Motion in two and three dimensions
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Transcript Motion in two and three dimensions
Motion in two and three
dimensions
• We now apply the concepts on motion and
vectors which were previously reviewed and
introduced.
• As me move our discuss into two and three
dimensions we start by introducing the
concept of the position vector.
• As me move our discuss into two and three
dimensions we start by introducing the
concept of the position vector.
• The position vector gives the coordinate (the
position) of an object relative to the origin.
r xiˆ yˆj zkˆ
y-axis
(Position vector)
z-axis
y
x
z
x-axis
Example
y-axis
• For example,
4
z-axis
2
5
r 5iˆ 4 ˆj 2kˆ
x-axis
y-axis
• The
displacement
in
unit
vector
notation
is,
r r2 r1 x2 x1 iˆ y2 y1 ˆj z2 z1 kˆ
r2
z-axis
r
r1
x-axis
y-axis
• The
displacement
in
unit
vector
notation
is,
r r2 r1 x2 x1 iˆ y2 y1 ˆj z2 z1 kˆ
r2
r
r1
x-axis
z-axis
• Alternatively,
r xiˆ yˆj zkˆ
• Example: Determine the displacement of a car
which is initially at r1 3.0miˆ 2.0m ˆj 5.0mkˆ and
later at r2 9.0miˆ 2.0m ˆj 8.0mkˆ .
• Example: Determine the displacement of a car
which is initially at r1 3.0miˆ 2.0m ˆj 5.0mkˆ and
later at r2 9.0miˆ 2.0m ˆj 8.0mkˆ .
• Sol: r 9 (3)iˆ 2 2 ˆj 8 5kˆ
12miˆ 0 ˆj 3mkˆ
Average and instantaneous velocity
• Average velocity in unit vector notation,
x y ˆ z ˆ
r
i
j
k
vavg
t
t
t
t
• Instantaneous velocity,
dr dx dy ˆ dz ˆ
v
i
j k
dt
dt
dt
dt
• Recall: the direction of the velocity is at a tangent to the
position of the particle.
• Example: The coordinates of a spy satellite
orbiting the Earth are given by the position
function (x & y in km):
x 0.3t 2 7t 12
y 0.2t 2 10t 20
• Find the velocity and acceleration at time t 15 s
in unit vector notation.
x 0.3t 2 7t 12
• Sol:
v
dx dy ˆ
i
j
dt
dt
vx 0.6t 7
vy 0.4t 10
y 0.2t 2 10t 20
x 0.3t 2 7t 12
y 0.2t 2 10t 20
• Sol:
v
dx dy ˆ
i
j
dt
dt
vx 0.6t 7
vy 0.4t 10
vx 15 0.615 7 2ms1
vy 15 0.415 10 16ms1
Thus, v 2ms1 i 16ms1 ˆj
vx 0.6t 7
vy 0.4t 10
• Continuing,
dvx dvy ˆ
a
i
j
dt
dt
ax 0.6
a y 0.4
2
a 0.6ms
i 0.4ms2 ˆj
• In addition the velocity and acceleration in
magnitude angle format.
v v x2 v y2 22 162 260ms1
v
and tan1 y
v
x
83
16
tan1
2
Similarly,
a ax2 a y2 0.62 0.42 0.72ms2
a
and tan1 y
a
x
34
0.4
tan1
0.6
Projectile Motion
A special case of motion in two
dimensions
• A projectile is an object where the only force
acting on it is gravity (which acts downward).
• A projectile is an object where the only force
acting on it is gravity (which acts downward).
• Consider a particle which is release with an
initial velocity v0.
vy
vy
vx
v
v
vx
vx
vx
vy
v
• The horizontal and vertical motion is
independent.
• The horizontal and vertical motion is
independent.
• What this means is that we can convert the
two dimensional projectile problem into two
one dimensional problems.
• ie: One 1D equation for the x-direction and
another for the y-direction. These equations
can be solved separately!
• For our analysis we assume that there is no air
resistance.
