n ewton ’ s l aws. i nertia newton’s 1 st law an object wants to keep on doing what it is...

NEWTON’S LAWS

INERTIA

INERTIA•Newton’s 1st Law•An object wants to keep on doing what it is already doing•In order to change, it needs a net force not equal to zero•sum of all forces

EXPLORING 1ST LAW: TORQUE•Force:•a push or pull [units: Newton (N)]•Translational motion: motion without rotation•Fulcrum: point of rotation (ex. hinge)•If a force is applied to an object with a fulcrum, you get rotational motion

Fulcrum

EXPLORING 1ST LAW: TORQUE•Force:•a push or pull•Translational motion: motion without rotation•Fulcrum: point of rotation (ex. hinge)•If a force is applied to an object with a fulcrum, you get rotational motion

Fulcrum

TORQUE•Rotational Force is most effective far from the fulcrum•Torque is the combination of force and distance from fulcrum (lever arm = r)

€

τ =Frr r

r = 0 Which lever arm has greatest torque?

A B

C

TORQUE•Only the perpendicular part of the force is applied to the torque.

•This requires a touch of trig.

€

τ =Frsinθ

€

θ

r

ROTATIONAL STABILITY•1st Law, rotational style:

•“In the absence of a net torque, an object not rotating will continue to not rotate and an object rotating will continue to rotate at a constant rotational velocity”

•This means that if a rotational system is in balance (not moving), the net torque is zero.

ROTATIONAL STABILITY•1st Law, rotational style:

•All torques add to zero•Convention: CW is positive, CCW is negative•Which dog has a positive torque?€

Στ =0

€

τ1 + τ 2 + τ 3...τ n = 0

CENTER OF MASS•Concentration of mass•Where the weight force of the object is located

•This weight force can add a torque if it is not at the fulcrum.

CENTER OF MASS•Concentration of mass

•This weight force can add a torque if it is not at the fulcrum.

What is applying the torque on the left side?

CENTER OF MASS•Where is the woman’s COM?

•In her upper body there are two main torques.

•To compensate she needs to lean back.

Check Yourself!

Go to pg. 349

NEWTON’S 2ND LAW: ACCELERATION•Recap of 1st law:• involves objects with no net force•When you do have a net force:•object will accelerate in same direction•This acceleration is proportional to the net force and inversely proportional to its mass

€

acceleration∝ force

€

acceleration∝1

mass

EXPLORING 2ND LAW

•Put the two together:

•mass units: kg•acceleration units: m/s2

•This makes force units: kg•m/s2

•Let’s call 1 kg•m/s2 a Newton (N)

€

a =Fnet

mtot

€

acceleration =Forcenet

masstotal

€

F = ma

FREEFALLING: 2ND LAW•Weight is a very common force•In freefalling the only force is weight•In freefall, the acceleration is -9.8 m/s2 (g)•So, if a = 9.8 m/s2 and m = mass of object

€

Fw = mg

2ND LAW•Let’s look back at the original eqn

•Pay close attention to the “net”

•This is a reminder you need to add all the forces present upon that object

•Let’s look at an example…

€

a =Fnet

mtot

2ND LAWHanging mass pulling a block on a frictionless tabletop

•How does the acceleration of m1 change by connecting it to m2?•Why?•Acceleration decreases because m2 is pulling more mass•More inertia

2ND LAW: EXAMPLEThe Batman, with a mass of 70-kg, rappels down a rope from his bat-copter with a downward acceleration of 3.5 m/s2. What vertical force does the rope exert on Batman?

2ND LAW: EXAMPLEThe Batman, with a mass of 70-kg, rappels down a rope from his bat-copter with a downward acceleration of 3.5 m/s2. What vertical force does the rope exert on Batman?

