Transcript University Physics - Erwin Sitompul

Lecture 5
Ch4. TWO- AND THREE-DIMENSIONAL MOTION
University Physics: Mechanics
Dr.-Ing. Erwin Sitompul
http://zitompul.wordpress.com
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.
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Solution of Homework 4: The Plane
→
Δr1
B
A
 r1   4 8 3iˆ k m
 r   966 ˆj km
 t1  45 min  0.75 h
 t 2  1.5 h
2
966 km,
1.5 h
483 km,
45 min
Δr→2
 rtotal   r1   r2   483iˆ  966 ˆj km
 t total   t1   t 2  0 .7 5  1 .5  2 .2 5 h
(a) the magnitude of its displacement
 rto tal 
(  4 8 3)  (  9 6 6 )
2
2
 1080.021 km
2 4 3 .4 3 5 
B
A
(b) the direction of its displacement
C
  rtotal  tan
1
  966 


  483 
 6 3 .4 3 5 
• Quadrant I
C
 243.435  • Quadrant III
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Solution of Homework 4: The Plane
(c) the magnitude of its average velocity
v avg 
v avg 
 rtotal

 t total
 483iˆ  966 ˆj km
  214.667 ˆi  429.333 ˆj km h
2.25 h
(  2 1 4 .6 6 7 )  (  4 2 9 .3 3 3)
2
2
 480 km h
(d) the direction of its average velocity
 v avg  tan
1
  429.333   2 4 3 .4 3 5  • Quadrant III


  214.667 
(e) its average speed
s avg 
to ta l d ista n ce tra ve le d
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to ta l tim e

483 km  966 km
 644 km h
2.25 h
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5/4
Average and Instantaneous Acceleration
→
→
 When a particle’s velocity changes from
v1 to v2 in a time
→
interval Δt, its average acceleration aavg during Δt is:
a v e ra g e  ch a n g e in v e lo city
a cce le ra tio n
tim e in te rv a l
a avg 
v 2  v1
t

v
t
 If we shrink Δt to zero, then→→
aavg approaches the
instantaneous acceleration a ; that is:
a 
dv
dt
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dv x ˆ dv y ˆ dv z ˆ
ˆ
ˆ
ˆ

