circular motion & gravitation
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Circular Motion & GravitationTRANSCRIPT
AP Physics Rapid Learning Series - 09
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Gravitation andGravitation and Circular Motion
Physics Rapid Learning Series
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Rapid Learning Centerwww.RapidLearningCenter.com/© Rapid Learning Inc. All rights reserved.
Wayne Huang, Ph.D.Keith Duda, M.Ed.
Peddi Prasad, Ph.D.Gary Zhou, Ph.D.
Michelle Wedemeyer, Ph.D.Sarah Hedges,Ph.D.
AP Physics Rapid Learning Series - 09
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Learning Objectives
Understand the nature of the gravitational force
By completing this tutorial, you will:
the gravitational force.
Calculate the gravitational force between objects.
Describe uniform circular motion.
C l l t t i t l f
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Calculate centripetal force and centripetal acceleration.
Describe simulated gravity situations.
Concept MapPhysics
Studies
Previous content
New content
Motion
F
Caused by
Circular M ti
Gravitation
Described by
Universal
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ForcesMotion
Centripetal Force
Universal Gravitational
Constant
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Gravitation
Previously, we learned that gravity accelerates falling objects at -9 8m/s2
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accelerates falling objects at -9.8m/s2. Now we will learn more about the origin of that acceleration.
Direction of Gravitation
Isaac Newton described this attractive force graivty that acts between all pieces of matter in the universe.
Gravity is always attractive.
There is no “repulsive” gravity. So far, antigravity is just science fiction.
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Law of Universal Gravitation
What factors do you think the gravitational attraction of two bodies would depend on?
mass of objects, and distance between objects
21mmF
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221
g dF →
Universal Gravitational Constant
The previous relationship only describes the factors that influence gravity. To get a numerically correct answer, with units, you need a constant included in the equation.
This constant is called the universal gravitational constant:
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G = 6.67 x 10 -11 N m2 /kg2
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Universal Law of Gravitation
Universal gravitational
constant
Two masses,
kg
221
g dmmGF =
Force from
constant
Distance between
bj t
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from gravity, N
objects, m
Because there is a quantity squared in the denominator of the fraction, this formula may be referred to as an inverse square law.
Gravitation Calculation Example
Calculate the gravitation force between the planet Mars, 6.4 x 10 23 kg, and the sun, 2 x 10 30 kg. Assume a distance of 2.0 x 10 11 m.
Fg
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Gravitation Example Solution
221
g dmmGF =
Substitute values, do math carefully!
(211
30232
211
g m)(2x10
kg)kg)(2x106.4x10)kgNm(6.67x10
F
−
=
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Fg= 2.13 x 1021 N
Notice how the units cancel leaving the correct force unit of Newtons.
Gravitational FieldsField lines show which way an object will move when exposed to some force.
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Thus, gravitational field lines on earth always point to the center of the earth.
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Thought QuestionImagine you are deep inside the earth in a cave. Would you weigh more, less, or the same as on the surface of the earth?
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Thought Question AnswerWhen on the surface of the earth, all of the mass of the earth is pulling you down.
When underground, the mass above you actually pulls you up, countering some of pull from the bulk of the planet. You weigh less!
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of the planet. You weigh less!
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Extension of Thought QuestionWhat would happen if you were at a cave at the center of the earth?
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Because the gravitational pull from the matter equally all around you would cancel out, you would be weightless! Fnet would be 0.
Planet DiscoveryGravity can have much more subtle effects.
The existence of Pluto and Neptune was predicted before they were ever observed.y
Their slight gravitational effects wobble the orbits of the other planets, betraying their presence before they were ever seen visually.
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UniformUniform Circular Motion
Obviously, not all objects move in a linear path Another common occurrence
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linear path. Another common occurrence is constant motion in a circular path.
Thought Question
Imagine you are swinging a ball on the end of a string. The string suddenly breaks when the ball is at the top.is at the top.
Snap!
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Which direction will the ball initially fly?
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Inertia in Action
When the string breaks, the ball will continue to move tangent to the circle. Linear inertia at work!
Snap!
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Obviously, if we were considering gravity, the ball would begin to fall also.
Circular Motion
In a line, we usually measure a speed or velocity in m/s, linear speed.
