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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
/
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