PHY2048
9/8/08
Position, velocity, acceleration vs. time plots (demo).
What causes acceleration?
Projectile motion
0
0
time (s)
time (s)
position (m)
time (s)
x(t) = x O + vt
velocity (m/s)
0
x(t) = x O = constant
xO
Constant velocity (wrt coord. syst.):
accel. (m/s2)
accel. (m/s2)
velocity (m/s)
position (m)
Object at rest
(wrt coord. syst.):
0
0
xO
time (s)
v(t) = v o = (constant)
time (s)
a=
0
dx
= v = (constant)
dt
dv
=0
dt
time (s)
accel. (m/s2)
velocity (m/s)
position (m)
Constant acceleration
0
dx(t)
= v(t) = (not constant) x(t) = x O + v O t + 12 at 2
x O dt
time (s)
dv(t) d 2 x(t)
=
a(t) =
dt
dt 2
v(t) = v O + at
0
vO
Note that:
dv(t)
= a(t)
dt
time (s)
This means that only a quadratic
(and lower) time dependence of x(t)
can give a constant acceleration.
E.g. suppose
x(t) = x O + v O t + αt 2
a = constant
0
time (s)
dx
= v O + αt
dt
d2x
a = 2 = 2α (time independent)
dt
x(t) has a parabolic form
Cart Motion Sensor demo
3-Dimensional Kinematics
1-Dimension
position:
G
x(t) = x(t)iˆ
velocity:
G
dx(t) ˆ
v(t) =
i
dt
3-Dimensions
G
r(t) = rx (t)iˆ + ry (t)ˆj + rz (t)kˆ
drx (t) ˆ dry (t) ˆ drz (t) ˆ
G
v(t) =
i+
j+
k
dt
dt
dt
G
v(t) = v x (t)iˆ + v y (t)ˆj + v z (t)kˆ
acceleration:
G
dv(t) ˆ
a(t) =
i
dt
G
dv x (t) ˆ dv y (t) ˆ dv z (t) ˆ
a(t) =
i+
j+
k
dt
dt
dt
G
a(t) = a x (t)iˆ + a y (t)ˆj + a z (t)kˆ
where:
ry
G
r
ˆj
î
k̂
rz
rx
HITT#1 (9/8/08)
G
r1
&
G
r2
are the position vectors of an object at
times t1 and t2 (t2 later than t1).
The displacement vector of the object during this time is:
G G G
A. Δ r = r1 + r2
Hint:
B.
G G G
Δ r = r1 − r2
C.
G G G
Δ r = r2 − r1
D.
G G G
Δ r = r1 ⋅ r2
E.
G G G
Δ r = r1 × r2
Draw the 3 vectors.
Recall position vectors
are always from the
coordinate system origin
to the object.
Displacement vectors
are from the objects initial
position to the later position.
A.
G G
Δr = r1 + r2
B.
G G
Δr = r1 − r2
C.
G G
Δr = r2 − r1
D.
G G
Δr = r1 ⋅ r2
E.
G G
Δr = r1 × r2
By definition the displacement
vector points from the earlier
to the later position vectors.
G
Δr
G
r2
(at t2)
G
r1 (at t1)
By the head to tail rules of
geometric vector addition
clearly,
G
G G
r1 + Δ r = r2
Hence
G G G
Δ r = r2 − r1
HITT#2 (9/8/08)
These position vectors are (in meters):
G
G
ˆ
ˆ
r1 = 2i + 2 j
r2 = 5iˆ − 2ˆj
G G G
So the magnitude of Δ r = r2 − r1 is:
A. Δr = 5 m
B. Δr = 8 m
C. Δr = 6 m
D. Δr = 2.6 m
E. Δr = 2 m
G G G
Δ r = r2 − r1
A. Δr = 5 m
B. Δr = 8 m
C. Δr = 6 m
G
r2 = 5iˆ − 2ˆj
G
r1 = 2iˆ + 2ˆj
G ˆ ˆ
Δ r = 3i − 4 j
Δr = (3 m) 2 + ( −4 m) 2 = 9 m 2 + 16 m 2
D. Δr = 2.6 m
E. Δr = 2 m
Δr = 25 m 2 = 5 m
In 1-D kinematics we derived that for constant acceleration
x = x o + v o t + 12 at 2
v = v o + at
It can similarly be shown that in 3 dimensions,
G G G
G2
1
r = ro + v o t + 2 at
G G G
v = v o + at
Where now each of these vector equations is really 3 equations,
one for the components along each coordinate direction:
x:
rx = rxo + v xo t + 12 a x t 2
v x = v xo + a x t
y:
ry = ryo + v yo t + 12 a y t 2
v y = v yo + a y t
z:
rz = rzo + v zo t + 12 a z t 2
v z = v zo + a z t
Also, in 1-D we developed the expression,
v 2 = v 2o + 2a ( x − x o )
In 3-D we have this becomes 3 equations,
x:
v 2x = v 2xo + 2a x ( rx − rxo )
y:
v 2y = v 2yo + 2a y ( ry − ryo )
z:
v 2z = v 2zo + 2a z ( rz − rzo )
These and the foregoing equations will let us handle (predict)
Projectile Motion in 3-dimensions.
