Gravity Notes physics

7/27/2019 Gravity Notes physics

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AP Physics Gravity

Sir Isaac Newton is credited with the discovery of gravity. Now, ofcourse we know that he didntreally discover the thing lets face it, people knew about gravity for as long as there have beenpeople. Gravity didnt have to be discovered for crying out loud! Your basic tiny little toddler

figures it out in the first few months of life. So what is the big deal with Newton and gravity?Well, what Newton did was to describe gravity, extend its effects from the surface of the earth(which everyone understood) out into space to explain the behavior of the planets (which nobodysuspected), and to formulate a mathematical law that accurately described the force of gravitybetween objects with mass.

Perhaps the greatest success of his theory of gravity was to successfully explain the motion of theheavens planets, moons, etc.

Newtons theory is very simple. Gravity is a force of attraction between any two objects that havemass. Two objects sitting on a desktop attract each other with a force that we call gravity. Theydont go flying together because gravity is a very weak force and is only significant when one or theother of the masses is enormous planet size. This is why we arent attracted to objects that wepass on our daily wanderings.

The size of the force of gravity is given by this equation:

1 2

2

m mF G

r= Newtons law of gravity

Gis the universal gravitational constant.

211

26.67 10

N mG x

kg

=

m1 is the mass of one of the bodies and m2 is the mass of the other body.

ris the distance between the two bodies.

The value ofGis the same everywhere throughout the universe.

On the old AP Physics Test the equation is written as:

1 2

2G

G m mF

r=

156


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Newtons law of gravity is an inverse-square Law. This means that the force of gravity getssmaller or larger by the square of the distance. The force is directly proportionalto the masses, soif the mass of one of the objects doubles, the force of gravity would double. But if the distancedoubled, the force of gravity would decrease by a factor of four. Thats because it decreases by thesquare of the distance. Inverse-square laws are very common in physics. Well see more of them inour explorations.

To give the particulars of the theory:

Universal Law of Gravitation

1. Gravitational force is a field force between two particles -- in all mediums.2. Force varies as the inverse square of the distance3. Force is proportional to mass of objects.4. The gravity force acts from the center of the two objects.5. The gravitational force is always attractive.6. The gravitational force cannot be shielded or canceled.

Solving gravity problems is quite simple. Lets do one.

A girl, Brandy (42.5 kg), sits 1.50 m from a boy (63.0 kg), George. What is the force of gravitybetween them? (This will tell us how attracted they are to each other.)

1 2

2

m mF G

r=

( ) ( )

( )

2

11

2 2

42.5 63.06.67 10

1.50

kg kg N mF x

kg m

=

11 8

7940 10 7.94 10F x N x N

= =You can see that this is a tiny force, one so small that Brandy will never,the Physics Kahuna is sure, notice its presence. George must generatesome other attractive force if there is to be a relationship between the twoof them.

One useful application of Newtons law of gravity was to weigh Earth this allowed physicists tomake an accurate determination of the earths mass. Lets do that.

Find the mass of the earth. re = 6.38 x 106 m. We will use a 10.0 kg mass in our solution (notthat it matters), the other mass will be that of the earth.

1

2

Em m

F Gr

= This is the force of gravity using Newtons law. But we know that the

force of gravity must also equal F= mg, so we set them equal to oneanother:

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1

1 2

Earthm m

m g Gr

= The mass of the object cancels out:2

Em

g Gr

=

Solve for the mass of the earth2

E

grm

G=

( )

( )

26

223 24

2

11

2 2

9.8 6.38 1059.8 10 5.98 10

6.67 10 10.0

E

m

x msm x kg x kg

kg m mx kg

s kg

= = =

23 2459.8 10 5.98 10E

m x kg x kg = =

Center of Gravity: A stone resting on the ground is acted upon by gravity. In fact, everyatom within the rock is attracted to the earth. The sum of all these forces is the thing that we callweight. Wouldnt it be a real pain to have to add up every single vector for every single atom in an

object in order to find out what it weighed? Well fortunately, we dont have to do that. Nature sortof does it for us. Basically the vectors all add together to give us one big vector, the weight vector,which is directed down and which acts at a point where all the weight appears to be concentrated.This point is called the center of gravity.

For an object like a uniform sphere, the center of gravity, orCG, is located at its geometric center.Irregular objects, like, say, your average baseball bat would have their CG located closer to the fatend of the bat rather than the center.

The dot represents the CG of the objects in the drawings below:

The weight of these objects, e.g., the vector we call weight, is applied at the CG.

