Physics 111: Lecture 4, Pg 1

Physics 111: Lecture 4Physics 111: Lecture 4

Today’s AgendaToday’s Agenda

Recap of centripetal acceleration Newton’s 3 laws

How and why do objects move? DynamicsDynamics

Physics 111: Lecture 4, Pg 2

Review: Centripetal AccelerationReview: Centripetal Acceleration

UCM results in acceleration:Magnitude: |a| = v2 / R = R Direction: - rr (toward center of circle)

R aa

^̂

Useful stuff:Useful stuff:

f = rotations / sec

T = 1 / f

ω = 2 / T = 2 f = rad/sec

v = R

Physics 111: Lecture 4, Pg 3

Isaac Newton (1643 - 1727) published Principia Mathematica in 1687

Physics 111: Lecture 4, Pg 4

DynamicsDynamics

In this work, he proposed three “laws” of motion:

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, FFNET = FF = maa

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

(For every action there is an equal and opposite reaction.)

Physics 111: Lecture 4, Pg 5

Newton’s First LawNewton’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 frameinertial 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!

1. Dishes2. Monkey

Physics 111: Lecture 4, Pg 6

Is Urbana a good IRF?Is Urbana a good IRF? Is Urbana accelerating? YES!

Urbana is on the Earth.The Earth is rotating.

What is the centripetal acceleration of Urbana? T = 1 day = 8.64 x 10 4 sec, R ~ RE = 6.4 x 10 6 meters .

Plug this in: aU = .034 m/s2 ( ~ 1/300 g) Close enough to 0 that we will ignore it. Urbana is a pretty good IRF.

av

RR

TRU

2 22

2

Icepuck

Physics 111: Lecture 4, Pg 7

Newton’s Second LawNewton’s Second Law

For any object, FFNET = FF = maa.

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

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

» This is the definition of mass.

» The mass of an object is a constant property of thatobject, and is independent of external influences.

Force has units of [M]x[L / T2] = kg m/s2 = N (Newton)

Physics 111: Lecture 4, Pg 8

Newton’s Second Law...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 like adding vectors.

FF1 FF2

aaFF1

FF2

aa

FFNET

FFNET = maa

Physics 111: Lecture 4, Pg 9

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

Components of FF = maa :

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 weeks:

tavv

ta21

tvxx

xx0x

2xx00

+=

++=

Physics 111: Lecture 4, Pg 10

Example: Pushing a Box on Ice.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 ii direction. If the box starts at rest, what is its speed v after being pushed a distance d = 10 m?

FF

v = 0

m a

ii

Physics 111: Lecture 4, Pg 11

Example: Pushing a Box on Ice.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 ii direction. If the box starts at rest, what is its speed v after being pushed a distance d = 10m ?

d

FF

v

m a

ii

Physics 111: Lecture 4, Pg 12

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

Start with F = ma.a = F / m.Recall that v2 - v0

2 = 2a(x - x0 ) (Lecture 1)

So v2 = 2Fd / m vFd

m

2

d

FF

v

m a

ii

Physics 111: Lecture 4, Pg 13

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

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

d

FF

v

m a

ii

vFd

m

2

Physics 111: Lecture 4, Pg 14

Lecture 4, Lecture 4, Act 1Act 1Force and accelerationForce 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.

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

Physics 111: Lecture 4, Pg 15

Lecture 4, Lecture 4, Act 1Act 1Force and accelerationForce and acceleration

Since a2 = 2a1 for the same applied force, m2 = (1/2)m1 !

m1+ m2 = 3m1 /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

Physics 111: Lecture 4, Pg 16

ForcesForces

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...

Action at a distance:» Gravity» Electricity

Physics 111: Lecture 4, Pg 17

Contact forces:Contact forces:

Objects in contact exert forces.

Convention: FFa,b means “the force acting on a due to b”.

So FFhead,thumb means “the force on the head due to the thumb”.

