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PH 221-1D Spring 2013
Motion in Two and Three Dimensions
Lectures 5,6,7
Chapter 4
(Halliday/Resnick/Walker, Fundamentals of Physics 9th 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 two- and 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 The position vector 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.
The position vectoExampl r in te : he f
r
igure is:
ˆˆ ˆr xi yj zk= + +
( )ˆˆ ˆ3 2 5r i j k m= − + +
P
3
t2
t1
Displacement Vector
1 2For a particle that changes postion vector from to we define the displacement vector as follows:
r rr∆
2 1r r r∆ = −
1 2The position vectors and are written in terms of components as:r r
1 1 1 1ˆˆ ˆr x i y j z k= + +
2 2 2 2ˆˆ ˆr x i y j z k= + +
( ) ( ) ( )2 1 2 1 2 1ˆ ˆˆ ˆ ˆ ˆr x x i y y j z z k xi yj zk∆ = − + − + − = ∆ + ∆ + ∆
2 1x x x∆ = −
2 1y y y∆ = −
2 1z z z∆ = −
The displacement r can then be written as:∆
4
5
(a) The position vector, according to is ˆ ˆ= ( 5.0 m) i + (8.0 m)jr − .
(b) The magnitude is 2 2 2 2 2 2| | + + ( 5.0 m) (8.0 m) (0 m) 9.4 m.r x y z= = − + + =
(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:
1 8.0 mtan 58 or 1225.0 m
θ − = = − ° ° −
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. (e) The displacement is r r r′∆ = −
where r is given in part (a) and ˆ (3.0 m)i.r′ = Therefore, ˆ ˆ(8.0 m)i (8.0 m)jr∆ = −
. (f) The magnitude of the displacement is 2 2| | (8.0 m) ( 8.0 m) 11 m.r∆ = + − =
(g) The angle for the displacement, using tanθ=ry/rx, is
1 8.0 mtan = 45 or 1358.0 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 fourth quadrant. A sketch of r∆ is shown on the right.
Problem 2. A watermelon seed has the following coordinates: 5.0 , 8.0 , 0 . Find its position vector (a) in unit-vector notation and as (b) a magnitude and (c) an angle relative to the positive
x m y m z m= − = = 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? ˆˆ ˆr xi yj zk= + +
t
t + Δt
Average and Instantaneous Velocity Following the same approach as in chapter 2 we define the average velocity as: displacementaverage velocity =
time interval
ˆ ˆˆ ˆ ˆ ˆavg
r xi yj zk xi yj zkvt t t t t
∆ ∆ + ∆ + ∆ ∆ ∆ ∆= = = + +∆ ∆ ∆ ∆ ∆
We define as the instantaneous velocity (or more simply the velocity) as the limit:
lim
0
r drvt dt
t
∆= =
∆∆ →
6
t
t + Δt
2 1
If we allow the time interval t to shrink to zero, the following things happen: 1. Vector moves towards vector and 0
2. The direction of the ratio (and thus )approachest avg
r r rr v
∆∆ →
∆∆
the direction
of the tangent to the path at position 13. avgv v→
( )ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆx y z
d dx dy dzv xi yj zk i j k v i v j v kdt dt dt dt
= + + = + + = + +
xdxvdt
=
ydyvdt
=
zdzvdt
=
The three velocity components are given by the equations:
drvdt
=
7
8
(a) Eq. leads to
2ˆ ˆ ˆ ˆ ˆ( ) (3.00 i 4.00 j + 2.00k) (3.00 m/s)i (8.00 m/s) jdv t t t tdt
= − = −
(b) Evaluating this result at t = 2.00 s produces ˆ ˆ = (3.00i 16.0j) m/s.v −
(c) The speed at t = 2.00 s is 2 2 | | (3.00 m/s) ( 16.0 m/s) 16.3 m/s.v v= = + − =
(d) The angle of v at that moment is
1 16.0 m/stan 79.4 or 1013.00 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.
2 ˆˆ ˆProblem 6. An electron position is given by 3.00 4.00 2.00 ,with in seconds and in meters. (a) In unit-vector notation, what is theelectron's velocity ( )? At 2.00 , what is (b)
r ti t j kt r
v t t s v
= − +
=
in unit-vector notationand as (c) a magnitude and (d) an angle relative to the positive direction of the x axis?
drvdt
=
Average and Instantaneous Acceleration The average acceleration is defined as:
change in velocityaverage acceleration = time interval
2 1avg
v v vat t− ∆
= =∆ ∆
We define as the instantaneous acceleration as the limit:
( )ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆlim
0
yx zx y z x y z
dvdvv dv d dva v i v j v k i j k a i a j a kt dt dt dt dt dt
t
∆= = = + + = + + = + +
∆∆ →
The three acceleration components are given by the equations:
xx
dvadt
= yy
dva
dt= z
zdvadt
=
Note: Unlike velocity, the acceleration vector does not have any specific relationship with the path.