• Consider the vertical motion:
2
1
• Recall: the equation of motion, y y0 v0t 2 at
• Through the motion,a g
g
2
1
y y0 v0 yt 2 gt
2
1
y
y
v
sin
t
gt
• Alt:
0
0
0
2
• Similarly, v y2 v0 sin 0 2 2 g y y0
+ve
direction
• Consider the horizontal motion:
• Again the general equation, x x0 v0t 12 at2
• However the acceleration in the x-dir is zero
and vx is constant.
+ve
2
direction
x x0 v0 xt 12 0t
g
• Thus, x x0 v0 xt
x x0 v0 cos0t
Other equations
• Equation of the path (trajectory):
gx2
y t an0 x
2
2v0 cos0
The trajectory gives the path of the projectile. For a given horizontal
distance we can calculate its height.
• Horizontal Range:
v02
R
sin 2 0
g
The range gives the maximum horizontal distance of the projectile for a given
initial velocity and launch angle.
Example
• A missile launcher on a cliff fires a missile at a
velocity of 25m/s at an angle of 53° above the
horizontal.
v
53
25m
Example
• A missile launcher on a cliff fires a missile at a
velocity of 25m/s at an angle of 53° above the
horizontal.
(a) The initial position of the object along
the x-axis is x0 0 .
v
53
+a
25m
g
Example
• A missile launcher on a cliff fires a missile at a
velocity of 25m/s at an angle of 53° above the
horizontal.
(a) The initial position of the object along
the x-axis is x0 0 .
v
53
+a
(b) The initial position of the object along
the y-axis is y0 0 .
25m
g
Example
• A missile launcher on a cliff fires a missile at a
velocity of 25m/s at an angle of 53° above the
horizontal.
(a) The initial position of the object along
the x-axis is x0 0 .
v
53
+a
(b) The initial position of the object along
the y-axis is y0 0 .
(c) Find the initial velocity along the xaxis and y-axis.
25m
g
Example
• A missile launcher on a cliff fires a missile at a
velocity of 25m/s at an angle of 53° above the
horizontal.
(a) The initial position of the object along
the x-axis is x0 0 .
v
53
+a
(b) The initial position of the object along
the y-axis is y0 0 .
(c) Find the initial velocity along the xaxis and y-axis.
Recall: v0 x v0 cos and v0 y v0 sin
v0 x 25cos53 15ms1
v0 y 25sin 53 20ms1
25m
g
Example
• A missile launcher on a cliff fires a missile at a
velocity of 25m/s at an angle of 53° above the
horizontal.
(d) Find the time at which it hits the
ground.
v
53
+a
25m
g
Example
• A missile launcher on a cliff fires a missile at a
velocity of 25m/s at an angle of 53° above the
horizontal.
(d) Find the time at which it hits the
ground.
v
53
+a
When the object hits the ground y=-25m.
Using, y y0 v0 yt 12 gt2
Then, 25 20t 12 9.81t 2
Solving gives us, t 5s and t 1s
The former is the solution as time can’t
be negative.
25m
g
Circular motion
• When an object moves in a circle or circular
arc at a constant speed we describe this
motion as uniform circular motion.
v
a
v
v
a
a
(a) The direction of the
velocity changes but not
its magnitude.
(b) The direction of the
acceleration is always
towards the centre.
• The acceleration is called the centripetal
acceleration.
v
a
v
v
a
a
(a) The direction of the
velocity changes but not
its magnitude.
(b) The direction of the
acceleration is always
towards the centre.
• For uniform motion:
v2
ˆ v2
ˆ
v2
which in unit vector notion is: a
j
a
cos
i
sin
r
r
r
and
T
2r
(The period)
v
v
a
v
v
a
a
Example
• A small object of mass m rotates anticlockwise
on a horizontal frictionless plane at the end of
a string of length r = 0.4 m with constant
speed v = 2.0 m/s. Draw the magnitude and
direction of (a) the velocity and (b) the
centripetal acceleration at points A, B, and C.
(c) Are the velocity and the acceleration
constant?
A
B
r
C
A
v
a
B
v
a
r
a
C
v
• The velocity is not constant.
• The magnitude of the acceleration is:
v2
a
r
2
2.0
a
0.4
10m s2
• The velocity is not constant.
• The magnitude of the acceleration is:
v2
a
r
2
2.0
a
0.4
10m s2
• The acceleration is not constant. The
magnitude is constant but the direction
changes.