However, when things travel in a circle we could describe a speed differently, rotational speed.
Example: 3 revolutions per second
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Example: 3 revolutions per second
5 revolutions per minute (rpm)
2π radians per second
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Linear Speed Example
If a tire on a car has a radius of .29m, and is being rotated at 830 rpm, what is the speed at the outer edge of the tire?
830 l ti
.29m
830 revolutions per minute
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Linear Speed Solution
Use a previous formula:
tdv =t
However, the distance is now in a circle. So we must find the distance around the circle…
tr π 2v =
Circumference of a circle
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25m/ssec 60
830 (.29m) 2πv ==
We also need to consider that it covers 830 circumferences in 1 minute
830 rpm
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Acceleration?
If an object is moving at a constant speed in a circular path, is it accelerating?
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YES. Although its speed isn’t changing, it still accelerates. Its direction is constantly changing, this means its velocity is constantly changing, thus it is accelerating.
Centripetal Acceleration
We can describe this type of acceleration in a circular path by the following relationship:
Li
rva
2
c =
Linear velocity,
m/s
radius of i lcentripetal
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circular path, m
centripetal acceleration
m/s2
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Accelerations Require ForcesIf an object is accelerating, it must have a net force applied to it!
This is the force that is responsible for making anThis is the force that is responsible for making an object travel in the circular path instead of a regular straight line.
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Instantaneous velocity
Centripetal force
Centripetal ForceCentripetal force is a “center seeking” force. It always points towards the center for an object moving in a circular path.
Centripetal force
Tangential or
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This isn’t some new force. Centripetal force can be provided by tension from a string, gravitational pull, friction on the road, etc.
ginstantaneous velocity
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Centrifugal Force?Often, it seems like there is a force pushing outward (not to the center).
There is no force like this. However, this apparent force has been given a name: centrifugal force.
This is an often misunderstood and misused term
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term.
Direction of Forces
Imagine you are driving along in a car, you make a hard right turn.
You feel as if you are pushed into the left side of your car. Actually, it is the left side of the car that is pushing into you!
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Nothing is pushing you out (centrifugal).
You are being pulled inward (centripetal).
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Fc Formula
Linear velocity,
m/sMass, kg
2
c rmvF =
CentripetalRadius of
i l
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Centripetal force, N
circular path, m
Here is the formula for centripetal force.
Similarities
Notice how the formula for centripetal force contains the formula for centripetal acceleration within itself.within itself.
2
rmvma =
cnet FF =Newton’s 2nd law:Fnet=ma
Mass cancels
out
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r
rva
2
c =
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Centripetal Force Example
Imagine you whirl a 0.1 kg rubber stopper attached to a string in a 1 m radius circle at 2 rev/sec. What is the centripetal force on that stopper?
1m
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Example Solution
First change 2 rev/sec into m/s:
=2 (2πr) /sec Circumference f i l 2= 2(2 π 1m) / sec
= 12.6 m/s
of circle =2πr
Linear velocity,
m/sThen, substitute into the Fc formula:
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Fc = .1kg (12.6 m/s)2 / 1mFc = 15.8 kg m/s2
=15.8 N
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Centripetal Acceleration Example
Imagine that a fighter pilot is traveling at 1000 km/hr, when he decides to make a sharp turn. He can withstand a maximum acceleration of 8g’s (8 timeswithstand a maximum acceleration of 8g s (8 times the normal acceleration from gravity). What is the smallest radius turn he can withstand?
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Centripetal Acceleration SolutionFirst, convert km/hr into m/s
277m/s3600sec
1hr1km
1000mhr
1000km=××
3600sec1kmhr
Next, convert the acceleration he can withstand
22
78m/s1g
9.8m/ss8g' =×
Finally, solve for the radius of the turn
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rva
2
c =
983m78m/s
(277m/s)avr 2
2
c
2
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Banked Turns for a CarSimilarly, when a car is driving in a circular path, on a banked road, the centripetal force comes from a component of the normal force.
This component of the normal force is Fc
normal force θ
i ht
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θ
These can be calculated using simple trigonometry.
weight
Example
At the Daytona 500 speedway, the turns in the oval track have a maximum radius of 316 m and are banked at an angle of 31°. Assuming no friction, how fast could the cars make the turn? Hint: use the previous diagram/example.
