Chapter 3: Kinematics in Two Dimensionsp

A scalar is a number with units It can be positiveVectors and Scalars

A scalar is a number with units. It can be positive, negative, or zero.

• Time: 100 s• Time: 100 s• Distance and speed are scalars, although they cannot be

negative because of the way they are definednegative because of the way they are defined.A vector is a quantity which specifies magnitude and

direction.direction.• Velocity: 4 m/s East• Acceleration: 10 m/s2 to the rightAcceleration: 10 m/s to the right

NotationNotationVector quantities are represented using boldq p g

type or an arrow above the symbol . The magnitude of a vector (a scalar) is g ( )represented using the absolute value symbol or without bold type:ypvelocity: = v = 5 m/s East

| v | = v = 5 m/sv

| v | = v = 5 m/s

Arrow Representation

Arrows are also used to represent vectors themselves.

• The length of the arrow is proportional to the magnitude.• The direction of the arrow indicates the direction of the vector.

4 m/s East 2 m/s North

Adding VectorsAdding Vectors

Only add vectors with the same units ForOnly add vectors with the same units. For example:

i OK bv3 = v1 + v2 is OK, butv3 = x1 + v2 does not make sense

Remember that a vector -A points opposite to A.Remember that a vector A points opposite to A.

A–A

Adding Vectors Graphicallyg p yTo add two vectors A + B, put the tail of B at

h i f A d h il f A hthe tip of A and connect the tail of A to the tip of B.

Use a ruler andUse a ruler and protractor to determine the magnitude and gdirection of C.

Subtracting Vectors, and Multiplication by a ScalarThe negative of a vector is a vector having the same length, but pointing in the opposite direction

So we can define the subtraction of one vector from another as

A vector can also be multiplied by a scalar c. We define this p yproduct so that cV has the same direction as V and a magnitude c|V|.

Components of a VectorComponents of a VectorA vector can be specified either by its magnitude and direction, or by its components in a coordinate system.

rθ r , rr,θ rx , ry

Be CarefulThe components of a vector are not always positive.

The angle is not always defined relative to the x axis.

Trigonometric Functions

sideopposite A

SOH CAH TOA

sideadjacenthypotnuse

sideoppositesin y

AAA

==θA

sideoppositehypotnuse

sideadjacent cos x

AAA

==θ Ay

22

sideadjacent sideoppositetan

x

y

AAA

AA

+

==θ

A

θ

yx AAA += Ax

Be careful with your calculator! A i d di ?Are you using degrees or radians?

ExampleExampleFind the x and y components of a vector of y pmagnitude 5, if its angle relative to the x axis is 200°.

( )cos 5cos 200 4.70xA A θ= = = −

( )sin 5sin 200 1.71yA A θ= = = −

Adding Vectors Using g gComponents

• Break vectors up into components.Add t• Add components.

• Find magnitude and direction of the sum from its components.

Example12. The vectors are shown in the figure below. Their magnitudes are given are given in arbitrary units. Determine the sum of the three vectors. Give the resultant in terms of (a) components, (b) magnitude and angle with x axis.

Unit VectorsUnit VectorsA unit vector is a vector with magnitude equal to 1

and pointing in a certain defined directionand pointing in a certain defined direction.Example: the unit vector in the x direction is usually

written: ˆˆ o rx iwritten:Now I can write a vector as its components

multiplied by the unit vector:

o rx i

p y

ˆ ˆx yA A x A y= +

orˆ ˆ

x yA A i A j= +

where Ax and Ay are scalars.x y j

ExampleWrite the vector in the figure in unit vector notation.

yy

θ

x

v 10v =x

50θ = °

Position and DisplacementPosition and Displacement Vectors

The displacementThe position vector rpoints from the origin to a

The displacement vector Δr indicates the change in position:points from the origin to a

particular location.g p

Δr = rf - ri

Writing of displacement in terms of components and unit vectors,

2 1 2 1 2 1ˆˆ ˆ( ) ( ) ( )r x x i y y j z z kΔ = − + − + −

We defined the instantaneous velocity as

0lim

t

r drvt dtΔ →

Δ= =

Δ

So we can write the instantaneous velocity as,

ˆ ˆˆ ˆ ˆ ˆx y z

dx dy dzv i j k v i v j v kdt dt dt

= + + = + +dt dt dt

Likewise for acceleration,,

2 1av

v vvat t

−Δ= =Δ Δt tΔ Δ

The instantaneous acceleration is given by

0lim v dva

t dtΔ

Δ= =

Δ0t t dtΔ → Δ

so

ˆ ˆˆ ˆ ˆ ˆyx zx y z

dvdv dva i j k a i a j a kdt dt dt

= + + = + +dt dt dt

Acceleration occurs whenever there is a change in velocity.g y

If a car travels in a circle at a constant speed of 50 mi/hr, is it accelerating?g

YES! Because the direction of the velocity vector is changingvector is changing.

