PH 221-3A Fall 2009

Motion in Two and Three DimensionsDimensions

Lectures 4, 5

Chapter 4 (Halliday/Resnick/Walker, Fundamentals of Physics 8th edition)

1

Chapter 4Motion in Two and Three DimensionsMotion 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 willrestriction 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-

d th di i l tiand three- dimensional motion:

DisplacementAverage and instantaneous velocity g yAverage and instantaneous acceleration

We will consider in detail projectile motion and uniform circular motion as l f i i di iexamples 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 withbetween two reference systems which move with respect to each other with constant velocity.

2

Position VectorTh iti t f ti l i d fi d t h t il i trThe 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.

r

pThe position vectoExampl r in te : he figure is:

ˆˆ ˆr xi yj zk= + +rr xi yj zk+ +

( )ˆ( )ˆˆ ˆ3 2 5r i j k m= − + +r

P

3

Displacement Vector

1 2For a particle that changes postion vector from to we define the displacement vector as follows:

r rrΔ

r r

r2 1r r rΔ = −

r r r

1 2The position vectors and are written in terms of components as:r rr r

1 1 1 1ˆˆ ˆr x i y j z k= + +

r2 2 2 2

ˆˆ ˆr x i y j z k= + +r

( ) ( ) ( )2 1 2 1 2 1ˆ ˆˆ ˆ ˆ ˆr x x i y y j z z k xi yj zkΔ = − + − + − = Δ + Δ + Δ

r

The displacement r can then be written as:Δr

2 1x x xΔ = −

2 1y y yΔ = −

t

2 1y y y

2 1z z zΔ = −

t2

t1

4

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( ) l l ti t th iti di ti ?

(a) The position vector, according to is ˆ ˆ= ( 5 .0 m ) i + (8 .0 m )jr −r .

(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= = − + + =

r (c) Many calculators have polar ↔ rectangular conversion capabilities which make this

(g) an angle relative to the positive x direction? ˆˆ ˆr xi yj zk= + +r

(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 mta n 5 8 o r 1 2 25 .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(d) The sketch is shown on the right. The vector is 122 counterclockwise from the +x direction. (e) The displacement is r r r′Δ = −

r r r where rr is given in part (a) and ˆ (3 .0 m ) i.r ′ =r Therefore, ˆ ˆ(8 .0 m ) i (8 .0 m )jrΔ = −

r . (f) The magnitude of the displacement is 2 2| | (8 .0 m ) ( 8 .0 m ) 1 1 m .rΔ = + − =

r (g) The angle for the displacement, using tanθ=ry/rx, is

1 8 .0 mta n = 4 5 o r 1 3 5− ⎛ ⎞ − ° °⎜ ⎟⎝ ⎠

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 fourth quadrant. A sketch of rΔ

r is shown on the right.

Average and Instantaneous VelocityFollowing the same approach as in chapter 2 we define the averageFollowing 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

Δ Δ + Δ + Δ Δ Δ Δ= = = + +Δ Δ Δ Δ Δ

rr

We define as the instantaneous velocity (or more simply the velocity) as the limit:

t

( p y y)

lim r drvt dt

Δ= =

Δ

r rr

t + Δt 0tΔ →

6

2 1

If we allow the time interval t to shrink to zero, the following things happen: 1. Vector moves towards vector and 0r r r

ΔΔ →

r r r2 1

2. The direction of the ratio (and thus )approachest avgr vΔΔ

rr the direction

f th t t t th th t iti 1 of the tangent to the path at position 13. avgv v→

r r

( )ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆd dx dy dzi j k i j k i j kr ( ) x y zyv xi yj zk i j k v i v j v k

dt dt dt dt= + + = + + = + +

r

The three velocity components are given by

tx

dxvdt

=

the equations:

t + Δt dt

ydyvdt

=drvdt

=r

r

zdzvdt

= 7

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

= − +

=

r

r r in unit-vector notation

(a) Eq. leads to

electron s velocity ( )? At 2.00 , what is (b) 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?

drv =r

r( ) q

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

= − = −

dt

(b) Evaluating this result at t = 2.00 s produces ˆ ˆ= (3.00i 16.0j) m/s.v −r

(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= = + − =

r (d) The angle of rv at that moment is

1 16.0 m/st 79 4 101− ⎛ ⎞−° °⎜ ⎟

1tan 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

8

281 counterclockwise from +x) since the signs of the components imply the vector is inthe fourth quadrant.

