Physics 111: Mechanics
Lecture 3
Dale Gary
NJIT Physics Department
Motion in Two Dimensions
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Reminder of vectors and vector algebra
Displacement and position in 2-D
Average and instantaneous velocity in 2-D
Average and instantaneous acceleration in 2-D
Projectile motion
Uniform circular motion
Relative velocity*
February 5-8, 2013
Vector and its components

 

A  Ax  Ay
The components are the
legs of the right triangle
whose hypotenuse is A
 Ax  A cos( )
 A  Ax2  Ay2
 Ay  A sin(  )
and

 A   A 2  A 2
x
y


Ay
Ay 
1 
or   tan  
tan   
Ax

 Ax 
 Ay 
  tan  
 Ax 
1
Or,
February 5-8, 2013
Vector Algebra
  
 Which diagram can represent
r  r2  r1 ?


r
r

A) 
B)
r1
C)

r1
r1

r2

r

r2

r2
 r1

r
D) 
r1

r2
February 5-8, 2013
Motion in two dimensions


Kinematic variables
 Position:
 Velocity:
 Acceleration:
in one dimension
x(t) m
v(t) m/s
a(t) m/s2
x
Kinematic variables in three dimensions

r (t )  xiˆ  yˆj  zkˆ m
 Position:

v (t )  vxiˆ  v y ˆj  vz kˆ m/s
 Velocity:

 Acceleration: a (t )  a x iˆ  a y ˆj  a z kˆ m/s2
y
j
i
x

All are vectors: have direction and
magnitudes
k
z
February 5-8, 2013
Position and Displacement

In one dimension
x  x2 (t2 )  x1 (t1 )
x1 (t1) = - 3.0 m, x2 (t2) = + 1.0 m
Δx = +1.0 m + 3.0 m = +4.0 m

  
r  r2  r1
In two dimensions
 Position: the position of an object is

described by its position vector r (t )
--always points to particle from origin.
  
 Displacement: r  r  r
2
1

r  ( x2iˆ  y2 ˆj )  ( x1iˆ  y1 ˆj )
 ( x2  x1 )iˆ  ( y2  y1 ) ˆj
 xiˆ  yˆj
February 5-8, 2013
Average & Instantaneous Velocity



r
 Average velocity v 
avg
t

x ˆ y ˆ
vavg 
i
j  vavg , x iˆ  vavg , y ˆj
t
t

Instantaneous velocity




r dr
v  lim vavg  lim

t 0
t 0 t
dt

 dr dx ˆ dy ˆ
v
 i
j  v x iˆ  v y ˆj
dt dt
dt

v is tangent to the path in x-y graph;
February 5-8, 2013
Motion of a Turtle
A turtle starts at the origin and moves with the speed of v0=10 cm/s in
the direction of 25° to the horizontal.
(a) Find the coordinates of a turtle 10 seconds later.
(b) How far did the turtle walk in 10 seconds?
February 5-8, 2013
Motion of a Turtle
Notice, you can solve the
equations independently for the
horizontal (x) and vertical (y)
components of motion and then
combine them!
  
v0  v x  v y



X components:
v0 x  v0 cos 25  9.06 cm/s
x  v0 x t  90.6 cm
v0 y  v0 sin 25  4.23 cm/s
y  v0 y t  42.3 cm
Y components:
Distance from the origin:
d  x 2  y 2  100.0 cm
February 5-8, 2013
Average & Instantaneous Acceleration
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Average acceleration


v
aavg 
t
v y

v
ˆj  aavg , xiˆ  aavg , y ˆj
aavg  x iˆ 
t
t

Instantaneous acceleration




v dv
a  lim aavg  lim

t 0
t 0 t
dt

 dv dvx ˆ dv y ˆ
a

i
j  a xiˆ  a y ˆj
dt
dt
dt
The magnitude of the velocity (the speed) can change
 The direction of the velocity can change, even though the
magnitude is constant
 Both the magnitude and the direction can change

