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Chapter 4
Motion in two and three
dimensions
Two principles for 2D and 3D motions:
1) The principle of independence of force
2) The principle of superposition of motion
F1
F3
F2
Section 4-1 Motion in three dimensions
with constant acceleration
Now we consider a particle move in three
dimensions with constant acceleration. We
can represent the acceleration as a vector:




a  ax i  a y j az k
The particle starts at t=0 with initial position




r0  x 0 i  y 0 j  z o k




and an initial velocity v0  v0 x i  v0 y j  v0 z k .

a , a constant
ax , a y , az , all constants
v v a t
v v a t
x
0x
y
0y
v v
z
0z
x
y
 az t



v  v0  a t
（4-1）
In a similar way:
1
2
x  x0  v0 x t  a x t
2
1
2
y  y  v0 y t  a y t
0
2
1
2
z  z 0  v0 z t  a z t
2



1 2
r  r0  v0 t  a t
2
（4-2）
Section 4-2 Newton’s laws in three
dimensional vector form


F  ma
(4-3)
Which includes the three component
equations
F
F
x
 max
z
 maz
F
y
 ma y
(4-4)
Sample problem
1. A crate of mass m=62 kg is sliding without
friction with an initial velocity of v0=6.4 m/s
along the floor. In an attempt to move it in a
different direction, Tom pushes opposite to
its initial motion with a constant force of a
magnitude F1=81N, while Jane pushes in a
perpendicular direction with a constant force
of magnitude F2=105N. If they each push
for 3.0s, in what direction is the crate
moving when they stop pushing?
Section 4-3 Projectile motion
Figure 4-4 shows the initial
motion of a projectile at the
instant of launch. Its initial
velocity is vo , directed at an
angle  from the horizontal.
y

vo

0
We choose suitable coordinate
system to make:
x0  0 y 0  0
o

mg
0
x
Fig 4-4 A particle is
launched with initial
velocity
The components of the initial velocity are
vox  v0 cos 0 voy  v0 sin 0 (4-6)
Gravity is the only force acting on the
particle, so the components of the net force
are  Fx  0
 Fy  mg (4-7)
(4-8)
a y  g
ax  0
v x  v0 x
v y  v 0 y  gt (4-9)
Position components:
1 2
x  v0 x t
y  v0 y t  gt
(4-10)
2
From Eqs. (4-10), we can eliminate t and obtain
the relationship between x and y (after considering
Eqs. (4-6)):
g
2
y  tan 0 x 
x
(4-13)
2(v0 cos 0 ) 2
which is the equation of a trajectory (轨线) of
the projectile, the equation of a parabola.
Hence the trajectory of a projectile is parabolic.
Fig 4-5 trajectory of a projectile
y
v
v
0y
v0
v0 x
y
v0 x
v
0x
o
v
v
0x
0x
v0 y
R
x
The “horizontal range R” of the projectile is
defined as the distance along the horizontal
where the projectile return to the level from
which it was launched.
Let y=0 in Eq(4-13), we obtain the range R:
2
2
2v0
v0
R
sin 0 cos 0 
sin 20 (4-14)
g
g
当子弹从枪口射出时，椰子刚好从树上由静止
自由下落 . 试说明为什么子弹总可以射中椰子 ?
Sample problem
4-3. In a contest to drop a package on a
target, one contestant’s plane is flying at a
constant horizontal velocity of 155km/h at an
elevation (海拔) of 225m toward a point
directly above the target. At what angle of
sight should the package be released to
strike the target?
Section 4-4 Drag forces and the effects on
motions



Drag force is a frictional force which
experienced by any object that moves
through a fluid medium, such as air or water.
Drag force must be taken into account in the
design of aircraft and seacraft.
Drag forces prevent the velocity from
increasing without limit in the nature.
Falling motion with drag force
We assume that the magnitude of the drag force
D depends linearly on the speed:


(4-17)
D  b v
We choose the y axis to be vertical and the positive
direction to be downward.
 Fy  mg  bv y and mg  bv y  ma y (4-18)
dv y
b
 ay  g  vy
dt
m
dv y
 dt
bv y
g
m
(4-20)
(4-19)
With v y  0 at time t=0, we integrate two
sides of Eq.(4-20)
vy
dv y

