M E C H A N I C S I

D R . B E N J A M I N C H A N

A S S O C I A T E P R O F E S S O R

P H Y S I C S D E P A R T M E N T

N O V E M B E R 2 0 1 3

PS 11 GeneralPhysics I for the Life Sciences

Definition

Mechanics is the study of motion and its causes

Before discussing the cause of motion, we need to know how to describe motion (kinematics)

first.

Kinematics

How would you describe a moving

object?

Locate the Object

Establish Reference Frame

Point of reference needed (origin)

Reference direction needed

Construct the position vector (magnitude and direction) from the origin to the object

We can also call this the displacement of the object from the origin

It consists of a magnitude (in meters) and direction

1-D Position Vector

Only two directions available

Can be represented by a signed scalar (magnitude)

O P

1 m

2-D Position Vector

Direction goes through a 360 angle

0 angle reference needed

Position vector = magnitude, angle

Polar Coordinates (r, ) are natural

r = magnitude, = direction

= r

= 0°

r

P

r

xy

yxr

ry

rx

/tan

sin

cos

1

22

Polar coordinates to cartesian coordinates and back:

Vector Components

Similar to polar coordinate transformation

x coordinate yields the x-component Ax of vector A

y coordinate yields the y-component Ay of vector A

xy

yx

y

x

AA

AAA

AA

AA

/tan

sin

cos

1

22

Second and Third Quadrant Adjustment

The direction is always measured from the +x axis

tan-1 (By/Bx) < 0 for quadrant II

tan-1 (Cy/Cx) > 0 for quadrant III

xy AA /tan180 1

3-D Position Vector

Direction consists of two angles

Choice for two angles

Geographer’s coordinates

Polar angle (longitude)

Angle from the horizon (latitude)

0 = horizontal view

Spherical coordinates

Polar angle

Azimuthal Angle

0 = view at the top (azimuth)

Geographer’s Coordinates

Spherical Coordinates

Exercise: Position Vector

2-D

Specify the position of the back door. Reference point:

Reference direction:

Magnitude:

Direction:

3-D

Specify the position of the projector in the classroom. Reference point:

Reference direction:

Magnitude:

Direction:

How fast is the object moving? (Velocity Vector)

Average velocity

Instantaneous velocity Velocity at any instant in time

Average velocity over a very short time interval

Slope of the position vs. time graph

Speed = magnitude of velocity

t

x

t

xx

elapsedtime

ntdisplacemev

if

ave

dt

xd

t

xv t

0lim

Example: Runner’s Average Velocity

During a 3.00 s time interval, a runner’s position changes from x1 = 50.0 m to x2 = 30.5 m towards you along a straight track. What is the runner’s average velocity?

Solution

sms

m

s

mm

t

xvave /50.6

00.3

5.19

00.3

0.505.30

Is it slowing down or picking up speed? (Acceleration Vector)

Average acceleration

Instantaneous acceleration

The acceleration at a particular point in time

The average acceleration over a very small time interval

The slope of the velocity vs. time graph

t

v

t

vva

if

ave

dt

vd

t

va t

0lim

Example: Accelerating Car

A car accelerates along a straight road from from rest to 75 kph in 5.0 s. What is the magnitude of the average acceleration?

Solution

2

2/2.4

3600

10001515

0.5

75sm

s

m

hs

km

s

kph

t

vva

if

ave

When we know the acceleration of an object we can figure out how it is moving!

Zero acceleration

At “rest” or moving with a constant velocity

Constant acceleration

Speed varies linearly (direction remains constant)

Position varies parabolically (in time)

Variable acceleration

Non-linear change in speed or changing direction

Motion Graphs

Position vs. time

Velocity vs. time

Velocity = time rate of change of position

Slope of position vs. time graph

Acceleration vs. time

Acceleration = time rate of change of velocity

Slope of velocity vs. time graph

Constant Velocity Graphs

x

t

v

t

a = 0

Constant Acceleration Graphs

Example: Graphical Analysis

The velocity of a motorcycle driven by a PNP officer is given by the graph below. How far does the officer go after 5 s? 9 s? 13 s?

Solution Consider the area under the

curve!

t=5 sx=(20 m/s)(5 s)=100m

t=9 sx=(20m/s)(9s)+(1/2 )(4s)(25m/s)

=180 m+50 m=230 m

t=13 sx=230 m+(1/2)(4s)(45m/s)=320m

Variable Acceleration

Uniform Circular Motion

Launching spaceships/satellites

Planetary Orbits

Motion in the very general sense

Questions and Problems for Contemplation

Giancoli (6th edition)

Chapter 2 Questions: 2, 4, 6, 8, 14, 17, 18, 21

Problems: 3, 7, 8, 20, 24, 26, 35, 39, 46, 50, 56

General Problems: 57, 58, 59, 60, 68, 76, 81

First Long Exam Wednesday, Dec. 4, 2013

Submit blue book by Dec. 2

Chapters 1, 2 and 3 including notes, assigned questions and problems

Free Fall

When an object falls through the air, how fast will it fall down?

