Chapter three
Motion in two dimension
Chapter Three
Motion in two dimension
3-1 The Displacement, velocity and acceleration
A particle moving in xy-plane is located with the position vector r
drawn from the origin to the particle. The displacement of the particle as
it moves from A to B in the time interval.
Dt = t - t is equal to the vector
f i
Dr = r - r
f i
y
ti
A
The instantaneous velocity v is given by:
Dr dr
=
Dt dt
Dt ®0
v = Lim
ri
rf
path of particle
Dr
B tf
x
The instantaneous acceleration a is :
Dv dv
=
Dt dt
Dt ®0
a = Lim
3-2 Two- Dimension motion with constant acceleration
The position vector for particle moving in the xy-plane can be
written:
r = x î + y ĵ
The velocity of the particle can be obtained from:
v=
dr d
dx dy
= ( x î + y ĵ ) =
î +
ĵ
dt dt
dt
dt
Chapter three
Motion in two dimension
v = v x î + v y ˆj
We can apple the eq's of motion to the x and y component of the
velocity vector, substituting
vx f = vxi + a x t
, vyf = vyi + a y t
We obtain the final velocity at any time t:
v f = ( v x i + a x t )î + ( v y i + a y ) ĵ
= ( v x i î + v y i ĵ ) + ( a x î + a y ĵ ) t
v f = vi + a t
Similarly, from equation x f = x i + v x i t +
1 2
a x t . We know that the x
2
and y coordinates of a particle moving with constant acceleration are:
1
1
x f = x i + v x i t + a x t 2 , yf = yi + v y i t + a y t 2
2
2
Obtain:
1
1
rf = ( x i + v x i t + a x t 2 ) î +( y i + v y i t + a y t 2 ) ĵ
2
2
1
2
= ( x i î + y i ĵ )+ ( v x i î + v y i ĵ ) t + ( a x î + a y ĵ ) t
2
1
rf = ri + v i t + a t 2
2
This equation tells us that the displacement vector Dr = rf + ri is the
vector sum of a displacement v i t arising from the initial velocity of the
particle and a displacement
acceleration of particle
1 2
a x t resulting from the uniform
2
Chapter three
Motion in two dimension
3-3 Projectile Motion
vi
vy
v
Vy=0
vxi C
D
B
vy i
h
vxi
v
vy
qi
A
vxi
R
E
The two assumption with projectile motion are:
(1) The free fell acceleration g is constant over the range of motion
and is direction downward.
(2) The effect of air resistance is negligible.
The path of projectile is a parabola.
At t=0, the projectile leaves the origin ( x i = y i = 0 ) with speed v i and
makes an angle qi with the horizontal
\ cos qi =
sin qi =
vxi
Þ v x i = v i cos qi .................(1)
vi
vyi
Þ v y i = v i sin qi
vi
...............( 2)
1
Q xf = xi + vxit + a x t 2
2
................(3)
1
yf = yi + v y i t + a y t 2
2
................( 4)
Substituting eq.(1) into eq.(3) with x i = 0 and a x = 0 , we find that:
x f = v x i t = ( v i cos qi ) t
..................( 5)
Chapter three
Substituting
Motion in two dimension
eq.(2) into eq.(4) and using y = 0 and a y = -g , we
obtain:
1
y f = ( v i sin qi ) t - g t 2
2
....................(6)
From eq.(5)obtain:
x f = v i cos qi t Þ t =
xf
v i cos qi
..................( 7)
Now substitute eq.(7) into eq.(6),this gives:
æ v i sin qi ö
ö
1 æ
x2
÷÷ x - g çç 2
÷÷ ................(8)
y = çç
2
v
cos
q
2
v
cos
q
è i
è i
i ø
i ø
æ
ö 2
g
÷÷ x
y = (tan qi ) x - çç 2
2
2
v
cos
q
è i
i ø
Where 0 < q <
................(9)
p
2
This equation is equivalent to parabola form y = ax - bx
2
From the expression above, we see that the projectile motion is the
superposition of two motion:
(1) Constant velocity motion in the horizontal direction.
(2) Free fall motion in the vertical direction.
From the diagram, the peak point c, which has Cartesian coordinates
(
R
, h ), and the point E, which has coordinates (R,0).
2
Chapter three
Motion in two dimension
The distance (R) is called the horizontal range of the projectile, and (h)
it's maximum height.
At the peak, v y 0 = 0 , then
Use v y f = v y i + a y t obtain:
0 = v i sin q i - g t c Þ t c =
v i sin qi
g
..............(10)
Substituting eq.(10) into eq.(4) and replacing y f = y c with (h) , we
obtain:
h = ( v i sin qi )
v i sin qi 1 v i sin qi 2
- g(
)
g
2
g
v i2 sin 2 q
h=
2g
.......................(11)
Now, use eq.(3) and setting R @ x f at t = t c , we find
R = ( v i cos qi ) 2t c = ( v i cos qi )
\R =
v 2 sin 2q
i
g
2 v i sin qi
g
.......................(12)
The maximum value of R from eq.(12) is
R max
v i2
=
g
where sin 2q = 1 when 2q = 90 0 therefore R
is max imum when qi = 450.