Kinematics
Review
Kinematics Review
a) Exam information
b) What kind of questions?
c) Review
Mechanics Lecture 4, Slide 1
First Midterm Exam on Friday
Covers Units 1-3 plus “Math”
One-dimensional Kinematics
Two-Dimensional Kinematics
Relative and Circular Motion
Unit conversion
Trigonometry and Algebra
50 minute duration
Multiple choice
Calculators Allowed/Needed
Three sheets of notes. (equivalent
font size>9)
(this is font size=9)
Mechanics Lecture 4, Slide 2
What type of questions?
 Interpretation and use of graphs
position (m)
position vs time
2
1.8
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
time (s)
Which of the following statements about the velocity of the cart is correct?
What is the velocity at t=?
What is the acceleration at t= ?
….
Mechanics Lecture 1, Slide 3
What type of questions?
Displacement, velocity and acceleration in 1 dimension
Given some combination of acceleration, initial velocity, final
velocity, displacement, elapsed time, etc. calculate an
unknown quantity using the kinematic equations….
Mechanics Lecture 1, Slide 4
What type of questions?
Motion with constant acceleration in 1-dimension
V0=0m/s
H=10m
V1
ground
Given some combination of acceleration, initial velocity, final
velocity, displacement, elapsed time, etc. calculate an
unknown quantity using the kinematic equations….
Mechanics Lecture 1, Slide 5
What type of questions?
Relative motion in 2 dimensions
Given some combination of relative velocity between frames
of reference answer questions related to the displacement
and elapsed times for certain actions…
Mechanics Lecture 1, Slide 6
What type of questions?
Projectile motion
vi
q
y
x
450 m
Given some combination of initial velocity, range , maximum
height, etc…calculate characteristics of projectile motion..
Mechanics Lecture 1, Slide 7
What type of questions?
Circular Motion
Top view
N
E
m
R
Given some combination of angular velocity, Radius, ac
calculate a characteristic of the motion of an object at
constant speed in a circle…
Mechanics Lecture 1, Slide 8
What type of questions?
 Any material from Units 1-3
 Calculations
 Unit conversion
 Algebra, trigonometry and vectors
 Conceptual Questions
 Similar to pre-lecture and checkpoint questions
Mechanics Lecture 1, Slide 9
What type of questions?
Vectors

A
q
 Decompose a vector into polar or Cartesian coordinates
 Vector Addition (Subtraction)
Mechanics Lecture 1, Slide 10
How to prepare?
Read the textbook
Review the video prelectures
Review the lecture slides
(Re-)Work homework problems
Prepare three sheets of notes
Basic Fundamental Equations
Derived Equations( and be able to derive them!)
Unit conversions
Algebra notes (quadratic equation)
Trigonometric relations
Vector concepts
Conceptual notes,…
Get enough sleep!!!
Mechanics Lecture 1, Slide 11
Main Points of Unit 1
Mechanics Lecture 1, Slide 12
Main Points of Unit 1
Mechanics Lecture 1, Slide 13
Vectors and 2d-kinematics – Main Points
Mechanics Lecture 2, Slide 14
Vectors and 2d-kinematics – Main Points
Mechanics Lecture 2, Slide 15
Vectors and 2d-kinematics
Important Equations
Mechanics Lecture 2, Slide 16
Relative and Circular Motion
a) Relative motion
b) Centripetal acceleration
Mechanics Lecture 3, Slide 17
Mechanics Lecture 3, Slide 18
Hyperphysics
Motion
Displacement vs timet
Velocity vs timet
Acceleration vs timet
Mechanics Lecture 1, Slide 19
Hyperphysics
Motion
Mechanics Lecture 1, Slide 20
1d-Kinematic Equations for constant acceleration
a ( t )  a0
v (t )  a0t  v0
1 2
x (t )  a0t  v0t  x0
2
2
2
( v (t ))  v0 )  2a ( x (t )  x0 )
 Basic Equations to be used for 1d –
kinematic problems.
 Need to apply to each object
separately sometimes with time
offset
 When acceleration changes from
one constant value to another say
a=0 The problem needs to be
broken down into segments
Mechanics Lecture 1, Slide 21
Maximum Height of ball thrown upward in gravitational field
What is the velocity of an object when it is at its maximum height?
vb f  vb ( xb  xbmax )  0m / s
Two ways to solve for maximum height…
Solve for position directly
Solve for time first
vb f  vb ( xb  xbmax )  0m / s
vb f  vb ( xb  xbmax )  0m / s
vb f  vb0  at
2a ( xb f  xb0 )  0m / s   vb20
 tf 
vb f  vb0
a
2

