma
y
 N rb  mg cos   0
31
Example: A 3.00 kg mass rests on a frictionless tabletop. This mass is
connected to a 5.00 kg mass by a light string as shown. Assume the pulley is
massless.
a. Draw free body diagrams for the
two masses and the pulley.
32
Example continued
b. Apply Newton’s Second law to the two masses.
33
Dynamics of Circular Motion
rt   r cos xˆ  r sin  yˆ
vt   r sin  xˆ  r cos yˆ
at   r 2 cos  xˆ  r 2 sin  yˆ
Rotational motion can be related to translational motion by converting between
Cartesian and polar coordinate systems.
34
The tangential velocity and radial acceleration of a body (constant
speed case).
y
v
ar
v
ar
ar
x
v
ar
v
35
Example: The Hubble Space Telescope orbits the Earth at an altitude of about
600 km with an orbit period of about 100 minutes, what is Hubble’s orbital
speed? (Assume a circular orbit.)
total distance r 2r
vav 


 7300 m/s
total time
t
T
r = Re + h = 6.98106 m
T = 6000 s
36
Example continued
(b) What is HST’s angular speed?
 2π
av 

 1.05 10 3 rad/sec
t
T
37
Example: What is the magnitude of the radial acceleration of HST?
v2
ar 
 7.63 m/s 2  0.78 g
r
38
Previously: Consider an object in uniform circular motion (speed=constant).
rt   r cos xˆ  r sin  yˆ
vt   r sin  xˆ  r cos yˆ
at   r 2 cos  xˆ  r 2 sin  yˆ
The magnitude of the (tangential) velocity is
v  r
The magnitude of the (radial) acceleration is
2
v
ar  r 2 
 v
r
39
Example: The rotor is an amusement park ride where people stand against the
inside of a cylinder. Once the cylinder is spinning fast enough the floor drops out.
(a) What force keeps the people from falling out the bottom of the cylinder?
y
fs;wp
Draw an FBD for a person with their
back to the wall:
Nwp
x
wep
It is the force of static friction.
40
Example continued
(b) If s = 0.40 and the cylinder has r = 2.5 m. what is the minimum angular
speed of the cylinder so that the people don’t fall out?
Apply Newton’s 2nd law:
From (2):
F  N
wp
 fs;wp  wep  ma
1  Fx  N wp  mar  m 2 r
2  Fy  fs;wp  wep  0
f s;wp  wep  mg
From (1)
s N wp  s m 2 r   mg
g


s r
9.8 m/s 2
 3.13 rad/s
0.402.5 m 
41
Example: A coin is placed on a record that is rotating at 33.3 rpm. If s = 0.1, how
far from the center of the record can the coin be placed without having it slip off?
Draw an FBD for the coin:
y
Nrc
Apply Newton’s 2nd law:
F  N
fs;rc
x
wec
rc
 fs;rc  wec  ma
1  Fx  f s;rc  maradial  m 2 r
2  Fy  N rc  wec  0
42
Example continued
From 1 : f s;rc  m 2 r
From (2)
f s;rc  s N rc  s mg   m 2 r
Solving for r:
s g
r 2

rev
  33.3
min
What is ?
 2 rad  1 min 


  3.5 rad/s
 1 rev  60 sec 
s g 0.19.8 m/s 2 
r 2 
 0.08 m
2

3.50 rad/s 
43
Example: What is the minimum speed for the car so that it maintains contact
with the loop when it is in the pictured position?
r
FBD for the car at the top
of the loop:
y
Apply Newton’s 2nd law:
F  N
x
F
y
Ntc
wec
tc
 w ec  ma
  N tc  wec  mar
v2
N tc  wec  m
r
44
Example continued
The apparent weight at the top of loop is
Ntc = 0 when
v2
N tc  mg  m
r

 v2

N tc  m  g 

 r
 v2

N tc  m  g   0
 r

v  gr
This is the minimum speed needed to make it around the loop.
45
Example continued
Consider the car at the bottom of the loop, how does the apparent weight
compare to the true weight?
FBD for the car at the bottom
of the loop:
y
Ntc
x
wec
Apply Newton’s 2nd law:
F  N
tc
 w ec  ma
v2
 Fy  N tc  wec  mar  m r
v2
N tc  mg  m
r
 v2

N tc  m  g 
 r

Here,
N  mg
46
Example continued
From (2) the normal force is
Using (1):
N rb  mg cos 
 f k;rb  mg sin   ma
  k N rb  mg sin   ma
  k mg cos    mg sin   ma
a  g sin    k cos    3.67 m/s 2
47
Example continued
c. What is the acceleration of the system?
d. What is the tension in the rope?
48
Work and Energy
49
Example: The extinction of the dinosaurs and the majority of species on Earth in
the Cretaceous Period (65 Myr ago) is thought to have been caused by an
asteroid striking the Earth near the Yucatan Peninsula. The resulting ejecta
caused widespread global climate change.
If the mass of the asteroid was 1016 kg (diameter in the range of 4-9 miles)
and had a speed of 30.0 km/sec, what was the asteroid’s kinetic energy?