dvadt
=
9
10
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
2.00ˆ ˆ ˆ ˆ [2.00(8) 5.00(2)]i + [6.00 7.00(16)] j (6.00 i 106 j) mtr = = − − = −
(b) Taking the derivative of the given expression produces 2 3ˆ ˆ( ) = (6.00 5.00) i 28.0 jv t t t− −
where we have written v(t) to emphasize its dependence on time. This becomes, at t = 2.00 s, ˆ ˆ = (19.0 i 224 j) m/s.v −
(c) Differentiating the v t( ) found above, with respect to t produces 2ˆ ˆ12.0 i 84.0 j,t t− which yields 2ˆ ˆ =(24.0 i 336 j) m/sa −
at t = 2.00 s. (d) The angle of v , measured from +x, is either
1 224 m/stan 85.2 or 94.819.0 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 the fourth quadrant.
3 4
Problem 11. The position of a particle moving in an xy plane is given byˆ ˆ(2.00 5.00 ) (6.00 7.00 ) , with in meters and in seconds.
In unit-vector notation, calculate (a) , (b) , (c
r
r t t i t j r tr v
= − + −
) for 2.00 s. (d) Whatis the angle between the positive direction of the axis and a line tangent to the particle at =2.00 s?
a tx
t
=
drvdt
=
xx
dvadt
=
Projectile Motion The motion of an object in a vertical plane under the influence of gravitational force is known as “projectile motion”
The projectile is launched with an initial velocity
The horizontal and vertical velocity components are:
ˆ ˆo ox oyv v i v j= +
cosox o ov v θ= sinoy o ov v θ=
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 x-axis has zero acceleration. Motion along the y-axis has uniform acceleration ay = -g
g
11
( )x 0 0
0 The velocity along the x-axis does not change (eqs.1) (eqs
Horizontal Motion:
Vertic.2)
al Motio Along the y-axiv c
n s os
tche projos
: e
x
y
o ox o o
a
av x x v t v t
gθ θ= −
=
= −
= =
( )
( ) ( )
2 2
0 0 0 0
220 0
ctile is in free fall
(eqs.3) (eqs.4)
If we eliminate between equations 3 an
sin sin2
d2
sin 2 4 we get:
y o oy
y o
gt gtv v gt y y v t v t
v v g y yt
θ θ
θ
= − − = − = −
= − −
Here and are the coordinates of the launching point. For many problems the launching point istaken at the origin. In this case
0 and 0 In this analysis of projectile
motiNote:
on we
o o
o o
x y
x y= =
neglect the effects ofair resistance
g
12
( ) ( )
( )( )
2
22
0 0
cos (eqs.2) sin (eqs.4)2
If we eliminate between equations 2 and 4 we get:
t
The equation of the p
an 2 cos
This equatio
a
e
:
d
th
noo
o
o
o
gy x
gtx v t y v
x
t
v
t
θθ
θ θ=
= −
= −
2
scribes the path of the motion
The path equations has the form: This is the equation of a parabolay ax bx= +
The equation of the path seems toocomplicated to be useful. Appearances candeceive: Complicated as it is, this equationcan be used as a short cut in many projectilemotion prob
Note:
lems
13
O A
R
t
( )
( )
x 0 0
2
0 0 0 0
v cos (eqs.1) cos (eqs.2)
sin (eqs.3) sin (eqs.4)2
The distance OA is defined as the horizantal range Horizontal RangAt point A
e:
o o
y
v x v t
gtv v gt y v t
R
θ θ
θ θ
= =
= − = −
( )2
0 0 0 0
we have: 0 From equation 4 we have:
sin 0 sin 0 This equation has two solutions:2 2
Solution 1. 0 This solution correspond to point O and is of no interest
Solution 2.
ygt gtv t t v
t
θ θ
=
− = → − =
=
0 0
0 0
sin 0 This solution correspond to point A2
2 sinFrom solution 2 we get: If we substitute in eqs.2 we get:
gtv
vt tg
θ
θ
− =
=
2 2
o2
max
2 sin cos sin 2
has its maximum value when 45
o oo o o
o
v vRg g
RvRg
θ θ θ
θ
= =
= °
=2sin cos sin 2A A A=
π/2
3π/2 ϕ
sinϕ
O
14
A t
H
g Maximum height H
2 2sin2
o ovHgθ
=
( ) ( )
0 0
0 00 0
220 0 0 0
0 0 0 0
2 2
The y-component of the projectile velocity is: sinsinAt point A: 0 sin
sin sin( ) sin sin2 2
sin2
y
y
o o
v v gtvv v gt t
g
v vgt gH y t v t vg g
vHg
θ
θθ
θ θθ θ
θ
= −
= → − → =
= = − = − →
= 15
( )2 2
2
We can calculate the maximum height using the third equation of kinematicsfor motion along the y-axis: 2
In our problem: 0 , , sin , 0 , and
2
y yo o
o yo o o y
yo
v v a y y
y y H v v v a g
v gH H
θ
= + −
= = = = = − →
− = − →2 2 2sin
2 2yo o ov vg g
θ= =
2 2sin2
o ovHgθ
=
Maximum height H (encore) A t
H g
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
21
We designate the given velocity ˆ ˆ(7.6 m/s)i (6.1 m/s) jv = + as v1 − as opposed to the
velocity when it reaches the max height v2 or the velocity when it returns to the ground v3 − and take v0 as the launch velocity, as usual. The origin is at its launch point on the ground. (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.
