PH 221-3A Fall 2010
Motion in Two and Three
Dimensions
Lectures 4, 5
Chapter 4
(Halliday/Resnick/Walker, Fundamentals of Physics 8th edition)
1
Chapter 4
Motion in Two and Three Dimensions
In this chapter we will continue to study the motion of objects without the
restriction we put in chapter 2 to move along a straight line. Instead we will
consider motion in a plane (two dimensional motion) and motion in space
(three dimensional motion). The following vectors will be defined for twoand three- dimensional motion:
Displacement
Average and instantaneous velocity
Average and instantaneous acceleration
We will consider in detail projectile motion and uniform circular motion as
examples of motion in two dimensions.
Finally we will consider relative motion, i.e. the transformation of velocities
between two reference systems which move with respect to each other with
constant velocity.
2
Position Vector
G
The position vector r of a particle is defined as a vector whose tail is at
a reference point (usually the origin O) and its tip is at the particle at
point P.
Example: The position vector in the figure is:
G
r = xiˆ + yjˆ + zkˆ
(
)
G
r = −3iˆ + 2 ˆj + 5kˆ m
P
3
Displacement Vector
G G
For a particle that changes postion vector from r1 to r2 we define the displacement
G
G G G
vector Δr as follows: Δr = r2 − r1
G
G
The position vectors r1 and r2 are written in terms of components as:
G
G
ˆ
ˆ
ˆ
r1 = x1i + y1 j + z1k
r2 = x2iˆ + y2 ˆj + z2 kˆ
G
The displacement Δ r can then be written as:
G
Δr = ( x2 − x1 ) iˆ + ( y2 − y1 ) ˆj + ( z2 − z1 ) kˆ = Δxiˆ + Δyjˆ + Δzkˆ
Δx = x2 − x1
Δy = y2 − y1
Δz = z2 − z1
t1
t2
4
Problem 2. A watermelon seed has the following coordinates: x = −5.0m, y = 8.0m, z = 0m. Find its position vector (a) in unitvector notation and as (b) a magnitude and (c) an angle relative to the positive direction of the x axis. (d) Sketch the vector on a right handed
coordinate system. If the seed is moved to the xyz coordinates (3.00m, 0m, 0m) what is its displacement
(e) in unit-vector notation as as (f) a magnitude and
(g) an angle relative to the positive x direction?
(a) The position vector, according to
G
(b) The magnitude is | r | =
x2 + y2 + z2
G
r = xiˆ + yjˆ + zkˆ
=
G
is r = ( − 5 .0 m ) ˆi + ( 8 .0 m )ˆj .
( − 5 .0 m ) 2 + (8 .0 m ) 2 + ( 0 m ) 2 = 9 .4 m .
(c) Many calculators have polar ↔ rectangular conversion capabilities which make this
computation more efficient than what is shown below. Noting that the vector lies in the
xy plane and using tanθ=ry/rx, we obtain:
⎛ 8 .0 m ⎞
⎟ = − 5 8° or 1 2 2°
⎝ − 5 .0 m ⎠
θ = ta n − 1 ⎜
where the latter possibility (122° measured counterclockwise from the +x
direction) is chosen since the signs of the components imply the vector is
in the second quadrant.
(d) The sketch is shown on the right. The vector is 122° counterclockwise
from the +x direction.
G
G
G
G
(e) The displacement is Δ r = r ′ − r where r is given in part (a) and
G
G
r ′ = (3 .0 m ) ˆi. Therefore, Δ r = (8 .0 m ) ˆi − (8 .0 m )ˆj .
G
(f) The magnitude of the displacement is | Δ r | =
( 8 .0 m ) 2 + ( − 8 .0 m ) 2 = 1 1 m .
(g) The angle for the displacement, using tanθ=ry/r x, is
⎛ 8 .0 m ⎞
ta n − 1 ⎜
⎟ = − 4 5 ° o r 1 3 5°
⎝ − 8 .0 m ⎠
where we choose the former possibility (−45°, or 45° measured clockwise
from +x) since the signs of the components imply the vector is in the
G
fourth quadrant. A sketch of Δ r is shown on the right.