0
bv y
g
m
then we obtain
ln(
mg  bv y
mg
t
 dt
0
b
) t
m
mg
vy 
(1  e
b
b
 t
m
)
(4-21)
(4-22)
• For large t, e
vy
b
 t
m
0
mg

b
The magnitude of the terminal speed approaches a
constant value, not increasing without limit.
• For small t,
b
t  1
m
So e
b
 t
m
b
 1 t
m
mg
b
v y (t ) 
[1  (1  t )]  gt (small t) (4-23)
b
m
At the beginning of the motion, it is nearly a freely
falling motion.
How is D changing with
time in general?
D
D
mg
mg
D
mg
(for cats)
mg
It is found that a cat is
much safer when it falls
from higher place. WHY?
Projectile motion with drag force
Y(m)
When the drag force

 is considered,
D  b v
the range R and the
maximum height H will
be reduced.
With drag
force
Without
drag force
  60 0
o
 60 0
 79
R
-
X (m)
The trajectory is also no longer symmetric about
the maximum; the descending motion is much
steeper than the ascending motion.
4-5 Uniform Circular Motion
In uniform circular motion, the particle
moves at constant speed in a circular path.
Since the direction of velocity
changes in the motion, it is an
acceleration motion.
How to find acceleration
from the constant speed for
uniform circular motion?
v
r
O
Fig 4-16
Find acceleration for the motion

v1x  v cos 
v1 y  v sin 
y
v
p1
v
v1 y
p
p2
1x
r 
v2 y
v2 x
v
x
O

v2 x  v cos 
(4-25)
v 2 y  v sin 
As the particle moves along
the arc from p1 to p 2 ,it
covers a distance of r  2 ,
and a time interval t  2r .
v
Acceleration:
Fig 4-16
ay 
v2 y  v1 y
t
v2 x  v1x
ax 
 0 (4-27)
t
 v sin   v sin 
v 2 sin 
(4-28)


(2r / v)
r 
In order to find the instantaneous acceleration,
we take t approaches zero, (then angle  goes
to zero) so that p1 and p 2 both approach p , which
gives
v 2 sin 
v2
sin 
v2
a y  lim [ (
)]   lim

(4-29)
 0
r 
r  0 
r
• Point p is an arbitrary point on the circle, so
Eq.(4-29) is a general result for the motion.
• The minus sign indicating that the acceleration
at p points toward the center of the circle.
• The acceleration a y is called centripetal acceleration
or radial acceleration. The corresponding force is
called centripetal force or seeking center force.
4-6 Relative Motion
If we have two inertial frames s and s’, how about
the relationship between the motions in them?
S’
y’
s, s’: inertial frame
P
s
y

rps'

r ps
0’

x’
rs 's
0
x



rps  rps'  rs ' s
Fig 4-17
(4-31)
Take the derivative with respect to time of
Eq.(4-31), we have



(4-32)
v ps  v ps'  vs 's
Eq.(4-32) is a law of the transformation of
velocities. It is often called the Galilean form
of the law of transformation of velocities.
It permits us to transform a velocity of
one frame of reference (s’) to another
frame of reference.
Differentiate Eq.(4-32), we have

d v ps

d v ps '

d vs ' s


dt
dt
dt
(4-33)
The last term of eq.(4-33) vanishes, because
the relative velocity of two reference frames
must be a constant. Thus

d v ps
dt

d v ps'
dt


or a ps  a ps'
(4-34)

The accelerations of P measured by two
observers are identical.

Eq.(4-34) indicates directly Newton’s second law
can be equally well applied in any inertial frames.
Sample problem
1. The compass in an airplane indicates that
it is headed due east; its air speed indicator
reads 215 km/h. A steady wind of 65 km/h is
blowing due north. (a) What is the velocity of
the plane with respect to the ground? (b) If
the pilot wishes to fly due east, what must
be the heading? That is, what must the
compass read?