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When a ball is thrown in the air, it follows a parabolic path. Actually, what follows this parabolicpath is the CG of the ball.

This is also true for irregular objects that are thrown. If you throw a crescent wrench so that it spinswhen you release it, it seems to follow a very erratic path. Its CG, however, traces out a nice

smooth parabola.

All irregular objects rotate about their CG when thrown.

Lets do some more gravity problems.

You weigh 567 N on earth. How much would you weigh on Mars?

First we must calculate the mass by using the weight on earth:

W mg=W

mg

=2

2

1567 57.86

9.8

kg mm kg

ms

s

= =

Now we need only find the force of gravity between our person and Mars. For the distance we usethe radius of Mars, since the person is on the surface of Mars and the force of gravity acts at thecenter of the planet.

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( ) ( )

( )

232

111 2

2 26

2

6.42 10 57.866.67 10 218

3.37 10

x kg kgm m N mF G x N

r kg x m

= = =

What is the acceleration of gravity on Venus?

We know: F ma= and 1 22

m mF G

r=

21 2

2

m mm a G

r=

( )

( )

242

11 1

2 2 2 26

2

4.88 106.67 10 0.886 10 8.86

6.06 10

x kgkg m m m ma x x

s kg s sx m

= = =

The acceleration on Venus would be slightly smaller than the acceleration on earth.

AP Question: Were revisiting one we already looked over, but this time we can analyze it onMars.

The Sojourner rover vehicle was used to explore the surface ofMars as part of the Pathfinder mission in 1997. Use the data in thetables below to answer the questions that follow.

Sojourner Data

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Mass of Sojourner vehicle: 11.5kgWheel diameter: 0.13 mStored energy available: 5.4 x l05 JPower required for driving under average conditions: 10 WLand speed: 6.7 x 10-3 m/s

Mars Data

Radius: 0.53 x Earth's radiusMass: 0.11 x Earth's mass

(a) Determine the acceleration due to gravity at the surface of Mars in terms ofg, the accelerationdue to gravity at the surface of Earth.

Force and mass are directly proportional to each other. Force is inversely proportional to the radiussquared.

We set up the proportions forg, the acceleration of gravity on earth.

( )2

MarsMars

Mars

mg g

r=

( )2 2 2

0.119.8 3.8

0.53Mars

m mg

s s= =

(b) Calculate Sojourner's weight on the surface of Mars.

211.5 3.84 44

mars

mw mg kg N

s

= = =

(c) Assume that when leaving the Pathfinder spacecraft Sojourner rolls down a ramp inclined at20 to the horizontal. The ramp must be lightweight but strong enough to support Sojourner.Calculate the minimum normal force that must be supplied by the ramp.

1. Draw a FBD:

cos 0 cosMars Mars

n mg n mg = =

211.5 3.8 cos20.0 41

mn kg N

s

= =

o

(d) What is the net force on Sojourner as it travelsacross the Martian surface at constant velocity?Justify your answer.

Zero. If it is traveling at constant velocity in the xdirection then there is no acceleration and no netforce. . The force of gravity is cancelled by the normal force resulting in no motion in the ydirection.

161

m g

m g s i n

n

M a r s

M a r s


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(e) Determine the maximum distance that Sojourner can travel on a horizontal Martian surfaceusing its stored energy.

5 415.4 10 5.4 10

10

W WP t x J x s

Jt P

s

= = = =

( )3 46.7 10 5.4 10 360

x mv x vt x x s m

t s

= = = =

(f). Suppose that 0.010% of the power for driving is expended against atmospheric drag asSojourner travels on the Martian surface. Calculate the magnitude of the drag force.

( ) ( )0.00010 10 0.001P W W= =

P Fv= ( )( )3

0.001 0.1496.7 10

WPF Nv m s

= = =

162

Dear Doctor Science,If, theoretically, you were at the exact center of the earth and you dropped

a ball, what would it do?

-- Rick McGuire from Birmingham, MI

Dr. Science responds:It would fly into your face at great velocity. Most probably, it would damage

one eye and, if your reflexes were slow enough, could permanently blind you.

That's why I can't urge you strongly enough to leave experiments like this to

trained scientists! We've been there before. We know what to expect in

situations like this. I wouldn't think of going to the center of the earth without

protective eyewear, nor would I tamper with terrestrial forces without first

having done my homework! The Center of the Earth, man! Didn't you see the

movie starring James Mason and Pat Boone? Sure, it had a happy ending -- allmovies did back then. But the reality of Pat Boone taking a shower surrounded

by glowing Technicolor rocks...the mind can only stand so much.


