1

Physics Chapter 6

Motion in Two Dimensions

2

Physics

Turn in Chapter 5 Homework & Worksheet & Lab

Lecture Q&A

3

1-D Review

Position:

t

dv

t

va

Displacement: Velocity:

– Instantaneous velocity = slope of Position-Time graph

Acceleration:

– Instantaneous acceleration = slope of Velocity-Time graph

dd = df – di

(unit: m)(unit: m)

(unit: m/s)

(unit: m/s2)

4

1D Review (2)

The 3 Great Equations of constant acceleration motion:

fv

fd

2fv

i fv at

21

2i i f fd v t at

2 2iv a d

v

d

2v

iv at

21

2i id v t at

2 2iv a d

5

Projectile Motion: 2-D

Choose coordinate so that the motion is confined in one plane 2-D motion

x

y

x y2 D

1 D

d yxv vx

vy

vi vix or vxi viy or vyi

a axay

6

Package Dropped From Airplane

7

Snapshots

Horizontal Direction: velocity is _______

Vertical Direction: velocity is __________

constant

increasing

8

Ball projected straight up from truck

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Snapshots

10

Projectile Motion: Horizontal and Vertical Displacements

11

Velocity Components of Projectile

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Independence of Motion

From observation:– The horizontal motion and the vertical motion are

independent of each other; that is, neither motion affects the other.

Connection:– Both horizontal and vertical motions are functions of

time. Time connects the two independent motions. – Though these two motions are independent, they

are connected to each other by time.

13

Initial Velocity

x

vi

y

i

ix

iy

v

v

Set up coordinate:

vix

viy

Initial velocity: vi at angle i with the horizontal

cosi iv

sini iv

Decompose vi into horizontal and vertical direction:

Horizontal: x direction Vertical: y direction

14

Projectile Motion Breakdown

Projectile: Object launched into the air Horizontal:

ix

x ix

x v t

v v

0netF 0netFa

m Constant velocity

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Projectile Motion Breakdown (2)

Similarly, Vertical: Constant acceleration (ay = g,

downward) ay = g if downward is defined as +y direction

ay = -g if upward is defined as +y direction

21

2i iy y

y iy y

y y v t a t

v v a t

16

Symmetry of Trajectory

Trajectory: Path of projectile Upward motion and downward motion are

symmetric at same height.– Upward total time is equal to downward total time if

landing point is at same height as initial point.– At the same height, speed is the same.

vx stays unchanged.

vy remains at the same magnitude but changes in direction. vy is upward when the projectile is going up and

vy is downward when coming down

17

Symmetry of Trajectory (2)

vx

vx

vy

vy

v

v

Velocity at any moment is tangent to the actual path. Velocity is horizontal at the top of the trajectory.

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Highest Point of Trajectory

y iyv v gt

2 2sin

2i iv

Hg

vy = 0

vx = vix = vi cos i

Maximum Height is

Minimum speed at top, but 0

0 t

y

siniy i iv v

g g

21

2iyv t gt 2

sin sin1sin

2i i i i

i i

v vv g

g g

2 2sin

2i iv

g

min cosi iv v

x

y

19

Horizontal Range

Two angles (complementary) with the same initial speed give the same range.

Horizontal range is maximum when the launch angle is 45o. Valid only when the landing point and initial point are at the same

height.

x 2

sin 2ii

v

g

y

cosi iv t

21sin

2i iv t gt 02 sin

0 or i ivt

g

2

sin 2ii

vR

g

2 sincos i ii i

vv

g

x

y

20

Equation of Path

No time involved. Can be used to find x or y when the other is given. Parabolic equation Trajectory is parabolic. Valid only when upward is defined as the +y direction and origin is

set at the initial point.

ix x cosi iv t iy y 21

sin2i iv t gt

cosi i

xtv

2

2tan2 cos

i

i i

gxy x

v

x

y

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Example: 150-1A stone is thrown horizontally at a speed of + 5.0 m/s from the top of a cliff 78.4 m high. a) How long does it take the stone to reach the bottom of the cliff? b) How far from the base of the cliff does the stone hit the ground? c) What are the horizontal and vertical components of the velocity of the stone just before it hits the ground?

vi = 5m/sx

y

0

y =

Set up coordinates as to the right. Then

ay =

y

x 0

a) t = ?

b) In x-direction:

g, yi = 0, xi = 0, vix = 5m/s, viy = 0

21

2 ya t

t

5.0 4.0 20.ix

mv t s m

s

78.4m

21

2i iy yy v t a t

In y-direction:

i ixx v t

2

2 2 78.44.0

9.8 /y

y ms

a m s

22

Continues …

c) Horizontal:

xv

yv

v

Vertical:

Speed:

Direction:

vi = 5m/sx

y

0

vvy

vx

5.0ix

mv

s

iyv ya t 29.8 4.0 39.m m

gt ss s

2 2 2 2(5.0 ) (39. ) 39.x y

m m mv v

s s s

1 139

tan tan 83 , below the horizontal5.0

y o

x

mv s

mvs

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Practice: A softball is tossed into the air at an angel of 50.0o with the vertical at an initial velocity of 11.0 m/s. What is its maximum height?

i

x

y

sin 11.0 sin 40.0 7.07oiy i i

m mv v

s s

yv t

maxy

Or Max. Height Eqn.

