2009 Physics 2111 Fundamentals of Physics Chapter 3 1 Fundamentals of Physics Chapter 3 Motion in 2...
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2009 Physics 2111 Fundamentals of Physics Chapter 3 1
Fundamentals of Physics
Chapter 3 Motion in 2 &3 Dimensions
1. Moving in 2 &3 Dimensions
2. Position & Displacement
3. Average & Instantaneous Velocity
4. Average & Instantaneous Acceleration
5. Projectile Motion
6. Projectile Motion Analyzed
7. Uniform Circular Motion
8. Relative Motion
Review & Summary
Questions
Exercises & Problems
2009 Physics 2111 Fundamentals of Physics Chapter 3 2
Position & Displacement
Position Vector:
A vector from a reference point (aka the origin) to the particle.
krjrirr zyx
kjir
523
The vector gives the position of the green ball.
2009 Physics 2111 Fundamentals of Physics Chapter 3 3
Position & Displacement
Displacement Vector: 12 rrr
kzzjyyixxr
121212
kzjyixr
Consider a particle moving along a path.
2009 Physics 2111 Fundamentals of Physics Chapter 3 4
Average & Instantaneous Velocity
Average Velocity = ratio of the displacement to the time interval:
Instantaneous Velocity = derivative of particle position wrt time:
A vector in the same direction as the straight line connecting the starting point to the end point.
A vector that is tangent to the particle’s path at the particle’s position.
t
rvavg
.
dt
rd
t
rv
t
lim0
2009 Physics 2111 Fundamentals of Physics Chapter 3 5
Velocity Components
dt
dzv
dt
dyv
dt
dxv
kvjvivv
dt
rdv
z
y
x
zyx
2009 Physics 2111 Fundamentals of Physics Chapter 3 6
Average & Instantaneous Acceleration
Acceleration of a particle need not point along the path of the particle.
Average Acceleration = ratio of the change in velocity to the time interval:
Instantaneous Acceleration = derivative of particle velocity wrt time:
If the velocity of a particle changes in either magnitude or direction, the particle is subject to an acceleration.
t
vaavg
.
kajaiadt
vda zyx
2009 Physics 2111 Fundamentals of Physics Chapter 3 7
Projectile Motion
The particle is launched with initial velocity vo:
For a projectile, the motions in the vertical and horizontal planes are independent of each other:
horizontal motion with zero acceleration
vertical motion with constant downward acceleration
sin
cos
00
00
000
vv
vv
jvivv
y
x
yx
gay
0xa
2-Dimensional Motion: a projectile moves in a vertical plane subject to the downward acceleration due to gravity.
2009 Physics 2111 Fundamentals of Physics Chapter 3 8
Projectile Motions
gay
0xajvivv yx
000
2009 Physics 2111 Fundamentals of Physics Chapter 3 9
Identical Vertical Motion
Two balls are released simultaneously
00 xv
00 xv
The vertical motions are identical.
2009 Physics 2111 Fundamentals of Physics Chapter 3 10
Projectile Motions
vx = v0x
vy = - v0yjvivv yx
000
g
22
100 tatvxx xx
000 cos
0
vvv
a
xx
x
tvxx 000 cos
22
100 tatvyy yy
000 sinvv
ga
y
y
22
1000 sin tgtvyy
tgvvy sin0
Equations that describe Projectile Motion
2009 Physics 2111 Fundamentals of Physics Chapter 3 11
Trajectory of a Projectile
Particle Trajectory:
22
100 sin tgtvy
tvx 00 cos
2
00
2
0cos2
tan
v
xgxy parabolic path
2009 Physics 2111 Fundamentals of Physics Chapter 3 12
Range of a Projectile
0
20 2sin
g
vR
22
1000 sin tgtvyy
00 yy
tvxx 000 cos
RANGE0 xx
cossin22sin
2009 Physics 2111 Fundamentals of Physics Chapter 3 13
Range equation
2009 Physics 2111 Fundamentals of Physics Chapter 3 14
Maximum Range of a Projectile
0
20 2sin
g
vR
R is a maximum for 0 = 45o.