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163

Dear Cecil:

Why are there high tides twice a day when the earth rotates beneath the moon

only once a day? In diagrams it appears the moon's gravity causes the earth's

oceans to bulge (creating a high tide) not only on the side toward the moon,

but also on the side away from the moon. I've heard some unconvincing

explanations for this, including: "the water on the far side is flung away fromthe earth" (why?); "the moon attracts the earth, and the water on the far side

is left behind" (why isn't the water on the far side attracted too?); and "the

earth and the moon both revolve around a common point" (I know that, but

what does that have to do with the question?). Please help.

--Kathleen Hunt, Brookline, Massachusetts

Cecil replies:Not to discourage you, Kathleen, but this makes 22 questions from you in threemonths. Think quality, not quantity. This isn't a scrap drive.The following homely metaphor is sometimes used to explain why there are twotides: the earth and moon, which are really dual planets, are like two figure skaters

spinning around one another while holding hands. Centrifugal force naturally tendsto pull them apart, but their clasped hands (i.e., their mutual gravitationalattraction) keep them together. Similarly, centrifugal force tends to fling the oceanoutward on the side of the earth away from the moon. On the near side, the water istugged moonward by lunar gravity.

There's just one problem with this explanation. It's wrong. Cecil has consulted withthe physics division of the Straight Dope Science Advisory Board and is satisfiedthat centrifugal force (OK, inertia, if you want to get technical) has nothing to dowith why there are two tides.The real reason is this. The pull of gravity drops off rapidly with distance. Lunargravity tugs on the side of the earth facing it a lot, on the earth itself a medium

amount, and on the opposite side of the earth relatively little. In short, the far-sidetide is a result of the moon attracting the earth, leaving the ocean behind. Which,looking back at your letter, I guess you already knew and didn't find convincing. Ifso, Kathleen, come on. Would I lie to you?


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AP Physics Centripetal Acceleration

All of our motion studies thus far have dealt with straight-line stuff. We havent dealt with thingschanging direction during their travel. This type of motion is called angular motion. A specialcase of angular motion is circular motion (when things travel around in circles).

We know that velocity and acceleration are both vectors. Recall that vectors have both magnitudeand direction. Change either one and you have changed the motion. We know from the first lawthat objects in motion stay in motion unless an outside force acts to change that motion. Wevestudied all about forces that change the magnitude of the velocity. Now well deal with forces thatchange the direction of a motion.

Circular Motion: Objects that have circular motion are moving in a circular path around acentral point which is called the axis orcenter of spin. If the axis is within the body, we say it isrotating. If the axis is outside the body, then the object is revolving. The earth rotates around itsaxis, which causes day and night. The earth revolves around the sun, which causes the four seasons

during the year (the time it takes the earth to revolve around the sun one time).

An object undergoing circular motion has an angular velocity and undergoes an angulardisplacement.

Angular displacement can be measured in several units: revolutions (the number of times the thingmakes a circle), degrees, or radians. These are the most common units. We will deal only withrevolutions.

The angular velocity is the rate of change in angular displacement.

angular displacementangular velocitytime

=

The Greek lower case letter omega is used as the symbol for angular velocity. Here is the equation:

t

=

t

=

The units for angular velocity we will use is: rpm min

rev.

A record is rotating with an angular velocity of 45 rpm. If it rotates for 35 seconds, how manyrotations does it make?

( )1min

45 35 26min 60

revt s rev

t s

= = =

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Linear Speed and Circular Motion:

When an object is rotating all points on the object have the same angular velocity. But at the sametime each point is also tracing out a circle. If you divide the time it takes to make one revolutioninto the circumference of the circle that is traced out you get the speed of the point. We call this thelinearortangential speed. The linear velocity is not constant as its direction changes every instant.

Calvins father is correct about the two points on a record. But, unlike Calvin, this should makeperfect sense to you, the sophisticated advanced student of physics. The further a point is from thecenter of spin, the larger will be its linear speed.

This is true for the earth as well. You have a much larger linear speed if you are at the equator thanyou would have if you were at one of the poles.

We can figure out the linear speed of a point using the angular velocity.

Speed is distance divided by time, the time it takes to make one revolution is:

tt

= = Let the angular displacement be 1 revolution:

1t

= = This is the period of the rotation, the time it takes to make

one revolution.

So we now have the time. Here is the equation for speed:x

vt

=

The distance is the circumference of the circular path, 2 r . We plug this in and also the value wefound for the time.

