Transcript Rotation and Newton`s Universal Law of Gravitation

Rotation and Newton’s
Universal Law of Gravitation
Chapter 7
Our friend Jupiter
Jupiter rotates once
every 10 hours
Of course Earth only
rotates once every 24
hours
How would we
calculate the speed of
a rotating object?
Objects in circular motion have a
tangential speed
What is tangential
speed?
– Tangential speed is
the instantaneous
linear speed of any
point rotating about an
axis
– r is radius of circle
– T is period (length of
time for one rotation)
2r
Vt 
T
So…how do we use that?
Let’s find the tangential speed of a point
along Jupiter’s equator. What information
do we need?
– Jupiter’s Radius at the equator: 7.15 x 107 m
– The time for one rotation: 10 hours= 36000 s
2r 2 (7.15 x107 m)
4 m
Vt 

 1.25 x10
 28,000 mph
T
36000s
s
Jupiter’s Speed at the Poles
Jupiter is not a solid sphere, so the radius at the
equator is not the same as the radius at the
poles.
– Jupiter’s Radius at the Poles: 6.69 x 107 m
– Time of rotation: 10 hours = 36000 s
2r 2 (6.69 x107 m)
4 m
Vt 

 1.17 x10
 26,208 mph
T
36000s
s
That’s a difference of 1,792 mph!
Compare that to Earth
Earth’s equatorial radius: 6.378 x 106 m
Earth’s rotational period: 24 hrs = 86,400s
2r 2 (6.378 x106 m)
m
Vt 

 463.82  1038 mph
T
86400s
s
Earth’s radius at the poles: 6.356 x 106 m
Earth’s rotational period: 24 hrs= 86,400 s
2r 2 (6.356 x106 m)
m
Vt 

 462.22  1035 mph
T
86400s
s
Does an object undergoing circular
motion always have the same
tangential velocity?
NO!
Direction of velocity: Right
A
Direction of velocity: Up
B Direction of velocity: Down
D
C
Direction of velocity: Left
To clarify things…
Points that are the same distance from the
center (i.e. that have the same radius)
have the same tangential speed but not
the same tangential velocity!
If velocity is changing then…
A change in velocity = ACCELERATION
Therefore, objects that are undergoing circular
motion must be accelerating!
 2r 


2
2
vt  T 
4 r
Centripeta l Accelerati on  ac 

 2
r
r
T
2
Sample Problem p.258 #2
A young boy swings a yo-yo horizontally
above his head so that the yo-yo has a
centripetal acceleration of 250 m/s2.If the
yo-yo’s string is 0.50 m long, what is the
yo-yo’s tangential velocity?
What do we have?
ac= 250 m/s2
r = 0.5 m
2
t
v
ac 
r
Rearrange the
equation and solve:
m
m
vt  ac r  (250 2 )(0.5m)  11.2
s
s
Follow up question
How long does it take
for the yo-yo to
complete one
revolution?
2r
Vt 
T
– In other words, what is
the period?
Rearrange the
tangential speed
equation to solve for
T
2r 2 (0.50m)
T

 .28s
m
vt
11.2
s
Another follow up question!
• How many rotations
per second does the
yo-yo make?
• In other words, what is
the frequency?
1
frequency  f 
T
1
1
revolution s
frequency  f  
 3.57
T 0.28s
second
What causes an acceleration?
A NET FORCE applied to an object
causes it to accelerate
Consequently, since an object that
undergoes circular motion is accelerating,
there must be a force that causes it to do
so
Centripetal Force
Remember Newton’s
Second Law says
F=ma
Therefore
mv
Centripeta l Force  Fc  ma c 
r
2
t
Centripetal Force
Centripetal Force is the force that
maintains circular motion
IT IS ALWAYS POINTED TOWARD THE
CENTER OF THE CIRCLE
Sample Problem p.261 #2
A bicyclist is riding at a tangential speed of
13.2 m/s around a circular track with a
radius of 40.0 m. If the magnitude of the
force that maintains the bike’s circular
motion is 377 N, what is the combined
mass of the bicycle and rider?
What do we know?
Vt= 13.2 m/s
r = 40m
Fc= 377 N
m=?
Rearrange the equation to solve for m.
Fc
Fc
377 N
m
 2 
 86.5 kg
2
ac  vt   