With no air friction

Constant acceleration

g = -9.8 m/s2

g = 9.8 m/s2

With air friction

Variable acceleration

Free Fall Demo

Falling Objects

Galileo’s experiment

Equations of Motion (No Friction)

a(t) = ao = -g = -9.8 m/s2

v(t) = -gt + vo

vo = initial velocity

x(t) = -(½)gt2 + vot + xo

xo = initial displacement

Reaction Time Calculation

Catch the falling meter stick!

vo = 0, xo = ?

g = 9.8 m/s2

v(t) = -gt

x(t) = -(½)gt2 + xo

Contest!

2 teams (3 persons each)

As quickly as you can, catch the falling meter stick between your thumb and pointing finger

Only one trial!

Throwing Objects Upwards

Equations with an initial velocity component

xmax , time of flight, or vo usually to be determined

xo = 0 point where object leaves hand

g = 9.8 m/s2

v(t) = -gt + vo

x(t) = -(½)gt2 + vot

Example

Calculate the initial velocity of an object thrown upward to a height of 2.0 m.

Solution Find vo. v(t)=0 at the highest point

0 = -(9.8 m/s2)t + vo

need to find t!

xmax = 2.0 m x(t) = -(½)gt2 + vot

2.0 m = -(4.9m/s2)t2 + vot

From the first equation, t = vo/(9.8 m/s2). Thus

2.0 m = -(4.9m/s2)(vo/(9.8 m/s2))2 + vo(vo/(9.8 m/s2))

Solve for vo

2.0 m = -vo2/(19.6 m/s2) + vo

2 /(9.8 m/s2)

Example continued

2.0 m = -vo2/(19.6 m/s2) + vo

2 /(9.8 m/s2)

2.0 m = vo2/(19.6m/s2)

vo2 = 39.2 m2/s2

vo= 6. 26099… m/s

vo= 6. 3 m/s

Questions and Problems for Contemplation

Giancoli (6th edition)

Chapter 3

Questions: 5, 6, 9, 11, 14, 16, 18

Problems: 2, 8, 18, 28, 35, 38, 47, 48, 49

General Problems: 53, 57, 63, 69, 75

Seatwork: Plot the motion graphs for a bouncing ball.

You may work in pairs

Motion Graphs for Bouncing Ball

2-D Motion

Projectile Motion

With No Air Resistance

Horizontal direction Constant velocity

Vertical direction Constant acceleration

Projectile launched with initial velocity vo at an angle from the horizontal vox = vo cos

voy= vo sin

Equations of Motion

Horizontal v(t) = vox

vox = initial velocity horizontal component

x(t) = voxt + xo xo = initial horizontal displacement

Vertical a(t) = ao = -g = -9.8 m/s2

vy(t) = -gt + voy

voy = initial velocity vertical component

y(t) = -½gt2 + voyt + yo yo = initial vertical displacement

Horizontal vo

Will you be able to jump across to the other building if your initial horizontal velocity is 10 m/s?

Solution

xo= 0, yo= 0 voy= 0, vox= 10 m/s

x(t) = (10 m/s)t

vy(t) = -(9.8 m/s2)t

y(t) = -(4.9 m/s2)t2

y(t)= -5.0 m = -(4.9 m/s2)t2

x(1.0s) = (10 m/s)(1.0s) = 10 m

You will fall short of the building!!

sst 0.19.4

0.5 2

mm 1210

vo

5.0m

12m

Exercise

What take-off velocity would you need to jump successfully to the other building?

If you cannot go any faster, will you succeed by just varying your take-off angle?

2-D Trajectory of the Projectile

Combine horizontal and vertical motion

2-D Trajectory

x(t) = voxt (1)

y(t) = -½gt2 + voyt (2)

From (1), t = x/vox

Substitute into (2),

2

22

2

2 cos2)(tan

2x

v

gxx

v

gx

v

vy

ooxox

oy

2BxAxyIsn’t this an equation for a parabola?!