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22

1

d rv r

t

= = =

So the linear speed is given by: 2v r =

Force and Circular Motion: A ball is attached to a string. The ball is whirled around andaround. You can picture this right? In the drawing, we are looking down on the path of the ballfrom above.

If the string breaks at the point shown, what will be the path of the ball? The Physics Kahuna willgive you three choices:

166

A B C

Dear Doctor Science,Why do you have to rotate the tires on your car? Don't they rotate when you

drive on them?

-- Kenneth Kuller from Minneapolis MN

Dr. Science responds:

No, Ken, they don't. What appears to be a spinning tire isactually an optical illusion. Car tires stay put and slide on aspecial greasy fluid they excrete as the rub along the road.Even wagon wheels turn backwards, as anyone who hasseen reruns of "Bonanza" can attest. Again, appearancesare deceiving, because in order to make the wagons appearto move forward, they had to run the film backwards


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In order for an object to undergo circular motion, a force must act. Picture an object that has somevelocity. What will happen to it if no forces act on it? Well, according to the first law, it willcontinue to move with a constant velocity. It will follow a straight-line path.

To make it change direction a force must act on it. In order to make it changes direction constantly,a force must act on it constantly.

What is the direction of the force needed to do this? Well, when you spin something in a circle,what do you have to do? You just pull it towards the center as you go around and around.

The object gets accelerated towards the center. We call this the centripetal acceleration. Theequation for the centripetal acceleration is:

2

c

va

r=

ac is the centripetal acceleration, v is the linear or tangential speed, and ris the radiusof the circular path.

This equation will be provided to you for the AP Physics Test.

A rotating object has a linear speed of 1.5 m/s. It undergoes a centripetal acceleration of 3.6m/s2. What is the radius of the mass's circular motion?

2

c

va

r

=2

c

vr

a=

2

2

11.5 0.62

3.6

mr m

ms

s

= =

The force that brings about this acceleration is called the centripetal force. Its direction is alsotowards the center of the circular path. Centripetal means "center seeking". The centripetal forcechanges the direction of the objects velocity vector. Without it, there would be no circular path.The centripetal force is merely a convenient name for the net force that is towards the center. It isalways caused by something it could be caused by the force of gravity, the reaction force betweenthe control surfaces of an airplane with the air, etc.

When you rotate a ball around your head in a circle, the centripetal force is supplied by the tension

in the string.

What is the source of the centripetal force that causes a racecar to travel in a circular path on theracetrack?

The force is brought about by the tires pushing on the racetrack. The friction between theroad and the tire is very important, so race tires are designed to maximize friction.

What is the source of the centripetal force required to make the earth revolve around the sun?

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This is where the apple falling on Newton story fits in. Before Newton no one could explain theorbits of the planets and moons. Newton, the story goes, was relaxing under an apple treepondering the problem of the moons orbit. He knew that there had to be a force acting on themoon to accelerating it towards the earth, but had no idea what was the source of the force. Then hesaw an apple fall and the simple solution struck him like the old thunderbolt. Just as the earthsgravity reached out and made the apple fall, so it reached out and made the moon fall. Thus, the

force that keeps the planets and moon following their orbital paths is gravity.

The AP Test equation sheet will notgive you the equation for centripetal force. It does give you theequation for centripetal acceleration. It also gives you the equation for the second law. Using thesetwo equations you can easily derive the formula for centripetal force. Heres how to do it:

F ma= so plug in the value of the centripetal acceleration:2C

va

r=

2 2

C

v mvF ma m F

r r

= = =

Thats all there is to it.

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Something Completely Different:

At the 1939 World's Fair, Dupont introduced nylon stockings to the world. Dupontchemists discovered nylon while trying to produce artificial silk. On May 15,1940, they were available in stores. Nineteen years later, Glen Raven Mills ofNorth Carolina introduced panty hose, eventually developing a seamless modeljust in time for the advent of the miniskirt in 1965.

In 1970, Hanes creatively packaged panty hose in plastic eggs in super markets,

drugstores, and convenience stores - places where they had never been availablebefore. Sheer Energy was introduced in 1973 as the first L'eggs panty hose madewith sheer spandex yarn.

The brand-name "L'eggs" is a combination of the words legs and eggs, with anapostrophe added to make the wordplay idiot-proof.

In 1991, L'eggs replaced the plastic egg with a cardboard package no doubt itwas cheaper. While the plastic eggs were recyclable and used for arts-and-craftsprojects, the new box, using 38 percent less material and made from recycledpaper, allows 33 percent more containers to fit into the store-display rack and isstill rounded at the top like an egg.