( v x i  v y j  v z k) 
i
j
k
dt
dt
dt
dt
d
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Average and Instantaneous Acceleration
 We can rewrite the last equation as
a  a x ˆi  a y ˆj  a z kˆ
→
where the scalar components of a are:
ax 
dv x
dt
, ay 
dv y
dt
, az 
dv z
dt
Acceleration of a particle does not have
to point along the path of the particle
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Average and Instantaneous Acceleration
→
A particle with velocity→v0 = –2i^ + 4j^ m/s at t = 0 undergoes a
constant acceleration a of magnitude a = 3 m/s2 at an angle
130° from the positive
direction of the x axis. What is the
→
particle’s velocity v at t = 5 s?
Solution:
v y  v0 y  a y t
v x  v0 x  a x t
v0 x   2 m s
v0 y  4 m s
a y  3  sin 130 
a x  3  cos130 
  1.928 m s
2
 2.298 m s
2
At t = 5 s,
v x   2  (  1.928)(5)
  1 1 .6 4 m s
v y  4  (2.298)(5)
 1 5 .4 9 m s
Thus, the particle’s velocity at t = 5 s is
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v   11.64iˆ  15.49 ˆj m s.
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Projectile Motion
 Projectile motion: a motion in a vertical plane, where
the
→
acceleration is always the free-fall acceleration g, which is
downward.
 Many sports involve the projectile motion of a ball.
 Besides sports, many acts also involve the projectile motion.
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Projectile Motion
 Projectile motion consists of horizontal motion and vertical
motion, which are independent to each other.
 The horizontal motion has no acceleration (it has a constant
velocity).
 The vertical motion is a free fall motion with constant
acceleration due to gravitational force.
ax  0
g  9.81 m s
2
ay  g
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Projectile Motion
ax  0
v 0  v 0 x ˆi  v 0 y ˆj
ay  g
v 0 x  v 0 cos  0
v 0 y  v 0 sin  0
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Projectile Motion
Two Golf Balls
v0 x  0
v0 x  0
• The vertical motions are quasiidentical.
• The horizontal motions are
different.
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University Physics: Mechanics
5/11
Projectile Motion Analyzed
The Horizontal Motion
x  x0  v0 x t
x  x 0  ( v 0 cos  0 ) t
The Vertical Motion
y  y 0  v0 y t 
1
gt
2
y  y 0  ( v 0 sin  0 ) t 
1
2
2
gt
2
v y  v 0 sin  0  gt
v y  ( v 0 sin  0 )  2 g ( y  y 0 )
2
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2
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Projectile Motion Analyzed
The Horizontal Range
x  x0  R
R  ( v 0 cos  0 ) t
y  y0  0
0  ( v 0 sin  0 ) t 
vx = v0x
vy = –v0y
1
2
gt
2
Eliminating t,
2
R 2
v0
g
sin  0 co s  0
• This equation is valid if the landing
height is identical with the launch height.
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5/13
Projectile Motion Analyzed
Further examining the equation,
2
v0
R 2
g
sin  0 co s  0
Using the identity
sin 2 0  2 sin  0 cos  0 ,
we obtain
2
R 
v0
g
sin 2 0
R is maximum when
sin2θ0 = 1 or θ0 =45°.
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• If the launch height and the
landing height are the same, then
the maximum horizontal range is
achieved if the launch angle is 45°.
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5/14
Projectile Motion Analyzed
• The launch height and the landing height differ.
• The launch angle 45° does not yield the
maximum horizontal distance.
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5/15
Projectile Motion Analyzed
The Effects of the Air
 Path I: Projectile movement if the air
resistance is taken into account
 Path II: Projectile movement if the air
resistance is neglected (as in a vacuum)
Our calculation along this chapter is based on
this assumption
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University Physics: Mechanics
5/16
Example: Baseball Pitcher
A pitcher throws a baseball at speed 40 km/h and at angle
θ = 30°.
h
30
(a) Determine the maximum height h
of the baseball above the ground.
v y  v 0 y  gt
0  5 .5 6  9 .8t
5.56
 0 .5 6 7 s
t
9.8
1
2
h  y  y 0  v 0 y t  2 gt
 (5.56)(0.567)  2 (9.8)(0.567)
 1.58 m
1
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4 0 k m h  1 1 .1 1 m s
v 0 y  v 0 sin 
 (11.11) sin 30 
 5.56 m s
v 0 x  v 0 co s 
 (1 1 .1 1) co s 3 0 
 9 .6 2 m s
x  x0  v0 xt
y  y0  v0 y t 
1
2
gt
2
v y  v 0 y  gt
2
v y  v0 y  2 g ( y  y 0 )
2
2
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5/17
Example: Baseball Pitcher
A pitcher throws a baseball at speed 40 km/h and at angle
θ = 30°.
30
d
(b) Determine the duration when the baseball is on the air.
t on air  t up  t dow n  0.567  0.567  1.134 s
(c) Determine the horizontal distance d
it travels.
d  x  x0  v0 x t
x  x0  v0 xt
y  y0  v0 y t 
 (9.62)(1.134)
 10.91 m
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1
2
gt
2
v y  v 0 y  gt
v y  v0 y  2 g ( y  y 0 )
2
2
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5/18
Example: Rescue Plane
A rescue plane flies at 198 km/h
and constant height h = 500 m
toward a point directly over a
victim, where a rescue capsule
is to land.
(a) What should be the angle Φ
of the pilot’s line of sight to
the victim when the capsule
release is made?
y  y 0  v0 y t 
1
2
1
gt
Released horizontally
2
d  x  x0
(  500)  (0)  (0) t  2 (9.8) t
2  500
2
t 
 102.041
9.8
t  1 0 .1 0 2 s
2
x  x0  v0 x t
 (55)(10.102)  5 5 5 .6 1 m
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h  y  y0
  tan
 tan
1
d
1
h
555.61
500
 48.016 
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5/19
Example: Rescue Plane
A rescue plane flies at 198 km/h
and constant height h = 500 m
toward a point directly over a
victim, where a rescue capsule
is to land.
(b) As the capsule reaches the
→
water, what is its velocity v
in unit-vector notation and in
magnitude-angle notation?
v y  v 0 y  gt
v y  (0)  (9.8)(10.102)
 99 m s
v x  v0 x
 55 m s
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Released horizontally
h  y  y0
d  x  x0
v  55iˆ  99 ˆj m s
Unit-vector notation
v  1 1 3 .2 5 2 m s   6 0 .9 4 5 
Magnitude-angle notation
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Example: Clever Stuntman
A stuntman plans a spectacular jump from
a higher building to a lower one, as can be
observed in the next figure.
Can he make the jump and safely reach
the lower building?
y  y 0  v0 y t 
1
2
1
gt
2
(  4.8)  (0)  (0) t  2 (9.8) t
2  4.8
2
t 
 0.98
9.8
t  0 .9 9 s
2
Time for the stuntman to fall 4.8 m
x  x0  v0 xt
x  x0  v0 x t
 (4.5)(0.99)
y  y0  v0 y t 
 4 .4 6 m
Horizontal distance jumped by the
stuntman in 0.99 s
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He cannot make
the jump
1
2
gt
2
v y  v 0 y  gt
v y  v0 y  2 g ( y  y 0 )
2
2
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5/21
Homework 5: Three Point Throw
New
A basketball player who is 2.00 m tall is standing on the floor
10.0 m from the basket. If he shoots the ball at a 40.0° angle
with the horizontal, at what initial speed must he throw so that it
goes through the hoop without striking the backboard? The
basket height is 3.05 m.
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University Physics: Mechanics
5/22