Average and Instantaneous AccelerationThe average acceleration is defined as:g

change in velocityaverage acceleration = time interval

2 1avg

v v vat t− Δ

= =Δ Δ

r r rr

We define as the instantaneous acceleration as the limit:

( )ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆlim yx zdvdvv dv d dva v i v j v k i j k a i a j a kΔ= = = + + = + + = + +

r rr ( )lim

0

x y z x y za v i v j v k i j k a i a j a kt dt dt dt dt dt

t

+ + + + + +Δ

Δ →

The three acceleration components are given by

Note: Unlike velocity, the acceleration vector does not have any specific relationship with the path.

p g ythe equations:

dv dv dv dvrxx

dvadt

= yy

dva

dt= z

zdvadt

= dvadt

=r

9

3 4

P rob lem 1 3 . T he po sition o f a p artic le m o ving in an xy p lane is g iven b yˆ ˆ( 2 .0 0 5 .00 ) (6 .0 0 7 .0 0 ) , w ith in m e te rs an d in secon ds.

In un it-vec to r no ta tio n , ca lcu la te (a ) , (b ) , (c

r

r t t i t j r tr v

= − + −

r

r r

r r ) fo r 2 .00 s . (d ) W hata t =r

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

, ( ) , ( ) , ( ) ( )is th e an gle b e tw een th e p ositive d irec tio n o f the ax is an d a lin e tangen t to th e p artic le a t = 2 .0 0 s?

xt drv

dt=

rr x

xdvadt

=

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 . 0 0ˆ ˆ ˆ ˆ [ 2 . 0 0 ( 8 ) 5 .0 0 ( 2 ) ] i + [ 6 .0 0 7 .0 0 ( 1 6 ) ] j ( 6 . 0 0 i 1 0 6 j) mtr = = − − = −

r

(b) Taking the derivative of the given expression produces 2 3ˆ ˆ( ) = ( 6 . 0 0 5 . 0 0 ) i 2 8 . 0 jv t t t− −

r

h h i ( ) h i i d d i Thi bwhere we have written v(t) to emphasize its dependence on time. This becomes, at t = 2.00 s, ˆ ˆ = ( 1 9 . 0 i 2 2 4 j ) m / s .v −

r (c) Differentiating the rv t( ) found above, with respect to t produces 2ˆ ˆ1 2 .0 i 8 4 . 0 j ,t t−

which yields 2ˆ ˆ = ( 2 4 . 0 i 3 3 6 j ) m / sa −r at t = 2.00 s.

(d) The angle of rv , measured from +x, is either

1 2 2 4 m / st a n 8 5 . 2 o r 9 4 . 81 9 . 0 m / s

− ⎛ ⎞−= − ° °⎜ ⎟

⎝ ⎠

10

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 inthe fourth quadrant.

Projectile MotionThe motion of an object in a vertical plane under the influence ofThe 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 ˆ ˆo ox oyv v i v j= +r

The horizontal and vertical velocity components are:

cosv v θ= sinoy o ov v θ=cosox o ov v θ oy o o

Projectile motion will be analyzed in a horizontal and a vertical motion

g

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-of each other. Motion along the xaxis has zero acceleration. Motion along the y-axis has uniform acceleration a = -gacceleration ay -g

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( )x 0 0

0 The velocity along the x-axis does not change (eqs.1) (eqs

Horizontal Motion:.2)v cos cos

x

o ox o o

av x x v t v tθ θ= −

=

= = ( )x 0 0 ( q ) ( qVertic

) al Motio Along the y-axin s the proj: ey

o ox o o

a g= −

( )2 2

ctile is in free fall

(eqs 3) (eqs 4)sin singt gtv v gt y y v t v tθ θ= − − = − = −( )

( ) ( )

0 0 0 0

220 0

(eqs.3) (eqs.4)