In the 1970s, when Peter Lynch, the most successful money manager in America,noticed his wife Carolyn bringing L'eggs panty hose home from the supermarket,his Fidelity Magellan fund bought stock in Hanes. The value of its shares rosenearly 600 percent.

L'eggs supported the women of the 1994 and 1996 United States Olympic Teams. L'eggs is the best-selling panty hose in America. 97 percent of all supermarkets, drugstores, and convenience stores in the United

States carry L'eggs panty hose.


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1.2 kg stone is attached to a 1.3 m line and swung in a circle. If it has a linear speed of 13 m/s,what is the centripetal force?

2

2 1.2 13

160

1.3C

mkg

mv sF N

r m

= = =

A car is traveling at a constant speed and makes a turn with a radius of 50.0 m. Its speed is15.0m/s. Find the minimum coefficient of friction needed tokeep the car traveling along the path.

Lets look at the FBD:

The frictional force must equal the centripetal force.

The centripetal force is given by:2

C

mvF

r=

We know that this must equal the frictional force. We alsoknow that the frictional force is:

s sf N=

Assume the road is flat, so n = mg

Set the two equal to each other and solve for the coefficient of friction:

2

s

mvmg

r =

2

s

v

gr =

2

2

115.0 0.459

9.8 50.0

s

m

s mm

s

= =

A child twirls a yo yo. If angle of the cord with the vertical is 30.0, find ac.

Look at the forces in they direction:

Fy = 0 cos 0cos

mgT mg T

= =

169

00

n

m g

fF c


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The horizontal component ofTis the centripetal force.

sinC

F T = Plug into equation forT:

sincos

C mgF = tanCF mg =

We know that: C CF ma=

so tanCma mg = tanCa g =

29.8 tan30.0

C

ma

s= o

25.66

C

ma

s=

Centrifugal force: You may have heard of the centripetal force

before you studied physics. It is possible. Most people dont use the term however. Instead theytalk about the centrifugal force. Just what the heck is that?

Okay, youve seen the word centrifugal force, now forget it! Its terribly bad form for a physicistsuch as yourself to use such a nave term. Heres the deal. The centrifugal force is the thing thatpeople blame for the feeling that things seem to be pushed away from the center of spin duringrotation. You place a coin on a turntable and then spin it really fast. What happens to the coin? Inyour mind you picture the coin flying straight away form the center of the record. The individualwho had not studied physics would say this was because of the centrifugal force.

Here is an important concept:

The centrifugal force is a fictional force. There is no actual force that is

pushing away from the center of a rotating system.

You feel this centripetal force when you are a passenger in a car that makes a turn. When the carenters the turn, you feel as if you have been pushed into the door, away from the center of thecircular path the car is making. So you think, Hey, Im being pushed into the door so there mustbe a force pushing me away from the center of the turn.

Thats certainly how you feel at any rate.

Remember the problem at the beginning of this distinguished paper? About the path a ball wouldfollow if the string were to break? You had you three of your basic choices:

170

m g

0

T


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The correct choice, youve hopefully (actually, the Physics Kahuna should say it is to be hoped,but that sounds very pompous, so we wont say that) figured out that the correct path isB. Why?Well at the point in the circle where the string breaks, the ball has a velocity that is tangent to thecircular path. The string is providing the centripetal force pulling the ball towards the center. Theball wants to follow the tangential path because of the first law, but the string wont let it. Thestring, via the tension it exerts, pulls the ball towards the center, changing the direction of its motionand making it follow the circular path. When the string breaks there is no longer any force to

change its direction, so the ball travels in a straight-line path that is tangent to the circle as in theBdrawing in accordance with the first law of motion.

This is sort of what is going on in the car with the passenger.

The passenger wants to travel in a straight-line path at a constant speed in accordance with the law,the first law to be exact. The car however has different ideas. It decides to go in a circular path.Its the tires pushing it toward the center exerting a force to make it all happen. So the car changesdirection, but you, the passenger, do not. No force is acting on you. So you go forward in youroriginal direction until you push into the door, which then pushes you toward the center and youthen go in a circle as well. The third law rears it head you push into the door and the door pushes

into you. You feel like you are being pushed into the door, even though there is no real force doingthis. Its just the reaction force to you pushing into the door. This is the so called centrifugal force,this sensation of being pushed away from the center.

