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
Erwin Sitompul
University Physics: Mechanics
5/2
Solution of Homework 4: The Plane
→
Δr1
B
966 km,
1.5 h
483 km,
45 min
Δr→2
A
r1  483iˆ km
r2  966jˆ km
rtotal
ttotal
t1  45 min  0.75 h
t2  1.5 h
 r1  r2  483iˆ  966jˆ km
 t1  t2  0.75  1.5  2.25 h
(a) the magnitude of its displacement
rtotal  (483) 2  (966) 2
 1080.021 km
243.435
B
A
(b) the direction of its displacement
C
 966 
rtotal  tan 1 

 483 
 63.435 • Quadrant I
C
 243.435 • Quadrant III
Erwin Sitompul
University Physics: Mechanics
5/3
Solution of Homework 4: The Plane
(c) the magnitude of its average velocity
vavg
rtotal
483iˆ  966jˆ km
 214.667iˆ  429.333jˆ km h


ttotal
2.25 h
vavg  (214.667) 2  (429.333) 2  480 km h
(d) the direction of its average velocity
 429.333   243.435 • Quadrant III
vavg  tan 1 

 214.667 
(e) its average speed
savg 
total distance traveled
Erwin Sitompul
total time

483 km  966 km
 644 km h
2.25 h
University Physics: Mechanics
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:
average  change in velocity
acceleration
time interval
aavg
v2  v1
v


t
t
 If we shrink Δt to zero, then→→
aavg approaches the
instantaneous acceleration a ; that is:
d
dv
dvx ˆ dv y ˆ dvz ˆ
ˆ
ˆ
ˆ
 (vx i  v y j  vz k) 
a
i
j
k
dt
dt
dt
dt
dt
Erwin Sitompul
University Physics: Mechanics
5/5
Average and Instantaneous Acceleration
 We can rewrite the last equation as
a  ax ˆi  ay ˆj  az kˆ
→
where the scalar components of a are:
dv y
dvx
dvz
ax 
, ay 
, az 
dt
dt
dt
Acceleration of a particle does not have
to point along the path of the particle
Erwin Sitompul
University Physics: Mechanics
5/6
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:
vx  v0 x  axt
v0 x  2 m s
ax  3  cos130
 1.928 m s2
vy  v0 y  ayt
v0 y  4 m s
ay  3  sin130
 2.298 m s2
At t = 5 s,
vx  2  (1.928)(5)
 11.64 m s
vy  4  (2.298)(5)
 15.49 m s
Thus, the particle’s velocity at t = 5 s is v  11.64iˆ  15.49jˆ m s.
Erwin Sitompul
University Physics: Mechanics
5/7
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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University Physics: Mechanics
5/8
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  9.81m s
2
ay   g
Erwin Sitompul
University Physics: Mechanics
5/9
Projectile Motion
ax  0
v0  v0 x ˆi  v0 y ˆj
v0 x  v0 cos0
v0 y  v0 sin 0
Erwin Sitompul
ay   g
University Physics: Mechanics
5/10
Projectile Motion
Two Golf Balls
v0 x  0
v0 x  0
• The vertical motions are quasiidentical.
• The horizontal motions are
different.
Erwin Sitompul
University Physics: Mechanics
5/11
Projectile Motion Analyzed
The Horizontal Motion
x  x0  v0xt
x  x0  (v0 cos0 )t
The Vertical Motion
y  y0  v0 yt  12 gt 2
y  y0  (v0 sin 0 )t  12 gt 2
vy  v0 sin 0  gt
vy2  (v0 sin 0 )2  2g ( y  y0 )
Erwin Sitompul
University Physics: Mechanics
5/12
Projectile Motion Analyzed
The Horizontal Range
x  x0  R
R  (v0 cos0 )t
y  y0  0
0  (v0 sin 0 )t  12 gt 2
vx = v0x
vy = –v0y
Eliminating t,
v02
R  2 sin 0 cos 0
g
• This equation is valid if the landing
height is identical with the launch height.
Erwin Sitompul
University Physics: Mechanics
5/13
Projectile Motion Analyzed
Further examining the equation,
v02
R  2 sin 0 cos 0
g
Using the identity
sin 20  2sin 0 cos0 ,
we obtain
v02
R  sin 20
g
R is maximum when
sin2θ0 = 1 or θ0 =45°.
Erwin Sitompul
• If the launch height and the
landing height are the same, then
the maximum horizontal range is
achieved if the launch angle is 45°.
University Physics: Mechanics
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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University Physics: Mechanics
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
Erwin Sitompul
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.
vy  v0 y  gt
0  5.56  9.8t
5.56
 0.567 s
t
9.8
h  y  y0  v0 yt  12 gt 2
1
 (5.56)(0.567)  2 (9.8)(0.567)2
 1.58 m
Erwin Sitompul
40km h  11.11 m s
v0 y  v0 sin 
 (11.11)sin 30
 5.56 m s
v0 x  v0 cos 
 (11.11) cos 30
 9.62 m s
x  x0  v0 xt
y  y0  v0 y t  12 gt 2
v y  v0 y  gt
v y2  v02y  2 g ( y  y0 )
University Physics: Mechanics
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.
ton air  tup  tdown  0.567  0.567  1.134 s
(c) Determine the horizontal distance d
it travels.
d  x  x0  v0xt
 (9.62)(1.134)
 10.91 m
Erwin Sitompul
x  x0  v0 xt
y  y0  v0 y t  12 gt 2
v y  v0 y  gt
v y2  v02y  2 g ( y  y0 )
University Physics: Mechanics
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?
Released horizontally
y  y0  v0 yt  12 gt 2
(500)  (0)  (0)t  12 (9.8)t 2
2  500
2
t 
 102.041
9.8
t  10.102 s
x  x0  v0xt
 (55)(10.102)  555.61 m
Erwin Sitompul
h  y  y0
d  x  x0
d
  tan
h
1 555.61
 tan
500
 48.016
1
University Physics: Mechanics
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?
vy  v0 y  gt
vy  (0)  (9.8)(10.102)
 99 m s
vx  v0 x
 55 m s
Erwin Sitompul
Released horizontally
h  y  y0
d  x  x0
v  55iˆ  99jˆ m s
Unit-vector notation
v  113.252 m s   60.945
Magnitude-angle notation
University Physics: Mechanics
5/20
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  y0  v0 yt  12 gt 2
(4.8)  (0)  (0)t  12 (9.8)t 2
2  4.8
2
t 
 0.98
9.8
t  0.99 s
He cannot make
the jump
x  x0  v0xt
 (4.5)(0.99)
 4.46 m
x  x0  v0 xt
y  y0  v0 y t  12 gt 2
v y  v0 y  gt
v y2  v02y  2 g ( y  y0 )
Time for the stuntman to fall 4.8 m
Horizontal distance jumped by the
stuntman in 0.99 s
Erwin Sitompul
University Physics: Mechanics
5/21