SPH4U: Lecture 1SPH4U: Lecture 1

Dynamics

How and why do objects move

Describing motion – so far… (from last Year)Describing motion – so far… (from last Year)

Linear motionLinear motion with const acceleration: with const acceleration:

0v v at

20 0

1

2x x v t at

a const

2 20 0

av 0

v v 2a(x x )

1v (v v)

2

What What aboutabout higher order rates of change? higher order rates of change?

If linear motion and circular motion are uniquely If linear motion and circular motion are uniquely determined by acceleration, do we ever need higher determined by acceleration, do we ever need higher derivatives?derivatives?

Certainly acceleration changes, so does that mean we need Certainly acceleration changes, so does that mean we need to find some “action” that controls the third or higher time to find some “action” that controls the third or higher time derivatives of position?derivatives of position?

NO. NO.

3

3

dt

rd

dt

adJ

Known as the “Jerk”

DynamicsDynamics Isaac Newton (1643 - 1727)Isaac Newton (1643 - 1727) publishedpublished Principia Mathematica Principia Mathematica

in 1687. In this work, he proposed three “laws” of motion:in 1687. In this work, he proposed three “laws” of motion: principiaprincipia

Law 1Law 1: : An object subject to no external forces is at rest or An object subject to no external forces is at rest or moves moves with a constant velocity if viewed from an inertial with a constant velocity if viewed from an inertial reference reference frame.frame.

Law 2Law 2: : For any objectFor any object, , FFNETNET = = FF = m = ma a (not m(not mvv!)!)

Law 3Law 3: : Forces occur in pairsForces occur in pairs: : FFA ,BA ,B = - = - FFB ,AB ,A

((For every action there is an equal and opposite reactionFor every action there is an equal and opposite reaction.).)These are the postulates of mechanicsThey are experimentally, not mathematically, justified.They work, and DEFINE what we mean by “forces”.

Newton’s First LawNewton’s First Law

An object subject to no external forces is at rest or moves An object subject to no external forces is at rest or moves with a constant velocity if viewed from anwith a constant velocity if viewed from an inertial reference inertial reference frameframe.. If no forces act, there is no accelerationIf no forces act, there is no acceleration..

The following statements can be thought of as the The following statements can be thought of as the definition of inertial reference frames.definition of inertial reference frames. An An IRFIRF is a reference frame that is not accelerating (or is a reference frame that is not accelerating (or

rotating) with respect to the “rotating) with respect to the “fixed starsfixed stars”.”. If one If one IRFIRF exists, infinitely many exist since they are exists, infinitely many exist since they are

related by any arbitrary constant velocity vector!related by any arbitrary constant velocity vector! If you can eliminate all forces, then anIf you can eliminate all forces, then an IRF IRF is a is a

reference frame in which a mass moves with a reference frame in which a mass moves with a constant velocityconstant velocity. . ((alternative definition of alternative definition of IRFIRF))

Is Waterloo a good IRF?Is Waterloo a good IRF?

Is Waterloo accelerating?Is Waterloo accelerating? YES!YES!

Waterloo is on the EarthWaterloo is on the Earth .. The Earth is rotating. The Earth is rotating.

What is the centripetal acceleration of Waterloo?What is the centripetal acceleration of Waterloo? T = 1 day = 8.64 x 10T = 1 day = 8.64 x 10 44 sec sec , , R ~ RR ~ R

EE = 6.4 x 10 = 6.4 x 1066 meters . meters .

Plug this in: Plug this in: aaUU = .034 m/s = .034 m/s 22 ( ~ 1/300 g) ( ~ 1/300 g) Close enough to Close enough to 00 that we will ignore it.that we will ignore it. Therefore Waterloo is a pretty good IRFTherefore Waterloo is a pretty good IRF ..

222 2

U

va R R

R T

Newton’s Second LawNewton’s Second Law

For any object, For any object, FFNETNET = = FF = m = maa..

The acceleration The acceleration aa of an object is proportional to the of an object is proportional to the net force net force FFNETNET acting on it.acting on it.

The constant of proportionality is called The constant of proportionality is called “mass”, “mass”, denoted denoted mm..

This is the definition of This is the definition of massmass andand force force.. TheThe mass mass of an object is a constant property of thatof an object is a constant property of that

object, and is independent of external influences.object, and is independent of external influences. The The forceforce is the external influence is the external influence The The accelerationacceleration is a combination of these two things is a combination of these two things

Force has units of [Force has units of [MM]x[]x[L / TL / T22] = ] = kg m/skg m/s22 == NN (Newton)(Newton)

Newton’s Second Law...Newton’s Second Law...