If we eliminate between equations 3 an

sin sin2

d2

sin 2 4 we get:

y o oy

y o

v v gt y y v t v t

v v g y yt

θ θ

θ

= = =

= − −

Here and are the coordinatesof the launching point. For many

o ox yg

problems the launching point istaken at the origin. In this case

0 d 00 and 0 In this analysis of projectile

motiNote:

on we

o ox y= =

neglect the effects ofo o we eg ec e e ec s oair resistance

12

( ) ( )2

0 0

cos (eqs 2) sin (eqs 4)

The equation of the pa :thgtx v t y v tθ θ= = −( ) ( )

( ) 2

0 0 cos (eqs.2) sin (eqs.4)2

If we eliminate between equations 2 and 4 we get:

Thi i d

oo

g

x v t y v

t

t

θ

θ θ

ib h h f h i( )( )

22tan

2 cos This equatio e dno

o o

gy x xv

θθ

= −

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 canNote:complicated to be useful. Appearances candeceive: Complicated as it is, this equationcan be used as a short cut in many projectilemotion problems

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( )

( )

x 0 0

2

0 0 0 0

v cos (eqs.1) cos (eqs.2)

sin (eqs 3) sin (eqs 4)

o ov x v t

gtv v gt y v t

θ θ

θ θ

= =

= − = − π/2

3π/2 ϕ

sinϕ

O( )0 0 0 0sin (eqs.3) sin (eqs.4)2

The distance OA is defined as the horizantal range Horizontal RangAt point A

e:

yv v gt y v t

R

θ θ

we have: 0 From equation 4 we have:y =

π/2

At point A

( )2

0 0 0 0

we have: 0 From equation 4 we have:

sin 0 sin 0 This equation has two solutions:2 2

ygt gtv t t vθ θ⎛ ⎞− = → − =⎜ ⎟

⎝ ⎠Solution 1. 0 This solution correspond to point O and is of no interest

Solution 2.

t =

0 0 sin 0 This solution correspond to point A2gtv θ − =

0 0

22 sinFrom solution 2 we get: If we substitute in eqs.2 we get:vt t

gθ

=

2 22

O A t

2 2

o

2 sin cos sin 2

has its maximum value when 45

o oo o o

v vRg g

R

θ θ θ

θ

= =

= °

R

o2

maxovRg

=2sin cos sin 2A A A= 14

Atg

Maximum height H

H 2 2sin2

o ovHgθ

=g

Th f h j il l i i i θ0 0

0 00 0

The y-component of the projectile velocity is: sin

sinAt point A: 0 sin

y

y

v v gt

vv v gt tg

θ

θθ

= −

= → − → =

( ) ( )22

0 0 0 00 0 0 0

sin sin( ) sin sin2 2

g

v vgt gH y t v t vg gθ θθ θ

⎛ ⎞= = − = − →⎜ ⎟

⎝ ⎠2 2sin

2o ovH

gθ

= 15

Maximum height H (encore)At

2 2sin2

o ovHgθ

=Hg

g

( )2 2

We can calculate the maximum height using the third equation of kinematicsfor motion along the y-axis: 2v v a y y= + −( )

2

for motion along the y axis: 2

In our problem: 0 , , sin , 0 , and

2

y yo o

o yo o o y

v v a y y

y y H v v v a gθ

+

= = = = = − →2 2 2sinyov v θ2 2yov gH H− = − →

sin2 2

yo o ovg g

θ= =

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 di tidirection

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 projectileb) The time to rise to the top of the trajectoryc) The total time of the projectile in the aird) The velocity of the projectile at the grounde) The range of the projectile

Problem 37. 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= +r

) W hat total horizontal distance does the ball travel? W hat are the (c) magnitude and (d) angle (below the horizontal) of the ball's velocity just before it hits the ground?We designate the given velocity ˆ ˆ( 7 . 6 m / s ) i ( 6 . 1 m / s ) jv = +

r as rv 1 − as opposed to the velocity when it reaches the max height rv 2 or the velocity when it returns to the groundrv 3 − and take rv 0 as the launch velocity, as usual. The origin is at its launch point on theground. (a) Different approaches are available, but since it will be useful (for the rest of theproblem) 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= − Δ ⇒ = −

which yields v0 y = 14.7 m/s. Knowing that v2 y must equal 0, we use Δy = h for the maximum height:

2 2 2 22 0 ( 1 4 7 m / s ) 2 ( 9 8 m / s )v v g h h⇒2 0 2 0 ( 1 4 . 7 m / s ) 2 ( 9 . 8 m / s )y yv v g h 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