5
Average and Instantaneous Velocity
Following the same approach as in chapter 2 we define the average
velocity as:
displacement
average velocity =
time interval
G
Δr Δxiˆ + Δyjˆ + Δzkˆ Δxiˆ Δyjˆ Δzkˆ
G
vavg =
=
=
+
+
Δt
Δt
Δt
Δt
Δt
We define as the instantaneous velocity
(or more simply the velocity) as the limit:
t
t + Δt
G
G
Δr dr
G
v = lim
=
Δt dt
Δt → 0
6
If we allow the time interval Δt to shrink to zero, the following things happen:
G
G
G
1. Vector r2 moves towards vector r1 and Δr → 0
G
G
Δr
2. The direction of the ratio
(and thus vavg )approaches the direction
Δt
of the tangent to the path at position 1
G
G
3. vavg → v
(
)
dx
dy ˆ dz ˆ
G d ˆ ˆ
v=
xi + yj + zkˆ = iˆ +
j + k = vx iˆ + v y ˆj + vz kˆ
dt
dt
dt
dt
The three velocity components are given by
the equations:
t
t + Δt
vx =
G
G dr
v=
dt
dx
dt
dy
vy =
dt
vz =
dz
dt
7
Problem 6. An electron position is given by r = 3.00tiˆ − 4.00t 2 ˆj + 2.00kˆ,
G
with t in seconds and r in meters. (a) In unit-vector notation, what is the
G
G
electron's velocity v ( t )? At t = 2.00 s, what is v (b) in unit-vector notation
and as (c) a magnitude and (d) an angle relative to the positive direction of the x axis?
(a) Eq.
G
G dr
v=
dt
v(t ) =
leads to
d
ˆ = (3.00 m/s)iˆ − (8.00t m/s) ˆj
(3.00tˆi − 4.00t 2 ˆj + 2.00k)
dt
G
ˆ m/s.
(b) Evaluating this result at t = 2.00 s produces v = (3.00iˆ − 16.0j)
G
(c) The speed at t = 2.00 s is v = |v |= (3.00 m/s) 2 + ( −16.0 m/s) 2 = 16.3 m/s.
G
(d) The angle of v at that moment is
⎛ −16.0 m/s ⎞
tan −1 ⎜
⎟ = −79.4° or 101°
3.00
m/s
⎝
⎠
where we choose the first possibility (79.4° measured clockwise from the +x direction, or
281° counterclockwise from +x) since the signs of the components imply the vector is in
the fourth quadrant.
8
Average and Instantaneous Acceleration
The average acceleration is defined as:
G
aavg
change in velocity
average acceleration =
time interval
G G
G
v − v Δv
= 2 1=
Δt
Δt
We define as the instantaneous acceleration as the limit:
G
G
dvx ˆ dv y ˆ dvz ˆ
Δv dv d
G
ˆ
ˆ
ˆ
a = lim
=
=
vx i + v y j + vz k =
i+
j+
k = ax iˆ + a y ˆj + az kˆ
Δt dt dt
dt
dt
dt
Δt → 0
(
)
Note: Unlike velocity, the acceleration vector does not have any specific relationship
with the path.
The three acceleration components are given by
the equations:
dv
ax = x
dt
ay =
dv y
dt
dvz
az =
dt
G
dv
G
a=
dt
9
G
P ro b le m 1 3 . T h e p o sitio n r o f a p a rtic le m o v in g in a n x y p la n e is g iv e n b y
G
G
r = ( 2 .0 0 t 3 − 5 .0 0 t ) iˆ + ( 6 .0 0 − 7 .0 0 t 4 ) ˆj , w ith r in m e te rs a n d t in se c o n d s.