0m / s  vb0
a
 xb f  xb0 
Remember a is negative
xb f  xb0 
Use tf to solve for maximum height…

xb f 
1 2
at f  vb0 t f  xb0
2
 vb20
2a
1 2
at f  vb0 t f
2
1 2
at f  vb0 t f  ( xb0  xb f )  0
2
 tf 
 vb0  vb20  2a ( xb0  xb f )
a
Mechanics Lecture 1, Slide 22
The meeting Problem
Define your problem; Identify useful equations Whenever possible solve symbolically.
Then plug in the numbers!!!! You can
object1  first _ launched
understand what is going on better if
td  0
you keep the equations organized.
x1 (t  tmeet )  x2 (t  tmeet )
1
1
a (tmeet  td ) 2  v10 (tmeet  td )  x10  a (tmeet ) 2  v20 (tmeet )  x20
2
2
Start solving…
1 2
1
1
1 2
atmeet  a ( 2)td tmeet  atd2  v10 tmeet  v10 td  x10  atmeet
 v20 tmeet  x20  0
2
2
2
2
Gather terms…Does the equation look reasonable?
1
(( atd  v10 )  v20 )tmeet  ( atd2  v10 td  x10 )  x20  0
2
Solve for desired quantity…
Position of object 1 when object 2 is launched
1
( atd2  v10 td  x10 )
1
2
( atd2  v10 td  x10 )  x20 x (t  t )
d
tmeet  2

(( atd  v10 )  v20 )
v (t  td )
This time is w.r.t object 2.
To obtain time w.r.t object 1
Speed of object 1 when object 2 is launched need to add delay time
( atd  v10 )
(tmeet )1  tmeet  td
Mechanics Lecture 1, Slide 23
Break problem into parts
Region 1: 0<t<1.5 s
Region 1 Region 2 Region 3
t  t f  ti  1.5s  0s
a  a1  v / t  (10m / s ) / 1.5s  6.667m / s 2
v (t  1.5s )  a1t  v0  a1 (t  0)  0  a1t  0  (6.667m / s 2 )(1.5s )  10.0m / s
1
1
x (t  1.5s )  x0  v0 t  a1 ( t ) 2  (6.667m / s 2 )(1.5s ) 2  7.5m
2
2
Region 2: 1.5s<t<3.0s
a  a2  v / t  (0m / s ) / 1.5s  0m / s 2
v (t )  a2t  v (t  1.5s )  0t  10m / s  10.0m / s
x (t  3.0s )  x (t  1.5s )  v (t  1.5s )(3.0s  1.5s )  (10.0m / s )(1.5s )  15.0m
x (t  3.0s )  x (t  1.5s )  15.0m  7.5m  15.0m  22.5m
v (t  3.0s )  0t  v (t  1.5s )  10.0m / s
Region 3: 3.0s<t<5.0s
t  t f  ti  5.0s  3.0s  2.0s
What happened in region 3?
Smartphysics Conclusion
a  a3  v / t  ( v f  vi ) / 2.0s  ( 10m / s  10.0m / s ) / 2.0s  10m / s 2
v (t  5s )  a2 t  v (t  3.0s )  ( 10m / s 2 )( 2 s )  10m / s  10.0m / s
1
1
a3 ( t ) 2  v (t  3.0s ) t  ( 10m / s 2 )( 2.0) 2  (10m / s )( 2.0s )
2
2
x (t  5.0s )  x (t  3.0s )  ( 20m )  ( 20m )  x (t  3.0s )  22.5m
x (t  5.0s )  x (t  3.0s ) 
v (t  5.0s )  a2 t  v (t  3.0s )  ( 10m / s 2 )( 2.0)  10m / s  10.0m / s
Mechanics Lecture 1, Slide 24
2d-Vectors

