Transcript 投影片 1

3. Motion in 2- & 3-D
1.
2.
3.
4.
5.
6.
Vectors
Velocity & Acceleration Vectors
Relative Motion
Constant Acceleration
Projectile Motion
Uniform Circular Motion
At what angle should this penguin leave the
water to maximize the range of its jump?
45
3.1. Vectors
Vectors:
1. Physics: Quantities with both magnitude & direction.
2. Mathematics : Members of a linear space.
Scalars: Quantities with only magnitude.
Displacement
 r  r
Position vector
r r
r   r  r  r
Vector addition:
1. Commutative: A + B = B + A
2.
Associative: (A + B) + C = A +( B + C )
(Free vectors)
a, b V  c   a   b V
Multiplication by scalar.
A
B  A
2A
z
Coordinate system.
A
Cartesian coordinate system.
Az k = Az
y
k
j
A
Ax i = Ax
Ay j = Ay
Ay = Ay j
i

j
i
x
x
Ax = Ax i
A  Ax  A y  Az
Vector components:
A  Ax  A y
A  Ax i  Ay j  Ax ˆi  Ay ˆj
Unit vectors:
Ax  A cos 
Ay  A sin 
A A A
tan  
2
2
x
2
y
y
Ay
Ax
A  Ax i  Ay j  Az k
 Ax ˆi  Ay ˆj  Az kˆ
Example 3.1. Taking a Drive
You drive to city 160 km from home, going 35 N of E.
Express your new position in unit vector notation, using an E-W / N-S coordinate system.
r  rx ˆi  ry ˆj
y (N)
rx  r cos  160 km cos35
city
 160  0.81915
km  131.06
km  131 km
r = 160 km
ry  r sin   160 km sin35
 = 35
j
x (E)
home
i
 160  0.57357
r  131 ˆi  92 ˆj km
km
 91.77
km  92 km
Vector Arithmetic with Unit Vectors
A  Ax ˆi  Ay ˆj
B  Bx ˆi  By ˆj
  Ax , Ay 
  Bx , By 
AB
 Ax  Bx & Ay  By
A  B   Ax  Bx  ˆi   Ay  By  ˆj   Ax  Bx , Ay  By 
3.2. Velocity & Acceleration Vectors
Average velocity
(Instantaneous) velocity
Average acceleration
(Instantaneous)
acceleration
v
r
xˆ yˆ

i+
j
t
t
t
v  lim
t 0
a
r
dr

t
dt

 x  y 

,


t
t 

dxˆ d yˆ
i+
j
dt
dt
d x d y 

,

d
t
d
t


 v  vx ˆ  vy ˆ   vx
 vy 

i+
j 
,

t
t
t

t

t


v
d vx ˆ d vy ˆ
 d vx d v y
dv

i
+
j


,

t 0  t
dt
dt
dt
d
t
dt

r


t 
d2 r
 d2 x
d2 x ˆ d2 y ˆ

 lim


i+
j  2
2
2
2
t 0
t
dt
dt
dt
dt
a  lim



d2 y 
,

d t2 
Velocity & Acceleration in 2-D
a  v  circular motion
3.3. Relative Motion
Motion is relative (requires frame of reference).
Man walks at v = 4 km/h down aisle to front of plane,
which move at V = 1000 km/h wrt (with respect to) ground.
Man’s velocity wrt ground is v = v + V.
Plane flies at v wrt air.
Air moves at V wrt ground.
Plane’s velocity wrt ground is v = v + V.
Example 3.2. Navigating a Jetliner
Jet flies at 960 km / h wrt air, trying to reach airport 1290 km northward.
Assuming wind blows steadly eastward at 190 km / h.
1. What direction should the plane fly?
2. How long will the trip takes?
Desired velocity
Wind velocity
V
Jet velocity
190
km/h
v  v  V
v
960
km/h

v  vy ˆj   0 , v y 
V  190 km / h i  190 , 0
 cos i  sin j 
  960 cos , 960 sin  
v  960 km / h
 0 , v    960 cos   190 , 960 sin  
y
v
cos   

190
960
vy  960 sin 
Trip time
t
1290 km
941 km / h
 190 
  101.4
960


  cos 1  
 941 km / h
 1.4 h
3.4. Constant Acceleration
Constant Acceleration:
2-D:
1
r  r0  v 0 t  a t 2
2
v  v0  a t
vx  v0 x  ax t
v y  v0 y  a y t
1
x  x0  vx 0 t  ax t 2
2
1
y  y0  v y 0 t  a y t 2
2
a  0 ,  g 
x  x0
y  y0 
x  x0  vx 0 t
1 2
gt
2
y  y0 
1 2
gt
2
Example 3.3. Windsurfing
You’re windsurfing at 7.3 m/s when a wind gust accelerates you
1
x  x0  vx 0 t  ax t 2
2
1
y  y0  v y 0 t  a y t 2
2
at 0.82 m/s2 at 60 to your original direction.
If the gust lasts 8.7 s, what is your net displacement?
r0   0 , 0  m
v0   7.3 , 0  m / s
2
a  0.82  cos60 , sin 60  m / s2   0.41 , 0.71  m / s
1
x  7.3 t  0.41 t 2
2
y
1
0.71 t 2
2
net displacement
r
x  79.0 m
y  26.9 m
x 2  y 2  84 m
3.5. Projectile Motion
2-D motion under constant gravitational acceleration
x  x0  vx 0 t
vx  v0 x
1
y  y0  v y 0 t  g t 2
v y  v0 y  g t
2
y ~ x2
parabola
vx  v0 x
Example 3.4. Washout
A section of highway was washed away by flood, creating a gash 1.7 m deep.
A car moving at 31 m/s goes over the edge.
v y  v0 y  g t
x  x0  vx 0 t
1
y  y0  v y 0 t  g t 2
2
How far from the edge does it land?
vx 0  31 m / s
x0  0
vy 0  0
y0  1.7 m
x   31 m / s  t
0  1.7 m 
t
1.7
s
4.9
1
9.8 m / s 2  t 2