1 2 1 16
K  mv  10 kg 30 103 m/s
2
2
 4.5 10 24 J

2
This is equivalent to ~109 Megatons of TNT.
50
Example: What is the net work done on a box of mass m that is being pushed
along a rough surface as shown?
F

x
An FBD for the box at left:
y
x
Nsb

fk;sb
web
x
Fpb
51
Example continued
The forces and displacement (in unit vector notation) are
N sb  N sb yˆ  mg  F sin   yˆ
w eb  web  yˆ   mg yˆ
f k;sb  f k;sb  xˆ     k mg  F sin   xˆ
Fpb  F cos  xˆ  F sin  yˆ
r  x xˆ
The work done by the pushing force is:
Wpb  Fpb  r  F cos  x
52
The work done by the Normal force is:
WN  Nsb  r  0
The normal force is perpendicular to the displacement.
The work done by gravity is:
Wg  w eb  r  0
The force of gravity is perpendicular to the displacement.
The work done by kinetic friction is:
Wkf  f k;sb  r    k mg  F sin  x
53
The net work done on the box is:
Wnet  Wpf  WN  Wg  Wkf
 F cos  x  0  0   k mg  F sin  x
 F cos    k mg   k F sin  x
54
Example: A ball is tossed straight up. What is the work done by the force of
gravity on the ball as it rises?
y
r
FBD for rising
ball:
x
web
Wg  w eb  r
 web y cos180  mgy
W < 0 and the KE of the ball decreases.
55
Example: An ideal spring has k = 20.0 N/m. What is the amount of work done
(by an external agent) to stretch the spring 0.40 m from its relaxed length?
0.40 m
0.40 m
1 2
W   F  dr   Fdx  kx
2
0
0
1
2
 20.0 N/m 0.40 m   1.6 J
2
56
Example continued
How much additional work must be done to stretch the spring to 0.80 m?
0.80 m
0.80 m
1 2
W   F  dr   Fdx  kx
2
0.40 m
0.40 m


1
2
2
 20.0 N/m 0.80 m   0.40 m   4.8 J
2
Note that the result is not 1.6 J.
57
Example: A 4.00 kg particle moves along the x-axis. Its position varies with time
according to x(t) = t + 2t3, where x is measured in meters and t is in seconds.
(a) What is the particle’s kinetic energy?
2

1 2 1  dx 
K  mv  m   2 1  6t 2
2
2  dt 

2
J
(b) What is the particle’s acceleration? What is the net force acting on the
particle?
d 2x
a  2  12t m/s 2
dt
Fnet  ma  48t N
58
Example continued
(c) The power delivered to the particle at time t?


P  Fv  48t  1  6t  48t  288t W
2
3
(d) The work done on the particle from t = 0 to t = 2 sec?
xf
tf
tf
dx
W   F  dr   Fdx   F dt   Fv dt
dt
xi
ti
ti
  48t  288t 3  dt  1248 J
2
0
59
Example: A box of mass m is towed up a frictionless incline at constant speed. The
applied force F is parallel to the incline. What is the net work done on the box?
y
F
Nrb
Fpb
x


web
Apply Newton’s 2nd
law:
F
F
x
 Fpb  web sin   0
y
 N rb  web cos   0
60
Example continued
The magnitude of F is:
Fpb  F  mg sin 
If the box travels along the ramp a distance of x then the work by the
force F is
WF  Fx cos 0  mgx sin 
The work by gravity is
Wg  web x cos  90  mgx sin 
61
Example continued
The work by the normal force is
WN  N rb x cos 90  0
The net work done on the box is
Wnet  WF  Wg  WN
 mgx sin   mgx sin   0
0
62
Example: What is the net work done on the box in the previous example if the box
is not pulled at constant speed?
 F  F  w sin   ma
 F  ma  w sin 
x
eb
eb
Proceeding as before:
Wnet  WF  Wg  WN
 ma  mg sin  x  mgx sin   0
 max  Fnet x
63
Example: A raindrop of mass 3.3510-5 kg falls vertically at constant speed
under the influence of the forces of gravity and drag. In falling through 100
m…
(a) What is the work done by gravity?
y
W   F  dr
Far
x

yf
  mg yˆ   dy yˆ 
yi
wer
 mgy  0.033 J
64
Example continued
(b) What is the work done by drag?
Wnet  Wdrag  Wg  0
Wdrag  Wg  mgy  0.033 Joules
65
Example: A 2.00 kg mass hangs suspended from a string. The ceiling is 3.00
m above the floor and the length of the string is 1.00 m.
ceiling
(a) What is the gravitational potential energy of
the mass if the ceiling is the reference point?
U g  mgy
floor
 2.00 kg 9.8 m/s 2  1.00 m 
 19.6 J
66
Example continued
(b) What is the gravitational potential energy of the mass if the floor is the
reference point?
U g  mgy
 2.00 kg 9.8 m/s 2 2.00 m 
 39.2 J
(c) What is the gravitational potential energy of the mass if the ball’s
location is the reference point?
U g  mgy
 2.00 kg 9.8 m/s 2 0 m 
0J
67
Example: A cart starts from position 4 with v = 15.0 m/s to the left. Find the speed
of the cart at positions 1, 2, and 3. Ignore friction.
E4  E3
U 4  K 4  U 3  K3
1 2
1 2
mgy4  mv4  mgy3  mv3
2
2
v3  v42  2 g  y4  y3   20.5 m/s
68
Example continued
E4  E2
U 4  K4  U 2  K2
Or use
E3=E2
1
1
mgy4  mv42  mgy2  mv22
2
2
v2  v42  2 g  y4  y2   18.0 m/s
E4  E1
U 4  K 4  U1  K1
1 2
1 2
mgy4  mv4  mgy1  mv1
2
2
Or use
E3=E1
E2=E1
v1  v42  2 g  y4  y1   24.8 m/s
69
The work-energy theorem can be used to solve problems instead of using
Newton’s laws.
Determine the speed of a mass m that starts from rest and
slides along a frictionless ramp for some distance d using (a)
Newton’s laws and (b) the work-energy theorem.
Answer:
v  2gd sin 
Repeat the calculation but take the coefficient of kinetic
friction to be k.
Answer:
v  2 gd sin   k cos 
70
Momentum and Collisions
Use conservation principles to calculate momentum and energy transfer between
two or more interacting bodies.
71
Example: Particle A is at the origin and has a mass of 30.0 grams. Particle B has a
mass of 10.0 grams. Where must particle B be located so that the center of mass
(marked with a red x) is located at the point (2.00 cm, 5.00 cm)?
y
xcm
ma xa  mb xb
mb xb