G
G
G
In u n it-v e c to r n o ta tio n , c a lc u la te (a ) r , (b ) v , (c ) a fo r t = 2 .0 0 s. (d ) W h a t
is th e a n g le b e tw e e n th e p o sitiv e d ire c tio n o f th e x a x is a n d a lin e ta n g e n t to th e
G
dv
dr
G
p a rtic le a t t = 2 .0 0 s?
ax = x
v=
dt
dt
In parts (b) and (c), we use
. For part (d), we
find the direction of the velocity computed in part (b), since that represents the asked-for
tangent line.
(a) Plugging into the given expression, we obtain
G
r
t = 2 .0 0
=
[ 2 . 0 0 ( 8 ) − 5 . 0 0 ( 2 ) ] ˆi + [ 6 . 0 0 − 7 . 0 0 ( 1 6 ) ] ˆj =
( 6 . 0 0 ˆi − 1 0 6 ˆj) m
(b) Taking the derivative of the given expression produces
G
v ( t ) = ( 6 . 0 0 t 2 − 5 . 0 0 ) ˆi − 2 8 . 0 t 3 ˆj
where we have written v(t) to emphasize its dependence on time. This becomes, at
G
t = 2.00 s, v = ( 1 9 . 0 ˆi − 2 2 4 ˆj ) m / s .
G
(c) Differentiating the v ( t ) found above, with respect to t produces 1 2 . 0 t ˆi − 8 4 . 0 t 2 ˆj ,
G
which yields a = ( 2 4 . 0 ˆi − 3 3 6 ˆj ) m / s 2 at t = 2.00 s.
G
(d) The angle of v , measured from +x, is either
ta n
−1
⎛ − 2 2 4 m /s ⎞
⎜
⎟ = − 8 5 .2 ° o r 9 4 .8 °
⎝ 1 9 .0 m /s ⎠
where we settle on the first choice (–85.2°, which is equivalent to 275° measured
counterclockwise from the +x axis) since the signs of its components imply that it is in
10
the fourth quadrant.
Projectile Motion
The motion of an object in a vertical plane under the influence of
gravitational force is known as “projectile motion”
G
The projectile is launched with an initial velocity vo = vox iˆ + voy ˆj
The horizontal and vertical velocity components are:
vox = vo cos θ o
g
voy = vo sin θ o
Projectile motion will be analyzed in
a horizontal and a vertical motion
along the x- and y-axes, respectively.
These two motions are independent
of each other. Motion along the xaxis has zero acceleration. Motion
along the y-axis has uniform
acceleration ay = -g
11
Horizontal Motion: ax = 0
v x = v0 cosθ0
(eqs.1)
Vertical Motion:
v y = v0 sin θ0 − gt
ay = − g
(eqs.3)
The velocity along the x-axis does not change
x − xo = vox t = ( vo cosθ o ) t
(eqs.2)
Along the y-axis the projectile is in free fall
gt 2
gt 2
y − yo = voy t −
= ( v0 sin θ0 ) t −
2
2
(eqs.4)
If we eliminate t between equations 3 and 4 we get: v 2y = ( v0 sin θ0 ) − 2 g ( y − yo )
2
g
Here xo and yo are the coordinates
of the launching point. For many
problems the launching point is
taken at the origin. In this case
xo = 0 and yo = 0
Note: In this analysis of projectile
motion we neglect the effects of
air resistance
12
The equation of the path:
gt 2
x = ( vo cos θ o ) t (eqs.2)
y = ( v0 sin θ 0 ) t −
2
If we eliminate t between equations 2 and 4 we get:
y = ( tan θ o ) x −
g
2 ( vo cos θ o )
2
x2
(eqs.4)
This equation describes the path of the motion
The path equations has the form: y = ax + bx 2 This is the equation of a parabola
Note: The equation of the path seems too
complicated to be useful. Appearances can
deceive: Complicated as it is, this equation
can be used as a short cut in many projectile
motion problems
13
v x = v0 cos θ 0
(eqs.1)
x = ( vo cos θ o ) t
(eqs.2)
sinϕ
gt 2
v y = v0 sin θ 0 − gt (eqs.3) y = ( v0 sin θ 0 ) t −
(eqs.4) O π/2
2
Horizontal Range: The distance OA is defined as the horizantal range R
At point A we have: y = 0 From equation 4 we have:
3π/2
gt 2
gt ⎞
⎛
sin
0
sin
v
t
−
=
→
t
v
−
θ
θ
(0
0)
0
⎜ 0
⎟ = 0 This equation has two solutions:
2
2⎠
⎝
Solution 1. t = 0 This solution correspond to point O and is of no interest
gt
= 0 This solution correspond to point A
2
2v0 sin θ 0
From solution 2 we get: t =
If we substitute t in eqs.2 we get:
g
2vo2
vo2
R=
sin θ o cos θ o = sin 2θ o
g
g
t
A
O
R has its maximum value when θ o = 45°
Solution 2.