2
 0.589 s
x  31 0.589 m  18 m
y0
Projectile Trajectory
1
x  x0  vx 0 t  ax t 2  x   v cos   t
0
0
0
2
1 2
1
y  y0  v y 0 t  a y t 2  y0   v0 sin 0  t  g t
2
2

t
x  x0
v0 cos  0
 x  x0  1  x  x0 
y  y0   v0 sin 0  
 g 

v
cos

2
v
cos

0 
0 
 0
 0
y  y0   x  x0  tan 0 
g
2
x

x


0
2 v02 cos2 0
2
Projectile trajectory:
parabola
Example 3.5. Out of the Hole
A construction worker stands in a 2.6 m deep hole, 3.1 m from edge of hole.
He tosses a hammer to a companion outside the hole.
Let the hammer leave his hand 1.0 m above hole bottom at an angle of 35.
1. What’s the minimum speed for it to clear the edge?
2.
How far from the edge does it land?
y  y0   x  x0  tan 0 
x0  0
x  3.1 m
y0  1.0 m
y  2.6 m
1.6  3.1tan 35 
minimum speed
1.6  x tan 35 
x
9.8
2 11 cos2 35
2
1
 0.70  0.33
0.12
x2
g
2
x

x


0
2 v02 cos2 0
9.8
2
3.1


2 v02 cos 2 35
v0  11 m / s
2
 0.060 x  0.70 x  1.6  0
8.7 m

 3.1 m

0  35
Lands at 5.5 m from edge.
The Range of a Projectile
y  y0   x  x0  tan 0 
g
2
x

x


0
2 v02 cos2 0
Horizontal range y = y0 :
0   x  x0  tan 0 

g
2
x

x


0
2 v02 cos 2 0
x  x0
2 v02
2 v02
v02
2
x  x0 
cos 0 tan 0 
cos 0 sin 0 
sin 20
g
g
g
Longest range at 0 = 45 = /4.
Prob 70: Range is same for 0 & /2  0.
Prob 2.77: Projectile spends 71% in upper half of trajectory.
Example 3.6. Probing the Atmosphere
After a short engine firing, a rocket reaches 4.6 km/s.
If the rocket is to land within 50 km from its launch site,
what’s the maximum allowable deviation from a vertical trajectory?
Short engine firing  y  0, v0 = 4.6 km/s.
g
2
0  x tan 0  2
x
2 v0 cos2 0
50 km 
 4.6 km / s 
g
2
x

x


0
2 v02 cos 2 0
v02
v02
x2
sin 0 cos 0 
sin 20
g
g
2
9.8 103 km / s 2
sin 20  0.0232

y  y0   x  x0  tan 0 
sin 2 0
 1.33
20  
180  1.33
 0.67
90  0.67
0  
 maximum allowable deviation from a vertical trajectory is 0.67.
3.6. Uniform Circular Motion
Uniform circular motion: circular trajectory, constant speed.
Examples:
Satellite orbit.
Planetary orbits (almost).
Earth’s rotation.
Motors.
Electrons in magnetic field.
⁞
r  r2  r1
v  v2  v1
v r


v
r

r  v t

v
v2
a

t
r
v v 2
a  lim

t  0 t
r
v2
a   rˆ
r
( centripetal )
Example 3.7. Space Shuttle Orbit
Orbit of space shuttle is circular at altitude 250 km, where g is 93% of its surface value.
Find its orbital period.
2 r
T
v
T
2 r
ar
 2
v2
a
r
 2
r
a
6.37 103 km  250 km
0.93  9.8 10 3 km / s 2
 5355 s
ISS: r ~ 350 km
15.7 orbits a day
 89 min
(low orbits)
Example 3.7. Engineering a Road
Consider a flat, horizontal road with 80 km/h (22.2 m/s) speed limit.
If the max vehicle acceleration is 1.5 m/s2,
what’s the min safe radius for curves on this road.
v2
a
r
2
 22.2 m / s   329 m
vmax


amax
1.5 m / s 2
2
rmin
Nonuniform Circular Motion
Nonuniform Circular Motion: trajectory circular, speed nonuniform
 a non-radial but ar = v2 / r
v
at
ar
a
GOT IT? 3.4.
Arbitrary motion:
ar = v2 / r
r = radius of curvature
If v1 = v4 , & v2 = v3 , rank ak.
Ans: a2 > a3 > a4 > a1