Kinematic Equations for Constant Acceleration in 2 Dimensions

x-component y-component

21x x v t a t= + +21

2o yo yy y v t a t= + +2

2 2 2 ( )

o xo x

x xo x

x x v t a tv v a t

v v a x x

= + +

= +

= + ( )

2

2 2 2

o yo y

y yo y

y y

v v a t

v v a y y

= +

= + −2 ( )x xo x ov v a x x= + − ( )2y yo y ov v a y y= +

Projectile MotionProjectile MotionMotion in the x direction is independent fromMotion in the x direction is independent from motion in the y direction. We use the same equations from Chapter 2 but for eachequations from Chapter 2, but for each dimension separately.

There are not really any new equations in this chapter.p

The equations you need.1 2

0 012x xx x v t a t= + +

20 0

12y yy y v t a t= + + NOTE

For projectile motion problems, a

0

2x x xv v a t= +

For projectile motion problems, ax(the horizontal componant of the acceleration) will usually be zero i ll th ill b

( )0

2 20 2

y y y

f

v v a t

v v a x x

= +

= + −

since usually there will be no acceleration in the x direction

( )( )

0

2 20

2

2

x x x f i

y y y f i

v v a x x

v v a y y

+

= + −( )y y y f

Projectile Motion

• Assume that acceleration of gravity is g yconstant, downward and has a magnitude of g = 9.81 m/s2

• Air resistance is ignored• The Earth’s rotation is ignoredg⇒Horizontal velocity is constant: because

a = 0ax 0⇒Vertical motion governed by the constant

acceleration of gravityacceleration of gravity

Motion of a Projectile Launched H i t llHorizontally

Th d hThe dots represents the position of the object every 0.05 s. t = 1 sy

y0 = 8 m; x0 = 0v0y = 0; v0x = 6 m/say = -9.81 m/s2; ax = 0

1. Verify the position of the object at t = 1s.

2. What would be the position of the object at t = 1 s if it were dropped (v0x = 0)?

210 t t t+ + + 20 0 00

21

x y yx v t y y v t a t= + = + +

( )216 1 6 m 8 9.8 1 3.1 m2

x y= × = = − × =

21__________________________________________________________

( )218 9.8 1 3.1 m2

y = − × =

0x =

ExamplepPitcher’s mounds are raised to compensate for the vertical drop of

h b ll i l 18 h hthe ball as it travels 18 m to the catcher.

(a) If a pitch is thrown horizontally with an initial speed of 32 m/s (71 mi/hr) how far does it drop by the time it reaches the(71 mi/hr), how far does it drop by the time it reaches the catcher?

(b) If the speed of the pitch is increased does the drop distance(b) If the speed of the pitch is increased, does the drop distance increase, decrease or stay the same? Explain.

(c) If this baseball game were to be played on the moon, would(c) If this baseball game were to be played on the moon, would the drop distance increase, decrease, or stay the same? Explain.

0( )18 32 0.56 s

x

reach catcher reach catcher

a x v tt t− −

=

= =

( )220 0 0

1 19.8 9.8 0.56 1.54 m2 2

1 54

yy y v t t y y= + − − = − × = −

0 1.54 my y− =

(b) If the speed of the pitch increases, the time gets smaller, so the drop distance is less

(c) On the moon “g” is less than 9.8 m/s2, so the drop distance is less.

General Launch AngleConsider an object launched from the origin at an angle θ with respect to the horizontal.g p

y

vo

xθ

What are the x and y components of the initial velocity vector?vector?

Example:On a hot summer day a young girl swings on a rope above the local swimming hole. When she lets go of the rope her initial velocity is 2 25 / l f 35 0° b h h i l If h i i fli h2.25 m/s at an angle of 35.0° above the horizontal. If she is in flight for 1.60 s, how high above the water was she when she let go of the rope?p

What is the girl’s minimumWhat is the girl s minimum speed during her flight?