A
q
Think of a vector as an arrow.
(An object having both magnitude and direction)
The object is the same no matter how we chose to describe it
Mechanics Lecture 2, Slide 25
Algebra and Trigonometry
Algebra
 Algebra is “fair”
 Quadratic Equation
ax 2  bx  c  0
 b  b2  4ac
x
2a
Trigonometry
Mechanics Lecture 2, Slide 26
Vector Addition
Add
Components!!!
AddTail to Head
Mechanics Lecture 2, Slide 27
Kinematics in 3D
Mechanics Lecture 2, Slide 28
Projectile Motion
Horizontal
Vertical
Boring
Mechanics Lecture 2, Slide 29
Ballistic Projectile Motion Quantities
Initial velocity
speed,angle
Maximum Height of trajectory, h=ymax
“Hang Time”
Time of Flight, tf
Range of trajectory, D
Height of trajectory at arbitrary x,t
Mechanics Lecture 2, Slide 30
Derived Projectile Trajectory Equations
Maximum height
v02 sin 2 q
h  y0 
2g
Time of Flight (“Hang Time”)
tf 
2v0 y
g

2v0 sin q
g
Range of trajectory
v02 sin 2q
D
g
Height of trajectory as f(t) , y(t)
y (t )  y0  v0 y t 
1 2
gt
2
Height of trajectory as f(x), y(x)
 x  1  x 
  g 

y ( x )  v0 sin q 
 v0 cos q  2  v0 cos q 
2
Mechanics Lecture 1, Slide 31
Relative Motion in 2 Dimensions
Speed relative to shore
Mechanics Lecture 3, Slide 32
Relative Motion in 2 Dimensions
Direction w.r.t shoreline
Mechanics Lecture 3, Slide 33
Centripetal Acceleration
Constant speed in circular path
Acceleration directed toward center
of circle
What is the magnitude of
acceleration?
Proportional to:
1. Speed
1. time rate of change of
angle or angular velocity
dq

dt
v = R
v2
ac 
R
Mechanics Lecture 3, Slide 34
Additional Slides
Mechanics Lecture 1, Slide 35
v = R
 is the rate at which the angle q changes:

dq
dt
q
Once around:
v  x / t = 2pR / T
  q / t = 2p / T
Mechanics Lecture 3, Slide 36
v = R
Another way to see it:
dq
R
v dt =R dq
dq
vR
dt
v = R
Mechanics Lecture 3, Slide 37
Two thrown balls problem #2
Use your equation in symbolic form and gather the values to be used….
1
( atd2  v10 td  x10 )  x20 x (t  t )
d
tmeet  2

(( atd  v10 )  v20 )
v (t  td )
a  9.81m / s; td  0.4 s; v10  1.2m / s; v20  23.8m / s; x10  26m; x20  1m
Solve for individual elements of equation…can check if things make sense
Position of object 1 at moment object 2 is launched
1
1
( atd2  v10 td  x10 )  ( ( 9.81m / s 2 )(0.4 s ) 2  ( 1.2m / s )(0.4 s )  26m  24.735m
2
2
Velocity of object 1 at moment object 2 is launched
( atd  v10 )  ( 9.81m / s 2 )(0.4 s )  ( 1.2m / s )  5.124m / s
Solve for displacement and relative velocity at time when object 2 launches
x (t  td )  24.735m  1m  23.735m
v (t  td )  5.124m / s  23.8m / s  28.94m / s
Note that acceleration term
nd
Divide to obtain time when objects meet after 2 object is released
drops out!!!
x ( t  t d )
tmeet 
 0.82 s
v ( t  t d )
tred  tmeet  td  1.22 s
Mechanics Lecture 1, Slide 38
Homework Hints – Stadium Wall
Calculate time to reach wall using vx:
twall  xwall / v0 x  xwall / v0 cos q 
Calculate y position at time to reach wall:
ywall  y0  v0 y twall 
1
2
g twall 
2
ywall  y0  v0 sin q  xwall / v0 cos q  
ywall  y0  xwall tan q 
1
2
g  xwall / v0 cos q 
2
1
2
g  xwall / v0 cos q 
2
Mechanics Lecture 1, Slide 39
Homework Hints – Stadium Wall
Calculate time to reach wall using vx:
twall  xwall / v0 x  xwall / v0 cos q 
Calculate y position at time to reach wall:
ywall  y0  v0 y twall 
1
2
g twall 
2
ywall  y0  v0 sin q  xwall / v0 cos q  
ywall  y0  xwall tan q 
1
2
g  xwall / v0 cos q 
2
1
2
g  xwall / v0 cos q 
2
Mechanics Lecture 1, Slide 40
Homework Solutions-Baseball Stadium
twall  xwall / v0 x  xwall / v0 cos q 
ywall  y0  v0 y twall 
1
2
g twall 
2
ywall  y0  v0 sin q  xwall / v0 cos q  
ywall  y0  xwall tan q 
1
2
g  xwall / v0 cos q 
2
1
2
g  xwall / v0 cos q 
2
x  565 ft;q  350 ; g  32.2 ft / s 2 ; y0  0
1
2
g  xwall / v0 cos q 
2
2
 3 ft  (565 ft )(.7002)  (16.1 ft / s 2 )565 ft / 176 ft / s  0.8192) 
ywall  y0  xwall tan q 
ywall
ywall  3 ft  395.61 ft  247.24 ft  151.37 ft
Mechanics Lecture 1, Slide 41
Plane Ride