February 5-8, 2013
Summary in two dimension

Position

r (t )  xiˆ  yˆj


r x ˆ y ˆ
v 

i
j  vavg , x iˆ  vavg , y ˆj
 Average velocity avg
t t
t
dx
dy
v

v

x
y
 Instantaneous velocity
dt
dt



r dr dx ˆ dy ˆ
v (t )  lim

 i
j  vx iˆ  v y ˆj
t 0 t
dt dt
dt

Acceleration
dv x d 2 x
ax 
 2
dt
dt
d2y
ay 
 2
dt
dt
dv y



v dv dvx ˆ dv y ˆ
a (t )  lim


i
j  a xiˆ  a y ˆj
t 0 t
dt
dt
dt




r (t), v (t ), and a (t ) are not necessarily same direction.
February 5-8, 2013
Motion in two dimensions
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Motions in each dimension are independent components
Constant acceleration equations
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Constant acceleration equations hold in each dimension
  
v  v0  at
vx  v0 x  axt
v y  v0 y  a y t
x  x0  v0 x t  12 a x t 2
y  y0  v0 yt  12 a yt 2
vx  v0 x  2ax ( x  x0 )
v y  v0 y  2a y ( y  y0 )
2



  
2
1
r  r  v0t  2 at
2
2
2
t = 0 beginning of the process;

a  a xiˆ  a y ˆj where ax and ay are constant;
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Initial velocity v0  v0 xiˆ  v0 y ˆj initial displacement r0  x0iˆ  y0 ˆj ;
February 5-8, 2013
Hints for solving problems
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Define coordinate system. Make sketch showing axes, origin.
List known quantities. Find v0x , v0y , ax , ay , etc. Show initial
conditions on sketch.
List equations of motion to see which ones to use.
Time t is the same for x and y directions.
x0 = x(t = 0), y0 = y(t = 0), v0x = vx(t = 0), v0y = vy(t = 0).
Have an axis point along the direction of a if it is constant.
vx  v0 x  axt
v y  v0 y  a y t
x  x0  v0 x t  12 a x t 2
y  y0  v0 yt  12 a yt 2
vx  v0 x  2ax ( x  x0 )
v y  v0 y  2a y ( y  y0 )
2
2
2
2
February 5-8, 2013
Projectile Motion
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2-D problem and define a coordinate
system: x- horizontal, y- vertical (up +)
Try to pick x0 = 0, y0 = 0 at t = 0
Horizontal motion + Vertical motion
Horizontal: ax = 0 , constant velocity motion
Vertical:
ay = -g = -9.8 m/s2, v0y = 0
Equations:
Horizontal
Vertical
vx  v0 x  axt
v y  v0 y  a y t
x  x0  v0 x t  12 a x t 2
y  y0  v0 yt  12 a yt 2
y f  yi  viyt  12 gt 2
vx  v0 x  2ax ( x  x0 ) v y 2  v0 y 2  2a y ( y  y0 )
2
2
February 5-8, 2013
Projectile Motion

X and Y motions happen independently, so
we can treat them separately
vx  v0 x
v y  v0 y  gt
x  x0  v0 xt
y  y0  v0 yt  12 gt 2
Horizontal

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Vertical
Try to pick x0 = 0, y0 = 0 at t = 0
Horizontal motion + Vertical motion
Horizontal: ax = 0 , constant velocity motion
Vertical:
ay = -g = -9.8 m/s2
x and y are connected by time t
y(x) is a parabola
February 5-8, 2013
Projectile Motion
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2-D problem and define a coordinate system.
Horizontal: ax = 0 and vertical: ay = -g.
Try to pick x0 = 0, y0 = 0 at t = 0.
Velocity initial conditions:




v0 can have x, y components.
v0x is constant usually. v0 x  v0 cos  0
v0y changes continuously. v0 x  v0 sin  0
Equations:
Horizontal
Vertical
vx  v0 x
v y  v0 y  gt
x  x0  v0 xt
y  y0  v0 y t  12 gt 2
February 5-8, 2013
Trajectory of Projectile Motion
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
Initial conditions (t = 0): x0 = 0, y0 = 0
v0x = v0 cosθ0 and v0y = v0 sinθ0
Horizontal motion:
x  0  v0 x t


t
Vertical motion:
x
v0 x
y  0  v0 y t  12 gt 2
 x  g x 
  

y  v0 y 
 v0 x  2  v0 x 
y  x tan  0 

2
g
2
x
2
2v0 cos 2  0
Parabola;

θ0 = 0 and θ0 = 90 ?
February 5-8, 2013
What is R and h ?

