Rotation and Newton’s Rotation and Newton’s Universal Law of GravitationUniversal Law of Gravitation

Chapter 7Chapter 7

Our friend Jupiter Our friend Jupiter

Jupiter rotates once Jupiter rotates once every 10 hoursevery 10 hours

Of course Earth only Of course Earth only rotates once every 24 rotates once every 24 hourshours

How would we How would we calculate the speed of calculate the speed of a rotating object?a rotating object?

Objects in circular motion have a Objects in circular motion have a tangential speedtangential speed

What is tangential What is tangential speed?speed?– Tangential speed is Tangential speed is

the instantaneous the instantaneous linear speed of any linear speed of any point rotating about an point rotating about an axisaxis

– r is radius of circler is radius of circle

– T is period (length of T is period (length of time for one rotation)time for one rotation)

T

rVt

2

So…how do we use that?So…how do we use that?

Let’s find the tangential speed of a point Let’s find the tangential speed of a point along Jupiter’s equator. What information along Jupiter’s equator. What information do we need?do we need?– Jupiter’s Radius at the equator: 7.15 x 10Jupiter’s Radius at the equator: 7.15 x 1077 m m– The time for one rotation: 10 hours= 36000 sThe time for one rotation: 10 hours= 36000 s

mph 000,281025.136000

)1015.7(22 47

s

mx

s

mx

T

rVt

Jupiter’s Speed at the PolesJupiter’s Speed at the Poles

Jupiter is not a solid sphere, so the radius at the Jupiter is not a solid sphere, so the radius at the equator is not the same as the radius at the equator is not the same as the radius at the poles.poles.– Jupiter’s Radius at the Poles: 6.69 x 10Jupiter’s Radius at the Poles: 6.69 x 1077 m m– Time of rotation: 10 hours = 36000 sTime of rotation: 10 hours = 36000 s

That’s a difference of 1,792 mph!That’s a difference of 1,792 mph!

mph 208,261017.136000

)1069.6(22 47

s

mx

s

mx

T

rVt

Compare that to EarthCompare that to Earth

Earth’s equatorial radius: 6.378 x 10Earth’s equatorial radius: 6.378 x 1066 m m

Earth’s rotational period: 24 hrs = 86,400sEarth’s rotational period: 24 hrs = 86,400s

Earth’s radius at the poles: 6.356 x 10Earth’s radius at the poles: 6.356 x 1066 m m

Earth’s rotational period: 24 hrs= 86,400 sEarth’s rotational period: 24 hrs= 86,400 s

mph 103882.46386400

)10378.6(22 6

s

m

s

mx

T

rVt

mph 103522.46286400

)10356.6(22 6

s

m

s

mx

T

rVt

Does an object undergoing circular Does an object undergoing circular motion always have the same motion always have the same

tangential velocity?tangential velocity?NO!NO!

Direction of velocity: Right

A

D

C

BDirection of velocity: Up

Direction of velocity: Left

Direction of velocity: Down

To clarify things…To clarify things…

Points that are the same distance from the Points that are the same distance from the center (i.e. that have the same radius) center (i.e. that have the same radius) have the same have the same tangential speedtangential speed but but notnot the same the same tangential velocitytangential velocity!!

If velocity is changing then…If velocity is changing then…

A change in velocity = ACCELERATIONA change in velocity = ACCELERATION

Therefore, objects that are undergoing circular Therefore, objects that are undergoing circular motion must be accelerating!motion must be accelerating!

2

2

2

2 42

onAccelerati lCentripetaT

r

rTr

r

va tc

Sample Problem p.258 #2Sample Problem p.258 #2

A young boy swings a yo-yo horizontally A young boy swings a yo-yo horizontally above his head so that the yo-yo has a above his head so that the yo-yo has a centripetal acceleration of 250 m/scentripetal acceleration of 250 m/s22.If the .If the yo-yo’s string is 0.50 m long, what is the yo-yo’s string is 0.50 m long, what is the yo-yo’s tangential velocity?yo-yo’s tangential velocity?

What do we have?What do we have?

aacc= 250 m/s= 250 m/s22

r = 0.5 mr = 0.5 m

Rearrange the Rearrange the equation and solve:equation and solve:

r

va tc

2

s

mm

s

mrav ct 2.11)5.0)(250(

2

Follow up questionFollow up question

How long does it take How long does it take for the yo-yo to for the yo-yo to complete one complete one revolution?revolution?– In other words, what is In other words, what is

the period?the period?

Rearrange the Rearrange the tangential speed tangential speed equation to solve for equation to solve for TT

T

rVt

2

s

smm

v

rT

t

28.2.11

)50.0(22

Another follow up question!Another follow up question!

• How many rotations per second How many rotations per second does the yo-yo make?does the yo-yo make?• In other words, what is the frequency? In other words, what is the frequency?

Tffrequency

1

second

srevolution 57.3

28.0

11sT

ffrequency

What causes an acceleration?What causes an acceleration?

A NET FORCE applied to an object A NET FORCE applied to an object causes it to acceleratecauses it to accelerate

Consequently, since an object that Consequently, since an object that undergoes circular motion is accelerating, undergoes circular motion is accelerating, there must be a force that causes it to do there must be a force that causes it to do soso

Centripetal ForceCentripetal Force

Remember Newton’s Remember Newton’s Second Law says Second Law says F=maF=ma

Therefore Therefore

r

mvmaF Force lCentripeta

2t

cc

Centripetal ForceCentripetal Force

Centripetal Force is the force that Centripetal Force is the force that maintains circular motionmaintains circular motion

IT IS ALWAYS POINTED TOWARD THE IT IS ALWAYS POINTED TOWARD THE CENTER OF THE CIRCLECENTER OF THE CIRCLE

Sample Problem p.261 #2Sample Problem p.261 #2

A bicyclist is riding at a tangential speed of A bicyclist is riding at a tangential speed of 13.2 m/s around a circular track with a 13.2 m/s around a circular track with a radius of 40.0 m. If the magnitude of the radius of 40.0 m. If the magnitude of the force that maintains the bike’s circular force that maintains the bike’s circular motion is 377 N, what is the combined motion is 377 N, what is the combined mass of the bicycle and rider?mass of the bicycle and rider?

