### 04_Kinema2D

```NEWTON’S LAWS
PHY1012F
KINEMATICS IN 2D
Gregor Leigh
[email protected]
MOTION IN A PLANE
NEWTON’S LAWS
PHY1012F
MOTION IN A PLANE
MOTION IN A PLANE
Learning outcomes:
At the end of this chapter you should be able to…
Extend the understanding, skills and problem-solving
strategies developed for kinematics problems in one
dimension to two dimensional situations.
Define the terms specific to projectile motion;
solve numerical problems involving projectile motion.
2
PHY1012F
NEWTON’S LAWS
MOTION IN A PLANE
KINEMATICS IN TWO DIMENSIONS
We shall apply vectors to the study of motion in 2-d.
How the position of a body, r , changes with time (t) is
determined by its acceleration, a , and depends on the
body’s initial position, r0 s , and initial velocity, v 0 s .
We use vectors and their components to represent these
directional quantities.
In this chapter we shall concentrate on motion in which
the x- and y-components are independent of each other.
Later (in circular motion) we shall investigate motion in
which the x- and y-components are not independent .
3
PHY1012F
NEWTON’S LAWS
MOTION IN A PLANE
KINEMATICS IN TWO DIMENSIONS
In one dimensional (1-d) motion…
Non-zero a and v can only be either…
parallel (body is speeding up), or
antiparallel (body is slowing down).
Or, in terms of components, vs and as can only either…
have the same sign (body is speeding up), or
have opposite signs (body is slowing down).
4
NEWTON’S LAWS
PHY1012F
MOTION IN A PLANE
KINEMATICS IN TWO DIMENSIONS
In two dimensional (2-d) motion…
v
a  0
a
0
a
a
can have
a
components
both parallel to
and perpendicular to v .
a
a
a
a
a
a
The parallel components alter the body’s speed;
The perpendicular components alter the body’s direction.
5
NEWTON’S LAWS
PHY1012F
MOTION IN A PLANE
POSITION and DISPLACEMENT IN 2-D
The curved path followed by an
object moving in a (2-d) plane is
called its trajectory.
Position vectors can be written:
r1  x 1 ˆi  y 1 ˆj
and r2  x 2 ˆi  y 2 ˆj
y
(x1, y1)
x
r
r1
r2
y
(x2, y2)
and the displacement vector as:
 r  r2  r1 
 x2 
x 1  ˆi 
 y2 
x
y 1  ˆj
i.e.  r   x ˆi   yˆj
WARNING: Distinguish carefully between y-vs-x graphs
(actual trajectories) and position graphs (x, or y-vs-t)!
6
NEWTON’S LAWS
PHY1012F
MOTION IN A PLANE
VELOCITY IN TWO DIMENSIONS
Average velocity is given by:
y ˆ
v a v g   r   x ˆi 
j
t
t
t
Instantaneous velocity by:
y
v
vy
dy ˆ
j
v  l i m  r  d r  d x ˆi 
d
t
d
t
dt
t 0  t
r

r
r
v x x
y
x
As seen in the diagram, as t0…
 r , and thus also v becomes tangent to the trajectory.
Since v  v x ˆi  v y ˆj , it follows that
v x  dx
dt
and
vy 
dy
dt
.
Motion in 2-d may be understood as the vector sum of two
simultaneous motions along the x- and y-axes.
7
NEWTON’S LAWS
PHY1012F
MOTION IN A PLANE
VELOCITY IN TWO DIMENSIONS
If v ’s angle is measured
relative to the positive x-axis,
its components are:
vx
and
 d x  v co s 
dt
y
v
vy

vx
x
dy
vy 
 v sin 
dt
where v 
2
vx  vy
2
is the body’s speed at that point.
Conversely, the direction of v is given by   tan
1
 vy 
v 
 x 
8
NEWTON’S LAWS
PHY1012F
MOTION IN A PLANE
VELOCITY IN TWO DIMENSIONS
Once again, distinguish carefully between position
graphs and trajectories…
s
y
v
t
Position graph:
Tangent gives the
magnitude of v .
x
Trajectory:
Tangent gives the
direction of v .
9
NEWTON’S LAWS
PHY1012F
MOTION IN A PLANE
A particle’s motion is described by the two equations:
x = 2t2 m and y = (5t + 5) m, where time t is in seconds.
(a) Draw the particle’s trajectory.
y (m)
t (s)
x (m)
y (m)
30
0
0
5
1
2
10
2
8
15
3
18
20
4
32
25
20
10
x (m)
0
0
10
20
30
10
NEWTON’S LAWS
PHY1012F
MOTION IN A PLANE
A particle’s motion is described by the two equations:
x = 2t2 m and y = (5t + 5) m, where time t is in seconds.