If the centrifugal force was real, i.e., there was a force pushing you away from the center of thecircular path, then if the door were to suddenly pop open you would fly straight away from thecenter of the circular path. But of course that does not happen. You would travel in a tangentialpath. The centripetal force is real, the centrifugal force is not.

Centripetal Force and Gravity: The Physics Kahuna did a silly demonstration involvinga bucket of water that was spinning in a vertical circle. The water stayed in the bucket and did not

fall out.

171

A B C


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So what was the deal? Does spinning somethingin a vertical circle somehow cancel out gravity?

Well, no, gravity is a force that cannot be stopped or canceled. It is always there, anytime you havethe appropriate masses.

The water does fall, it falls but the bucket falls with it and catches it.

This only works if the bucket is moving fast enough to catch the water. If the bucket is too slow,then the waterwillfall out of it.

The minimum linear speed for this is called the critical velocity.

Critical velocity minimum velocity for an object to travel in vertical circleand maintain its circular path against the force of gravity.

The same thing is needed for satellites in orbit around the earth or planets in orbit around the sun.They too must travel at the critical velocity.

The critical velocity formula is not provided on the AP Test, but it is very simple to figure out.

You just set the centripetal force equal to the weight of the object that is in circular motion. If thetwo forces are equal, then the object wont be able to fall out of the bucket.

2

C

mvF

r= and F mg= Set them equal to each other:

2mvmg

r= 2v gr= v gr=

So here is the critical velocity v gr=

A carnival ride travels in a vertical circle. If the ride has a radius of 4.5 m, what is the criticalvelocity?

2

2 24.5 9.8 44.1 6.6

m m mv rg m

ss s

= = = =

AP Test Question Time:

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A heavy ball swings at the end of a string as shown here, withnegligible air resistance. PointPis the lowest point reached bythe ball in its motion, and point Q is one of the two highestpoints.

(a) On the following diagrams draw and label vectors that could represent the velocity andacceleration of the ball at pointsPand Q. If a vector is zero, explicitly state this fact. Thedashed lines indicate horizontal and vertical directions.

(b) After several swings, the string breaks. The mass of the string and air resistance arenegligible. On the following diagrams, sketch the path of the ball if thebreak occurs when the ball is at pointPor point Q. In each case, brieflydescribe the motion of the ball after the break.

i. PointP

ii. Point Q

Okay, here we go. Lets answer the questions.

(a) On the following diagrams draw and label vectors that could represent the velocity andacceleration of the ball at pointsPand Q. If a vector is zero, explicitly state this fact. The

dashed lines indicate horizontal and vertical directions.

Many foolish AP Physics students forgot about the centripetal acceleration when they

i. PointP

ii. Point Q

i. PointP

173

a t P

a t Q

v

a c


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labeled the ball at point P. You woulnd forget something like that would you?

At point Q the ball has come to rest (for an instant) and so its velocity iszero. The only acceleration acting on it is the acceleration of gravity.

(d) After several swings, the string breaks. The mass of the string and air resistance arenegligible. On the following diagrams, sketch the path of the ball if the break occurs whenthe ball is at pointPor point Q. In each case, briefly describe the motion of the ball after thebreak.

i. PointP

At point P the ball is at its lowest point in the path. When the centripetal force disipears (thebroken string, right?) the ball continues to move with the velocity it had at this point, which ishorizontal. So its path looks like the projectiles we studied that started out with a horizontalvelocity.

ii. Point Q

At point Q the ball is at rest so when the string breaks it simply falls straight down.

ii. Point Q

174

a t P

a t Q

a t Q


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175

Dear Doctor Science,If the earth spins on its axis while it rotates around the sun, why don't Iget

dizzy more often?

-- Star Wholethumb from Eugene OR

Dr. Science responds:You can only get so dizzy and then you break through theother side, and progress towards undizziness. This processusually reaches its peak in the first year of life, which is whybabies fall down so often. If the earth were to stop spinningand circling the sun, even newborn babies would be able towalk just fine, but the rest of us who have adapted to thisamusement park ride would suffer horribly from motionsickness. So it's a trade-off, us or them. Until newborns get


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176

Dear Straight Dope:

What is down? If the Earth were a perfect sphere, then a line perpendicular to the tangent

of any point on the Earth's surface would go through the core and would certainly be

down. But since the Earth is not a perfect sphere, is down the perpendicular to the tangentor the most direct line to the core? While you're at it, what's up?

--Andrew Mattison, Waltham, Massachusetts

SDSTAFF Dex replies:For this one we turned to our physicist/penguinist friend Karen. Her reply:

Let's examine the two possibilities. If I drop an object, that object goes down. Therefore, Idefine down as the direction an object goes when dropped.