Initial conditions (t = 0): x0 = 0, y0 = 0
v0x = v0 cosθ0 and v0x = v0 sinθ0, then
0  0  v0 yt  12 gt 2
x  0  v0 xt
t
2v0 y
g

h
2v0 sin  0
g
2v cos  0v0 sin  0 v0 sin 2 0
R  x  x0  v0 xt  0

g
g
2
t
g
t


2
h  y  y0  v0 y t h  12 gt h  v0 y   
2 2 2
2
v0 sin 2  0
h
2g
2
v y  v0 y  gt  v0 y  g
2v0 y
g
 v0 y
Horizontal
Vertical
vx  v0 x
v y  v0 y  gt
x  x0  v0 xt
y  y0  v0 y t  12 gt 2
February 5-8, 2013
Projectile Motion
at Various Initial Angles

Complementary
values of the initial
angle result in the
same range


v0 sin 2
R
g
2
The heights will be
different
The maximum range
occurs at a projection
angle of 45o
February 5-8, 2013
Uniform circular motion
Constant speed, or,
constant magnitude of velocity
Motion along a circle:
Changing direction of velocity
February 5-8, 2013
Circular Motion: Observations

Object moving along a
curved path with constant
speed






Magnitude of velocity: same
Direction of velocity: changing
Velocity: changing
Acceleration is NOT zero!
Net force acting on the
object is NOT zero
“Centripetal force”


Fnet  ma
February 5-8, 2013
Uniform Circular Motion

Centripetal acceleration
v r
vr

so, v 
v
r
r
v r v v 2


t t r r
v v 2
ar 

t
r

vi
A
vf
vi
Δv = vf - vi
y
Δr
ri
B
vf
R
rf
O
Direction: Centripetal
February 5-8, 2013
x
Uniform Circular Motion

Velocity:



ac 
v2
r
Acceleration:



Magnitude: constant v
The direction of the velocity is
tangent to the circle


ac  v
v2
ac 
Magnitude:
r
directed toward the center of
the circle of motion
Period:

time interval required for one
complete revolution of the
2r
T
particle
v
February 5-8, 2013
Summary

Position

r (t )  xiˆ  yˆj


r x ˆ y ˆ
v 

i
j  vavg , x iˆ  vavg , y ˆj
 Average velocity avg
t t
t
dx
dy
v

v

x
y
 Instantaneous velocity
dt
dt



r dr dx ˆ dy ˆ
v (t )  lim

 i
j  vx iˆ  v y ˆj
t 0 t
dt dt
dt

Acceleration
dv x d 2 x
ax 
 2
dt
dt
d2y
ay 
 2
dt
dt
dv y



v dv dvx ˆ dv y ˆ
a (t )  lim


i
j  a xiˆ  a y ˆj
t 0 t
dt
dt
dt




r (t), v (t ), and a (t ) are not necessarily in the same direction.
February 5-8, 2013
Summary

If a particle moves with constant acceleration a, motion
equations are
  
2
rf  ri  vit  12 at

rf  x f iˆ  y f ˆj  ( xi  vxi t  12 axi t 2 )iˆ  ( yi  v yi t  12 a yi t 2 ) ˆj
  
v  vi  at

v f (t )  v fxiˆ  v fy ˆj  (vix  axt )iˆ  (viy  a y t ) ˆj

Projectile motion is one type of 2-D motion under constant
acceleration, where ax = 0, ay = -g.
February 5-8, 2013