Plotting the Parabola

Roots: y = 0

0 = x(A - Bx)

Two roots

x = 0 (take-off point)

x = A/B (landing point, range R)

Vertex: x = A/2B

x coordinate of ymax

ymax = A2/2B – A2/4B = A2/4B

g

vR o 2sin2

g

vy o

2

sin 22

max

tanA 22 cos2 ov

gB

Time of flight

Time from launching point to landing point

Time to reach maximum height ymax

This is just half the time of flight!

g

v

v

gv

v

R

v

xt o

o

o

oox

sin2

cos

2sin

cos

2

g

vt o sin

gtvoy0

Range of the Projectile

Varying the projection angle

Azkal’s Football Kick

vo = 20.0 m/s, 37.0

Calculate

Maximum height

Time of flight

Range

Velocity at the maximum height

Velocity as it hits the ground

Solution

Resolve initial velocity into its components

At maximum height, vy = 0

With yo = 0

smsmvv

smsmvv

ooy

oox

/0.12)602.0)(/0.20(0.37sin

/0.16)799.0)(/0.20(0.37cos

ssm

sm

g

vt

oy22.1

/80.9

/0.122

mssmssm

gttvy oy

35.7)22.1)(/90.4()22.1)(/0.12(

2

22

2

Solution

Time of flight Time to go up done

Time to go up = time to go down

Time of flight = 2x time to go up = 2.44s

Alternatively,

])/90.4()/0.12[(0

)/90.4()/0.12(2

0

2

222

tsmsmt

tsmtsmgt

tvoy

ssm

smt 45.2

/90.4

/0.122

Solution

Range = how far will it go horizontally Horizontal displacement at t = tflight

mssmtvx flightox 2.39)45.2)(/0.16(

Solution

Velocity at the maximum point vy = 0

vx = 16.0 m/s

v = 16.0 m/s, 0 (horizontal)

Velocity as the football hits the ground Velocity at t = tflight

vx = 16.0 m/s

vy = ?smsmssmvy /0.12/0.12)45.2)(/8.9( 2

smsmsmv /0.20)/0.16()/0.12( 22

9.36/0.16

/0.12tan 1

sm

sm

“Hang-Time”

Calculate the fraction of time the football spends on the upper half of its flight.

Solution

From y=(1/2) ymax to highest point, ymax, back to y=(1/2) ymax

ymax = 7.35 m

2

2gttvy oy

22 )/9.4()/0.12(2

35.7tsmtsm

m

22 )/9.4()/0.12(675.30 tsmtsmm

st 359.01 st 090.22

ssstt 731.1359.0090.212%6.70706.0

45.2

73.112

s

s

t

tt

flight

Jumping From Building A to Building B

Jump at an angle of 15 from the horizontal, 10.0 m/s

Time to go up to max height

Time to go down

Determine ymax first

Now determine tdown (free fall from a height of 5.00m +0.34m = 5.34m)

Time of flight = tup+ tdown = 0.264s + 1.044s = 1.308s

Horizontal distance covered

ssm

sm

g

vt o 264.0

/8.9

15sin)/10(sin2

msm

sm

g

vy o 342.0

)/8.9(2

15sin)/10(

2

sin2

2222

max

2

2gty s

sm

mtdown 044.1

/8.9

)34.5(22

mssmtvx flightox 6.12)308.1(15cos)/0.10( You’ll makethe jump!

Basketball Exercise

There are two ways to shoot the ball given the same initial velocity

High arc

Low arc

Determine the angles of projection for the two shots mentioned above for your favorite player

Reminders

LONG TEST 1 on December 4, 2013 (Wednesday)

Chapters 1, 2 and 3

Submit Blue Book on Dec. 2 Monday

Relative Velocity (1-D)

The velocity with respect to a particular reference frame The woman

The train

The road

The bike rider

Woman’s velocity relative to the train is 1.0 m/s

Train’s velocity relative to bike rider is 3.0 m/s

What is the woman’s velocity with respect to the bike rider?

ABBPAP vvv ///

Example

You are driving north on a straight road at a constant velocity of 88 kph. A truck is traveling at a constant velocity of 104 kph on the opposite lane.

Relative velocity of truck with respect to you

Your relative velocity with respect to the truck

Relative velocities don’t change after the truck has passed you!

kphkphvvv EYETYT 88104///

kphv YT 192/

EYYTET vvv ///

kphvv YTTY 192//

Relative Velocity (2-D)

Vector addition required

Woman is walking at an angle with respect to the train’s displacement

Train is moving at an angle with respect to the normal to the bike rider’s line of sight

Position vector Velocity Vector

ABBPAP rrr ///

ABBPAP vvv ///

Example

An airplane is headed north at 240 kph. If there is a wind of 100 kph from west to east, determine the resultant velocity of the airplane with respect to the ground.

P=plane, A=air, E=earth

From the diagram

Westduekphv AP 240/

Eastduekphv EA 100/

EAAPEP vvv ///

kphkphkphv EP 26010024022

/

NofEkph

kph23

240

100tan 1

Correcting Flight Path

In what direction should you fly the plane so that its resultant direction is northwards?

From the diagram,

unknowndirectionkphv AP 240/

Eastduekphv EA 100/

EAAPEP vvv ///

NofWkph

kph25

240

100sin 1

kphkphkphv EP 21810024022

/