Download Physics 106P: Lecture 5 Notes

SPH4U: Lecture 1
Dynamics
How and why do
objects move
Describing motion – so far… (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 3r
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