Transcript Physics 106P: Lecture 5 Notes

SPH4U: Lecture 1
Dynamics
How and why do
objects move
Course Info & Advice

Course has several components:
 Lecture: (me talking, demos and you asking questions)
 Discussion sections (tutorials, problem solving, quizzes)
 Homework Web based
 Labs: (group exploration of physical phenomena)
 What happens if you miss a lab or class test




Read notes from online
What if you are excused?? (What you need to do.)
That topic of your exam will be evaluated as your test
The first few weeks of the course should be review, hence the pace is
fast. It is important for you to keep up!
 Then,
watch out….
Fundamental Units


How we measure things!
All things in classical mechanics can be expressed in terms of the
fundamental units:




Length: L
Mass : M
Time : T
For example:
 Speed has units of L / T (e.g. miles per hour).
 Force has units of ML / T2 etc... (as you will learn).
Units...

SI (Système International) Units:
 mks: L = meters (m), M = kilograms (kg), T = seconds (s)
 cgs: L = centimeters (cm), M = grams (gm), T = seconds (s)

British Units:
 Inches, feet, miles, pounds, slugs...

We will use mostly SI units, but you may run across some
problems using British units. You should know where to look to
convert back & forth.
Ever heard of Google
Converting between different systems of units

Useful Conversion factors:
 1 inch
= 2.54 cm
 1m
= 3.28 ft
 1 mile
= 5280 ft
 1 mile
= 1.61 km

Example: convert miles per hour to meters per second:
mi
mi
ft
1 m
1 hr
m
1
 1  5280


 0.447
hr
hr
mi 3.28 ft 3600 s
s
Dimensional Analysis

This is a very important tool to check your work
 It’s also very easy!

Example:
Doing a problem you get the answer distance
d = vt 2 (velocity x time2)
Units on left side = L
Units on right side = L / T x T2 = L x T

Left units and right units don’t match, so answer must be
wrong!!
Dimensional Analysis

The period P of a swinging pendulum depends only on
the length of the pendulum d and the acceleration of
gravity g.
 Which of the following formulas for P could be
correct ?
(a) P = 2
(dg)2
(b)
d
P  2
g
(c)
P  2
Given: d has units of length (L) and g has units of (L / T 2).
d
g
Solution


Realize that the left hand side P has units of
time (T )
Try the first equation
2
(a)
(a)
L 
L4

L 2   4  T
 T  T
P  2  dg 
2
(b)
Not Right!
P  2
d
g
(c) P  2
d
g
Solution


Realize that the left hand side P has units of
time (T )
Try the second equation
(b)
(a)
L
T2 T
L
T2
P  2  dg 
2
Not Right!
(b)
P  2
d
g
(c) P  2
d
g
Solution


Realize that the left hand side P has units of
time (T )
Try the first equation
(c)
(a)
L
 T2 T
L
T2
P  2  dg 
2
(b)
Dude, this is it
P  2
d
g
(c) P  2
d
g
Time to Drop an Apple
Want to ask myself a question: If I drop an Apple from a certain height,
h, what then happens to the time, t, it takes for the apple to fall?
The time t, must be proportional to the height to the power of some
value. The time t may be proportional to the mass of the Apple to the
power of some value. The time t, may be proportional to the
acceleration due to gravity, g to some power


th m g

1

  L 
T

C
L
M
 
     2
T 
 0
There is no L on Left side:     0
There is no M on Left side:
There is T to power 1 on Left side: 1  2
1
1





Therefore
2
2

Dimensional Analysis
to the Rescue
1
2
t  Ch m 0 g
C
 h
h
g

1
2
Time for Apple to Drop
t h
Dimensional Analysis tells us, that if we increase the height by a factor
of 100, then to time will increase by the square root of 100, or by a
factor of 10
Let’s verify this by dropping an apple from 3 m then from 1.5 m and
compare the times
 x   y 
e  E     
 x   y 
2
2
2
 2   2 
 1.417 
 

 781   551 
 0.006
2
Since we double the height, the time
should be
2  1.4142 longer.
The experiment shows: 1.417±0.006
781ms
 1.417
551ms
Vectors


A vector is a quantity that involves both magnitude and direction.
 55 km/h [N35E]
 A downward force of 3 Newtons
A scalar is a quantity that does not involve direction.
 55 km/h
 18 cm long
Vector Notation

Vectors are often identified with arrows in graphics and labeled
as follows:
We label a vector with a variable.
This variable is identified as a vector either by an arrow
above itself :
A
Or
By the variable being BOLD:
A
Displacement

