Transcript Slide 1

Lecture 4
Ch4. TWO- AND THREE-DIMENSIONAL MOTION
University Physics: Mechanics
Dr.-Ing. Erwin Sitompul
http://zitompul.wordpress.com
Solution of Homework 3: The Beetles
2nd run, ?
New
location
40
30
Starting 1st run, 0.5 m
point
Erwin Sitompul
University Physics: Mechanics
4/2
Solution of Homework 3: The Beetles
→
D
A  B  0.5iˆ  (0.8cos30ˆi  0.8sin30ˆj)
 1.19iˆ  0.4jˆ
C  1.6cos50ˆi  1.6sin 50ˆj
 1.03iˆ  1.23jˆ
New
location
40
30
Starting
point
→
A
A B  C  D
D  A B C
ˆ  (1.03iˆ 1.23j)
ˆ
 (1.19iˆ  0.4j)
 0.16iˆ  0.83jˆ
Thus, the second run of the
green beetle corresponds to the
vector D  0.16iˆ  0.83jˆ m.
Erwin Sitompul
University Physics: Mechanics
4/3
Solution of Homework 3: The Beetles
(a) The magnitude of the second run?
D  0.16iˆ  0.83jˆ m
D  (0.16)2  (0.83)2  0.85 m
N
(b) The direction of the second run?
 0.83 
  tan 
  79.09
 0.16 
1
W
E
79.09
S
0.85 m
→
D
The direction of the second run is
79.09° south of due east or
10.91° east of due south.
Erwin Sitompul
University Physics: Mechanics
4/4
Moving in Two and Three Dimensions
 In this chapter we extends the material of the preceding
chapters to two and three dimensions.
 Position, velocity, and acceleration are again used, but they
are now a little more complex because of the extra
dimensions.
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University Physics: Mechanics
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Position and Displacement
 One general
way of locating a particle is with a position
→
vector r,
r  xˆi  yˆj  zkˆ
 The coefficients x, z, and y give the particle’s location along
the coordinate axes and relative to the origin.
 The following figure shows a particle with position vector
r  (3 m)iˆ  (2 m)jˆ  (5 m)kˆ
 In rectangular coordinates, the position
is given by (–3 m, 2 m, 5 m).
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University Physics: Mechanics
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Position and Displacement
 As a particle moves, its position vector changes in a way that
the vector always extends from the origin to the particle.
→
→
 If the position vector changes from r1 to r2, then the particle’s
displacement delta is:
r  r2  r1
ˆ  ( x ˆi  y ˆj  z k)
ˆ
r  ( x2ˆi  y2ˆj  z2k)
1
1
1
r  ( x  x )iˆ  ( y  y )jˆ  ( z  z )kˆ
2
1
2
1
2
1
r  xˆi  yˆj  zkˆ
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University Physics: Mechanics
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Average Velocity and Instantaneous Velocity
→
 If a particle moves through a displacement Δr in a time
interval Δt, then its average velocity →
vavg is:
average velocity 
vavg
displacement
time interval
r

t
 The equation above can be rewritten
in vector components as:
xˆi  yˆj  zkˆ
vavg 
t
x ˆ y ˆ z ˆ

i
j
k
t
t
t
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University Physics: Mechanics
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Average Velocity and Instantaneous Velocity
→
 The particle’s instantaneous velocity v is the velocity of the
particle at some instant.
dr
v
dt
 The direction of instantaneous velocity of a particle is always
tangent to the particle’s path at the particle’s position.
Erwin Sitompul
University Physics: Mechanics
4/9
Average Velocity and Instantaneous Velocity
 Writing the last equation in unit-vector form:
d ˆ ˆ ˆ
dx ˆ dy ˆ dz ˆ
v  ( xi  yj  zk)  i 
j k
dt
dt
dt
dt
 This equation can be simplified by rewriting it as:
v  vx ˆi  vy ˆj  vz kˆ
→
where the scalar components of v are:
dx
dy
dz
vx  , v y 
, vz 
dt
dt
dt
 The next figure shows a velocity
vector →
v and its scalar x→and y
components. Note that v is tangent
to the particle’s path at the particle’s
position.
Erwin Sitompul
University Physics: Mechanics
4/10
Average Velocity and Instantaneous Velocity
The figure below shows a circular path taken by a particle.
If
the instantaneous velocity of the particle at a certain time is
→
v = 2i^ – 2j^ m/s, through which quadrant is the particle
v
currently moving when it is traveling
(a) clockwise
v
(b) counterclockwise
2
y
x
–2
v  2ˆi  2ˆj m s
First
quadrant
Third
quadrant
(a) clockwise
Erwin Sitompul
(b) counterclockwise
University Physics: Mechanics
4/11
Homework 4: The Plane
A plane flies 483 km west from city A to city B in 45 min and
then 966 km south from city B to city C in 1.5 h.
From the total trip of the plane, determine:
(a) the magnitude of its displacement;
(b) the direction of its displacement;
(c) the magnitude of its average velocity;
(d) the direction of its average velocity;
(e) its average speed.
Erwin Sitompul
University Physics: Mechanics
4/12
Homework 4
New
A turtle starts moving from its original position with the speed
10 cm/s in the direction 25° north of due east for 1 minute.
Afterwards, it continues to move south for 2 m in 8 s.
From the total movement of the turtle, determine:
(a) the magnitude of its displacement;
(b) the direction of its displacement;
(c) the magnitude of its average velocity;
(d) the direction of its average velocity;
(e) its average speed.
Erwin Sitompul
University Physics: Mechanics
4/13