From gravitational physics, we know that two objects will gravitationally attract each otheraccording to their centers of mass, therefore, down is directly towards the center of mass of theearth, which for our purposes is the geometrical center of the earth. (The "core" of the earth is

quite large.)

Now what if you define down as the direction perpendicular to the tangent to the earth? (Ingeometry this is called the "normal.") Won't work. First, as you rightly point out, typicallythere's an angle between the direction of gravity and the normal. Therefore, when you drop anobject, it will not fall "down" but at some angle to "down." That's too weird.

Second, I can think of no instance where anyone would need to know the direction of thenormal. If you do care about the surface of the globe and its curvature, you are lots moresensitive to the local terrain--for example, a contractor building a house on the side of a hill.

Third, that contractor is mostly concerned with the effects of gravity tending to pull his housedown. So if all he knew were the normal, he would have to get some paper and a calculator and

figure out his latitude and calculate the direction of "gravitational down," similar to therigmarole people go through to convert magnetic north to true north. And why bother? It is veryeasy to measure "gravitational down" (even on a hillside) with a plumb bob, and it's the mostrelevant concept anyway. (As opposed to magnetic/true north where you only measure magneticnorth and you only care about true north.)

OK, so that's the theory. The real Earth is another story. When people hear the earth is an oblatespheroid, they tend to think "hamburger bun" when they should be thinking "billiard ball." Theearth shrunk to the size of a billiard ball would be out-of-round by +/-0.004", whereas thetolerance for billiard balls according to the Billiard Congress of America is +/-0.005"(http://www.bca-pool.com/rules/equip.htm). I think you'll agree that the press has really blownthe flattening of the earth's poles all out of proportion. The normal misses the earth's geometrical

center by only 21.5 km (earth's inner core has a diameter of 1200 km.) If you had two poles astall as the Empire State building, and you took them to latitude 45 degrees (where the differencebetween ""gravitational down"" and the normal is greatest) and the bottoms of the poles both hitthe earth at the same spot, but one pointed along ""gravitational down"" and one pointed alongthe normal, the tops of the poles would be separated by only 5 feet.

Fine, Karen. Blame the press foreverything.--SDSTAFF DexStraight Dope Science Advisory Board
http://www.bca-pool.com/rules/equip.htmhttp://www.bca-pool.com/rules/equip.htmhttp://www.bca-pool.com/rules/equip.htm


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AP Physics Satellite Stuff

One of Sir Isaac Newtons greatest claims to fame is his explanation of how the planets orbit thesun. That and the ability to compute the orbits, orbital speeds, orbital periods, etc. BeforeNewton, the motion of the heavens was a mystery. After Newton, the motion of the heavens

was an explainable physical phenomenon.

Lets do us a mind experiment. This is an experiment where you think instead of do. Anyway,picture a cannon that is set to fire horizontally. What does the path of the projectile look like?

The projectile will follow a curved path. This is because it is being accelerated downwards by theforce of gravity. The greater the velocity of the projectile, the farther it will go before it strikes the

Earth.

The Earth, however, is not flat, although over short distances we can pretend that it is. So whatactually is happening is that the projectile moves over and falls to the ground on a curved surface.So we have a curved path and a curved surface. We have to take the curvature of the Earth intoaccount when firing long range projectiles. Possible paths would look like this:

Again, the greater the velocity of the projectile, the greater the range. Newton showed that if thevelocity was great enough, the curving path of the falling projectile would match the curved surfaceof the earth and the falling projectile would never actually hit the Earth. Here is a drawing of this.

Newton imagined a mountain on the earth that was so high that its summit was outside of theearths atmosphere (this eliminates friction with air). On top of the mountain is a powerful cannon.The cannon fires a projectile horizontally. The projectile follows a curved path and eventually hitsthe earth. Now we add more gunpowder to the charge and fire another cannonball. This cannonballwill travel a greater distance before it too hit the surface of the earth. We keep firing the gun with abigger and bigger charge. The cannonball goes further and further before it strikes the earth.

177

P a t h o f s h o r t r a n g e p r o j e c t i l e

P a t h o f l o n g r a n g e p r o j e c t i l e

P a t h o f p r o j e c t i l e w i t h s a m e c u r v a t u r e

a s s u r f a c e o f t h e e a r t h


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Eventually the velocity is great enough so that the curved path of the projectile matches the curvedsurface of the earth and the cannonball never gets closer to the planets surface. It keeps fallingforever.