ma  mb
ma  mb
ycm
ma ya  mb yb
mb yb


ma  mb
ma  mb
x
A
x
72
Example continued
xcm 
10.0 g xb
10.0 g  30.0 g
xb  8.00 cm
ycm 
10.0 g  yb
30.0 g  10.0 g
yb  20.0 cm
 2.00 cm
 5.00 cm
rb  8.00 xˆ  20.0 yˆ cm
73
Example: The positions of three particles are (4.00 m, 0.00 m); (2.00 m, 4.00 m);
and (-1.00 m, -2.00 m). The masses are 4.00 kg, 6.00 kg, and 3.00 kg
respectively. What is the location of the center of mass?
y
2
1
x
3
74
Example continued
xcm
m1 x1  m2 x2  m3 x3

m1  m2  m3

4.00 kg 4.00 m   6.00 kg 2.00 m   3.00 kg  1.00 m 

4.00  6.00  3.00 kg
 1.92 m
ycm
m1 y1  m2 y2  m3 y3

m1  m2  m3

4.00 kg 0.00 m   6.00 kg 4.00 m   3.00 kg  2.00 m 
4.00  6.00  3.00 kg
 1.38 m
75
Example: In a railroad freight yard an empty freight car of mass m rolls along a
straight level track at 1.00 m/s and collides with an initially stationary, fully loaded,
boxcar of mass 4.00m. The two cars couple together upon collision.
(a) What is the speed of the two cars after the collision?
pi  p f
p1i  p2i  p1 f  p2 f
m1v1  0  m1v  m2 v  m1  m2 v
 m1 
v1  0.20 m/s
v  
 m1  m2 
76
Example continued
(b) Suppose instead that both cars are at rest after the collision with what
speed was the loaded boxcar moving before the collision if the empty one had
v1i = 1.00 m/s.
pi  p f
p1i  p2i  p1 f  p2 f
m1v1i  m2 v2i  0  0
 m1 
v2i   v1i  0.25 m/s
 m2 
77
Example: Body 1 of mass M has an original velocity of 6.00 m/s in the +xdirection toward a stationary body 2 of the same mass. After the collision, body
1 has vx=+1.00 m/s and vy=+2.00 m/s. What is the magnitude of body 2’s
velocity after the collision?
Final
Initial
1
1
v1i
2
2
78
Example continued
y momentum:
x momentum:
pix  p fx
piy  p fy
p1ix  p2ix  p1 fx  p2 fx
p1iy  p2iy  p1 fy  p2 fy
m1v1ix  0  m1v1 fx  m2 v2 fx
Solve for v2fy:
Solve for v2fx:
v2 fx 
0  0  m1v1 fy  m2 v2 fy
m1v1ix  m1v1 fx
m2
 v1ix  v1 fx
v2 fy 
 m1v1 fy
m2
 v1 fy
 2.00 m/s
 5.00 m/s
The mag. of v2 is
v2 f  v22 fy  v22 fx  5.40 m/s
79
Example: Head-on 1D elastic collision. The values of v1i and v2i are known.
Determine the velocity of the pucks after the collision.
Initial
1
v1i
Final
v2i
v2f
1
v1f
2
2
Take Fext = 0 so that pi = pf.
pi  p f
p1i  p2i  p1 f  p2 f
m1v1i  m2 v2i  m1v1 f  m2 v2 f
m1 v1i  v1 f   m2 v2 f  v2i 
(1)
80
Example continued
The collision is elastic so Ki = Kf..
Ki  K f
K1i  K 2i  K1 f  K 2 f
m1v12i  m2v22i  m1v12f  m2v22 f
(2)
Rewrite (2):
m1v12i  m1v12f  m2 v22 f  m2 v22i

v
m1 v  v
(3)
m1 v1i  v1 f
2
1i
2
1f
1i
 v1 f
  m v
  m v
2
2
2f
v
2
2i
 v2 f
2
2i

v
2i
 v2 f 
81
Example continued
Divide (3) by (1) to get
Take
v1i  v1 f  v2i  v2 f
v2 f  v1i  v1 f  v2i
substitute into (1) and solve for v1f
 m1  m2 
 2m2 
v1i  
v2i
v1 f  
 m1  m2 
 m1  m2 
Then solve for v2f:
v2 f
 2m1 
 m2  m1 
v1i  
v2i
 