Start all force questions with a diagram showing the forces

Pick the positive direction(make direction of motion positive)

€

Frope

€

Fw

€

m1 = 70kg

Identify all givens with symbols

€

a = 3.5m /s2

2ND LAW: EXAMPLE

Given:

€

Frope

€

Fw

€

m1 = 70kg

€

a = 3.5m /s2€

m = 70kg

€

a = 3.5m /s2

€

Fw =

€

mg = (70kg)(9.8m /s2)

€

Fw = 686N

Want:

€

Frope

The Batman, with a mass of 70-kg, rappels down a rope from his bat-copter with a downward acceleration of 3.5 m/s2. What vertical force does the rope exert on Batman?

2ND LAW: EXAMPLECalculations:

€

Frope

€

Fw€

Fnet = mtota

€

Fnet = ΣF

€

=Fw − Frope

€

=mtota

€

Frope = Fw − ma

€

Frope = (686N) − (70kg)(3.5 m /s2)

€

Frope = 441N

€

=441N

Lab

Go to pg. 359

2ND LAW LAB PREPTo be able to complete the lab you need to be able to find acceleration in the situation below:

2ND LAW LAB PREPDraw diagram of the moving “body”(in this case: both masses connected by string)

m1

m2

2ND LAW LAB PREPDraw diagram of the moving “body”(in this case: both masses connected by string)

Draw in all forces acting on the body Which forces affect the motion?

m1

m2

m2g

2ND LAW LAB PREP

Use Newton’s 2nd Law

m1

m2

m2g€

Fnet =

€

mtota

Begin Lab

Go to pg. 359

FREE BODY DIAGRAMS•Identifies all forces acting upon an object•Shows direction and relative sizesScenario 1:

10 kg block in freefall (no air resistance)

€

Fw

€

a =Fnet

m

€

a =mg

m

€

=mg

€

a = g = 9.8m /s2

10 kg

FREE BODY DIAGRAMSScenario 2:

10 kg block in freefall with air resistance

€

Fw

€

Fair

€

a =Fnet

m

€

=Fw − Fair

m

10 kg

FREE BODY DIAGRAMSScenario 3:10 kg block pulled across a

frictionless floor by a string

€

mg

€

FN

Just like projectiles, we treat x and y separately

10 kg

Does it fall through the floor?

Normal Force• perpendicular () to the surface• equal to the force acting on the opposite side of surface

Y

FREE BODY DIAGRAMSScenario 3:10 kg block pulled across a

frictionless floor by a string

€

mg

€

N

€

ay =Fnet

m

€

=FN − Fw

m= 0

10 kg

X

€

T

net force only in x direction

€

ax =Fs

m

3 Forces


10 kg block on a frictionless ramp at 30°

10 kg



€

N

10 kg

€

mg

There are 2 forces



10 kg

Choose x and y to line up with movement

xy

€

N

€

mg

There are 2 forces



€

N

10 kg

€

mg

Choose x and y to line up with movement

There are 2 forcesWeight has both an x and y component

FREE BODY DIAGRAMS

€

mg

xy

€

θ

€

Fy

€

Fx

€

=Fw cosθ

€

=Fw sinθ

€

θ

The angle between the weight and Fy is the same as the angle between the ramp and the ground



€

FN

10 kg

€

Fw

€

a =Fnet

m€

Fwy

€

Fwx

The y forces cancel out (block does not move in the y direction)

Only force left is:the x component of the weight

€

=Fx

m

€

=Fw sinθ

m

€

=(98N)(sin30°)

10kg

€

=4.9m

s2

AIR RESISTANCE•Two things determine amount of air resistance•surface area (shape)•speed

TIME TO PRACTICETurn to pg. 371

2ND LAW: ROTATIONAL STYLE•Translational motion takes you from place to place•Rotational motion moves about an axis (spinning)•for every type of translational measurement there is rotational one

ROTATIONAL SYMBOLS•Measurements•The angles (θ) are in radians

RADIANS•s = arc length (sector)•r = radius•radians are just the ratio of arc length to the radius