2 2 2 2 21 0 02 (6.1 m/s) 2(9.8 m/s )(9.1 m)y y yv v g y v= − ∆ ⇒ = −
which yields v0 y = 14.7 m/s. Knowing that v2 y must equal 0, we use ∆y = h for the maximum height:
2 2 2 22 0 2 0 (14.7 m/s) 2(9.8 m/s )y yv v gh 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
20 0
10 ,2y xv t gt R v t= − =
which leads to 0 02 / .x yR v v g= Noting that v0x = v1x = 7.6 m/s, we plug in values and obtain
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 y = –14.7 m/s, we have
2 2 2 23 3 3 (7.6 m/s) ( 14.7 m/s) 17 m/s.x yv v v= + = + − =
(d) The angle (measured from horizontal) for v3 is one of these possibilities:
1 14.7 mtan 63 or 1177.6 m
− − = − ° °
where we settle on the first choice (–63°, which is equivalent to 297°) since the signs of its components imply that it is in the fourth quadrant.
Problem 43. A ball is shot from the ground into the air. At a height of 9.1 m, its velocity is ˆ ˆ ˆ ˆ(7.6 6.1 ) m.s, with horizontal and upward. (a) To what maximum height does the
ball rise? (bv i j i j= +
) What total horizontal distance does the ball travel? What are the (c) magnitude and (d) angle (below the horizontal) of the ball's velocity just before it hits the ground?
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” 2. Its magnitude a is given by the equation:
2var
=
C P
R
Q
r
r r
The time T it takes to complete a full revolution is known as the “period”. It is given by the equation:
2 rTvπ
=22
A
P
C C
( ) ( )ˆ ˆ ˆ ˆsin cos sin cos
Here and are the coordinates of the rotating particle
P Px y
P P
y xv v i v j v i v jr r
x y
θ θ θ θ= + = − + = =
ˆ ˆ ˆ ˆ Acceleration = P P P Py x dv v dy v dxv v i v j a i jr r dt r dt r dt
= − + = − +
We note that: cos and sinP Py x
dy dxv v v vdt dt
θ θ= = = = −
( ) ( )2 2 2 2
2 22 2ˆ ˆ cos sin cos sinx yv v v va i j a a ar r r r
θ θ θ θ
= − + − = + = + =
( )( )
2
2
/ sintan tan points towards C
/ cosy
x
v raa
a v r
θφ θ φ θ
θ
−= = = → = →
−
sinxv v θ= −cos yv v θ=
( ) ( )2 2cos sin 1θ θ+ =23
24
We apply to solve for speed v and 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
v rT
= =×
= ×2 2 7 01 10
98 0 607 49 10
63π π .
. / min.
mmin s
m / s.c h
b gb g
(b) The magnitude of the acceleration is
a vr
= =×
×=
2 3 2
6
7 49 107 01 10
8 00.
.. .
m / sm
m / s2c h
Problem 56. 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?
2var
=2 rTvπ
=
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.
Here is the coordinate of B with respect to APA PB BA BAx x x x= +
( ) ( ) ( )We take derivatives of the above equation: PA PB BAd d dx x xdt dt dt
= + →
PA PB BAv v v= + If we take derivatives of the last equation and take
into account that 0BAdvdt
= → PA PBa a=
Even though observers A and Bmeasure different velocities for P,they measure the same acceler
Note:
ation25
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
and , respectively. PA PBr r
We take the time derivative of both sides of the equationPA PB BAr r r= +
PA PB BA PA PB BAd d dr r r v v vdt dt dt
= + → = + PA PB BAv v v= +
If we take the time derivative of both sides of the last equation we have:
If we take into account that 0 BAPA A BB AP PB P
d d d dvv v vdt dt d
at dt
a= + = =→
As in the one dimensional case, even though observers A and Bmeasure different velocities for P,they measure the same acceler
Note:
ation26
27
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,
(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^ . (a) The magnitude of agv is 2 2
ag| | ( 171 km/h) ( 70.0 km/h) 185 km/h.v = − + − =
(b) The direction of agv is
1 70.0 km/htan 22.3 (south of west).171 km/h
θ − − = = ° −
Problem 76. 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 20.0 east of due north to fly there directly . The plane arrives in2.00 h. What were the (a) magnitude and (b) direction of the wind velocity?
pgv
pav
agv
