Physics 111: Lecture 4 Today’s Agenda

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Physics 111: Lecture 4, Pg 1 Physics 111: Lecture 4 Today’s Agenda Recap of centripetal acceleration Newton’s 3 laws How and why do objects move? Dynamics

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Physics 111: Lecture 4 Today’s Agenda. Recap of centripetal acceleration Newton’s 3 laws How and why do objects move? Dynamics. Review: Centripetal Acceleration. UCM results in acceleration: Magnitude : a = v 2 / R =   R Direction : - r (toward center of circle). ^. v =  R. - PowerPoint PPT Presentation

Transcript of Physics 111: Lecture 4 Today’s Agenda

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Physics 111: Lecture 4, Pg 1

Physics 111: Lecture 4

Today’s Agenda

Recap of centripetal acceleration Newton’s 3 laws

How and why do objects move? Dynamics

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Review: Centripetal Acceleration

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

R a

^

Useful stuff:

f = rotations / secT = 1 / fω = 2 / T = 2 f = rad/sec

v = R

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Dynamics

Isaac Newton (1643 - 1727) published Principia Mathematica in 1687. 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, FNET = F = ma

Law 3: Forces occur in pairs: FA ,B = - FB ,A

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

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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!

1. Dishes2. Monkey

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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 104 sec, R ~ RE = 6.4 x 106 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

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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.» The mass of an object is a constant property of that

object, and is independent of external influences.

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

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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.

F1 F2

aF1

F2

a

FNET

FNET = ma

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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 weeks:

tavv

ta21

tvxx

xx0x

2xx00

+=

++=

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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 i direction. If the box starts at rest, what is its speed v after being pushed a distance d = 10 m?

F

v = 0

m a

i

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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 i direction. If the box starts at rest, what is its speed v after being pushed a distance d = 10m ?

d

F

v

m a

i

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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

F

v

m a

i

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Example: Pushing a Box on Ice...

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

d

F

v

m a

i

vFd

m

2

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Lecture 4, Act 1Force 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) 2/3 a1 (b) 3/2 a1 (c) 3/4 a1

F a1

m1 F a2 = 2a1

m2

F a = ? m1 m2

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Lecture 4, Act 1Force and acceleration

Since a2 = (1/2) a1 for the same applied force, m2 = (1/2)m1 !m1+ m2 = 3m2 /2

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

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

So a = (2/3)F / m1but F/m = a

a = 2/3 a1

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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...

Action at a distance:» Gravity» Electricity

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Contact forces:

Objects in contact exert forces.

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

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

Fhead,thumb

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Action at a distance

Gravity:

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Gravitation(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: |FMm |= GMm / R2

R RE

amoong

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

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Gravity...

The magnitude of the gravitational force F12 exerted on an object having mass m1 by another object having mass m2 a distance R12 away is:

The direction of F12 is attractive, and lies along the line connecting the centers of the masses.

F Gm m

R12

1 2

122

R12

m1 m2F12 F21

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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 Fg

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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

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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

Fg The force that gravity exerts on any object is called its Weight W = 539 N

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Lecture 4, Act 2Force 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) 0.75 W (b) 1.5 W

(c) 2.25 W

E

X

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Lecture 4, Act 2Solution

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

Mm

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

.

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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:

F F121 2

122 21 G

m m

R

R12

m1m2

F12 F21

Newton’sSailboard

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Newton's Third Law...

FA ,B = - FB ,A. is true for contact forces as well:

Fm,w Fw,m

Fm,f

Ff,m

2 Skateboards

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Example of Bad Thinking

Since Fm,b = -Fb,m, why isn’t Fnet = 0 and a = 0 ?

a ??Fm,b Fb,m

ice

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Example of Good Thinking

Consider only the box as the system!Fon box = mabox = Fb,m

Free Body Diagram (next time).

abox

Fm,b Fb,m

ice

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Lecture 4, Act 3Newton’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

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Lecture 4, Act 3Solution:

(c) 4

FE,a

Fa,E

a

b

FE,b

a

bFb,E

Fb,a

Fa,b

a

bFg,b

Fb,g

a

b

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Recap of today’s lecture

Newton’s 3 Laws: (Text: 4-1 to 4-5)

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. (Text: 4-1)

Law 2: For any object, FNET = F = ma (Text: 4-2 & 4-3)

Law 3: Forces occur in pairs: FA ,B = - FB ,A. (Text: 4-4 & 4-5)

Look at Textbook problems Chapter 4: # 3, 5, 7, 19

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