€

Angle,θ€

Radius, r€

Arc length, s

€

θ =s

r

RADIANS•For a whole circle, s is just the circumference

•One whole circle makes 2π radians•2π = 360°

€

Angle,θ€

Radius, r€

Arc length, s

€

θ =2πr

r= 2π

€

s = 2πr

CONVERTING RADIANS

€

Angle (degrees) =360°

2π radians

⎛

⎝ ⎜

⎞

⎠ ⎟• Angle (radians)

€

Angle (radians) =2π radians

360°

⎛

⎝ ⎜

⎞

⎠ ⎟• Angle (degrees)

ROTATIONAL EQUATIONS•Each translational equation has a rotational version

€

ω ≡

€

α ≡

€

τ ≡

€

I ≡

€

rotational velocity

€

rotational

€

acceleration

€

torque

€

rotational inertia

ROTATIONAL INERTIA•α = τnet / I•Which shapes accelerate faster?•The ones with smaller I

RELIEF

ROTATIONAL MOTIONExample: Two dumbbells are spun about their center axis with a torque of 5.0 N•m. Both consist of a 50 cm long rod with a mass of 0.20 kg and two 1.0 kg spheres. Calculate rotational acceleration of each.

#1

#2

ROTATIONAL MOTION

•First, we need to find the moment of inertia.•The I of the dumbbell will be the sum of the I of the rod and the I of the two spheres

#1

#2

I OF ROD

•Find I for a rod spinning around center axis

•So, for both dumbbells the I of the rod is:

€

I =1

12ML2

€

I =1

12(0.20kg)(0.50m)2

€

=4.2 ×10−3 kg⋅m2

I OF SPHERES

•Find I for a mass rotating in a circle with a radius R

• shape of path makes a hollow loop

•Each dumbbell has a different R, so I will be different also:

€

I = MR2

€

I1 = 2 • MR2

€

=2(1.0kg)(0.25m)2

€

=0.125 kg⋅m2

€

I2 = 2 • MR2

€

=2(1.0kg)(0.05m)2

€

=0.005kg⋅m2

OF DUMBBELLS•Now we get to use Newton’s Second Law to calculate the rotational acceleration

•Dumbbell #2 will spin almost 14 times for every single spin of #1!

€

α1 =τ net

I1

€

=5.0N⋅m

(4.2 ×10−3 + 0.125)kg⋅m2

€

=39 rads2

€

α2 =τ net

I2

€

=5.0N⋅m

(4.2 ×10−3 + 0.005)kg⋅m2

€

=543 rads2

€

vα

NEWTON’S 3RD LAW•Equal and Opposite

•For every force that one object exerts on a second object, the second object exerts an equal (in size) and opposite (in direction) reaction force on the first object.

NEWTON’S 3RD LAW SUBTLETIES•Forces only come in pairs (you cannot create a single force)

•Does action cancel out reaction?

•NO! Each force acts on a different object.

•Each force has different effect on their object! (ex. gun and bullet)

FRICTION•Friction is everywhere there is motion on a surface or within a fluid

•Friction is a resistive force (a force that opposes motion)

•Friction evidence?

FRICTION•Let’s examine sandpaper

400 grit40 grit

FRICTION•Even “smooth” sandpaper is rough

400 grit40 grit

FRICTION•Slide the surfaces of two pieces together

•Atoms at the peaks bond with atoms at the peaks of the other•It requires force to break these bonds

•opposing force is friction

TRIBOLOGY•The microscopic effect of friction was discovered by tribologists in the 1950’s

•a tribologist is someone who studies friction

•Test friction between two pieces of metal

TRIBOLOGY•Which would experience more friction, normal surfaces or very polished surfaces?

•Polished! There are way more atoms that come into contact and make bonds.

TRIBOLOGY•If surface is polished smoothly enough, it can create a strong bond known as cold weld

•Almost impossible to break

FORCE OF FRICTION•Friction between solid surfaces depends on:

1. The perpendicular (Normal) force between the surfaces in contact

2. The nature of the surfaces in contact

•together, these create a net frictional force

PERPENDICULAR FORCE•Rub your hands together

•What happens to friction if you press harder?