 m1  m2 
 m1  m2 
82
Example: An elastic collision in 2D. Take particles A and B to be protons and vAi
= 3.50105 m/s and vBi = 0. After the collision, particle A travels at an angle  =
37 with the x-axis and particle B travels at an unknown angle . Determine the
velocities of the protons after the collision and also the value of .
Final
Initial
A
A
vAi
B
B
83
Take Fext = 0 so that pi = pf.
pix  p fx
x-component:
p Ai, x  pBi, x  p Af , x  pBf , x
m Av Ai  0  m Av Af cos   mB vBf cos 
piy  p fy
y-component:
p Ai, y  pBi, y  p Af , y  pBf , y
0  0  m Av Af sin   mB vBf sin 
Ki  K f
Kinetic energy:
K Ai  K Bi  K Af  K Bf
2
2
m Av Ai
 mAv Af
 mB vBf2
84
Example continued
Summary, so far:
v Ai  v Af cos   vBf cos 
(1)
0  v Af sin   vBf sin 
(2)
v  v v
2
Ai
Rewrite (1) and (2):
2
Af
2
Bf
(3)
v Ai  v Af cos   vBf cos 
v Af sin   vBf sin 
Square each and add together (and use a trig identity):
2
2
vAi
 vAf
 2vAivAf cos  vBf2
85
Example continued
vAf  vAi cos  2.80 10 m/s
5
Use (3) to simplify to:
Now use (3) to determine vBf:
vBf  v  v
2
Ai
2
Af


2
 v Ai
1  cos 2   v Ai sin   2.11105 m/s
Now use (2) to solve for :
sin  
v Af
vBf
  53
sin   0.8
86
Example: The Ballistic Pendulum. A bullet of mass mb is fired at a block of wood
of mass mp. The bullet embeds in the wood and the block rises a height h before
coming to rest. What was the speed of the bullet as it entered the block?
This is in inelastic collision.
pi  p f
pbi  p pi  pbf  p pf
h
vb
mb
mp
mb vb  0  mb  m p v
 mb  m p 
v
vb  
 mb 
87
Example continued
What is v?
Even though energy is not conserved during the collision, it will
be conserved during the upward swing.
Ei  E f
Ki  U i  K f  U f
1
mb  m p v 2  0  mb  m p  gh
2
v  2 gh
88
Example continued
How much kinetic energy is lost in the collision?
K  K f  K i
1
1
2
 mb  m p v  mb vb2
2
2
2
1
1  mb  m p  2
2
 v
 mb  m p  v  mb 
2
2  mb 
 mp  2
1
v
  mb  m p  
2
 mb 
89
Example: A 3.00 gram particle is moving toward a 7.00 gram particle with a
speed of 3.00 m/s.
(a) With what speed does each particle approach the center of mass?
vcm
v1m1  v2 m2

 0.900 m/s
m1  m2
v1  v1  vcm 
 3.00 m/s  0.900 m/s  2.10 m/s
v2  v2  vcm
 0.00 m/s  0.900 m/s  0.900 m/s
90
Example continued
(b) What is the momentum of each particle relative to the center of mass?
p1  m1v1  0.0063 kg m/s
p2  m2v2  0.0063 kg m/s
Note: the total momentum relative to the center of mass is zero.
91
Momentum in the center of mass frame
ptotal  p1  p2  m1v1  m2v2
 m1 v1  vcm   m2 v2  vcm 
m1m2
m1m2
v1  v2  
v2  v1   0

m1  m2
m1  m2
The momentum of a system in the center of
mass frame is zero.
92
Rotational Momentum and Inertia
Systems of particles that involve rotation require separate equations to account
for total system energy and momentum.
93
Example: A centrifuge in a medical laboratory rotates at an angular speed of
3600 rev/min. When switched off, it rotates 50 times before coming to rest. Find
the constant angular acceleration of the centrifuge.
  3600
rev  2 rad  1 min 


  120 rad/sec
min  1 rev  60 sec 
 2 rad 
  50 rev
  100 rad
 1 rev 
 2  02  2  0
02
 
 72 rad/sec 2
2 
94
Example: A pulsar is a rapidly spinning neutron star that emits a radio beam like a
lighthouse emits a light beam. We receive a radio pulse for each rotation of the
star. The period T of rotation is found by measuring the time between pulses. The
pulsar in the Crab Nebula has a period of T = 0.033 sec that is increasing at the
rate of 1.2610-5 sec/year.
(a) What is the pulsar’s angular acceleration?
d d  2

 
dt dt  T
2 dT

9
2




2

10
rad/sec

2
T
dt

dT
1 year

5 sec 
-13 sec
 1.26 10

  4 10
7
dt
year  3.15 10 sec 
sec
95
Example continued
(b) If the pulsar’s angular acceleration is constant, how many years from now
will the pulsar stop rotating?
   0  t  0
2
0
T0
t

 9  1010 sec  2800 years


(c) The pulsar originated in a supernova explosion in the year 1054 A.D. What
was the initial period for the pulsar?
t  953 years  3.0 1010 sec
2
 0    t 
 t  250 rad/sec
T
2
T
 0.025 sec  25 msec
0
96
Example continued
(d) What is the rotational kinetic energy of the pulsar?
K rot
1 2
 I
2
Assume the pulsar to be a solid sphere of
uniform density. M = 1.4Msun and R = 10
km.
The moment of inertia for a sphere is
2
I  MR 2
5
 11038 kg m 2
  190 rad/sec
K rot
1 2
 I  2 10 42 J
2
97
The KE of the pulsar decreases over time. The rate of energy loss is
dE d  1 2 
d
P
  I   I
 4  1031 W
dt dt  2
dt