Displacement is an object’s change in position.
Distance is the total length of space traversed
by an object.
1m
6.7m
3m
Start
= 500 m
Finish
5m
Displacement:
 6m   3m
2
2
 6.7m
Distance: 5m  3m  1m  9m
Displacement = 0 m
Distance = 500 m
Vector Addition
A
R
E
B
R
E
C
B A
D
D
E
D
R
C
A
C
B
A + B + C + D + E = Distance
R = Resultant = Displacement
Rectangular Components
Quadrant II
R  A2  B2
R sin  
y
A
R

-x
A opp
sin   
R hyp
B adj
cos   
R hyp
A opp
tan   
B adj
Quadrant I
R cos  
Quadrant III
B
Quadrant IV
-y
x
Vectors...

The components (in a particular coordinate system) of r,
the position vector, are its (x,y,z) coordinates in that
coordinate system
 r = (rx ,ry ,rz ) = (x,y,z)

Consider this in 2-D (since it’s easier to draw):
where r = |r |
 rx = x = r cos 
 ry = y = r sin 
(x,y)
y
arctan( y / x )
r

x
Vectors...

The magnitude (length) of r is found using the
Pythagorean theorem:
r
y
r r
x2  y 2
x

The length of a vector clearly does not depend on its direction.
Vector Example



Vector A = (0,2,1)
Vector B = (3,0,2)
Vector C = (1,-4,2)
What is the resultant vector, D, from adding A+B+C?
(a) (3,5,-1)
(b) (4,-2,5)
(c) (5,-2,4)
Resultant of Two Forces
• force: action of one body on another;
characterized by its point of application,
magnitude, line of action, and sense.
• Experimental evidence shows that the
combined effect of two forces may be
represented by a single resultant force.
• The resultant is equivalent to the
diagonal of a parallelogram which
contains the two forces in adjacent legs.
• Force is a vector quantity.
Vectors
• Vector: parameters possessing magnitude and direction
which add according to the parallelogram law. Examples:
displacements, velocities, accelerations.
• Scalar: parameters possessing magnitude but not
direction. Examples: mass, volume, temperature
P
Q
• Vector classifications:
- Fixed or bound vectors have well defined points of
application that cannot be changed without affecting
an analysis.
- Free vectors may be freely moved in space without
changing their effect on an analysis.
- Sliding vectors may be applied anywhere along their
line of action without affecting an analysis.
P
-P
• Equal vectors have the same magnitude and direction.
• Negative vector of a given vector has the same magnitude
and the opposite direction.
Addition of Vectors
• Trapezoid rule for vector addition
• Triangle rule for vector addition
P
• Law of cosines,
Q
R 2  P 2  Q 2  2 PQ cos B
Q
R  PQ
P
P
Q
• Law of sines,
sin A sin B sin C


Q
R
A
• Vector addition is commutative,
PQ  Q P
-Q
Q
P-Q
P
• Vector subtraction
 
P  Q  P  Q
Addition of Vectors
Q
S
• Addition of three or more vectors through
repeated application of the triangle rule
P
Q
S
P
• The polygon rule for the addition of three or
more vectors.
• Vector addition is associative,



PQ S  PQ  S  P Q S
P
2P
-1.5P

• Multiplication of a vector by a scalar
increases its length by that factor (if scalar
is negative, the direction will also change.)
Resultant of Several Concurrent Forces
• Concurrent forces: set of forces
which all pass through the same
point.
A set of concurrent forces applied
to a particle may be replaced by
a single resultant force which is
the vector sum of the applied
forces.
• Vector force components: two or
more force vectors which,
together, have the same effect as
a single force vector.
Sample Problem
• Graphical solution - construct a
parallelogram with sides in the
same direction as P and Q and
lengths in proportion.
Graphically evaluate the
resultant which is equivalent in
direction and proportional in
magnitude to the diagonal.
Two forces act on a bolt at A.
Determine their resultant.
• Trigonometric solution - use the
triangle rule for vector addition
in conjunction with the law of
cosines and law of sines to find
the resultant.
Sample Problem Solution
R
Q

P
R  98 N   35
• Graphical solution - construct a
parallelogram with sides in the same
direction as P and Q and lengths in
proportion. Graphically evaluate the
resultant which is equivalent in
direction and proportional in
magnitude to the diagonal.
Sample Problem Solution
• Trigonometric solution
From the Law of Cosines,
R 2  P 2  Q 2  2 PQ cos B
  40N    60N   2  40N  60N  cos155
2
2
R  97.73N
From the Law of Sines,
sin A sin B

Q
R
sin A  sin B
Q
R
 sin155
A  15.04
  20  A
 35.04
60N
97.73N
Dynamics
How and why do
objects move
Describing motion – so far… (review from last Year)

Linear motion with const acceleration:
x  x0  v0t 
v  v0  at
a  const
1 2
at
2
v 2  v 02  2a(x  x 0 )
v av 
1
(v 0  v)
2
What about higher order rates of change?