•So,

•or

•When the object is on a surface parallel to Earth’s surface, N = mg

€

F ∝ F⊥

€

F ∝ N

NATURE OF SURFACES•Rub your hands together again

•What would happen to friction if you added some oil to your hands?

•What would happen to friction if you added some dry glue?

•This condition of a surface is given as a quantity known as the coefficient of friction

€

(μ)

NATURE OF SURFACES•Each situation has two coefficients

•μs = static friction Surfaces μs μk

steel on steel 0.74 0.57

glass on glass

0.94 0.40

tire on dry road

1.0 0.80

tire on icy road

0.30 0.015

bone joints 0.010 0.0030

•when an object is being pushed, but hasn’t moved yet

•μk = kinetic friction

•the friction once the object starts moving

FRICTIONAL FORCE•Friction is the product of the normal force and the coefficient of friction

•when μ = 0, surface contact is frictionless

•when μ = 1, force required to move object is equal to its weight

€

f = μN

FRICTION EXAMPLE•Let’s compare the friction force of a 2,000-pound car skidding on a dry road compared to an icy one:

€

fdry = μ k N

€

=(0.8)(2,000lbs)

€

=1,600lbs

€

f icy = μ k N

€

=(0.015)(2,000lbs)

€

=30lbs

•That’s why it takes so long to stop on ice

TIME FOR LABTurn to pg. 394

CIRCULAR MOTION•On the Gravitron ride you feel like you are being pushed outwards

•Centrifugal Force•does not exist•It is an apparent force

•Which way would you fly out if centrifugal force existed?

CENTRIPETAL FORCE•In reality, what you feel is the force of the wall pushing you in

•Centripetal Force = force pulling you inward when you move in a circle

•If circular force was centripetal which way would you fly out?

€

Fc

CENTRIPETAL FORCE•But, what do you feel?

•Imagine if you drove into a wall

•What would you feel?

CENTRIPETAL FORCE•What “threw” the driver over wall?

•inertia

•Hard to know source when you feel

€

va

CENTRIPETAL FORCE•Inertia is at work in a turn also

CENTRIPETAL FORCE•The “centrifugal” force you feel is your inertia

•If you removed the gravitron wall you would fly out tangent to the circular path

CENTRIPETAL ACCELERATION•Imagine twirling a ball on a string

•What affects its acceleration?1. Speed of the twirl2. Length of the string

€

centripetal acceleration =(speed)2

distance to point of rotation

CALCULATING CENTRIPETAL FORCE

•Now let’s mix it with Newton’s Second Law

€

ac =v 2

r

€

Fc = mac

€

Fc = mv 2

r

CIRCULAR VELOCITY•Easiest to find when using the period (T)

•Average speed can be found with

•For a circle, the distance in one period (the time) is the circumference, 2πr

€

v =d

t

€

v =2πr

T

CENTRIPETAL FORCE•Plugging the new v into centripetal force gives:

•note:•velocity always tangent to path

€

Fc =4π 2mr

T 2

CENTRIPETAL FORCE•Example: imagine twirling a .50 kg ball on a 0.80 m long string. The ball moves in a circle 20 times in 16 seconds. How much force is required to keep the ball moving this way?given:

€

m =

€

0.50kg

€

r =

€

0.80 m

€

T =

€

16s

20

€

=0.80 s

want:

€

Fc

CENTRIPETAL FORCE•plug ‘em in

€

=4π 2(0.50kg)(0.80m)

(0.80s)2

€

Fc =4π 2mr

T 2

€

=24.7N

VERTICAL CIRCULAR MOTION•When moving in a vertical circle, gravity becomes important.

•Net force at each part of ride varies.

BOTTOM OF THE RIDE

•At the bottom, the Fc pushes the seat up.

•This pulls the seat against your weight (up).