For comparison, the luminosity of the sun
is 3.861026 W.
98
Example: What is the moment of inertia of a thin disk when you add a small
mass near the rim? The disk has a rotation axis though its center and points out
of the page.
R
I new  I disk  mr 2
1
2
2
 M disk R  mR
2
99
Example: (a) Find the moment of inertia of the system below. The masses are m1
and m2 and they are separated by a distance r. Assume the rod connecting the
masses is massless.

m1
r1
r2
r1 and r2 are the distances between
mass 1 and the rotation axis and
mass 2 and the rotation axis (the
dashed, vertical line) respectively.
m2
100
Example continued
Take m1 = 2.00 kg, m2 = 1.00 kg, r1= 0.33 m ,
and r2 = 0.67 m.
2
I   mi ri  m r  m r  0.67 kg m
2
i 1
2
1 1
2
2 2
2
(b) What is the moment of inertia if the axis is moved so that is passes through
m1 ?
2
I   mi ri 2  m2 r22  m2 r 2  1.00 kg m 2
i 1
101
Parallel Axis theorem
I new  I cm  md 2
Where m is the mass of the object and d is the distance the axis
is moved. Note: the new axis must be parallel to the original
axis!
102
Example: Rank the I1, I2, I3, and I4 from greatest to smallest.
a
2
1
I new  I cm  md
2
3
4
I 3  I cm
b
a
I 4  I cm  m 
2
2

1
 m a 2  b2
12

 a  2  b  2 
I 2  I cm  m      
 2   2  
I1  I cm  mL
2
Ranking: I1 > I2 > I4 > I3
103
Example: A centrifuge in a medical laboratory rotates at an angular speed of
3600 rev/min. When switched off, it rotates 50 times before coming to rest. Find
the constant angular acceleration of the centrifuge.
  3600
rev  2 rad  1 min 


  120 rad/sec
min  1 rev  60 sec 
 2 rad 
  50 rev
  100 rad
 1 rev 
 2  02  2  0
02
 
 72 rad/sec 2
2 
104
Example: A pulsar is a rapidly spinning neutron star that emits a radio beam like a
lighthouse emits a light beam. We receive a radio pulse for each rotation of the
star. The period T of rotation is found by measuring the time between pulses. The
pulsar in the Crab Nebula has a period of T = 0.033 sec that is increasing at the
rate of 1.2610-5 sec/year.
(a) What is the pulsar’s angular acceleration?
d d  2

 
dt dt  T
2 dT

9
2




2

10
rad/sec

2
T
dt

dT
1 year

5 sec 
-13 sec
 1.26 10

  4 10
7
dt
year  3.15 10 sec 
sec
105
Example continued
(b) If the pulsar’s angular acceleration is constant, how many years from now
will the pulsar stop rotating?
   0  t  0
2
0
T0
t

 9  1010 sec  2800 years


(c) The pulsar originated in a supernova explosion in the year 1054 A.D. What
was the initial period for the pulsar?
t  953 years  3.0 1010 sec
2
 0    t 
 t  250 rad/sec
T
2
T
 0.025 sec  25 msec
0
106
Example continued
(d) What is the rotational kinetic energy of the pulsar?
K rot
1 2
 I
2
Assume the pulsar to be a solid sphere of
uniform density. M = 1.4Msun and R = 10
km.
The moment of inertia for a sphere is
2
I  MR 2
5
 11038 kg m 2
  190 rad/sec
K rot
1 2
 I  2 10 42 J
2
107
The KE of the pulsar decreases over time. The rate of energy loss is
dE d  1 2 
d
P
  I   I
 4  1031 W
dt dt  2
dt

For comparison, the luminosity of the sun
is 3.861026 W.
108
Top view of door
F
Hinge
end

r

Line of action
of the force
Lever arm
r
sin  
r
r  r sin 
The torque is:
  r F
 rF sin 
Same as
before
109
Example: Calculate the torque due to the three forces shown about the left end of
the bar (the red X). The length of the bar is 4m and F2 acts in the middle of the
bar.
F2=30 N
30
F3=20 N
X
10
45
F1=25 N
110
Example continued
Lever arm
for F2
F2=30 N
30
F3=20 N
10
X
45
F1=25 N
Lever arm
for F3
The lever arms are:
r1  0
r2  2m sin 60  1.73 m
r3  4m sin 10  0.695 m
111
Example continued
The torques are:
1  0
 2  1.73 m 30 N   51.9 Nm
 3  0.695 m 20 N   13.9 Nm
The net torque is + 65.8 Nm and is the sum of the above results.
112
Example: A bicycle wheel (a hoop) of radius 0.300 m and mass 2.00 kg is rotating
at 4.00 rev/sec. After 50.0 sec the wheel comes to a stop because of friction. What
is the magnitude of the average torque due to frictional forces?
2


I


MR


rev  2 rad 
i  4.00

  25.1 rad/sec
sec  1 rev 
f  0
  f  i
 av 

 0.503 rad/s 2
t
t
 av  MR 2 av  9.05 102 Nm
113
Example: A flywheel of mass 182 kg has a radius of 0.62 m (assume the flywheel is
a hoop).
(a) What is the torque required to bring the flywheel from rest to a speed of 120
rpm in an interval of 30 sec?
 f  120
rev  2 rad  1 min 