(b) Draw a speed-vs-time graph for the particle.
v x  d x  4t m / s
dt
2
2
dy
and v y 
 5 m /s
dt
2
v 
vx  vy 
16 t  25 m /s
t (s)
v (m/s)
0
5.0
16
1
6.4
12
2
9.4
3
13.0
8
4
4
16.8
0
v (m/s)
0
1
2
3
4
t (s)
11
NEWTON’S LAWS
PHY1012F
MOTION IN A PLANE
ACCELERATION IN TWO DIMENSIONS
a avg   v
t
where  v  v 2  v 1 is the
change in instantaneous
velocity during the interval t.
As we approach the limit
t0…
y
v1 v
v2
a
v2
v2
v
 v1
a  lim  v  d v
dt
t 0  t
x
…the instantaneous acceleration is found at the same
point on the trajectory as the instantaneous velocity.
12
PHY1012F
NEWTON’S LAWS
MOTION IN A PLANE
ACCELERATION IN TWO DIMENSIONS
Instantaneous acceleration
can be resolved…
We can resolve it into
components parallel to and
perpendicular to the
instantaneous velocity…
y
v
a
a
a
Where…
a alters the speed, and
a alters the direction.
x
a and a, however, are constantly changing direction,
so it is more practical to resolve the acceleration into…
13
NEWTON’S LAWS
PHY1012F
MOTION IN A PLANE
ACCELERATION IN TWO DIMENSIONS
x- and y-components…
y
v
a  a x ˆi  a y ˆj
ax
dv x ˆ dv y ˆ
d
v
a 

i
j
dt
dt
dt
Hence a x 
dv x
dt
and a y 
ay
dv y
dt
.
a
x
We now have the following parametric equations:
vfx = vix + axt
vfy = viy + ayt
xf = xi + vixt + ½ax(t)2
yf = yi + viyt + ½ay(t)2
14
NEWTON’S LAWS
PHY1012F
MOTION IN A PLANE
ACCELERATION IN ONE DIRECTION ONLY
Let us consider a special case in the xy-plane, in which a
particle experiences acceleration in only one direction…
xf = xi + vixt + ½ax(t)2
yf = yi + viyt + ½ay(t)2
vfx = vix + axt
vfy = viy + ayt
y
Letting ax = 0, and starting at the
origin at t = 0 with v0 making an angle
of  with the x-axis, we get:
vx = v0 cos
vy = v0 sin + ayt
x = v0 cos t
y = v0 sin t + ½ayt2
v0
v0y

v0x
x
15
NEWTON’S LAWS
PHY1012F
MOTION IN A PLANE
ACCELERATION IN ONE DIRECTION ONLY
Since the equations are parametric, we can eliminate t:
x = v0 cos t
t 
Substituting in
we get
i.e.
x
v 0 cos 
y = v0 sin t + ½ayt2,
y  v 0 sin 
x

v 0 cos 
y  x tan  
ayx

1
x


a

2 y
v
cos

 0

2
2
2 v 0 cos 

2
Any object for which one component of the acceleration
is zero while the other has a constant non-zero value
follows a parabolic path.
16
PHY1012F
NEWTON’S LAWS
MOTION IN A PLANE
A particle with an initial velocity of v  5.00 ˆi m /s experiences
a constant acceleration a   1.63 ˆj m /s 2.
(a) Draw a physical representation of the particle’s motion.
y (m)
v
a
x (m)
17
NEWTON’S LAWS
PHY1012F
MOTION IN A PLANE
A particle with an initial velocity of v  5.00 ˆi m /s experiences
a constant acceleration a   1.63 ˆj m /s 2.
(b) Draw the particle’s trajectory.
y  x tan  
ayx
2

2 v 0 cos 

2
y (m)
0
 y   1 .6 3 x
50
2
x (m)
y (m)
–0.1
0
0.000
–0.2
1
–0.033
–0.3
2
–0.130
–0.4
3
–0.293
–0.5
4
–0.522
1
2
3
4
x (m)
–0.6
18
PHY1012F
NEWTON’S LAWS
MOTION IN A PLANE
PROJECTILE MOTION
A projectile is an object upon which the only force acting
is gravity. I.e. a projectile is an object in free fall.
The start of a projectile’s motion is called its launch.
The angle above the x-axis at which a projectile is
launched is called the launch angle, or elevation.
vix = vi cos , viy = vi sin, and ay = –g.
y
vi  0, although vix and viy may be < 0.
The horizontal distance travelled by
a projectile before it returns to its
original height is called its range, R.
vi
v iy

v ix
R
x
19
NEWTON’S LAWS
PHY1012F
MOTION IN A PLANE
PROJECTILE MOTION
Since ax = 0, and ay = –g , the kinematic equations for
projectile motion are:
xf = xi + vix t
yf = yi + viy t – ½g (t)2
vfx = vix = constant
vfy = viy – g t
Notes:
t is the same for the x- and y-components.
We determine t using one component and then
use that value for the other component.
Although the x- and y-components are linked in this
way, the motion in one direction is completely
independent of the motion in the other direction.