•Feels heavierFc

mg

NORMAL FORCE OF THE SEAT•We can find the net force acting on you using Newton’s 2nd law.

•Let’s call the normal force at the bottom FB

•The normal force at the top FT

€

a =Fnet

m

NORMAL FORCE AT THE BOTTOM•At the bottom of the ride:

•Notice FB gets stronger the faster the ride goes.

FB

€

a =Fnet

m

€

⇒v 2

r=

FB − mg

m

€

⇒ FB = m g + v 2

r( )

€

⇒ FB = mg + mv 2

r

mg

NORMAL FORCE AT THE TOP•At the top of the ride the acceleration points down (negative):

•Opposite situation at the top

FT

€

⇒ −v 2

r=

FT − mg

m

€

⇒ FT = m g − v 2

r( )

TIME FOR LABTurn to pg. 404

GRAVITY•Not everything thrown up must come down!

•The higher something goes the smaller the effect of earth’s gravity on it.

•If you throw it hard enough, “g” becomes too small to slow it to a stop.

•This is called the escape velocity

•escape velocity from earth = 25,000 mph

GRAVITY•Gravity is not only caused by large planets, moons and stars.

•Everything with mass attracts everything else with mass no matter the distance

•Unless the mass is really large, you will not notice it

ANTHROCENTRIC UNIVERSE•Aristotle reasoned that all objects wanted to be at the center of the universe…

•The Earth!

•all objects circle Earth

HELIOCENTRIC UNIVERSE•Copernicus noticed that the planets did not stay the same distance from the earth.

•He used observations to come up with a new theory.

•All objects circle the sun

TYCHO AND KEPLER•Tycho Brahae spent 20 years trying to proves Copernicus wrong with very careful measurements of the planets.

•Compromise

•Planets circle both the Earth and sun.

TYCHO AND KEPLER•Johannes Kepler wanted to carefully analyze Tycho’s data.

•Kepler got his chance when Tycho died.

•Using Tyco’s data and mathematical theories, Kepler developed the Three Laws of Planetary Motion.

KEPLER’S LAWS1. Each planet moves around the Sun

in an elliptical orbit with the Sun at one focus of the ellipse.

KEPLER’S LAWS2. The line from the Sun to any planet

sweeps out equal areas of space in equal time intervals.

KEPLER’S LAWS3. The squares of the periods of the

planets are proportional to the cubes of their average distances from the Sun

€

T 2 ∝ R3

UNIVERSAL LAW OF GRAVITY•Isaac Newton took Kepler’s ideas further.

•He noticed that the force making the moon go around the earth was the same as the force causing an apple to fall.

•“Between any two masses there exists an attractive force of gravity that is proportional to the product of the masses and inversely porportional to the square of the distance between their centers.”

UNIVERSAL LAW OF GRAVITY•This gives the expression:

•or, the formula:

•G = universal gravitational constant

€

FG ∝m1m2

d2

€

FG = Gm1m2

d2

GRAVITATIONAL CONSTANT•Newton never was able to calculate “G”

•In order to calculate, first rearrange:

•Measuring each of the variables is easy until you get to force.

•The force between materials on earth was too small for Newton to be able to detect.

€

G =FGd2

m1m2

MEASURING “G”•Henry Cavendish used an apparatus made by John Michell.

MEASURING “G”•Used hanging lead balls.

•Force of gravity can be found by measuring the tiny angle the small balls moved

€

G = 6.67 ×10−11 N⋅m2

kg2

MASS OF THE EARTH•Cavendish’s main reason to calculate G was to use it to find the mass of the Earth.

•Imagine any object on the Earth’s surface. It’s force due to gravity (FG) is mg.

•Since height of any object on Earth is negligible, R is just Earth’s radius.

