Download Chapter 4 – 2D and 3D Motion

Physics 2048
Spring 2008
Lecture #4 Chapter 4
motion in 2D and 3D
Chapter 4 – 2D and 3D Motion
I.
Definitions
II. Projectile motion
III. Uniform circular motion
IV. Relative motion
I. Definitions
Position vector: extends from the origin of a coordinate system to the particle.

r  xiˆ  yˆj  zkˆ
(4.1)
Displacement vector: represents a particle’s position change during a certain
time interval.
  
r  r2  r1  ( x2  x1 )iˆ  ( y2  y1 ) ˆj  ( z2  z1 )kˆ
Average velocity:


r x ˆ y ˆ z ˆ
vavg 

i
j
k
t t
t
t
(4.3)
(4.2)
Instantaneous velocity:


d
r
dx
dy ˆ dz ˆ
v  v xiˆ  v y ˆj  v z kˆ 
 iˆ 
j k
dt dt
dt
dt
(4.4)
-The direction of the instantaneous velocity of a
particle is always tangent to the particle’s path at
the particle’s position
 

v2  v1 v


t
t
(4.5)

dv y

dv
d
v
ˆj  dvz kˆ
a  a xiˆ  a y ˆj  a z kˆ 
 x iˆ 
dt
dt
dt
dt
(4.6)
Average acceleration: 
aavg
Instantaneous acceleration:
II. Projectile motion
Motion of a particle launched with initial velocity, v0 and free fall acceleration
g.
The horizontal and vertical motions are independent from each other.
- Horizontal motion: ax=0  vx=v0x= cte
x  x0  v0 xt  (v0 cos 0 )t
(4.7)
Range (R): horizontal distance traveled by a
projectile before returning to launch height.
- Vertical motion: ay= -g
y  y0  v0 y t 
1 2
1
gt  (v0 sin  0 )t  gt 2
2
2
v y  v0 sin  0  gt
(4.9)
(4.8)
v y  (v0 sin  0 ) 2  2 g ( y  y0 )
2
(4.10)
- Trajectory: projectile’s path.
x0  y0  0
2

x
x
1 
x

( 4.7)  ( 4.8)  t 
 y  v0 sin  0
 g
 
v0 cos  0
v0 cos  0
2 
v
cos

0 
 0
gx 2
y  (tan  0 ) x 
2(v0 cos  0 ) 2
( 4.11)
- Horizontal range: R = x-x0; y-y0=0.
R  (v0 cos  0 )t  t 
R
v0 cos  0
2

1 2
R
1 
R
1
R2
  R tan  0  g 2
0  (v0 sin  0 )t  gt  (v0 sin  0 )
 g 

2
v0 cos  0 2  v0 cos  0 
2 v0 cos 2  0
2 sin  0 cos  0 2 v02
R
v0  sin 2 0
g
g
(4.12) (Maximum for a launch angle of 45º )
Overall assumption: the air through which the projectile moves has no effect
on its motion  friction neglected.
## In Galileo’s Two New Sciences, the author states that “for elevations (angles of projection) which exceed
or fall short of 45º by equal amounts, the ranges are equal…” Prove this statement.
y
  45
1  45  
v
v02
Range : R  sin 2 0  d max at h  0
g
 2  45  
0
θ=45º
x=R=R’?
x








v02
v02

R  sin 2 45    sin 90  2
g
g
v02
v02

R '  sin 2 45    sin 90  2
g
g
sin( a  b)  sin a cos b  cos a sin b
sin( a  b)  sin a cos b  cos a sin b




v02
v02


R
sin 90 cos( 2 )  cos 90 sin( 2 )  cos( 2 )
g
g
v02
v02


R' 
sin 90 cos( 2 )  cos 90 sin( 2 )  cos( 2 )
g
g
III. Uniform circular motion
Motion around a circle at constant speed.
Magnitude of velocity and acceleration constant.
Direction varies continuously.
-Velocity: tangent to circle in the direction of motion.
v2
a
r
- Acceleration: centripetal
- Period of revolution: T  2r
v
v0y
v0x
(4.13)
(4.14)
  v  y p ˆ  v  x p  ˆ

i  
 j
v  v x iˆ  v y ˆj  (v sin  )iˆ  (v cos  ) ˆj  
r
r

 


   v2

 dv   v dy p  ˆ  v dx p  ˆ   v  ˆ  v  ˆ   v 2
ˆ



 ˆj




a

i

j

v
i

v
j

cos

i

sin





y
x







dt  r dt   r dt 
 r
 r 
 r
  r

v2
v2
2
2
a

cos   sin  
r
r
a y sin 

a directed along radius  tan  

 tan 
a x cos 
a x2
 a 2y
1- A cat rides a merry-go-round while turning with uniform circular motion. At time t1= 2s, the cat’s velocity
is: v1= (3m/s)i+(4m/s)j, measured on an horizontal xy coordinate system. At time t2=5s its velocity is:
v2= (-3m/s)i+(-4m/s)j. What are (a) the magnitude of the cat’s centripetal acceleration and (b) the cat’s
average acceleration during the time interval t2-t1?
v
1
x
In 3s the velocity is reversed  the cat reaches the opposite
side of the circle
v2
y
v  32  42  5m / s
2r
r
 3s 
 r  4.77m
v
5m / s
v 2 25m 2 / s 2
ac 

 5.23m / s 2
r
4.77m
T
 

v2  v1 (6m / s)iˆ  (8m / s) ˆj
aavg 

 (2m / s 2 )iˆ  (2.67m / s 2 ) ˆj
t
3s

aavg  3.33m / s 2
IV. Relative motion
Particle’s velocity depends on reference frame
vPA  vPB  vBA
(4.15)
1D
Frame moves at constant velocity
0
d
d
d
(vPA )  (vPB )  (vBA )  aPA  aPB
dt
dt
dt
(4.16)
Observers on different frames of reference measure the same acceleration
for a moving particle if their relative velocity is constant.