20
NEWTON’S LAWS
PHY1012F
MOTION IN A PLANE
RANGE
y
The horizontal distance travelled by
a projectile before it returns to its
original height is called its range.
v0
v0y

v0x
i.e. R = x – x0 if and only if y – y0 = 0
x
R
Starting at the origin at t0 = 0,
y = v0 sin t – ½gt2 = 0
 t (v0 sin – ½gt) = 0
R = x = v0 cos t  v 0 cos 
R 
2
v 0 sin  2 
 t = 0 (trivial) or t 
2 v 0 sin 
g
v 0 2 sin  cos 
2

2 v 0 sin 
g
g
g
21
NEWTON’S LAWS
PHY1012F
RANGE
R 
2
v 0 sin  2 
g
MOTION IN A PLANE
y
v0
80°
70°
50°
40°
20°
10°

R
Notes:
x
R is a maximum when sin(2) = 1
i.e. 2 = 90° i.e.  = 45° for maximum range.
Since sin(180° – ) = sin, it follows that
sin(2(90° – )) = sin(2)…
I.e. a projectile fired at an elevation of (90° – ) will
have the same range as a projectile launched at
angle .
22
NEWTON’S LAWS
PHY1012F
MOTION IN A PLANE
A small plane flying at 50 m/s, 60 m above the ground, comes
up behind a bakkie travelling in the same direction at 30 m/s
and drops a package into it. At what angle to the horizontal
should the “bomb sights” (a straight sighting tube) be set?
vP
a yP
vB
23
NEWTON’S LAWS
PHY1012F
MOTION IN A PLANE
A small plane flying at 50 m/s, 60 m above the ground, comes
up behind a bakkie travelling in the same direction at 30 m/s
and drops a package into it. At what angle to the horizontal
should the “bomb sights” (a straight sighting tube) be set?
y
x0P, y0P, t0
vxP, v0yP

x1P, y1P, t1
vxP, v1yP
ayP

x0B, t0, vxB
x0P = t0 = v0yP = 0
y0P = +60 m
vxP = +50 m/s
ayP = –g = –9.8 m/s2
x
x1B, t1, vxB
y1P = 0 m
x0B = ?
vxB = +30 m/s
x1B = x1P = ?
t1 = t = ?
24
NEWTON’S LAWS
PHY1012F
MOTION IN A PLANE
A small plane flying at 50 m/s, 60 m above the ground, comes
up behind a bakkie travelling in the same direction at 30 m/s
and drops a package into it. At what angle to the horizontal
should the “bomb sights” (a straight sighting tube) be set?
y1P = y0P + v0yPt – ½gt2
0 = 60 + 0 – ½  9.8 t2
x1P = x0P + v0xPt + ½axPt2 = 0 + 50  3.5 + 0
 t = 3.5 s
 x1P = x1B = 175 m
x1B = x0B + v0xBt + ½axBt2 175 = x0B + 30  3.5 + 0  x0B = 70 m
y0P
  tan

60 m
1
 
60
70
  = 40.6°

70 m
x0B
25
PHY1012F
NEWTON’S LAWS
MOTION IN A PLANE
IN REALITY…
Air resistance does affect the motion of
projectiles, destroying the symmetry of their
trajectories. The less dense the projectile,
the more noticeable the drag effect.
y
x
Although vertical speed is independent of horizontal speed, the
action of running helps to increase the initial vertical component
of an athlete’s jump speed.
If the landing is lower than the launch height, the “range equation”
is not valid. Here, angles less than 45° produce longer ranges.
When projecting heavy objects (e.g. shot puts), the greater the
elevation, the greater the force expended against gravity, and
the slower the launch speed. Thus maximum range for heavy
projectiles thrown by humans is attained for angles less than 45°.
26
NEWTON’S LAWS
PHY1012F
MOTION IN A PLANE
HOW FAR WILL IT GO?
Any projectile, irrespective of mass, launch angle or launch
speed, loses 5t2 m of height every second in free fall (from rest).
5m
trajectory
without gravity
5m
1s
1s
20 m
20 m
5m
20 m
2s
1s
2s
2s
So, no matter how great a projectile’s initial horizontal
speed, it must eventually hit the ground…
27
PHY1012F
NEWTON’S LAWS
MOTION IN A PLANE
HOW FAR WILL IT GO?
…except that the Earth is not flat!
Because of its curvature,
the Earth’s surface drops
a vertical distance of 5 m
every 8 000 m tangent to the surface.
8 000 m
5m
So theoretically, in the absence of air resistance, tall
buildings, etc, an object projected horizontally at 8 000 m/s
at a height of, say, 1 m will have the Earth’s surface
dropping away beneath it at the same rate it falls and will
consequently get no closer to the ground…
The object will be a satellite, in orbit around the Earth!
28
NEWTON’S LAWS
PHY1012F
MOTION IN A PLANE
MOTION IN A PLANE
Learning outcomes:
At the end of this chapter you should be able to…
Extend the understanding, skills and problem-solving
strategies developed for kinematics problems in one
dimension to two dimensional situations.
Define the terms specific to projectile motion;
solve numerical problems involving projectile motion.
29
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