Chapter 3 – 2D Motion
description
Transcript of Chapter 3 – 2D Motion
Chapter 3 – 2D MotionChapter 3 – 2D Motion
I.I. DefinitionsDefinitions
II.II. Projectile motionProjectile motion
Find the horizontal and vertical components of the d = 140 m displacement of a superhero who flies from the top of a tall building following the path shown in the Figure where = 30.0°.
Superperson
2 Dimensional Motion
• We will consider motion the the x-y plane.
• Positions now have (x,y) coordinates so we need to use vectors.
• There are two types of problems we need to consider– Throw or drop an object at an angle to the
horizontal– Make something go around in a circle
Position vector:Position vector: extends from the origin of a coordinate system to the particle. extends from the origin of a coordinate system to the particle.
jyixr ˆˆ
jyyixxrrr ˆ)(ˆ)( 121212
I.I. DefinitionsDefinitions
Average velocity:
jt
yit
x
t
rvavg ˆˆ
Displacement vector: Displacement vector: represents a particle’s position change during a certainrepresents a particle’s position change during a certain time interval.time interval.
Instantaneous velocity:Instantaneous velocity:
jdt
dyi
dt
dx
dt
rdkvjvivv zyx
ˆˆˆˆˆ
-The direction of the instantaneous velocity of a The direction of the instantaneous velocity of a particle is always tangent to the particle’s path at particle is always tangent to the particle’s path at the particle’s positionthe particle’s position
Instantaneous accelerationInstantaneous acceleration:
Average acceleration:Average acceleration:t
v
t
vvaavg
12
jdt
dvi
dt
dv
dt
vdjaiaa yx
yxˆˆˆˆ
Instantaneous velocity:Instantaneous velocity:
jdt
dyi
dt
dx
dt
rdjvivv yx
ˆˆˆˆ
-The direction of the instantaneous velocity of a The direction of the instantaneous velocity of a particle is always tangent to the particle’s path at particle is always tangent to the particle’s path at the particle’s positionthe particle’s position
Instantaneous accelerationInstantaneous acceleration:
Average acceleration:Average acceleration:t
v
t
vvaavg
12
jdt
dvi
dt
dv
dt
vdjaiaa yx
yxˆˆˆˆ
Projectile Motion
RANGE (R)
height (h)
What is its velocity here?It’s acceleration?How long did it take to get here?
here?
II. Projectile motionII. Projectile motion
Motion of a particle launched with initial velocity, vMotion of a particle launched with initial velocity, v00 and free fall acceleration and free fall acceleration
g.g.
- - Horizontal motion:Horizontal motion: aaxx=0 =0 v vxx=v=v0x0x= cte= cte
- - Vertical motion:Vertical motion: aayy= -g = -g
Range (R):Range (R): horizontal distance traveled by a horizontal distance traveled by a projectile before returning to launch height.projectile before returning to launch height.
tvtvxx x )cos( 0000
200
200 2
1)sin(
2
1gttvgttvyy y
gtvvy 00 sin
The horizontal and vertical motions are independent from each other.The horizontal and vertical motions are independent from each other.
- Trajectory: - Trajectory: projectile’s path.projectile’s path.
200
2
0
2
000000
00
)cos(2)(tan
cos2
1
cossin
cos
v
gxxy
v
xg
v
xvy
v
xt
000 yx
- Horizontal range: - Horizontal range: R = x-x R = x-x00; y-y; y-y00=0.=0.
0
202
000
022
0
2
0
2
000000
200
0000
2sincossin2
cos2
1tan
cos2
1
cos)sin(
2
1)sin(0
cos)cos(
g
vv
gR
v
RgR
v
Rg
v
Rvgttv
v
RttvR
(Maximum for a launch angle of 45º )
Overall assumption:Overall assumption: the air through which the projectile moves has no effect the air through which the projectile moves has no effect on its motion on its motion friction neglected. friction neglected.
In Galileo’s Two New Sciences, the author states that “for elevations (angles of In Galileo’s Two New Sciences, the author states that “for elevations (angles of projection) which exceed or fall short of 45projection) which exceed or fall short of 45º by equal amounts, the ranges are equal…” º by equal amounts, the ranges are equal…” Prove this statement.Prove this statement.
45
45
45
2
1
290sin452sin'
290sin452sin
20
20
20
20
g
v
g
vR
g
v
g
vR
x
v0
x=R=R’?
y
θ=45º
02sin: max0
20 hatdg
vRRange
bababa
bababa
sincoscossin)sin(
sincoscossin)sin(
)2cos()2sin(90cos)2cos(90sin'
)2cos()2sin(90cos)2cos(90sin
20
20
20
20
g
v
g
vR
g
v
g
vR