FFhead,thumb

Physics 111: Lecture 4, Pg 18

Action at a distanceAction at a distance

Gravity:

Physics 111: Lecture 4, Pg 19

GravitationGravitation(Courtesy of Newton)(Courtesy of Newton)

Newton found that amoon / g = 0.000278 and noticed that RE

2 / R2 = 0.000273

This inspired him to propose the Universal Law of Gravitation:Universal Law of Gravitation: |FMm |= GMm / R2

R RE

amoong

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

Physics 111: Lecture 4, Pg 20

Gravity...Gravity...

The magnitude of the gravitational force FF12 exerted on an object having mass m1 by another object having mass m2 a distance R12 away is:

The direction of FF12 is attractive, and lies along the line connecting the centers of the masses.

F Gm m

R12

1 2

122

R12

m1 m2FF12 FF21

Physics 111: Lecture 4, Pg 21

Gravity...Gravity...

Near the Earth’s surface:R12 = RE

» Won’t change much if we stay near the Earth's surface.

» i.e. since RE >> h, RE + h ~ RE.

RE

m

M

h2E

Eg R

mMGF FFg

Physics 111: Lecture 4, Pg 22

Gravity...Gravity...

Near the Earth’s surface...

2

E

E2E

Eg R

MGm

RmM

GF

So |Fg| = mg = ma

a = g

All objects accelerate with acceleration g, regardless of their mass!

22E

E s/m81.9RM

Gg Where:

=g

Leaky Cup

Physics 111: Lecture 4, Pg 23

Example gravity problem:Example gravity problem:

What is the force of gravity exerted by the earth on a typical physics student?

Typical student mass m = 55kgg = 9.8 m/s2.Fg = mg = (55 kg)x(9.8 m/s2 )

Fg = 539 N

FFg The force that gravity exerts on any object is

called its Weight

W = 539 N

Physics 111: Lecture 4, Pg 24

Lecture 4, Lecture 4, Act 2Act 2Force and accelerationForce and acceleration

Suppose you are standing on a bathroom scale in 141 Loomis and it says that your weight is W. What will the same scale say your weight is on the surface of the mysterious Planet X ? You are told that RX ~ 20 REarth and MX ~ 300 MEarth.

(a)(a) 00.75.75 W (b)(b) 1.5 W

(c)(c) 2.25 W

E

X

Physics 111: Lecture 4, Pg 25

Lecture 4, Lecture 4, Act 2Act 2SolutionSolution

The gravitational force on a person of mass m by another object (for instance a planet) having mass M is given by:

F GMm

R

2

WW

FF

X

E

X

E

Ratio of weights = ratio of forces:

G

M m

R

GM m

R

X

X

E

E

2

2

MM

RR

X

E

E

X

2

WW

X

E

300120

752

.

Physics 111: Lecture 4, Pg 26

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

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

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

We have already seen this in the case of gravity:

F F121 2

122 21 G

m m

R

R12

m1m2

FF12 FF21

Newton’sSailboard

Physics 111: Lecture 4, Pg 27

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

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

FFm,w FFw,m

FFm,f

FFf,m

2 Skateboards

Physics 111: Lecture 4, Pg 28

Example of Bad ThinkingExample of Bad Thinking

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

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

ice

Physics 111: Lecture 4, Pg 29

Example of Good ThinkingExample of Good Thinking Consider only the box only the box as the system!

FFon box = maabox = FFb,m

Free Body Diagram (next time).

aaboxbox

FFm,b FFb,m

ice

Physics 111: Lecture 4, Pg 30

Lecture 4, Lecture 4, Act 3Act 3Newton’s 3rd LawNewton’s 3rd Law

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

(a) 2

(b) 3

(c) 4

a

b

Physics 111: Lecture 4, Pg 31

Lecture 4, Lecture 4, Act 3Act 3Solution:Solution:

(c) 4

FFE,aE,a

FFa,Ea,E

a

b

FFE,bE,b

a

bFFb,Eb,E

FFb,ab,a

FFa,ba,b

a

b

FFg,bg,b

FFb,gb,g

a

b

Physics 111: Lecture 4, Pg 32

Recap of today’s lectureRecap of today’s lecture

Newton’s 3 Laws:

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, FFNET = FF = maa

Law 3: Forces occur in pairs: FFA ,B = - FFB ,A.

ExtinguisherCart