2009 Physics 2111 Fundamentals of Physics Chapter 3 15
Range of a Projectile
2009 Physics 2111 Fundamentals of Physics Chapter 3 16
Projectile Motion Applet
Launch Internet Explorer Browser.lnk
http://physics.uwstout.edu/physapplets/virginia/www.phys.virginia.edu/classes/109n/more_stuff/applets/ProjectileMotion/enapplet.html
2009 Physics 2111 Fundamentals of Physics Chapter 3 17
The Ball Hits the Can Every Time!
Magnet (M) releases the can just as the projectile leaves the blow gun (G).
During the time-of-flight, both the projectile and the can fall the same distance, h, under the constant acceleration –g.
2009 Physics 2111 Fundamentals of Physics Chapter 3 18
Projectiles
2009 Physics 2111 Fundamentals of Physics Chapter 3 19
Object Falling from a Moving Plane
released horizontally
h = 500m, v0 = 55 m/s
At what value of should the payload be released?
ght
tgh
tgtvy
/2
sin2
21
22
100
00
tvx
tvx
tvx x
0
00
0
cos
h
xtan
tgv
vv
y
x
0
2009 Physics 2111 Fundamentals of Physics Chapter 3 20
Cannon Firing at Pirate Ship
V0 cannonball = 82 m/sPirate ship is 560 m offshorea) What for cannonball to hit shipb) Safe range for the ship?
2009 Physics 2111 Fundamentals of Physics Chapter 3 21
Great Zacchini
The “flight of Emanuel Zacchini” over 3 ferris wheels
v0 = 26.5 m/s, 0 = 530, h0 = 3.0 mDoes he clear first Ferris wheel ?If hmax is above second Ferris Wheel, by how much does he clear it?
2009 Physics 2111 Fundamentals of Physics Chapter 3 22
Uniform Circular Motion
The velocity is changing, but the speed is not.Hence, the particle is accelerating.
The acceleration is always directed radially inward.
Centripetal Acceleration:
r
va
2
v
rTperiod
2
The velocity is always tangent to the circular path.
Let’s prove this.
2009 Physics 2111 Fundamentals of Physics Chapter 3 23
Circular Motion
r
r
v
v
r
tv
v
v
r
v
t
va
2
2009 Physics 2111 Fundamentals of Physics Chapter 3 24
“Reference Frame B moves @ constant velocity wrt Reference Frame A”
Relative Motion in One Dimension
Differentiate wrt time
BAPBPA vvv
Differentiate wrt time
PBPA aa
2009 Physics 2111 Fundamentals of Physics Chapter 3 25
Relative Motion in 2 Dimensions
Two observers watching the motion of particle P from the origins of reference frames A and B, while B moves at constant velocity vba relative to A.
Observers in different reference frames that move at constant velocity relative to each other will measure the same acceleration.
BAPBPA vvv
BAPBPA rrr
PBPA aa
Differentiate wrt time
Differentiate wrt time
2009 Physics 2111 Fundamentals of Physics Chapter 3 26
Relative Motion
WGPWPG vvv
vpw = 215 km/h, south of east
vwg = 65.0 km/hr 200 E of N
What is vpg and
2009 Physics 2111 Fundamentals of Physics Chapter 3 27
Example
y – y0 = v0 sin 0 t - ½ g t2
22
128.960sin60200 ttm
smo
sm
081.4060.102 tt
storst 00.36.13
R = x – x0 = v0 cos 0 t
sR os
m 6.1360cos60
mR 408
2009 Physics 2111 Fundamentals of Physics Chapter 3 28
Example: How fast should she be going?
22
10 tgyy
22
128.95 tm
sm
st 01.1
tvxx 00
sm
s
m
t
xxv 8.23
01.1
2400
hmi
hkmv 53850
2009 Physics 2111 Fundamentals of Physics Chapter 3 29
Example: Can He Make the Jump?
x – x0 = v0 cos0 t
y – y0 = v0 sin 0 t - ½ g t2
0 = 0
v0
y0
y = 0
He can’t make it!
t = 0.990 s
x – x0 = 4.5 m