€

mg =GMearthm

Rearth2

€

⇒ g =GMearth

Rearth2

MASS OF THE EARTH•The radius of the Earth had been calculated way back in 200 BC

€

Mearth =gRearth

2

G

€

=(9.8m /s2)(6.37 ×106 m)2

6.67 ×10−11 kg⋅m

s2€

Rearth = 6.37 ×106 m

€

Mearth = 5.96 ×1024 kg

SATELLITE MOTION•Imagine standing on a really tall building and throwing a ball really hard.•The ball would follow a parabolic path and land in front of the building.

SATELLITE MOTION•Now throw it harder.

•Same thing except balls goes a little farther.

SATELLITE MOTION•If you throw it at 8,000 m/s, something interesting happens.

•Once the ball goes out 8,000 m it will have fallen by 4.9 m.

•The curvature of the earth happens to fall at the same rate.

SATELLITE MOTION•The ball is freefalling, but never gets closer to the Earth’s surface!

•As long as the speed doesn’t change it will freefall forever.

•The ball is now a satellite.

SATELLITE CALCULATIONS•What is acting as the centripetal force?

•The force of gravity.

€

Fc = FG

€

⇒mv 2

r=

GmME

r2

€

⇒v 2

r=

GME

r2

€

⇒ v =GME

r

ME is the mass of the Earth and r is the distance from the center of the Earth to the satellite

SATELLITE CALCULATIONS•Let’s do the same thing in terms of period now (T)

€

Fc = FG

€

⇒4π 2mr

T 2 =GmME

r2

€

⇒ T =4π 2r3

GME

Remember:the units for period should always be seconds

Works for other planets if you change ME to the planet’s mass.

WEIGHTLESS ASTRONAUTS•Why do astronauts feel weightless in space?

•Do they really have almost no weight?

•Lets see…

WEIGHTLESS ASTRONAUTS•First, let’s calculate the weight on Earth of a 70 kg astronaut:

€

Fg =Gmast ME

r2

€

=6.67 ×10−11 N⋅m2

kg2

⎛

⎝ ⎜

⎞

⎠ ⎟ 5.96 ×1024 kg( )(70kg)

6.37 ×106 m( )2

given:

€

=685 N

€

mast = 70kg

€

ME = 5.96 ×1024 kg

€

rE = 6.37 ×106 m

WEIGHTLESS ASTRONAUTS•Next, let’s calculate the astronaut’s weight in orbit on the space station (380 km high)

€

Fg =Gmast ME

r2

€

=6.67 ×10−11 N⋅m2

kg2

⎛

⎝ ⎜

⎞

⎠ ⎟ 5.96 ×1024 kg( )(70kg)

6.75 ×106 m( )2

given:

€

=541 N

€

mast = 70kg

€

ME = 5.96 ×1024 kg

€

r = 6.37 ×106 m + 3.80 ×105 m = 6.75 ×106 m

WEIGHTLESS ASTRONAUTS•685 N vs 541 N

• Astronaut on space station still has about 80% of his weight.

•Astronauts are actually freefalling with the space station.

•Imagine an amusement park ride that never stops falling. Yup, that’s what it’s like in space. (Vomit Comet)

GEOSYNCHRONOUS ORBIT•In order for communication satellites to be useful they have to be accessible

•They are always above the same point on Earth and appear motionless

•Actually, they are freefalling with the same period as the spin of the Earth (24 hours)

DARK MATTER•Remember, we can find the mass of something by observing the speed and radius of its orbiting satellites

•So, if we observed a star in a galaxy we could calculate the mass of the galaxy.€

M =v 2r

G

DARK MATTER•You can also get the mass of a galaxy using brightness.

•Brightness can be converted to number of sun-sized stars and then mass.

•This was done in 1967

DARK MATTER•The stars in the Andromeda galaxy move way too fast compared to its brightness.

DARK MATTER•Scientists concluded that there must be some undetectable dark matter.

•Other galaxies give the same phenomenon.

•Over 90% of the universe appears to be made of this dark matter.

DARK MATTER•The dark matter appears to stretch much further than the galaxies themselves

n ewton ’ s l aws. i nertia newton’s 1 st law an object wants to keep on doing what it is...

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