What is aWhat is a force force?? A Force is a A Force is a pushpush or a or a pullpull.. A Force hasA Force has magnitude magnitude & & directiondirection ((vectorvector)).. Adding forces is just adding force vectors.Adding forces is just adding force vectors.

FF1 FF2

aaFF1

FF2

aa

FFNET FFNET = maa

Newton’s Second Law...Newton’s Second Law...

Components of Components of FF = m = maa : :

FFX X = ma= maXX

FFYY = ma = maYY

FFZ Z = ma = maZZ

Suppose we know Suppose we know mm and and FFXX , we can solve for , we can solve for aaXX and apply the and apply the

things we learned about kinematics over the last few lectures: (if things we learned about kinematics over the last few lectures: (if the force is constant)the force is constant)

20 0

0

1

2x x

x x x

x x v t a t

v v a t

Example: Pushing a Box on Ice.Example: Pushing a Box on Ice.

A skater is pushing a heavy box (mass A skater is pushing a heavy box (mass mm = 100 kg = 100 kg) across a sheet ) across a sheet of ice (horizontal & frictionless). He applies a force of of ice (horizontal & frictionless). He applies a force of 50 N 50 N in the in the xx direction. If the box starts at rest, what is its speed direction. If the box starts at rest, what is its speed vv after being after being pushed a distance pushed a distance d = d = 10 m10 m??

FF

v = 0

m a

xx

Example: Pushing a Box on Ice.Example: Pushing a Box on Ice.

A skater is pushing a heavy box (A skater is pushing a heavy box (mass mass mm = 100 kg = 100 kg) across a sheet ) across a sheet of ice (horizontal & frictionless). He applies a force of of ice (horizontal & frictionless). He applies a force of 50 N 50 N in the in the xx direction. If the box starts at rest, what is its speed direction. If the box starts at rest, what is its speed vv after being after being pushed a distance pushed a distance d = d = 10m 10m ??

d

mFF

v

a

xx

Example: Pushing a Box on Ice...Example: Pushing a Box on Ice...

Start with Start with FF = m = maa.. aa = = FF / m / m.. Recall that Recall that vv22 - v - v00

22 = 2a(x - x = 2a(x - x0 0 )) (Last Yeat)(Last Yeat)

So So vv22 = 2Fd / m = 2Fd / m 2Fdv

m

d

FF

v

m a

ii

Example: Pushing a Box on Ice...Example: Pushing a Box on Ice...

Plug in Plug in F = F = 50 N50 N, d = , d = 10 m10 m, m = , m = 100 kg:100 kg: Find Find v = v = 3.2 m/s3.2 m/s..

d

FF

v

m a

ii

2Fdv

m

QuestionQuestionForce and accelerationForce and acceleration

A force A force FF acting on a mass acting on a mass mm11 results in an acceleration results in an acceleration aa11..

The same force acting on a different mass The same force acting on a different mass mm22 results in an acceleration results in an acceleration aa22 = 2= 2aa11..

If m1 and m2 are glued together and the same force F acts on this combination, what is the resulting acceleration?

(a)(a) 2/3 a1 (b)(b) 3/2 a1 (c)(c) 3/4 a1

F a1

m1 F a2 = 2a1

m2

F a = ? m1 m2

SolutionSolutionForce and accelerationForce and acceleration

Since Since aa2 2 = = 22aa11 for the same applied force, for the same applied force, mm22 = (1/2)m = (1/2)m11 !!

mm11+ m+ m22 = 3m= 3m1 1 /2/2

(a)(a) 2/3 a1 (b) (b) 3/2 a1 (c)(c) 3/4 a1

F a = F / (m1+ m2)m1 m2

So a = (2/3)F / m1 but F/m1 = a1

a = 2/3 a1

ForcesForces

We will consider two kinds of forces:We will consider two kinds of forces: Contact forceContact force::

This is the most familiar kind.This is the most familiar kind. I push on the desk.I push on the desk. The ground pushes on the chair...The ground pushes on the chair... A spring pulls or pushes on a massA spring pulls or pushes on a mass A rocket engine provides some number of Newtons of A rocket engine provides some number of Newtons of

thrust (1 lb of thrust = mg = 2.205*9.81 = 21.62 Newtons)thrust (1 lb of thrust = mg = 2.205*9.81 = 21.62 Newtons)

Action at a distanceAction at a distance:: GravityGravity ElectricityElectricity MagnetismMagnetism

Contact forces:Contact forces:

Objects in contact exert forces.Objects in contact exert forces.

