motion in a plane

Physics Chapter 3 1

Motion in a Plane

Chapter 3

Physics Chapter 3 2

Position Vector

• Goes from the origin to the object – point p

x

y

rP

Physics Chapter 3 3

Average Velocity Vector• Change in displacement over change in

time

trvav

Physics Chapter 3 4

Instantaneous velocity vector

trv

t

0lim

Physics Chapter 3 5

Instantaneous velocity

• In two dimensions

22yx vvvv

x

y

vv

tan

Physics Chapter 3 6

Average acceleration vector

tvaav

Physics Chapter 3 7

Instantaneous acceleration vector

tva

t

0lim

Physics Chapter 3 8

Example

• A particle has x and y coordinates (4.0 m, 2.0 m) at time t1 = 2.0 s and coordinates (7.0 m, 6.0 m) at time t2 = 2.5 s. Find the components of the average velocity and the magnitude and direction of the average velocity during this time interval.

Physics Chapter 3 9

Projectile Motion

• Projectile – any body given an initial velocity which then follows a path (trajectory) based on gravitational acceleration and air resistance

• Thrown ball• Bullet• Dropped package

Physics Chapter 3 10

Projectile motion

• We will neglect air resistance• We will also neglect curvature and rotation

of the earth


Projectile Motion

• Consists of both horizontal and vertical motion

• We will break these problems into x and y components to make them easier to solve


Projectile motion

• Two dimensional – gravity only acts vertically

• We assign y as the vertical direction• We assign x as the horizontal direction

0xa gay


Projectile Motion

200 2

1 tatvxx xx tavv xxx 0

xx vv 0 tvxx x 0

gtvv yy 0 200 2

1 gttvyy y

0 0


Projectile Motion

• If we express initial velocity in terms of its magnitude and angle with the x-axis

000 cosvv x

000 sinvv y


Projectile Motion

• We can calculate the projectile’s speed and the direction of its velocity

22yx vvv

x

y

vv

tan


Trajectory shape

• Projectiles always travel in parabolas


Example

• A policeman chases a thief across city rooftops. They are both running at 5 m/s when they come to a gap between buildings that is 4 m wide and has a drop of 3 m.

• The thief leaps at 5 m/s at an angle of 45°. Does he clear the gap?

• The policeman leaps at 5 m/s horizontally. Does he clear the gap?


Example0x

0y


Example – the thief

tvxx x 0

tvx x

45 @ sm 50 v

00 x

45cossm 5

xv

tx

45cos

sm 5



200 2

1 gttvyy y

20 2

1 gttvy y

0

00 y

45 @ sm 50 v 45sin

sm 50

yv



0m 345sinsm 5

sm 9.4 2

2

tt

021

02 ytvgt y

m 3roof, thehits hewhen

y


Quadratic formula

02 cbxax

aacbbx

242



0m 345sinsm 5

sm 9.4 2

2

tt

2

2

2

sm 9.42

m 3sm 9.4445sin

sm 545sin

sm 5

t



2

2

2

2

2

sm 8.9

sm 8.58

sm 5.12

sm3.54

t

2sm 8.9

sm8.44

sm3.54

t



22 sm 8.9

sm 9.4

or

sm 8.9

sm 98.11

tt

s 22.1t



• Yes, he clears the gap

tx

45cos

sm 5

s 22.145cossm 5

x

m 32.4x


Example – the policeman

• Everything is the same, except that

0 @ sm 50 v

sm 5xv

00 yv



tx

sm 5

g

yt

21

2

2

21 gty



2

2

sm 9.4

m 3t

m 3roof, thehits hewhen

y

22 s 61.0t

s 78.0t



• No, he does not clear the gap

tx

sm 5

s 78.0sm 5

x

m 9.3x


Example 2• A rookie quarterback throws a football

with an initial upward velocity component of 16.0 m/s and a horizontal component of 20.0 m/s.

a) How much time is required for the football to reach the highest point of the trajectory?


Example 2 a)

• At the highest point, vy must be zero.• Before that, it was moving up, and after

that it is moving down, so it must momentarily stop at that point.

gtvv yy 0

0yv sm 0.160 yv


Example 2 a)

t

2s

m 8.9sm 0.160

2sm 8.9

sm 0.16

t

s 6.1t


Example 2 b)

• b)How high is this point?

200 2

1 gttvyy y

00 ysm 0.160 yv

s 6.1t


Example 2 b)

22 s 6.1

sm 8.9

21s 6.1

sm 16

y

m 5.12m 6.25 y

m 13.1y


Example 2 c)

• c) How much time is required for the ball to return to its original height?

200 2

1 gttvyy y

00 y

sm 0.160 yv

0y


Example 2 c)

20 2

10 gttv y

021

02 tvgt y

021

0

yvgtt


Example 2 c)

0t 021or 0

yvgt

yvgt 021

gv

t y02


Example 2 c)

2sm 8.9

sm 162

t

s 2.3t


Example 2 d)

• d) How does the answer to part c) compare to the answer to part a)?

• It is double. When something is thrown upward, it always takes the same amount of time to go up as it does to come down.

• So the total flight time (back to the original height) is always twice the time to the highest point.


Example 2 e)

• e) How far has the football traveled horizontally?

tvxx x 0

00 xsm 20xv

s 2.3t


Example 2 e)

s 2.3sm 20

x

m 64x


Normal acceleration

• Normal means perpendicular• The component of acceleration that is

perpendicular to the path• Shows change in direction of velocity

normaa or


Tangential acceleration

• Parallel to the path• Shows change in magnitude of velocity

(change in speed)

tanaa or


Uniform Circular Motion

• When an object moves around a circle with constant speed, the acceleration vector points towards the center of the circle.

• The velocity is tangent to the circle. In order for the speed to remain constant, the acceleration must be towards the center of the circle.

• We call this centripetal acceleration.


Uniform circular motion

• See Fig. 3-11 on page 63

Rs

vv

1s

Rvv 1

tRsv

tvaav

1

ts

Rvat

0lim

Rva

2


Uniform circular motion• The period of any motion is the time it takes to

make one complete cycle. It is denoted with the letter T.

• In this case it’s the time to go around the circle once.

tdvav

T

Rv 2

2

24TRa


Example

• The radius of the earth’s orbit around the sun is 1.49 x 1011 m, and the earth travels around the sun in 365.25 days.

• What is the earth’s orbital speed in m/s?• What is the acceleration of the earth

toward the sun in m/s2?

motion in a plane

Documents

example physics chapter

thief physics chapter

velocity physics chapter

time physics chapter

gap physics chapter

policeman physics chapter

dimensions physics chapter

b physics chapter