ˆ
ˆ
v plane, ground  v plane,air  vair , ground
v plane,air  140i  0 j

vair , ground  ( 35 * sin 30)iˆ  (35 * cos 30) ˆj


ˆ
ˆ
v
ˆ
v plane, ground  (140  35 * sin 30)i  (35 * cos 30) j
j
plane , ground

q
v plane, ground  (122.5)iˆ  (30.31) ˆj


iˆ
v plane,air
v plane, ground  (122.5) 2  (30.31) 2  126.19m / s
N

vair , ground
E

v plane, ground  (122.5)iˆ  (30.31) ˆj
v

 30.31 
0
  tan 1  plane, ground , y   tan 1 
  13.89
 122.5 
 v plane, ground , x 
q  90    90  13.89  76.100 East of north
x  (122.5m / s ) * (3600s / hr ) * 1hr  441,000m
Mechanics Lecture 3, Slide 42
Car on Curve
v 2 22m / s 
ac 

 2.547m / s 2
R
190m
2
ac , x  ( 2.547m / s 2 ) * cos(1800  160 )  2.4483m / s 2
ac , y  ( 2.547m / s 2 ) * sin(1800  160 )  0.702m / s 2
Mechanics Lecture 3, Slide 43
Maximum Height of Trajectory
v y (t )  v0 y  gt
Height of trajectory,h=ymax
ymax  v y  0  t ymax 
v0 y
g
h  y (t ymax )
h  y0  v0 y t ymax 
1 2
gt ymax
2
2
v02 y
1v
 0y
h  y0 
g 2 g
h  y0 
v02 y
2g
v02 sin 2 q
h  y0 
2g
Mechanics Lecture 2, Slide 44
Time of Flight, “Hang Time”
1 2
gt
2
t  t f  y ( t  t f )  y0
y (t )  y0  v0 y t 
1 2
gt f
2
1
1

v0 y t f  gt 2f  t f  v0 y  gt f
2
2

2v0 y 2v0 sin q
t f  0; t f 

g
g
y0  y0  v0 y t f 

0

Mechanics Lecture 2, Slide 45
Range of trajectory
x (t )  x0  v0 x t
t  t f  y ( t  t f )  y0
D  x (t  t f )  0  v0 x t f
D
2v0 x v0 y
g
2v02 cos q sin q

g
v02 sin 2q
D
g
Mechanics Lecture 2, Slide 46
Angle for Maximum Range
MAXIMUM range OCCURS AT 450
f (q )  sin(2q )
df (q )
 2 cos(2q )
dq
df (q )
 0  cos(2q )  0
dq
 2q  900
 q  450
Mechanics Lecture 2, Slide 47
Height of Trajectory at time t or position x
Height of trajectory, y(t)
1 2
y (t )  y0  v0 y t  gt
2
Height of trajectory, y(x)
x  x0  v0 x t
t
x  x0
v0 x
 x  x0  1  x  x0 
  g 

y ( x )  y0  v0 y 
v
2
v
 0x 
 0x 
x0  0; y0  0
 x  1  x 
  g 

y ( x )  v0 y 
v
2
v
 0x 
 0x 
2
2
 x  1  x 
  g 

y ( x )  v0 sin q 
v
cos
q
2
v
cos
q
 0

 0

2
Mechanics Lecture 2, Slide 48
Launch Velocity-Given R and q
Mechanics Lecture 1, Slide 49
Launch Angle
Mechanics Lecture 1, Slide 50
Launch Velocity –Given R and h
Mechanics Lecture 1, Slide 51