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.

Chapter 4Motion 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 that move with respect to each other with constant velocity. (4-1)

Position VectorThe position vector of a particle is defined as a vector whose tail is at

a reference point (usually the origin ) and its tip is at the particle at

point .

The position vecExamp tor il ne: the f

r

O

P

igure is

ˆ ˆ ˆi j kr x y z

ˆ ˆ ˆ3i 2 j 5kr m

(4-2)

P

t2

t1

Displacement Vector

1 2For a particle that changes position vector from to we define the displacement

vector as follows:

r r

r

2 1.r r r

1 2The position vectors and are written in terms of components asr r

1 1 1 1ˆ ˆ ˆi j kr x y z

2 2 2 2

ˆ ˆ ˆi j kr x y z

2 1 2 1 2 1ˆ ˆ ˆ ˆ ˆ ˆi j k i j kr x x y y z z x y z

(4-3)

2 1x x x

2 1y y y

2 1z z z

The displacement r can then be written as

t

t + Δt

Average and Instantaneous Velocity

Following the same approach as in Chapter 2 we define the average velocity as displacement

average velocity = time interval

avg

ˆ ˆ ˆ ˆ ˆ ˆi j k i j kr x y z x y zv

t t t t t

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

lim

0

r drv

t dtt

(4-4)

t

t + Δt

2 2

avg

If we allow the time interval to shrink to zero, the following things happen:

1. Vector moves toward vector and 0.

2. The direction of the ratio (and thus ) approaches th

t

r r r

rv

t

avg

e direction

of the tangent to the path at position 1.

3. v v

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆi j k i j k i j kx y z

d dx dy dzv x y z v v v

dt dt dt dt

(4-5)

x

dxv

dt

y

dyv

dt

z

dzv

dt

The three velocity components are given by the equations

drv

dt

Average and Instantaneous Acceleration The average acceleration is defined as:

change in velocityaverage acceleration =

time interval

2 1avg

v v va

t t

We define the instantaneous acceleration as the limit:

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆlim i j k i j k i j k

0

yx zx y z x y z

dvdv dvv dv da v v v a a a

t dt dt dt dt dtt

The three acceleration components are given by the equations

xx

dva

dt y

y

dva

dt z

z

dva

dt

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

(4-6)

dva

dt

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:

0.v

0 0 0cosxv v 0 0 0sinyv v

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

(4-7)

0 0 0 0

0 The velocity along the -axis does not change:

(eq. 1)

Horizontal Motion:

Vertical Motion:

(eq. 2)

Alo

cos

ng the -axis the projec l

os

t

c

ix o

x

y

v v x x v t

a x

a g y

2

0

2

0 0 0

20 0

e is in free fall

(eq. 3) (eq. 4)

If we eliminate between equations 3 and 4 we get

sin

sin 2

2

.

siny o

y ot v

gtv v

v

gt y y v t

g y y

Here and are the coordinates

of the launching point. For many

problems the launching point is

taken at the origin. In this case

0 and 0.

In this analysis of projectile

mot

Note

on e

:

i w

o o

o o

x y

x y

neglect the effects of

air resistance.

g

(4-8)

2

0 0 0

20 2

0 0

0

cos (eq. 2) sin (eq. 4)2

If we eliminate between equations 2 and 4 we

ta

ge

The equation of the p

t:

This equatin . 2 co

ath

son descri

:

gtx v t y v t

t

gy x x

v

2

bes the path of the motion.

The path equations has the form: . This is the equation of a parabola.y ax bx

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 shortcut in many projectile

motion probl

Note:

ems.

(4-9)

O A

R

t

0 0 0 0

2

0 0 0 0

cos (eq. 1) cos (eq. 2)

sin (eq. 3) sin (eq. 4)2

The distance is defined as the horizontal ranHorizontal Range: ge

At point

x

y

v v x v t

gtv v gt y v t

OA R

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

Solution 1. 0. This solution corresponds to point and is of no interest.

Soluti

y

gt gtv t t v

t O

0 0

0 0

on 2. sin 0. This solution corresponds to point .2

2 sinFrom solution 2 we get . If we substitute in eq. 2 we get

gtv A

vt t

g

2 20 0

0 0 0

0

20

max

2sin cos sin 2 .

has its maximum value when 45 :

v vR

g g

R

vR

g

(4-10)2sin cos sin 2A A A

/2

3/2

sin

O

At

H

gMaximum Height H

2 20 0sin

2

vH

g

0 0

0 00 0

220 0 0 0

0 0 0 0

2 20 0

The -component of the projectile velocity is sin .

sinAt point : 0 sin

sin sin( ) sin sin

2 2

sin

2

y

y

y v v gt

vA v v gt t

g

v vgt gH y t v t v

g g

vH

g

(4-11)

2 2

0 0 0

22

We can calculate the maximum height using the third equation of kinematics

for motion along the -axis: 2 .

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

2

y yo o

yo y

yoyo

y v v a y y

y y H v v v a g

vv gH H

2 20 0sin

.2 2

v

g g

2 20 0sin

2

vH

g

At

H

g

(4-12)

Maximum Height H (encore)

Uniform Circular Motion:

A particle is in uniform circular motion if 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 toward the center C of the circular path, thus the name “centripetal.” 2. Its magnitude a is given by the equation

2

.v

ar

C P

R

Q

r

rr

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

2.r

Tv

(4-13)

A

P

C C

ˆ ˆ ˆ ˆi j sin i cos j sin cos

Here and are the coordinates of the rotating particle.

P Px y

P P

y xv v v v v

r rx y

ˆ ˆ ˆ ˆi j. Acceleration = i j.P P P Py x dy dxdv v vv v v a

r r dt r dt r dt

We note that cos and sin .P Py x

dy dxv v v v

dt dt

2 2 2 2

2 22 2ˆ ˆ cos i sin j cos sinx y

v v v va a a a

r r r r

2

2

/ sintan tan points toward .

/ cosy

x

v raa C

a v r

(4-14)

sinxv v cos yv v

2 2cos sin 1

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. Observers A and B determine the coordinates of particle P to be xPA and xPB , respectively.

. Here is the coordinate of with respect to .PA PB BA BAx x x x B A

We take derivatives of the above equation: PA PB BA

d d dx x x

dt dt dt

PA PB BAv v v If we take derivatives of the last equation and take

into account that 0BAdv

dt PA PBa a

Even though observers and

measure different velocities for ,

they measure the same accel

Note

erat

:

ion.

A B

P

(4-15)

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 and determine the position vector of particle to be

and , respectively. PA PB

A B P

r r

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

PA PB BA PA PB BA

d d dr r r v v v

dt 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 PBA

PA A PBPB BA

dvd d dv v v

dta

dt dta

dt

As in the one-dimensional

case, even though observers and

measure different velocities for ,

they measure the same accel

Note:

eration.

A B

P

(4-16)