If linear motion and circular motion are uniquely
determined by acceleration, do we ever need higher
derivatives?
 da d 3 r
Known as the “Jerk”
J
 3
dt dt

Certainly acceleration changes, so does that mean we need
to find some “action” that controls the third or higher time
derivatives of position?

NO.
Dynamics


Isaac Newton (1643 - 1727) published Principia Mathematica
in 1687. In this work, he proposed three “laws” of motion:
principia
Law 1: An object subject to no external forces is at rest or moves
with a constant velocity if viewed from an inertial reference
frame.
Law 2: For any object, FNET = F = ma (not mv!)
Law 3: Forces occur in pairs: FA ,B = - FB ,A
(For every action there is an equal and opposite reaction.)
These are the postulates of mechanics
They are experimentally, not mathematically, justified.
They work, and DEFINE what we mean by “forces”.
Newton’s First Law

An object subject to no external forces is at rest or moves
with a constant velocity if viewed from an inertial reference
frame.
 If no forces act, there is no acceleration.

The following statements can be thought of as the
definition of inertial reference frames.
 An IRF is a reference frame that is not accelerating (or
rotating) with respect to the “fixed stars”.
 If one IRF exists, infinitely many exist since they are
related by any arbitrary constant velocity vector!
 If you can eliminate all forces, then an IRF is a
reference frame in which a mass moves with a
constant velocity. (alternative definition of IRF)
Is Waterloo a good IRF?


Is Waterloo accelerating?
YES!
 Waterloo is on the Earth.
 The Earth is rotating.

What is the centripetal acceleration of Waterloo?
 T = 1 day = 8.64 x 104 sec,
2
2
v
 2 
aU    2 R  
 R ~ RE = 6.4 x 106 meters .
 R
R
 T 

Plug this in: aU = .034 m/s2 ( ~ 1/300 g)
Close enough to 0 that we will ignore it.
Therefore Waterloo is a pretty good IRF.


Newton’s Second Law

For any object, FNET = F = ma.



The acceleration a of an object is proportional to the
net force FNET acting on it.
The constant of proportionality is called “mass”, denoted
m.
 This is the definition of mass and force.
 The mass of an object is a constant property of that
object, and is independent of external influences.
 The force is the external influence
 The acceleration is a combination of these two things
Force has units of [M]x[L / T2] = kg m/s2 = N (Newton)
Newton’s Second Law...

What is a force?
 A Force is a push or a pull.
 A Force has magnitude & direction (vector).
 Adding forces is just adding force vectors.
a
a
F1
F1
FNET
F2
F2
FNET = ma
Newton’s Second Law...

Components of F = ma :
FX = maX
FY = maY
FZ = maZ

Suppose we know m and FX , we can solve for aX and apply the
things we learned about kinematics over the last few lectures: (if
the force is constant)
x  x0  v0 x t 
vx  v0 x  ax t
1
axt 2
2
Example: Pushing a Box on Ice.

A skater is pushing a heavy box (mass m = 100 kg) across a sheet
of ice (horizontal & frictionless). He applies a force of 50 N in the x
direction. If the box starts at rest, what is its speed v after being
pushed a distance d = 10 m?
v=0
F
m
a
x
Example: Pushing a Box on Ice.

A skater is pushing a heavy box (mass m = 100 kg) across a sheet
of ice (horizontal & frictionless). He applies a force of 50 N in the x
direction. If the box starts at rest, what is its speed v after being
pushed a distance d = 10m ?
v
F
m
a
x
d
Example: Pushing a Box on Ice...