What is her acceleration at the top of her trajectory?

20 0

1sin 9.812

y y v t tθ= + −

20

210 2.25sin 35 9.812

y t t= + −

( ) ( )20

10 2.25sin 35 1.6 9.81 1.62

10 49 m

y

y

= + −

=0 10.49 my =

Girl’s minimum speed is at the maximum in her trajectory!C h ?Can you see why?

Girl’s acceleration is 9 81 m/s2 at the top of her trajectoryGirl s acceleration is 9.81 m/s at the top of her trajectory.What is her acceleration at other points on her trajectory?

RangeRangeThe range R of a projectile is the horizontal distance it g p jtravels before landing.

2⎟⎞

⎜⎛ v (BE CAREFUL!, This equation only

θ2sin0⎟⎟⎠

⎞⎜⎜⎝

⎛=

gvR

(BE CAREFUL!, This equation only works when the initial and final elevation are the same)

You should try to derive this equation! It’s not difficult. Just remember that x=R when the projectile hits the ground.

What angle θ results in the maximum range? Remember 0°≤ θ≤90°.

What if we do not ignore air resistance?

ExampleExampleA projectile is fired at 200 m/s at an angle of 45 degrees above

2 ⎞⎛

p j g gthe horizontal. How far does it travel?

θ2sin20⎟⎟⎠

⎞⎜⎜⎝

⎛=

gvR

2

2

(200m/s) sin(2(45 )) 4081m9 8m/s

oR⎛ ⎞

= =⎜ ⎟⎝ ⎠9.8m/s⎝ ⎠

Maximum HeightThe maximum height (and therefore the “hang time”) of aThe maximum height (and therefore the hang time ) of a projectile depends only on the vertical component of its initial velocity.y

At ymax, the vertical velocity vy is zero.

ygv

yavv yy

)(2sin0

2

max22

0

20

2

θ −+=

Δ+=

gvy

2sin22

0max

θ=

ExampleExample

From the previous example how high does the projectile go?From the previous example, how high does the projectile go?

sin22v θ2sin

max =o

gvy θ

( ) m1020)89(2

)45(sin/200 22

max ==osmy

)8.9(2maxy

Uniform Circular MotionUniform Circular Motionv1

C id ti l i i i lConsider a particle moving in a circle.

Is the particle accelerating?

lim v dvad

Δ= =

Δ0t t dtΔ → Δ

v2- The speed of the particle is constant- The direction of the ball is changing

What is the direction of the acceleration?

Magnitude of Radial Acceleration

This type acceleration is called centripetal or

Magnitude of Radial Acceleration

This type acceleration is called centripetal or radial acceleration. The magnitude of the acceleration is given byacceleration is given by,

2

Rvar

=

Period and FrequencyThe amount of time that it takes the particle to make one full revolution is called the period , T.

The number of times that the particle goes around the circle per unit time is called the frequency, f. q y, f

1Tf

=f

The speed of a particle moving around in a circle is given by,

distance 2 2time

rv rfTπ π= = =

Example

33 A h t tt th th th h t ( 7 3 k ) ith33. A shotputter throws throws the shot (mass = 7.3 kg) with an initial speed of 14.0 m/s at a 40° angle to the horizontal. Calculate the horizontal distance traveled by the shot if it leaves the athlete’s hand at a height of 2.2 m above the ground.

Examplep

18. The position of a particular particle as a function of time is18. The position of a particular particle as a function of time is given by,

2 ˆˆ ˆ(7 60 8 85 )r ti j t k m= +

(a) Determine the particle’s velocity and acceleration as a

(7.60 8.85 )r ti j t k m= + −

function of time.(b) Describe the motion in each direction.

Examplep

Chapter 2 #40. A space vehicle accelerates uniformly from 65 m/s at t=0 to 162 m/s at 10.0 s. How far did it move b t t 2 0 d t 6 0 ?between t=2.0s and t=6.0 s?

Example

A baseball is hit at an angle of 50° relative to the horizontalA baseball is hit at an angle of 50 relative to the horizontal with an initial speed of 120 miles/hour.

( ) H hi h d it ?(a) How high does it go?(b) How long does it spend in the air?(c) How far does it travel?