  12.6 rad/sec
min  1 rev  60 sec 
  
  rF  r ma  rmr   mr 

 t 
 f  i 
f 
2
2
  mr 
  29.4 Nm
 mr 
 t 
 t 
2
114
Example continued
(b) How much work is done in this 30 sec period?
W     av t 
 i   f
  
 2

f
t   

 2

t  5600 J

115
Rolling
An object that is rolling combines translational motion (its center of mass
moves) and rotational motion (points in the body rotate around the center of
mass).
For a rolling object
K tot  K T  K rot
1 2 1 2
 mvcm  I
2
2
If the object rolls without slipping then vcm = R.
116
Pure
Translation
Pure Rotation
Rolling
117
Example: Two objects (a solid disk and a solid sphere) are rolling down a ramp.
Both objects start from rest and from the same height. Which object reaches the
bottom of the ramp first?
h

The object with the largest linear velocity (v) at the bottom of the
ramp will win the race.
118
Example continued
Apply conservation of mechanical energy:
Ei  E f
U i  Ki  U f  K f
1 2 1 2 1 2 1 v
mgh  0  0  mv  I  mv  I  
2
2
2
2 R
2
1
I 
mgh   m  2 v 2
2
R 
Solving for v:
v
2mgh
I 

m


2 
R 

119
Example continued
1
I disk  mR 2
2
2
I sphere  mR 2
5
The moments of inertia are
For the disk:
For the sphere:
vdisk 
4
gh
3
Since Vsphere> Vdisk the sphere
wins the race.
10
vsphere 
gh
7
Compare these to a box sliding down the ramp.
vbox  2 gh
120
Example: Consider a billiard ball that has been hit “dead center”.
y
FBD:
Nsb
   f
F   f
F  N
s ;ro
x
fs;sb
x
y
R   I
s ; so
ro
 macm
 weo  0
web
Initially, x(t0) = 0, (t0) = 0, and (t0) = 0; v(t0) = v0.
121
Example continued
The “dead center” hit means  = 0 so the ball will slide (not roll!) initially.
But static friction will slow the ball until v = R and the ball will roll without
slipping.
f s ;ro R  I
2
 s N   s mg  mR 2
5
5
 R   s g
2
Knowing , then
5
R   s gt
2
122
Example continued
What is the velocity of the center of mass?
 f s ;so  macm
dvcm
  s mg  m
dt
vcm  v0   s gt
123
Example continued
When does v = R?
How far does the ball slide in
this time?
What is vcm at this time?
5
vcm  R   s gt
2
5
v0   s gt   s gt
2
2v0
t
7s g
12v02
1 2
x  at  v0t 
2
49 s g
vcm
5
 v0   s gt  v0
7
124
Example: What is the angular momentum of the Earth about its own rotation
axis?
2
L  I  MR 2
5
2
2  2 
33
2
 MR 

7
.
15

10
kg
m
/s

5
T 
Example: What is the angular momentum of the Earth revolving around the sun?
L  rp  rMv tan
 2
 rM r   Mr   Mr 
 T
2
2

40
2

2
.
7

10
kg
m
/s


125
Example: A skater is initially spinning at a rate of 10.0 rad/sec with I=2.50 kg m2
when her arms are extended. What is her angular velocity after she pulls her
arms in and reduces I to 1.60 kg m2?
On ice so we can ignore external torques.
Li  L f
I ii  I f  f
 Ii
 f  
 If

 2.50 kg m 2 
i  
10.0 rad/sec   15.6 rad/sec
2


 1.60 kg m 

126
Example: A child of mass 25.0 kg stands on the edge of a rotating platform of
mass 150 kg and radius 4.00 m. The platform with the child on it rotates with an
angular speed of 6.20 rad/sec. The child jumps off in the radial direction, what
happens to the angular speed of that platform?
The moment of inertia for the platform with the child is
1
I i  I p  mc r  Mr 2  mc r 2
2
2
Assume there are no external torques so that Li = Lf.
Li  L f
I ii  I f  f
1
2
2
 Mr  mc r i  I p f  Lc
2

127
Example continued
What is the angular momentum of the
child after the step off the platform?
ptan
r
r
L  rp
From before:
pr
Lc  rptan  rmcvtan  rmc r f   mc r 2 f
Li  L f
1
1
2
2
2
2
Mr

m
r


Mr


m
r


c
i
f
c f
2
2

i   f
128
Example continued
What happens to the platform if, a little while later, the child, starting from
rest, jumps back on the platform?
Li  L f
1
2
2
I pi   Mr  mc r  f
2

1
Mr 2
i
2
f 
i 
 4.65 rad/sec
1
2m
Mr 2  mc r 2
1 c
2
M
129
Example: A sign is supported by a uniform horizontal boom of length 3.00 m and
weight 80.0 N. A cable, inclined at a 35 angle with the boom, is attached at a
distance of 2.38 m from the hinge at the wall. The weight of the sign is 120.0 N.
What is the tension in the cable and what are the horizontal and vertical forces
exerted on the boom by the hinge?
130
Example continued
y
FBD for the bar:
FHby
Tcb
X