Start with F = ma.
 a = F / m.
 Recall that v2 - v02 = 2a(x - x0 )

So v2 = 2Fd / m
(Last Yeat)
v
2Fd
m
v
F
m
a
i
d
Example: Pushing a Box on Ice...
2Fd
m
v

Plug in F = 50 N, d = 10 m, m = 100 kg:
 Find v = 3.2 m/s.
v
F
m
a
i
d
Question
Force and acceleration

A force F acting on a mass m1 results in an acceleration a1.
The same force acting on a different mass m2 results in an
acceleration a2 = 2a1.
m1
F

a1
F
m2
a2 = 2a1
If m1 and m2 are glued together and the same force F acts
on this combination, what is the resulting acceleration?
F
(a)
2/3 a1
m1
m2
(b) 3/2 a1
a=?
(c)
3/4 a1
Solution
Force and acceleration
m1
F


m2
a = F / (m1+ m2)
Since a2 = 2a1 for the same applied force, m2 = (1/2)m1 !
 m1+ m2 = 3m1 /2
So a = (2/3)F / m1
but F/m1 = a1
a = 2/3 a1
(a)
2/3 a1
(b) 3/2 a1
(c)
3/4 a1
Forces

We will consider two kinds of forces:
 Contact force:
 This is the most familiar kind.





I push on the desk.
The ground pushes on the chair...
A spring pulls or pushes on a mass
A rocket engine provides some number of Newtons of
thrust (1 lb of thrust = mg = 2.205*9.81 = 21.62 Newtons)
Action at a distance:
 Gravity
 Electricity
 Magnetism
Contact forces:

Objects in contact exert forces.
The Force

Convention: Fa,b means acting on a due to b”.

So Fhead,thumb means “the force on
the head due to the thumb”.
Fhead,thumb
Action at a Distance

Gravity:
Burp!
Gravitation
(Courtesy of Newton)


Newton found that amoon / g = 0.000278
and noticed that RE2 / R2 = 0.000273
Hey, I’m
in UCM!
amoon
g
R

We will discuss
these concepts
later
RE
This inspired him to propose the
Universal Law of Gravitation:
where G = 6.67 x 10 -11 m3 kg-1 s-2
FMm
GMm

R2
And the force is attractive along a line between the 2 objects
Understanding
If the distance between two point particles is doubled, then
the gravitational force between them:
A)
B)
C)
D)
E)
Decreases by a factor of 4
Decreases by a factor of 2
Increases by a factor of 2
Increases by a factor of 4
Cannot be determined without knowing the masses
Newton’s Third Law:

Forces occur in pairs: FA ,B = - FB ,A.


For every “action” there is an equal and opposite “reaction”.
We have already seen this in the case of gravity:
m1
m2
F12  G
F12
F21
R12
m1m2
 F21
2
R12
We will discuss
these concepts in
more detail later.
Newton's Third Law...

FA ,B = - FB ,A. is true for contact forces as well:
Fm,w
Force on me from wall is
equal and opposite to the
force on the wall from the me.
Fw,m
Force on me from the floor is
equal and opposite to the force
on the floor from the me.
Ff,m
Fm,f
Example of Bad Thinking

Since Fm,b = -Fb,m, why isn’t Fnet = 0 and a = 0 ?
Fm,b
a ??
block
ice
Fb,m
Example of Good Thinking

Consider only the box as the system!
 Fon box = mabox = Fb,m
 Free Body Diagram (next power point).
Fm,b
Fb,m
abox
block
ice
No ice
Friction force
Add a wall that stops the motion of the block

Now there are two forces acting (in the horizontal direction)
on block and they cancel
 Fon box = mabox = Fb,m + Fb,w = 0
 Free Body Diagram (next power point).
Fb,w
Fw,b
Fm,b
block
abox
ice
Fb,m
Newton’s 3rd Law Understanding

Two blocks are stacked on the ground. How many action-reaction
pairs of forces are present in this system?
(a) 2
a
b
(b) 3
(c) 4
(d) 5
Solution:
gravity
gravity
a
a
Fa,E
gravity
very tiny
contact
a
a
Fa,b
a
b
b
Fb,a
b
Fb,a
b
b
Fb,E
Fa,b
contact
Fg,b
Fb,g
FE,a
FE,b
5
Understanding
A moon of mass m orbits a planet of mass 100m. Let the
strength of the gravitational force exerted by the planet on the
moon be denoted by F1, and let the strength of the
gravitational force exerted by the moon on the planet be F2.
Which of the following is true?
A)
B)
C)
D)
E)
F1=100F2
F1=10F2
F1=F2
F2=10F1
F2=100F1
Newton’s Third Law
Flash: Newton’s 1St Law
Flash: Newton’s 2nd Law
Flash: Newton’s 3rd Law
Flash: Applications of Newton