v0 sin θ 0 −
R
2sin A cos A = sin 2 A
Rmax
vo2
=
g
14
ϕ
tA
g
Maximum height H
vo2 sin 2 θ o
H=
2g
H
The y-component of the projectile velocity is: v y = v0 sin θ 0 − gt
At point A: v y = 0
→ v0 sin θ 0 − gt → t =
H = y (t ) = ( v0 sin θ 0 ) t −
vo2 sin 2 θ o
H=
2g
v0 sin θ 0
g
2
v sin θ 0 g ⎛ v0 sin θ 0 ⎞
gt
= ( v0 sin θ 0 ) 0
− ⎜
⎟ →
g
g ⎠
2
2⎝
2
15
Maximum height H (encore)
tA
g
H
vo2 sin 2 θ o
H=
2g
We can calculate the maximum height using the third equation of kinematics
for motion along the y-axis: v y 2 = v 2yo + 2a ( y − yo )
In our problem: yo = 0 , y = H , v yo = vo sin θ o , v y = 0 , and a = − g →
vo2 sin 2 θ o
=
− v = −2 gH → H =
2g
2g
2
yo
v 2yo
16
Problem : A car drives straight off the edge of a cliff that is 54m high. The police at the scene of the
accident note that the point of impact is 130m from the base of a cliff. How fast was the car
traveling when it went over the cliff?
During this time the car travels a horizontal distance of 130m
Problem : A diver springs upward from a board that is three meters above the water. At the instant
she contacts the water her speed is 8.90 m/s and her body makes an angle of 75.0 degrees with
respect to the horizontal surface of the water. Determine her initial velocity, both magnitude and
direction
The motion of a ballistic missile can be regarded as the motion of a projectile because along the
greatest part of its trajectory the missile is in free fall. Suppose that a missile is to strike a target
1000 km away. What minimum speed must the missile have at the beginning of its trajectory?
What maximum height does it reach when launched with this minimum speed? How long does it
take to reach the target? For these calculations assume that g = 9.8 m/s2 everywhere along the
trajectory.
~7.5 min
Problem: A projectile is fired at an initial velocity of 35.0 m/s at an angle of 30.0 degrees above the
horizontal from the roof of a building 30.0 m high, as shown. Find
a) The maximum height of the projectile
b) The time to rise to the top of the trajectory
c) The total time of the projectile in the air
d) The velocity of the projectile at the ground
e) The range of the projectile
Problem 37. A ball is shot from the ground into the air. A t a height of 9.1 m , its velocity is
G
v = (7.6 iˆ + 6.1 ˆj ) m .s, w ith iˆ horizontal and ˆj upw ard. (a) To w hat m axim um height does the
ball rise? (b ) W hat total horizontal distance does the ball travel? W hat are the (c) m agnitude and
(d) angle (below the horizontal) of the ball's velocity just before it hits the ground?
We designate
velocity when
G
v 3 − and take
ground.