What do we know?What do we know?VVtt= 13.2 m/s= 13.2 m/s

r = 40mr = 40m

FFcc= 377 N= 377 N

m = ?m = ?

Rearrange the equation to solve for m.Rearrange the equation to solve for m.

kg 5.86

40

2.13

37722

msm

N

rv

F

a

Fm

t

c

c

c

Gravity!Gravity!We all know that the gravitational force is We all know that the gravitational force is the mutual force of attraction between the mutual force of attraction between particles of matterparticles of matter

Credit & Copyright: Robert Gendler

Newton’s Law of Universal Newton’s Law of Universal GravitationGravitation

Every object in the Universe exerts a Every object in the Universe exerts a gravitational force on every other gravitational force on every other object in the universeobject in the universe

If that’s true, why aren’t things always If that’s true, why aren’t things always flying together?flying together?

Gravitational LensingGravitational Lensing

http://occamsmachete.com/bling/gravitational-lensing.jpg

Newton’s Law of Universal Newton’s Law of Universal GravitationGravitation

FFgg is the gravitational is the gravitational forceforce

G is a constantG is a constant

mm11 and m and m22 are the are the masses of the objectsmasses of the objects

r is the distance r is the distance between the objectsbetween the objects

221

r

mGmFg

2

2111067.6kg

NmxG

Newton’s Law of Universal Newton’s Law of Universal GravitationGravitation

The law of universal gravitation is an The law of universal gravitation is an inverse-square lawinverse-square law– The force between two masses decreases as The force between two masses decreases as

the distance between them increasesthe distance between them increases

The gravitational force between two The gravitational force between two objects is proportional to the product of the objects is proportional to the product of the objects’ massesobjects’ masses

Sample Problem (not in book)Sample Problem (not in book)

The average distance between the Sun and the The average distance between the Sun and the Earth is 1.5 x 10Earth is 1.5 x 101111 m (93 million miles). The m (93 million miles). The average distance between the Sun and Pluto is average distance between the Sun and Pluto is 5.9 x 105.9 x 101212 m (3.66 x 10 m (3.66 x 1099 miles). The mass of the miles). The mass of the Sun is 1.99 x 10Sun is 1.99 x 103030 kg, the mass of the Earth is kg, the mass of the Earth is 5.98 x 105.98 x 102424 kg and the mass of Pluto is 1.31 x kg and the mass of Pluto is 1.31 x 10102222 kg. Find the gravitational force between the kg. Find the gravitational force between the Sun and the Earth, the gravitational force Sun and the Earth, the gravitational force between the Sun and Pluto and the gravitational between the Sun and Pluto and the gravitational force between the Earth and Pluto. force between the Earth and Pluto.

Force between the Sun and the Force between the Sun and the EarthEarth

NF

mx

kgxkgxkgNm

x

r

mGmF

g

g

22

211

24302

211

221

10 x 53.3

105.1

)1098.5)(1099.1(1067.6

Force between the Sun and PlutoForce between the Sun and Pluto

NF

mx

kgxkgxkgNm

x

r

mGmF

g

g

16

212

22302

211

221

10 x 5

109.5

)1031.1)(1099.1(1067.6

Force between Pluto and EarthForce between Pluto and Earth

NxF

mx

kgxkgxkgNm

x

r

mGmF

g

g

11

212

22242

211

221

1058.1

109.5

)1031.1)(1098.5(1067.6

Calculating gCalculating g

We’ve been using the value for We’ve been using the value for acceleration due to gravity all year without acceleration due to gravity all year without knowing where it comes from.knowing where it comes from.

Now is the time to find out!Now is the time to find out!

Calculating gCalculating g

Let’s say we want to know the gravitational Let’s say we want to know the gravitational force between the Earth and a person force between the Earth and a person standing on the surface of the Earth.standing on the surface of the Earth.

What do we knowWhat do we know

Mass of the person = mMass of the person = mpp

Mass of the Earth= MMass of the Earth= MEE

RREE= radius of Earth= radius of Earth

– Note that for problems like this we pretend like Note that for problems like this we pretend like all the mass of the Earth is at its center. all the mass of the Earth is at its center. Therefore the distance between the two Therefore the distance between the two objects is equal to the radius of the Earth.objects is equal to the radius of the Earth.

Relationship between weight and Relationship between weight and FgFg

We know that an object’s weight is equal We know that an object’s weight is equal to the gravitational force acting on it from to the gravitational force acting on it from the Earththe Earth

Therefore FTherefore Fgg=m=mppgg

Calculating gCalculating g

Using Newton’s law of Using Newton’s law of gravitation Fgravitation Fgg = mg = mg

becomes:becomes:

So the expression for So the expression for g isg is

gmR

mGMp

pE

E

2

2

ER

GMg E

Plug in numbersPlug in numbers

226

242

211

281.9

10378.6

)1098.5(1067.6

s

m

mx

kgxkgNm

x

R

GMg

E

E

It works for all spherical bodiesIt works for all spherical bodies

2object

objectobject R

GMg

Calculate the acceleration due to Calculate the acceleration due to gravity on Plutogravity on Pluto

226

222

211

268.

1013.1

)1031.1(1067.6

s

m

mx

kgxkgNm

x

R

GMg

pluto

plutopluto