iy yv a t

iy 222

1 17.07 0.721 9.8 0.721 2.55

2 2iy y

m mv t a t s s m

s s

2

0 7.07 /0.721

9.8 /y iy

y

v v m ss

a m s

H

22

2 2

2

11.0 sin 40.0sin

2.552 2 9.8

o

i i

mv s

mmgs

11.0 / , 0, 0,i i iv m s x y

,max

max

0, , 0

?

x y ya a g v

y

90.0 50.0 40.0 ,o o o

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Practice: 152-10A tennis ball is thrown out a window 28 m above the ground at an initial velocity of 15.0 m/s and 20.0o below the horizontal. How far does the ball move horizontally before it hits the ground?

x

y

0, 0, 28 , 0, ,

15.0 / , 20.0 , ?

i i x y

oi i

x y y m a a g

v m s x

cos 15.0 / cos 20.0 14.1 /oix i iv v m s m s

sin 15.0 / sin 20.0 5.13 /oiy i iv v m s m s

y

229.8 /

28 5.132

m m sm t t

s

x

1.92 or 2.97t s s

0

iy21

2i y yv t a t

ix 14.1 1.92 27.1ix

mv t s m

s

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Uniform Circular Motion

Uniform Circular Motion:

v

v

Magnitude of velocity:

Circular path or circular arc Uniform = Constant speed (Constant velocity? why?)

Direction of velocity:

Velocity:

constant

changing

changing

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Centripetal Acceleration

Direction of acceleration is always toward the center of circle (or circular arc)

Centripetal

2va

r

v: speed of particle

r: radius of circle or circular arc, where

a

a

a

v

v

v

for uniform circular motion at any time.a v

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Direction of Acceleration

a in the same direction as v:

a opposite to v:

a v:

Speed: increases

Speed: decreases

Speed: does not change

Direction of velocity: changing

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Uniform Circular Motion

Period: Time for one complete cycle

2 rT

v

t

1f

T

Unit: f

Frequency: Number of cycles per unit time

Hertz Hz

for uniform circular motion2

vf

r

D

v

2 r

v

1

T

1

s

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Centripetal Force

Centripetal force is in general not a single physical force; rather, it is in general the net force.

Do not draw centripetal force on force diagram (Free Body Diagram)

2

c c

vF ma m

r

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Example: 156-12A runner moving at a speed of 8.8 m/s rounds a bend with a radius of 25 m. a) Find the centripetal acceleration of the runner. b) What agent exerts force on the runner (to round the bend)?

) 8.8 , 25 , ?c

ma v r m a

s

ca

b)

The friction the ground giving to the shoes provides the centripetal force.

22

2

8.8 /3.1 , towards the center

25

m sv m

r m s

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Practice: 156-13A car racing on a flat track travels at 22 m/s around a curve with a 56-m radius. Find the car’s centripetal acceleration. What minimum coefficient of static friction between the tires and road is necessary for the car to round the curve without slipping?

) 22 , 56 , ?c

ma v r m a

s

ca

b) The friction provides the centripetal force, so

22

2

22 /8.6 , towards the center

56

m sv m

r m s

, and ?cf ma

N

f

W mg

N

f

N cma

mg

2

2

8.60.88

9.8

c

ma s

mgs

W

N

f

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What if?

If centripetal force is not provided or not large enough, the object will not be able to move in the circle it intends to move.

Centripetal force disappears. Insufficient Centripetal force.

• •

2

c

vF m

r

33

Relative Velocity

Velocity of A relative to B Velocity of A measured by B Velocity of A at reference frame B

VA,B

, ,A C A Bv v

Bv

,C

1D or 2D 1D: Define positive direction 2D: vector addition (head-tail or parallelogram)

10m

s

+

2m

s

12m

s 2 10

m m

s s

,p gv ,t gv,p tv

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Example:159-22You are riding in a bus moving slowly through heavy traffic at 2.0 m/s. You hurry to the front of the bus at 4.0 m/s relative to the bus. What is your speed relative to the street?

Let direction bus moving = ”+” direction, also let

you = y, bus = b, street = s,

,y sv

then vb,s = 2.0m/s, Vy,b = 4.0 m/s, Vy,s =?

, ,y b b sv v 4.0 2.0 6.0m m m

s s s

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Example: 167-71A weather station releases a balloon to measure cloud conditions that rises at a constant 15 m/s relative to the air, but there is also a wind blowing at 6.5 m/s toward the west. What are the magnitude and direction of the velocity of the balloon?

Let west = +x, up = +y.

Balloon = b, air/wind = a, ground = g.

vb,a = 15 m/s, va,g = 6.5 m/s, vb,g = ?, = ?

x

y

vb,a

va,g

vb,g

,b gv

2 2, ,b a a gv v

2 2

15 6.5 16m m m

s s s

,1

,

tan b a

a g

v

v 1

15tan 67

6.5

o

msms

36

Practice:159-24A boat is rowed directly upriver at a speed of 2.5 m/s relative to the water. Viewers on the shore see that the boat is moving at only 0.5 m/s relative to the shore. What is the speed of the river? Is it moving with or against the boat?

Let upriver = ”+” direction, also let

boat = b, river/water = w, shore = s,

,b sv then vb,w = 2.5 m/s, Vb,s = 0.5 m/s, Vw,s =?

, ,b w w sv v

,w sv , , 0.5 2.5 2.0b s b w

m m mv v

s s s against

,Or w sv , ,w b b sv v

, ,b w b sv v 2.5 0.5 2.0m m m

s s s