m

   13.2  
s 
 r  
 40m 




Gravity!
We all know that the gravitational force is
the mutual force of attraction between
particles of matter
Credit & Copyright: Robert Gendler
Newton’s Law of Universal
Gravitation
Every object in the Universe exerts a
gravitational force on every other
object in the universe
If that’s true, why aren’t things always
flying together?
Gravitational Lensing
http://occamsmachete.com/bling/gravitational-lensing.jpg
Newton’s Law of Universal
Gravitation
Fg is the gravitational
force
G is a constant
m1 and m2 are the
masses of the objects
r is the distance
between the objects
Gm1m2
Fg 
2
r
2
Nm
G  6.67 x1011
kg 2
Newton’s Law of Universal
Gravitation
The law of universal gravitation is an
inverse-square law
– The force between two masses decreases as
the distance between them increases
The gravitational force between two
objects is proportional to the product of the
objects’ masses
Sample Problem (not in book)
The average distance between the Sun and the
Earth is 1.5 x 1011 m (93 million miles). The
average distance between the Sun and Pluto is
5.9 x 1012 m (3.66 x 109 miles). The mass of the
Sun is 1.99 x 1030 kg, the mass of the Earth is
5.98 x 1024 kg and the mass of Pluto is 1.31 x
1022 kg. Find the gravitational force between the
Sun and the Earth, the gravitational force
between the Sun and Pluto and the gravitational
force between the Earth and Pluto.
Force between the Sun and the
Earth
Gm1m2
Fg 

2
r
2
Nm
30
24
6.67 x10 11
(
1
.
99
x
10
kg
)(
5
.
98
x
10
kg)
2
kg
Fg  3.53 x 10 22 N
1.5x10 m
11
2
Force between the Sun and Pluto
Gm1m2
Fg 

2
r
Fg  5 x 1016 N
2
Nm
6.67 x10 11 2 (1.99 x1030 kg)(1.31x10 22 kg)
kg
5.9 x10 m
12
2
Force between Pluto and Earth
Gm1m2
Fg 

2
r
2
Nm
24
22
6.67 x10 11
(
5
.
98
x
10
kg
)(
1
.
31
x
10
kg)
2
kg
Fg  1.58 x1011 N
5.9 x10 m
12
2
Calculating g
We’ve been using the value for
acceleration due to gravity all year without
knowing where it comes from.
Now is the time to find out!
Calculating g
Let’s say we want to know the gravitational
force between the Earth and a person
standing on the surface of the Earth.
What do we know
Mass of the person = mp
Mass of the Earth= ME
RE= radius of Earth
– Note that for problems like this we pretend
like all the mass of the Earth is at its center.
Therefore the distance between the two
objects is equal to the radius of the Earth.
Relationship between weight and
Fg
We know that an object’s weight is equal
to the gravitational force acting on it from
the Earth
Therefore Fg=mpg
Calculating g
Using Newton’s law of
gravitation Fg = mg
becomes:
So the expression for
g is
GM E m p
2
RE
 mp g
GM E
g
2
RE
Plug in numbers
2

11 Nm 
24
 6.67 x10

(
5
.
98
x
10
kg)
2 
kg 
GM E 
m
g

 9.81 2
2
2
RE
s
6.378 x106 m


It works for all spherical bodies
g object 
GM object
2
object
R
Calculate the acceleration due to
gravity on Pluto
g pluto 
GM pluto
2
R pluto
2


Nm
11
22
 6.67 x10

(
1
.
31
x
10
kg)
2 
kg 
m


 .68 2
2
6
s
1.13x10 m