Digression - Chapt. 5 preview
Force ⇔ acceleration
(Law of nature)
Both are vector quantities
If a net force acts on an object
that net force has a direction
The object will accordingly accelerate
in that direction
Mathematically,
G G
a∝F
as a vector relationship
(we’ll see what the proportionality constant is later)
E.g. in free space (no gravity)
y
G
F = (2 N)iˆ
if:
object will accelerate along î direction.
The time dependent (increasing)
G
displacement will be Δ r = ΔrX ˆi
x
y
z
Or if:
G G
Since a ∝ F a
a constant force
causes a constant
acceleration
z
G
F = (2 N)iˆ + (2 N)ˆj
x
object will accelerate along line
in which the time dependent
G
displacement is Δ r = ΔrX ˆi + ΔrY ˆj
with ΔrX = ΔrY
Survey HITT (not for credit)
Given that
Force ⇔ acceleration
(Law of nature)
In the absence of any other forces, what happens if a constant
force acts for some time and then instantly drops to zero (i.e.
stops acting)?
A. Object immediately stops.
B. Object continues moving in the same direction
but immediately begins slowing until it stops.
C. Object continues moving with the last velocity it had.
A. Object immediately stops.
B. Object continues moving in the same direction
but immediately begins slowing until it stops.
C. Object continues moving with the last velocity it had.
(note that velocity is a vector so implicit here is that
it maintains the same direction)
This is codified in Newton’s first law of motion:
If no net force acts on a body the bodies velocity can not change,
that is, the body can not accelerate.
VPython simulation
In projectile motion an object is launched, with some initial velocity
(i.e. speed in a particular direction).
To simplify we initially assume that no force other than gravity acts
(after the launch).
Importantly this means that if we align our coordinate axes so that
the force of gravity lies along one of the axes, e.g. the y axis, then
acceleration of the projectile only occurs along that direction.
Since no forces act along the other axes, the velocity along
the x and z axes is constant.
By rotating our coordinate system to make all the initial velocity
in say the x, y plane, we simplify further in that the z component
drops out of the problem (without an initial velocity along the z
direction and no acceleration along that direction the z component
of the object remains zero throughout the trajectory).
G
a = − gjˆ
Since there is no acceleration
in the x direction the
x component of the velocity
is the same throughout
the trajectory (vx(t) = vxo is the
same length everywhere in
the trajectory).
G
vO
θο
vyo
vxo
G
v
vy
vx
Due to the force
of gravity in the
y direction there
Gis acceleration
a = − gjˆ causing
the initially upward y component, vy
to shrink to zero at the
top of the trajectory and
then grow again in the
opposite direction.
G
a = − gjˆ
Since the magnitude
of the velocity is
G
vO
G
v = v = v 2x + v 2y
θο
This vector initially shrinks as
v y → 0 and then grows
again.
vyo
vxo
G
v
entirely
v x = v xo
at the top of
the trajectory
vy
vx
Now consider the trajectory.
The general expression for the position along the x direction is,
rx = rxo + v xo t + 12 a x t 2
but the launch occurs from rx = 0 and with no force along x, ax = 0
so this reduces to,
rx = v xo t
For the y direction,
ry = ryo + v yo t + 12 a y t 2
but the launch occurs from ry = 0 and a y = − g so we have,
y:
ry = v yo t − 12 gt 2
&
x:
rx = v xo t
Since rx is proportional to time, the trajectory reflects the
quadratic ry dependence as an inverted parabola (trajectory is
parabolic).
To be definitive consider the case in which the magnitude of the
the initial velocity is vo = 35 m/s and the launch angle is θο = 65o.
We ask for the distance travelled along x when the projectile is at
it’s maximum height.
