+ All Categories
Home > Education > Circular Motion & Gravitation

Circular Motion & Gravitation

Date post: 14-Jun-2015
Category:
Upload: timothy-welsh
View: 283 times
Download: 2 times
Share this document with a friend
Description:
Circular Motion & Gravitation
Popular Tags:
28
AP Physics Rapid Learning Series - 09 © Rapid Learning Inc. All rights reserved. - http://www.RapidLearningCenter.com 1 Rapid Learning Center Chemistry :: Biology :: Physics :: Math Rapid Learning Center Presents … Rapid Learning Center Presents Teach Yourself AP Physics in 24 Hours 1/55 *AP is a registered trademark of the College Board, which does not endorse, nor is affiliated in any way with the Rapid Learning courses. Gravitation and Gravitation and Circular Motion Physics Rapid Learning Series 2/55 Rapid Learning Center www.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.
Transcript
Page 1: Circular Motion & Gravitation

AP Physics Rapid Learning Series - 09

© Rapid Learning Inc. All rights reserved. - http://www.RapidLearningCenter.com 1

Rapid Learning CenterChemistry :: Biology :: Physics :: Math

Rapid Learning Center Presents …Rapid Learning Center Presents …

Teach Yourself AP Physics in 24 Hours

1/55*AP is a registered trademark of the College Board, which does not endorse, nor is

affiliated in any way with the Rapid Learning courses.

Gravitation andGravitation and Circular Motion

Physics Rapid Learning Series

2/55

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.

Page 2: Circular Motion & Gravitation

AP Physics Rapid Learning Series - 09

© Rapid Learning Inc. All rights reserved. - http://www.RapidLearningCenter.com 2

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

3/55

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

4/55

ForcesMotion

Centripetal Force

Universal Gravitational

Constant

Page 3: Circular Motion & Gravitation

AP Physics Rapid Learning Series - 09

© Rapid Learning Inc. All rights reserved. - http://www.RapidLearningCenter.com 3

Gravitation

Previously, we learned that gravity accelerates falling objects at -9 8m/s2

5/55

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.

6/55

Page 4: Circular Motion & Gravitation

AP Physics Rapid Learning Series - 09

© Rapid Learning Inc. All rights reserved. - http://www.RapidLearningCenter.com 4

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

7/55

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:

8/55

G = 6.67 x 10 -11 N m2 /kg2

Page 5: Circular Motion & Gravitation

AP Physics Rapid Learning Series - 09

© Rapid Learning Inc. All rights reserved. - http://www.RapidLearningCenter.com 5

Universal Law of Gravitation

Universal gravitational

constant

Two masses,

kg

221

g dmmGF =

Force from

constant

Distance between

bj t

9/55

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

10/55

Page 6: Circular Motion & Gravitation

AP Physics Rapid Learning Series - 09

© Rapid Learning Inc. All rights reserved. - http://www.RapidLearningCenter.com 6

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

=

11/55

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.

12/55

Thus, gravitational field lines on earth always point to the center of the earth.

Page 7: Circular Motion & Gravitation

AP Physics Rapid Learning Series - 09

© Rapid Learning Inc. All rights reserved. - http://www.RapidLearningCenter.com 7

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?

13/55

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!

14/55

of the planet. You weigh less!

Page 8: Circular Motion & Gravitation

AP Physics Rapid Learning Series - 09

© Rapid Learning Inc. All rights reserved. - http://www.RapidLearningCenter.com 8

Extension of Thought QuestionWhat would happen if you were at a cave at the center of the earth?

15/55

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.

16/55

Page 9: Circular Motion & Gravitation

AP Physics Rapid Learning Series - 09

© Rapid Learning Inc. All rights reserved. - http://www.RapidLearningCenter.com 9

UniformUniform Circular Motion

Obviously, not all objects move in a linear path Another common occurrence

17/55

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!

18/55

Which direction will the ball initially fly?

Page 10: Circular Motion & Gravitation

AP Physics Rapid Learning Series - 09

© Rapid Learning Inc. All rights reserved. - http://www.RapidLearningCenter.com 10

Inertia in Action

When the string breaks, the ball will continue to move tangent to the circle. Linear inertia at work!

Snap!

19/55

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

20/55

Example: 3 revolutions per second

5 revolutions per minute (rpm)

2π radians per second

Page 11: Circular Motion & Gravitation

AP Physics Rapid Learning Series - 09

© Rapid Learning Inc. All rights reserved. - http://www.RapidLearningCenter.com 11

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

21/55

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

22/55

25m/ssec 60

830 (.29m) 2πv ==

We also need to consider that it covers 830 circumferences in 1 minute

830 rpm

Page 12: Circular Motion & Gravitation

AP Physics Rapid Learning Series - 09

© Rapid Learning Inc. All rights reserved. - http://www.RapidLearningCenter.com 12

Acceleration?

If an object is moving at a constant speed in a circular path, is it accelerating?

23/55

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

24/55

circular path, m

centripetal acceleration

m/s2

Page 13: Circular Motion & Gravitation

AP Physics Rapid Learning Series - 09

© Rapid Learning Inc. All rights reserved. - http://www.RapidLearningCenter.com 13

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.