FHbx
x
web
Apply the conditions for
equilibrium to the bar:
Fsb
(1)  Fx  FHbx  Tcb cos   0
(2)  Fy  FHby  web  Fsb  Tcb sin   0
L
(3)    web    Fsb L   Tcb sin  x  0
2
131
Example continued
Equation (3) can be solved for T:
L
web    Fsb L 
2
T
x sin 
 352 N
Equation (1) can be solved for Fx:
FHbx  Tcb cos   288 N
Equation (2) can be solved for Fy:
FHby  web  Fsb  Tcb sin 
 2.00 N
132
Example: Find the force exerted by the biceps muscle in holding a one liter milk
carton with the forearm parallel to the floor. Assume that the hand is 35.0 cm
from the elbow and that the upper arm is 30.0 cm long. The elbow is bent at a
right angle and one tendon of the biceps is attached at a position 5.00 cm from
the elbow and the other is attached 30.0 cm from the elbow. The weight of the
forearm and empty hand is 18.0 N and the center of gravity is at a distance of
16.5 cm from the elbow.
133
Example continued
Fb
“hinge”
(elbow
joint)
Fea
  F x  F
b 1
Fca
x  Fca x3  0
ea 2
Fea x2  Fca x3
Fb 
 130 N
x1
134
Periodic Motion
135
Example: The displacement of a particle is given by the
expression x(t) = (4.00 m) cos (3t + ) where t is in
seconds.
(a) What is the frequency and period of the motion?
1  3 rad/sec
f  

 1.5 Hz
T 2
2
1
T   0.67 sec
f
(b) What is the amplitude of the motion?
A = 4.00 m
136
Example continued
(c) What is the displacement of the particle at t = 0.25
sec.
xt   4.00 m  cos3t   
xt  0.25 sec   4.00 m  cos3 0.25 sec    
 2.83 m
137
Example: A particle moving with simple harmonic motion
travels a total distance of 20.0 cm in each cycle of its motion
and its maximum acceleration is 50.0 m/s2.
(a) What is the angular frequency of the particle’s motion?
xt   A cost   
x t    A sin t   
xt    A cost   
2
amax  A 2
amax
50 m/s 2


 31.6 rad/s
A
0.05 m
138
Example continued
(b) What is the maximum speed of the particle?
vmax  A  1.58 m/s
139
Example: A mass-spring system oscillates with amplitude
3.50 cm. The spring constant is 250 N/m and the mass is
0.500 kg.
(a) Determine the mechanical energy if the system.
E t   U max
1 2
 kA  0.153 J
2
(b) What is the maximum speed of the mass?
1 2
E  K max  mvmax
2
2E
vmax 
 0.783 m/s
m
140
Example continued
(c) What is the maximum acceleration of the mass?
2
amax
 k 
Ak
2


 A  A


17
.
5
m/s

m
m


2
141
Example: A clock has a pendulum that performs one full
swing every 1.0 sec. The object at the end of the string
weighs 10.0 N. What is the length of the pendulum?
L
T  2
g
Solving for L:


gT 2 9.8 m/s 2 1.0 s 
L

 0.25 m
2
2
4
4
2
142
Example: The gravitational potential energy of a pendulum is
U = mgy. Taking y = 0 at the lowest point of the swing, show
that y = L(1 – cos).

Lcos
L
L
y  L(1  cos  )
y=0
143
Example: Consider a rod of length L and mass m that is
pivoted about one end. What is the period of the oscillations
for this physical pendulum?
X
An FBD for
the rod
Apply N2L in rotational form:
L

  mg  2 sin    I

wer
144
Example continued
2
L
1
d



2
  mg  2 sin    I  3 ML dt 2
d 2
3g

sin 
2
dt
2L
Assume small amplitude oscillations so that  << 1 rad.
d 2
3g
3g

sin    
2
dt
2L
2L
Which now is the equation for SHM with
3g
 
2L
2
145
Example continued
2
3L
 2
The period of oscillations is T 

2g
146
Gravitation
147
Example: Three masses are arranged as shown. What is the force of mass 1
on mass 3?
y
b
M2
M3
a

r̂13
M1
GM1M 3
F13  
rˆ
2
r
x
r  x3  x1  xˆ   y3  y1  yˆ
 b xˆ  a yˆ
r b xˆ  a yˆ
rˆ  
 sin  xˆ  cos  yˆ
2
2
r
a b
148
Example Continued
GM1M 3
F13   2
a  b2



b

  a 2  b2

 
a
xˆ  
  2
2
a

b
 
 
yˆ 
 
 
149
Example: Three masses are arranged as shown. What is the net force on mass
2?
M2
b
r̂32
M3
y
x
a
r̂12
M1
Fnet  F12  F32
GM 3 M 2
GM1M 2

rˆ12 
rˆ32
2
2
r12
r32
GM 3 M 2
GM1M 2
yˆ   2  xˆ 

2
a
b
150
Let M1 = mass of the Earth.
 GM E
F  2
 r

M 2

Here F = the force the Earth exerts on mass M2. This is the force known as
weight, w.
 GM E
w   2
 rE

 M 2  gM 2 .


