Chapter 3

Motion in Two and Three Dimensions

Types of physical quantities

• In physics, quantities can be divided into such general categories as scalars, vectors, matrices, etc.

• Scalars – physical quantities that can be described by their value (magnitude) only

• Vectors – physical quantities that can be described by their value and direction

Vectors

• Vectors are labeled either a or

• Vector magnitude is labeled either |a| or a

• Two (or more) vectors having the same magnitude and direction are identical

a

Vector sum (resultant vector)

• Not the same as algebraic sum

• Triangle method of finding the resultant:a) Draw the vectors “head-to-tail”b) The resultant is drawn from the tail of A to the head of B

A

B

R = A + B

Addition of more than two vectors

• When you have many vectors, just keep repeating the process until all are included

• The resultant is still drawn from the tail of the first vector to the head of the last vector

Commutative law of vector addition

A + B = B + A

Associative law of vector addition

(A + B) + C = A + (B + C)

Negative vectors

Vector (- b) has the same magnitude as b but opposite direction

Vector subtraction

Special case of vector addition: A - B = A + (- B)

Multiplying a vector by a scalar

• The result of the multiplication is a vector

c A = B

• Vector magnitude of the product is multiplied by the scalar

|c| |A| = |B|

• If the scalar is positive (negative), the direction of the result is the same as (opposite to that) of the original vector

Vector components

• Component of a vector is the projection of the vector on an axis

• To find the projection – drop perpendicular lines to the axis from both ends of the vector – resolving the vector

Vector components

x

y

yx A

AAAA tan

22

inAAAA yx s cos

Unit vectors

• Unit vector:A) Has a magnitude of 1 (unity)B) Lacks both dimension and unitC) Specifies a direction

• Unit vectors in a right-handed coordinate system

Adding vectors by components

In 2D case:

jbibb

jaiaa

yx

yx

ˆˆ

ˆˆ

bar

yyy

xxx

bar

bar

Chapter 3Problem 42

Vector A has magnitude 1.0 m and points 35° clockwise from the x-axis. Vector B has magnitude 1.8 m. Find the direction of B such that A + B is in the y-direction.

Position

The position of an object is described by its position vector,

kzjyixr ˆˆˆ

r

Displacement

The displacement vector is defined as the change in its position,

if rrr

)ˆˆˆ()ˆˆˆ( kzjyixkzjyixr iiifff

kzzjyyixx ifififˆ)(ˆ)(ˆ)(

kzjyixr ˆˆˆ

r

Velocity

• Average velocity

• Instantaneous velocity

t

kzjyix

t

rv

ˆˆˆ

kdt

dzj

dt

dyi

dt

dx

dt

rdv ˆˆˆ

kvjvivv zyxˆˆˆ

Instantaneous velocity

• Vector of instantaneous velocity is always tangential to the object’s path at the object’s position

Acceleration

• Average acceleration

• Instantaneous acceleration

t

v

t

vva if

kajaiadt

vda zyx

ˆˆˆ

Acceleration

• Acceleration – the rate of change of velocity (vector)

• The magnitude of the velocity (the speed) can change – tangential acceleration

• The direction of the velocity can change – radial acceleration

• Both the magnitude and the direction can change

Chapter 3Problem 23

What are (a) the average velocity and (b) the average acceleration of the tip of the 2.4-cm-long hour hand of a clock in the interval from noon to 6 PM? Use unit vector notation, with the x-axis pointing toward 3 and the y-axis toward noon.

Projectile motion

• A special case of 2D motion

• An object moves in the presence of Earth’s gravity

• We neglect the air friction and the rotation of the Earth

• As a result, the object moves in a vertical plane and follows a parabolic path

• The x and y directions of motion are treated independently

Projectile motion – X direction

• A uniform motion: ax = 0

• Initial velocity is

• Displacement in the x direction is described as

iixi vv cos

tvxx iii )cos(

Projectile motion – Y direction

• Motion with a constant acceleration: ay = – g

• Initial velocity is

• Therefore

• Displacement in the y direction is described as

iiyi vv sin

2

2

1)sin( gttvyy iii

gtvv iiy sin

Projectile motion: putting X and Y together

constvv iix cos

2

2

1)sin( gttvyy iii

gtvv iiy sin

tvxx iii )cos(

Projectile motion: trajectory and range

ii

g

vR 2sin

2

2

2

)cos(2)(tan

iii v

gxxy

ii

g

vh 2

2

sin2

Projectile motion: trajectory and range

ii

g

vR 2sin

2

2

2

)cos(2)(tan

iii v

gxxy

ii

g

vh 2

2

sin2

Chapter 3Problem 33

A carpenter tosses a shingle horizontally off an 8.8-m-high roof at 11 m/s. (a) How long does it take the shingle to reach the ground? (b) How far does it move horizontally?

Uniform circular motion

• A special case of 2D motion

• An object moves around a circle at a constant speed

• Period – time to make one full revolution

• The x and y directions of motion are treated independently

v

rT

2

Uniform circular motion

• Velocity vector is tangential to the path

• From the diagram

• Using

• We obtain

jvivjvivv yxˆ)cos(ˆ)sin(ˆˆ

r

xpcosr

y psin

jr

vxi

r

vyv pp ˆˆ

Centripetal acceleration

dt

vda

j

dt

dx

r

vi

dt

dy

r

v pp ˆˆ

jvr

viv

r

v ˆ)sin(ˆ)cos(

)sinˆcosˆ(2

jir

v

22yx aaa

r

v2

x

y

a

atan

cos)/(

sin)/(2

2

rv

rv

tan

222

sincos r

v

jdt

dvi

dt

dv yx ˆˆ

jr

vxi

r

vyv pp ˆˆ

jvivjvivv yxˆ)cos(ˆ)sin(ˆˆ

Centripetal acceleration

During a uniform circular motion:

• the speed is constant

• the velocity is changing due to centripetal (“center seeking”) acceleration

• centripetal acceleration is constant in magnitude (v2/r), is normal to the velocity vector, and points radially inward

Chapter 3Problem 38

How fast would a car have to round a 75-m-radius turn for its accelerationto be numerically equal to that of gravity?

Relative motion

• Reference frame: physical object and a coordinate system attached to it

• Reference frames can move relative to each other

• We can measure displacements, velocities, accelerations, etc. separately in different reference frames

Relative motion

• If reference frames A and B move relative to each other with a constant velocity

• Then

• Acceleration measured in both reference frames will be the same

BAPBPA vvv

BAPBPA rrr

PBPA aa

BAv

Questions?

Answers to the even-numbered problems

Chapter 3

Problem 28:

196 km/h

Answers to the even-numbered problems

Chapter 3

Problem 30:

3.6 ˆi − 4.8 ˆj m/s2

6.0 m/s2; 53°

Answers to the even-numbered problems

Chapter 3

Problem 34:

1.5 m