210 v t g t R v t0 00 ,2y xv t g t 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 andobtain

R = 2(7.6 m/s)(14.7 m/s)/(9.8 m/s2) = 23 m. ( ) Si 7 6 / d 14 7 / h(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 ) ( 1 4 . 7 m / s ) 1 7 m / s .x yv v v= + = + − =

(d) The angle (measured from horizontal) for rv 3 is one of these possibilities:

1 4 7⎛ ⎞

21

1 1 4 . 7 mt a n 6 3 o r 1 1 77 . 6 m

− −⎛ ⎞ = − ° °⎜ ⎟⎝ ⎠

where we settle on the first choice (–63°, which is equivalent to 297°) since the signs ofits 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 p pradius 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 g p p g p ychanges 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 . ts vecto po ts towa ds t e ce te C o t e c cu a pat , t us t e a e“centripetal” 2. Its magnitude a is given by the equation:

2var

=

Q

rThe time T it takes to complete a full revolution is known as the “period”. It is given by theC P

Rr

r known as the period . It is given by the equation:

2 rπR 2 rTvπ

=22

( ) ( )ˆ ˆ ˆ ˆsin cos sin cos

Here and are the coordinates of the rotating particle

P Px y

y xv v i v j v i v jr r

x y

θ θ θ θ= + = − + = =r

Here and are the coordinates of the rotating particleP Px y

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

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − + = − +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

rr r

⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

We note that: cos and sinP Py x

dy dxv v v vdt dt

θ θ= = = = −

2 2 2 2v v v v⎛ ⎞ ⎛ ⎞ ( ) ( )2 22 2ˆ ˆ cos sin cos sinx yv v v va i j a a ar r r r

θ θ θ θ⎛ ⎞ ⎛ ⎞

= − + − = + = + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

r

( )2 / sinv ra θ−( )( )2

tan tan points towards C/ cos

y

x

aa

a v r

θφ θ φ θ

θ= = = → = →

−r

sinxv v θ= −cosv v θ=

A

P

C C

cos yv v θ=

( ) ( )2 2cos sin 1θ θ+ =23

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?

2v2 rπWe 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

var

=2 rTvπ

=

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

v ×2 3 27 49 10

8 00. m / s

/ 2c h

24

ar

= =×

=67 01 108 00

.. .

mm / s2c h

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 k t t l it ith t t b A Ob A d B d t iknown 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

= + →

v v v+ If t k d i ti f th l t ti d t kPA 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,Note:

they measure the same acceleration25

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-planerespect to observer A in the xy-plane.

Observers A and B determine the position vector of particle P to be and , respectively. PA PBr rr r

We take the time derivative of both sides of the equationPA PB BAr r r= +r r r

d d dr r r r r r +r r r

PA PB BA PA PB BAr r r v v vdt dt dt

= + → = +r r r r r r

PA PB BAv v v= +

If we take the time derivative of both sides of the last equation we have:d d d dr If we take into account that 0 BA

PA A BB AP PB Pd d d dvv v vdt dt d

at dt

a= + = =→r r

rr r r

As in the one dimensional case, even though observers A and Bmeasure different velocities for P

Note:

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 20.0 east of due north to fly there directlyo . The plane arrives in

The destination is D→

= 800 km j^ where we orient axes so that +y points north and +x

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 xpoints east. This takes two hours, so the (constant) velocity of the plane (relative to theground) 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 ofth i ( i d) l ti t th d

→ Ththe air (wind) relative to the ground vag . Thus,

(400 km/h) j^ = (500 km/h) cos70º i^ + (500 km/h) sin70º j^ + vag

→

agvuur

which yields vag →

=( –171 km/h)i^ –( 70.0 km/h)j^ . ( ) Th it d f r i 2 2| | ( 171 k /h) ( 70 0 k /h) 185 k /hr pgv

uuuruuur

ag

(a) The magnitude of agv is 2 2ag| | ( 171 km/h) ( 70.0 km/h) 185 km/h.v = − + − =

(b) The direction of agvr is

pgpav

27

1 70.0 km/htan 22.3 (south of west).171 km/h

θ − −⎛ ⎞= = °⎜ ⎟−⎝ ⎠