GM E
2
where g 

9
.
8
N/kg

9
.
8
m/s
2
rE
M E  5.98 1024 kg
rE  6400 km
Near the surface of the
Earth.
151
Note that
In general,
F
g
m
is the gravitational force per unit mass. This is
called the gravitational field strength. It is often
referred to as the acceleration due to gravity.
GM
g   2 rˆ
r
Where the unit vector points from the source of the
gravitational field (location of mass M) to the point
you wish to know g at.
152
Example: What is the weight of a 100 kg astronaut on the surface of the Earth
(force of the Earth on the astronaut)? How about in low Earth orbit? This is an
orbit about 300 km above the surface of the Earth.
w  mg  980 N
On Earth:
In low Earth orbit:
 GM E 
  890 N
w  mg (h)  m
2 
 RE  h  
Their weight is reduced by about 10%. The
astronaut is NOT weightless!
153
Example: If the mass of Mars is 0.108 ME and its radius is 0.6 RE, what is the
gravitational field strength at the surface of Mars?
g mars
GM mars
 2
rmars
G 0.108M E 

0.6 RE 2
 0.108  GM E 
2



 

0
.
30
g

2
.
9
m/s
E
2 
2 
 0.6   RE 
154
Escape Speed
With what speed does an object of mass m need to be launched from the
surface of a body with mass M so that it can just make it to r = ?
Ei  E f
Ki  U i  K f  U f
1 2 GmM
mvesc 
 00
2
R
2GM
vesc 
R
For the Earth vesc = 11.2 km/sec.
155
Example: What is the size of an object with the mass of the sun (21030 kg)
and an escape velocity equal to the speed of light?
2GM
vesc 
c
R
2GM
R  2  3 km
c
This size is called the Schwarzchild radius.
156
Consider m1
Gm1m2 4 2 m1r1
4 2 m1m2
 F1  a 2  T 2  T 2 m  m  a
1
2
GT 2 m1  m2   4 2 a 3
157
For planets in the solar system m1+m2  Msun and
GT 2 M sun  4 2 a 3
Can use the Earth to rewrite Kepler’s third law as T2=a3 if
the period is in years and a is in AUs.
1 AU = 1.5108 km
158
When Kepler formulated these laws they were empirical. It
was not until much later when they were found to have a
physical justification (Newtonian physics).
Planet
a(AU)
T(years)
a3
T2
a3/T2
Mercury
0.39
0.24
0.0593
0.576
1.03
Venus
0.72
0.62
0.373
0.384
0.971
Earth
1.0
1.0
1.0
1.0
1.0
Mars
1.52
1.88
3.51
3.53
0.994
Jupiter
5.20
11.86
141
141
1.0
Saturn
9.54
29.42
868
865
1.003
Uranus
19.19
83.75
7066
7014
1.01
Neptune
30.07
163.7
27189
26798
1.01
Pluto
39.48
248.0
61536
61504
1.000
159
Is there a way to detect the presence of a planet short of
direct imaging? Consider the figure on slide 10.
Gm1m2
v12
2
F


m

m

 1 r  r 2 1 r 1 1 r1
1
1
2
Gm1m2
v22
2
F


m

m

 2 r  r 2 2 r 2 2 r2
2
1
2
Assume that 1 = 2 = 
The center of mass is located at
r2 m2  r1m1
rcm 
0
m1  m2
r1m1  r2 m2
160
Take m1 to be the Sun and m2 to be Jupiter, then r1+r2 = d
is Jupiter’s distance from the Sun.
 m1 
 m1 

d  r1  r2  r1  r1    r1 1 
 m2 
 m2 
d  778.4 106 km
m1
 1050
m2
Are measured values.
Use these values to determine that r1 = 7.40105 km. This
is the size of the Sun’s orbit around the solar system’s
barrycenter. (This value is approximately the radius of the
Sun.)
161
The speed of the Sun in its orbit is
2r1
v1 
 12.4 m/s
T
The period is 11.86 years which is the Jupiter’s orbital
period (assumed 1 = 2 = ).
A distant astronomer viewing the Sun would be able to
detect this motion of the Sun on the sky by noting the
Doppler shift of features in the spectrum of the Sun. The
astronomer would be able to deduce a estimates for the
mass of Jupiter, its orbital period, eccentricity, and orbit
radius.
162
Rocket Motion
Accelerating systems where the mass of the accelerating body changes with
time. Momentum is still conserved.
163
Example: A rocket has an initial mass of 7104 kg and upon
firing its fuel burns at a rate of 250 kg/s. The exhaust
velocity is 2500 m/s. If the rocket has a vertical ascent from
resting on Earth, how long after the engines fire will the
rocket lift off?
Apply N2L
FBD for
the
rocket
y
Fthrust
F  F
thrust
x
wer
 wer  ma
What is the magnitude of the thrust
and weight force on the rocket?
dm
Fthrust   u
 6.25 105 N
dt
wer  mg  6.86 105 N
164
Example continued
Since the thrust is less than the weight force the rocket
cannot take off immediately.
When does the thrust equal the weight force?
Fthrust  wer
Fthrust  mg
Fthrust
m
 63800 kg
g
165
Example continued
 dm 
The mass of the rocket is determined by mt   m0   t
 dt 
When does m(t) = 63 800 kg?
t
mt   m0
 25 sec
dm
dt
166
Example: To perform a rescue a lunar landing craft needs
to hover just above the surface of the moon which has a
gravitational acceleration of g/6. The exhaust velocity is
2000 m/s, but fuel amounting to only 20% of the mass may
be used. How long can the landing craft hover?
FBD for
the
rocket
y
Fthrust
Apply N2L
x
F  F
thrust
 wmr  ma  0
wmr
167
Example continued
What function describes
the fuel consumption?
Fthrust  wmr
dm
u
 mg m
dt
m
t

u dm

  dt

g m m0 m 0
u
ln m  ln m0   t

gm
 g mt 
mt   m0 exp  

 u 
168
Example continued
Since only 20% of the mass can be spent, at what
time is m(t) = 0.8m0?
 g mt 
mt   m0 exp  

 u 
 g mt 
0.8m0  m0 exp  

 u 
u
t
ln 0.8  273 sec
gm
169