25/55

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

26/55

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

Page 14: Circular Motion & Gravitation

AP Physics Rapid Learning Series - 09

© Rapid Learning Inc. All rights reserved. - http://www.RapidLearningCenter.com 14

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

27/55

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!

28/55

Nothing is pushing you out (centrifugal).

You are being pulled inward (centripetal).

Page 15: Circular Motion & Gravitation

AP Physics Rapid Learning Series - 09

© Rapid Learning Inc. All rights reserved. - http://www.RapidLearningCenter.com 15

Fc Formula

Linear velocity,

m/sMass, kg

2

c rmvF =

CentripetalRadius of

i l

29/55

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

30/55

r

rva

2

c =

Page 16: Circular Motion & Gravitation

AP Physics Rapid Learning Series - 09

© Rapid Learning Inc. All rights reserved. - http://www.RapidLearningCenter.com 16

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

31/55

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:

32/55

Fc = .1kg (12.6 m/s)2 / 1mFc = 15.8 kg m/s2

=15.8 N

Page 17: Circular Motion & Gravitation

AP Physics Rapid Learning Series - 09

© Rapid Learning Inc. All rights reserved. - http://www.RapidLearningCenter.com 17

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?

33/55

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

34/55

rva

2

c =

983m78m/s

(277m/s)avr 2

2

c

2

===

Page 18: Circular Motion & Gravitation

AP Physics Rapid Learning Series - 09

© Rapid Learning Inc. All rights reserved. - http://www.RapidLearningCenter.com 18

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

35/55

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.

36/55

Page 19: Circular Motion & Gravitation

AP Physics Rapid Learning Series - 09

© Rapid Learning Inc. All rights reserved. - http://www.RapidLearningCenter.com 19

Banked Turn Diagram

Use the right triangle to find the Fc:

adjopptanθ =

WFtanθ c=

Ftanθ c=

Fc

normal force

37/55

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

38/55

2vtanθ g r =

43m/s))(tan31s316m(9.8m/

tanθ g rv2 ==

=

out, it is irrelevant

Page 20: Circular Motion & Gravitation

AP Physics Rapid Learning Series - 09

© Rapid Learning Inc. All rights reserved. - http://www.RapidLearningCenter.com 20

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.

39/55

Simulated Gravity

When objects are moving in a circle, the centripetal force applied may mimic the

40/55

centripetal force applied may mimic the usual gravitational force.

Page 21: Circular Motion & Gravitation

AP Physics Rapid Learning Series - 09

© Rapid Learning Inc. All rights reserved. - http://www.RapidLearningCenter.com 21

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.

41/55

Gravity Changes Direction

Fcit

Tangential velocity

gravity

42/55

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.

Page 22: Circular Motion & Gravitation

AP Physics Rapid Learning Series - 09

© Rapid Learning Inc. All rights reserved. - http://www.RapidLearningCenter.com 22

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

43/55

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).

44/55

Don’t just plug in the distance above the surface of the earth!

Page 23: Circular Motion & Gravitation

AP Physics Rapid Learning Series - 09

© Rapid Learning Inc. All rights reserved. - http://www.RapidLearningCenter.com 23

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.

45/55

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!

46/55

Page 24: Circular Motion & Gravitation

AP Physics Rapid Learning Series - 09

© Rapid Learning Inc. All rights reserved. - http://www.RapidLearningCenter.com 24

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.

47/55

Artificial Gravity

When astronauts live for long time periods in space, it impacts their bodies. Bones may weaken, muscles may lose mass, etc.

48/55

In the future, humans may design space ships that create “artificial” gravity.

Page 25: Circular Motion & Gravitation

AP Physics Rapid Learning Series - 09

© Rapid Learning Inc. All rights reserved. - http://www.RapidLearningCenter.com 25

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.

49/55

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.

50/55

Page 26: Circular Motion & Gravitation

AP Physics Rapid Learning Series - 09

© Rapid Learning Inc. All rights reserved. - http://www.RapidLearningCenter.com 26

Normal

Calculation of Linear Speed

rva

2

c =Centripetal

force formula

Normal acceleration from gravity

on Earth.rav c=

)(15m)(9.8m/sv 2=

51/55

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

52/55

revolutionsX 7.7 rpm=

srevolutionXr π 2

s)12.1m/s(60=

Page 27: Circular Motion & Gravitation

AP Physics Rapid Learning Series - 09

© Rapid Learning Inc. All rights reserved. - http://www.RapidLearningCenter.com 27

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.

53/55

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

Rapid Learning Center

Page 28: Circular Motion & Gravitation

AP Physics Rapid Learning Series - 09

© Rapid Learning Inc. All rights reserved. - http://www.RapidLearningCenter.com 28

Rapid Learning Center

Wh t’ N t

Chemistry :: Biology :: Physics :: Math

What’s Next …

Step 1: Concepts – Core Tutorial (Just Completed)

Step 2: Practice – Interactive Problem Drill

Step 3: Recap Super Review Cheat Sheet

55/55

Step 3: Recap – Super Review Cheat Sheet

Go for it!

http://www.RapidLearningCenter.com


Recommended