G
the given velocity v = ( 7
it reaches the max height
G
v 0 as the launch velocity,
G
as opposed to the
. 6 m / s ) ˆi + ( 6 . 1 m / s ) ˆj as v 1 −
G
v 2 or the velocity when it returns to the ground
as usual. The origin is at its launch point on the
(a) Different approaches are available, but since it will be useful (for the rest of the
problem) to first find the initial y velocity, that is how we will proceed.
v 12 y = v 02 y − 2 g Δ y
⇒
( 6 . 1 m / s ) 2 = v 02 y − 2 ( 9 . 8 m / s 2 ) ( 9 . 1 m )
which yields v0 y = 14.7 m/s. Knowing that v2
maximum height:
2
2 y
v
= v
2
0 y
− 2 g h
⇒
0 = (1 4 . 7
y
must equal 0, we use Δy = h for the
m /s )
2
− 2 (9 .8 m /s
2
)h
which yields h = 11 m.
(b) Recalling the derivation of Eq. 4-26, but using v0 y for v0 sin θ0 and v0x for v0 cos θ0,
we have
1
0 = v0 yt −
g t 2 ,
R = v0 xt
2
which leads to R
obtain
= 2 v
0 x
v
0 y
/ g .
Noting that v0x = v1x = 7.6 m/s, we plug in values and
R = 2(7.6 m/s)(14.7 m/s)/(9.8 m/s2) = 23 m.
(c) Since v3x = v1x = 7.6 m/s and v3y = – v0
v
3
=
v
2
3 x
+ v
2
3 y
=
( 7 .6
y
= –14.7 m/s, we have
m /s )
(d) The angle (measured from horizontal) for
ta n
− 1
⎛ − 1 4 .7 m
⎜
7 .6 m
⎝
2
+ ( − 1 4 .7
G
v
3
m /s )
2
= 1 7
m /s .
is one of these possibilities:
⎞
⎟ = − 6 3 °
⎠
o r
1 1 7 °
where we settle on the first choice (–63°, which is equivalent to 297°) since the signs
21 of
its components imply that it is in the fourth quadrant.
Uniform circular Motion
A particles is in uniform circular motion it moves on a circular path of
radius r with constant speed v. Even though the speed is constant, the
velocity is not. The reason is that the direction of the velocity vector
changes from point to point along the path. The fact that the velocity
changes means that the acceleration is not zero. The acceleration in uniform
circular motion has the following characteristics:
1. Its vector points towards the center C of the circular path, thus the name
“centripetal”
v2
2. Its magnitude a is given by the equation: a =
r
Q
r
C r
P
The time T it takes to complete a full revolution is
known as the “period”. It is given by the
equation:
r
R
2π r
T=
v
22
G
v = vx iˆ + v y ˆj = ( −v sin θ ) iˆ + ( v cos θ ) ˆj
yP
xP
sin θ =
cosθ =
r
r
Here xP and yP are the coordinates of the rotating particle
G
G ⎛ y P ⎞ ˆ ⎛ xP ⎞ ˆ
G dv ⎛ v dyP ⎞ ˆ ⎛ v dxP
v = ⎜ −v ⎟ i + ⎜ v ⎟ j Acceleration a =
= ⎜−
⎟i + ⎜
r ⎠ ⎝ r ⎠
dt ⎝ r dt ⎠ ⎝ r dt
⎝
dyP
dxP
We note that:
= v y = v cos θ and
= vx = −v sin θ
dt
dt
⎞ ˆ ⎛ v2
⎞ˆ
G ⎛ v2
a = ⎜ − cos θ ⎟ i + ⎜ − sin θ ⎟ j
⎝ r
⎠ ⎝ r
⎠
tan φ =
vx = −v sin θ
v y = v cos θ
ay
ax
=
− ( v 2 / r ) sin θ
− ( v 2 / r ) cos θ
2
v
a = ax2 + a y2 =
r
⎞ˆ
⎟j
⎠
( cos θ ) + ( sin θ )
2
2
v2
=
r
G
= tan θ → φ = θ → a points towards C
P
C
A
C
( cos θ ) + ( sin θ )
2
23
2
=1
Problem 60. An earth satellite moves in a circular orbit 640 km above Earth's surface with a period of 98.0 min.