Convention:Convention: FFa,ba,b means acting on means acting on aa due to due to bb”.”.

So So FFhead,thumbhead,thumb means “ means “the force on the force on

the head due to the thumbthe head due to the thumb”.”.

FFhead,thumb

The Force

Action at a DistanceAction at a Distance

Gravity:Gravity:

Burp!Burp!

GravitationGravitation(Courtesy of Newton)(Courtesy of Newton)

Newton found that Newton found that aamoonmoon / / gg = 0.000278 = 0.000278 and noticed that and noticed that RREE

22 / R / R22 = 0.000273= 0.000273

This inspired him to propose the This inspired him to propose the Universal Law of Gravitation:Universal Law of Gravitation:

R RE

amoon g

where G = 6.67 x 10 -11 m3 kg-1 s-2

And the force is attractive along a line between the 2 objects

Hey, I’m in UCM!

2Mm

GMmF

R

We will discuss these concepts

later

UnderstandingUnderstanding

If the distance between two point particles is doubled, then the gravitational force between them:

A)Decreases by a factor of 4B)Decreases by a factor of 2C)Increases by a factor of 2D)Increases by a factor of 4E)Cannot be determined without knowing the masses

Newton’s Third Law:Newton’s Third Law:

Forces occur in pairs: Forces occur in pairs: FFA ,BA ,B = - = - FFB ,AB ,A..

For every “action” there is an equal and opposite “reaction”.For every “action” there is an equal and opposite “reaction”.

We have already seen this in the case of gravity:We have already seen this in the case of gravity:

1 212 212

12

mmGR

F F

R12

m1m2

FF12 FF21

We will discuss these concepts in more detail later.

Newton's Third Law...Newton's Third Law...

FFA ,BA ,B = - = - FFB ,AB ,A. . is true for contact forces as wellis true for contact forces as well::

FFm,w FFw,m

FFm,f

FFf,m

Force on me from wall is equal and opposite to the force on the wall from the me.

Force on me from the floor is equal and opposite to the force on the floor from the me.

block

Example of Bad ThinkingExample of Bad Thinking

Since Since FFm,b m,b = -= -FFb,mb,m, why isn’t , why isn’t FFnetnet = 0 = 0 and and aa = 0 = 0 ??

a ??a ??FFm,b FFb,m

ice

block

Example of Good ThinkingExample of Good Thinking

Consider Consider only the box only the box as the system!as the system! FFon boxon box = = mmaaboxbox == FFb,mb,m

Free Body Diagram (next power point).Free Body Diagram (next power point).

aaboxbox

FFm,b FFb,m

ice No ice

Friction force

block

Add a wall that Add a wall that stopsstops the motion of the block the motion of the block

Now there are two forces acting (Now there are two forces acting (in the horizontal in the horizontal directiondirection) on block and they cancel) on block and they cancel FFon boxon box = = mmaabox box = = FFb,mb,m + + FFb,wb,w = = 00

Free Body Diagram (next power point).Free Body Diagram (next power point).

aaboxbox

FFm,b FFb,m

ice

FFb,w FFw,b

Newton’s 3rd Law UnderstandingNewton’s 3rd Law Understanding

Two blocks are stacked on the ground. How many action-reaction pairs of Two blocks are stacked on the ground. How many action-reaction pairs of forces are present in this system?forces are present in this system?

(a) 2

(b) 3

(c) 4

(d) 5

a

b

Solution:Solution:

FFE,aE,a

FFa,Ea,E

a

b

FFE,bE,b

a

bFFb,Eb,E

FFb,ab,a

FFa,ba,b

a

b

FFb,gb,g

FFg,bg,b

a

b

FFa,ba,b

FFb,ab,a

a

b

contactgravity gravity contactgravityvery tiny

5

UnderstandingUnderstanding

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)F1=100F2

B)F1=10F2

C)F1=F2

D)F2=10F1

E)F2=100F1

Newton’s Third Law

Flash: Newton’s 1Flash: Newton’s 1StSt Law Law

Flash: Newton’s 2Flash: Newton’s 2ndnd Law Law

Flash: Newton’s 3Flash: Newton’s 3rdrd Law Law

Flash: Applications of NewtonFlash: Applications of Newton