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Banked Turn Diagram
Use the right triangle to find the Fc:
adjopptanθ =
WFtanθ c=
Ftanθ c=
Fc
normal force
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31o
weightmg
tanθ mgFc =
Banked Turn Calculation
Next, substitute our expression for Fc back into the formula to find v:
mv2
rmvF
2
c =
rmvtanθ mg
2
=
2t θ
Since the mass cancels out it is
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2vtanθ g r =
43m/s))(tan31s316m(9.8m/
tanθ g rv2 ==
=
out, it is irrelevant
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Reality Check
This number represents how fast the car could go without any friction. Obviously there is friction so the actual maximum speed on the track is much higher.
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Simulated Gravity
When objects are moving in a circle, the centripetal force applied may mimic the
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centripetal force applied may mimic the usual gravitational force.
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Orbits
When in a circular orbit, an object is continually falling ( under the influence of the earth’s gravity).
However, it is continuing to move tangent to the earth, so it continues in a circular path at a constant speed.
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Gravity Changes Direction
Fcit
Tangential velocity
gravity
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Notice that gravity does not pull the satellite forward or backward. Gravity simply acts as the centripetal force to keep it going in a circular orbit.
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Gravity Equals Centripetal ForceSince the centripetal force is provided by gravity, we can equate the two forces:
G FF = cG FF
2
2 rmv
rmMG =
2MG
Notice the mass of the satellite cancels
out!
The speed of an i l biti
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2vr
G =
rGMv =
circular orbiting satellite depends only on the radius, gravitational constant and mass of the earth!
Correct Distance Value
The radius used in the previous equation is measured from the center of the orbit ( center of the earth).
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Don’t just plug in the distance above the surface of the earth!
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Importance of MassThis means that satellites of any mass will have the same orbital speed for any particular radius.
A giant satellite will have the same speed as a tinyA giant satellite will have the same speed as a tiny satellite in the same orbit.
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However, it can be much more difficult to get that large satellite into orbit in the first place.
Energy Costs
It turns out that it takes 62,000,000 J of energy to put 1kg outside of the Earth’s orbit. (62 MJ)
This is a large amount of energy, which is why it is so costly and difficult to put people and objects into space!
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Weight and Weightlessness
You can feel weightless even though gravity is acting on you.acting on you.
Astronauts in free fall are still being pulled around the earth by gravity.
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Artificial Gravity
When astronauts live for long time periods in space, it impacts their bodies. Bones may weaken, muscles may lose mass, etc.
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In the future, humans may design space ships that create “artificial” gravity.
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Rotating Space HabitatsWe can’t create a gravitational force, but we can use centripetal force to act like gravity.
If a round space ship is large enough, and spins at p p g g , pthe correct rate, the centripetal force would simulate gravity.
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Rotating Spaceship Example
If we want to simulate regular earth gravity in a circular spaceship of radius 15m, how fast must it rotate? Give m/s and rpm.rotate? Give m/s and rpm.
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Normal
Calculation of Linear Speed
rva
2
c =Centripetal
force formula
Normal acceleration from gravity
on Earth.rav c=
)(15m)(9.8m/sv 2=
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This gives the linear speed (in m/s) at the edge.
12.1m/sv =
Next, consider the number of revolutions it makes in 1 minute ( 60 sec).
d
Calculation of Rotational Speed
Circumference
Linear speed
previously found
60secX r π 2v srevolution=
tdv =
s)12 1m/s(60
C cu e e ceof a circle
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revolutionsX 7.7 rpm=
srevolutionXr π 2
s)12.1m/s(60=
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221
g dmmGF =
Rotational speed refers
to motion in a
Rotational speed refers
to motion in a
In many instances, the
In many instances, the
Learning Summary
2g dG = 6.67 x 10 -11
N m2 /kg2
2 Centripetal forceCentripetal force
to motion in a circular path,
rev/sec.
to motion in a circular path,
rev/sec.
,mass variable cancels out.
,mass variable cancels out.
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rva
rmvF
2
c
2
c
=
=Centripetal force
points inward and forces objects to
maintain a circular motion.
Centripetal force points inward and forces objects to
maintain a circular motion.
Congratulations
You have successfully completed the core tutorial
Gravitation and Circular MotionMotion
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