What are the (a) speed and (b) magnitude of the centripital acceleration of the satellite?
2π r to solve for speed v and
We apply T =
v
v2
a=
r
to find acceleration a.
(a) Since the radius of Earth is 6.37 × 106 m, the radius of the satellite orbit is
r = (6.37 × 106 + 640 × 103 ) m = 7.01 × 106 m.
Therefore, the speed of the satellite is
c
h
2π 7.01 × 106 m
2πr
v=
=
= 7.49 × 103 m / s.
T
98.0 min 60 s / min
b
gb
g
(b) The magnitude of the acceleration is
a=
2
v
r
7.49 × 10
c
=
3
h
m/s
7.01 × 10 m
6
2
= 8.00 m / s2 .
24
Relative Motion in One Dimension:
The velocity of a particle P determined by two different observers A and B varies from
observer to observer. Below we derive what is known as the “transformation
equation” of velocities. This equation gives us the exact relationship between the
velocities each observer perceives. Here we assume that observer B moves with a
known constant velocity vBA with respect to observer A. Observer A and B determine
the coordinates of particle P to be xPA and xPB , respectively.
xPA = xPB + xBA
Here xBA is the coordinate of B with respect to A
d
d
d
We take derivatives of the above equation:
=
+
x
x
( PA )
( PB )
( xBA ) →
dt
dt
dt
vPA = vPB + vBA
If we take derivatives of the last equation and take
dvBA
aPA = aPB
=0→
into account that
dt
Note: Even though observers A and B
measure different velocities for P,
they measure the same acceleration
25
Relative Motion in Two Dimensions:
Here we assume that observer B moves with a known constant velocity vBA with
respect to observer A in the xy-plane.
Observers A and B determine the position vector of particle P to be
G
G
rPA and rPB , respectively.
G
G
G
rPA = rPB + rBA We take the time derivative of both sides of the equation
G
G
G
d G
d G
d G
G
G
G
rPA = rPB + rBA → vPA = vPB + vBA
vPA = vPB + vBA
dt
dt
dt
If we take the time derivative of both sides of the last equation we have:
G
d G
d G
d G
dvBA
G
G
vPA = vPB + vBA If we take into account that
= 0 → aPA = aPB
dt
dt
dt
dt
Note: As in the one dimensional
case, even though observers A and B
measure different velocities for P,
they measure the same acceleration
26
Problem 74. A light plane attain an airspeed of 500 km/h. The pilot sets out for a destination 800 km due north
but discovers that the plane must be headed 20.0Deast of due north to fly there directly. The plane arrives in
2.00 h. What were the (a) magnitude and (b) direction of the wind velocity?
→
^
The destination is D = 800 km j where we orient axes so that +y points north and +x
points east. This takes two hours, so the (constant) velocity of the plane (relative to the
→
^
ground) is vpg = (400 km/h) j . This must be the vector sum of the plane’s velocity with
respect to the air which has (x,y) components (500cos70º, 500sin70º) and the velocity of
→
the air (wind) relative to the ground vag . Thus,
^
^
^
→
JJG
vag
(400 km/h) j = (500 km/h) cos70º i + (500 km/h) sin70º j + vag
which yields
→
^
^
vag =( –171 km/h)i –( 70.0 km/h)j .
G
G
(a) The magnitude of vag is | vag |= (− 171 km/h) 2 + (− 70.0 km/h) 2 = 185 km/h.
JJJG
v pg
JJJG
v pa
G
(b) The direction of vag is
⎛ − 70.0 km/h ⎞
⎟ = 22.3 ° (south of west).
⎝ − 171 km/h ⎠
θ = tan − 1 ⎜
27