G
a = − gjˆ
G
vO
θο
vyo
vxo
If we find the time taken for the projectile to reach it’s maximum
height we can then plug that into the rx = vxot equation to get the
desired x distance travelled. For the need time to get to the max
height use,
v y = v yo + a y t
v y = v yo − gt
Since vy = 0 at the top of the trajectory,
0 = v yo − gt
gt = v yo
t=
v yo
g
Then the x distance travelled at that time is found by using this
time in,
rx = v xo t
⎛ v yo ⎞ v xo v yo
rx = v xo ⎜
=
⎟
g
⎝ g ⎠
We must now determine vxo and vyo from the information given.
G
v
Resolving o into its
x and y components,
G
a = − gjˆ
G
vO
vox = vocosθo
θο
voy = vosinθo
vxo
vyo
So,
(v o cos θo )(v o sin θo )
rx =
=
g
g
m 2
(35
)
v 2o
s cos(65o )sin(65o )
rx = cos θo sin θo =
m
g
9.8 2
s
v xo v yo
rx = 48 m
What is the maximum height reached by the projectile?
Since,
v 2y = v 2yo + 2a y ( ry − ryo )
(use vy = 0 at top)
0 = v 2yo − 2g ( ry max − 0 )
2gry max = v 2yo
ry max =
v 2yo
2g
v o sin θ )
(
=
ry max = 51 m
2g
2
⎛ m
o⎞
35
sin
65
⎜
⎟
s
⎠
=⎝
m
2(9.8 2 )
s
2
What is the total horizontal distance travelled by the projectile?
To answer this we first find the vy component with which the
projectile hits the ground. To do this we again use,
v 2y = v 2yo + 2a y ( ry − ryo )
But we now take the top of the trajectory as the initial point
that gets the zero subscript. In that case,
vyo = 0
&
ryo = rymax
So,
v 2y = 0 − 2g ( 0 − ry max )
v 2y = 2gry max
But we found previously that, ry max =
v 2yo
2g
(where vyo here is the y component of the original velocity)
Using this above,
v 2y = 2g
Evidently,
v 2yo
2g
= v 2yo
v y = ± v yo
Since the velocity points downward when the projectile
strikes the ground the – sign is correct.
We now use this to find the time of the fall from the top of
of the trajectory. Applying,
v y = v yo − gt
− v yo = 0 − gt
t=
v yo
g
Which is the same time it took to get from the launch point to the
top of the trajectory. The total travel time is then twice this and
recall that,
rx = v xo t
Then,
v xo v oy
⎛ v oy ⎞
rx = v xo ⎜ 2
=2
⎟
g
⎝ g ⎠
Making the substitutions we made above for the velocity
components gives the range equation,
v 2o
rx = 2 cos θo sin θo
g
Which gives the horizontal distance travelled for given launch
speed and angle (careful: only valid when the landing is at the
same height as the launch).
We found half the distance above to be 48 m so,
rx max = 2(48 m) = 96 m
We can make the additional observation that the flight is symmetric
about the top of the trajectory i.e. we have half the time going up,
half the time coming down and the launch and landing speeds
are equal.
Another example: For a movie stunt, a driverless car is rigged to
accelerate from a stand still, at 4 m/s2, to the edge of a 60 m high cliff,
100 m away. How far from the edge of the cliff does the car hit?
G
a x = 4m / s2
100 m
60 m
rxmax ?
Once the car just comes off the edge of the cliff, the drive wheels
come out of contact with the road, so it stops accelerating
horizontally.
From that point on it retains whatever lateral velocity it had acquired.
That velocity is obtained
by applying eqn. 2-16
to the accelerated part of
the motion,
G
a x = 4.0m / s2
100 m
60 m
v 2x = v 2xo + 2a x (rx − rxo )
v 2x = 0 + 2(4m / s 2 )(100m)
v x = 28.3 m / s
rxmax ?
This is now the horizontal velocity retained throughout the rest of the
car’s motion independent of what it does along the vertical axis.
To find the distance it goes we have,
rx = rxo + v xo t
rx = 0 + (28.3 m / s)t + 0
rx = (28.3 m / s)t
but we don’t know the time, t when
it hits the ground.
To find t, we solve for the time taken to fall using,
ry = ryo + v yo t − 12 gt 2
0 = 60 m + 0t − 12 (9.8 m / s 2 )t 2
t = 3.5 s
Then,
rx max = (28.3 m / s)t = (28.3 m / s)(3.5 s) = 99 m
Note that if the launch point is inclined at say 30o upwards, we
can resolve the velocity at that point into its horizontal
component, v xo and its vertical component, v yo (it then has
an initial velocity upwards). So we can still tackle the problem.