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