That basically is your orbit.

The earth is not flat. It is a sphere and its surface has a fairly constant curvature. The surface drops

4.9 m in 8 000 m of horizontal travel.

If we launch a cannonball with a velocity of 8 000 m/s, it will fall a distance of 4.9 m and travel ahorizontal distance of 8 000 m in one second. This means that it will stay at the same height abovethe earths surface throughout its path. Of course, if we did this near the surface, wed have the airslowing the projectile down. Wed also have to worry about the cannonball running into houses and

178

P a t h o f p r o j e c t i l e f i r e dw i t h l a r g e r a n d l a r g e r c h a r g e s

W i t h j u s t t h e r i g h t v e l o c i t y ,t h e p r o j e c t i l e n e v e r r e a c h e st h e e a r t h s s u r f a c e

E a r t h E a r t h


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mountains and trees and so forth. Above the atmosphere, however, all these impediments areeliminated.

The orbit of the everyday celestial object is described by a combination of the law of gravity,Newtons three laws, and the stuff we just learned about, circular motion.

179

Sherlock Holmes is the most portrayed character on film,having been played by 72 actors in 204 films. The historicalcharacter most represented in films is Napoleon Bonaparte,with 194 film portrayals. Abraham Lincoln is the U.S.

President to be portrayed most on film, with 136 filmsfeaturing actors playing the role.


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Orbital Equations: Let us assume that the orbit of a satellite about the earth (or any othermassive body) is a circle. Most orbits, the Physics Kahuna must point out, are not circles butare instead ellipses. This was discovered by Johanes Keppler in the 1600s. But lets keep itsimple and look at a circular orbit.

In order to have a circular path, a centripetal force is required. This is supplied by the force of

gravity between the two bodies.

So we can set the centripetal force equal to Newtons law of gravity:

1 2

2

m mF G

r= gravity 2

2

C

m vF

r= centripetal force

Set them equal to one another:

2

1 2 2

2

m m m vG

rr=

Notice how the mass of the object canceled out.

2 1Gm

vr

= 1Gmvr

=

This gives us an equation for the orbital velocity:

Gmv

r=

The mass, m, in the equation is the mass of the body being orbited. If we are talking about a planetorbiting the sun, then the mass we would use would be that of the sun. The mass of the satellitecancels out, so its mass is not involved in the orbital velocity equation at all.

The equation for the orbital velocity will not be given you on the AP Physics Test. So be preparedto derive it if you need it.

Period of satellite: This is another simple derivation job. The period of a satellite is T, thetime to make one orbit. What would be the period of the earth around the sun?

Lets develop the equation for the period of a satellite. Well use the equation for distance andsolve it for the time:

xv

t=

xt

v=

d, the distance traveled is the circumference of the orbit. We know that it would be:

180


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2x r=

So we can plug that in to the equation we solved for time:

2 rt

v

=

but v is also given by the equation we just derived for the orbital velocity:

Gmv

r=

If we plug the orbital velocity into our working equation, i.e., put them together, we get:

2 rt

Gmr

=

Square both sides:

2 22 4 rt

Gm

r

=

Clean up everything up nice and neatlike using our potent algebra skills:

2 2 2

4r

t rGm

=2 3

2 4 rtGm=

2 34 rt

Gm

=

3

2r

tGm

=

And we end up with an equation for the period of a satellite. Again the mass in the thing is the massof the body being orbited:

3

2 rtGm

=

Now lets do some exciting problems.

What is orbital velocity of the earth around the sun? The sun has a mass of 1.99 x 1030kg, themean distance from the earth to the sun is 1.50 x 1011 m.

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( )2

11 30

2 2 11

16.67 10 1.99 10

1.50 10

kg m mGmv x x kg

r s kg x m

= =

2

8 42

8.85 10 2.97 10m mv x xss

= =

A satellite is in a low earth orbit, some 250 km above the earth's surface. rearth is 6.37 x 106 mand mearth = 5.98 x 10

24 kg. Find the period of the satellite in minutes.

ris the radius of the earth plusy, the height of the satellite. earthr r y= +3

610250 0.25 101

my km x m

km

= =

6 6 66.37 x 10 m + 0.25 x 10 m = 6.62 x 10 mr=

( )

( )

36

3

2

11 24

2 2

6.62 102 2

6.67 10 5.98 10

x mrT

kg mGm mx x kg

s kg

= =

182 5 2

13

290.12 102 2 7.273 10

39.89 10

xT s x s

x = =

4 2 22 72.73 10 53.57 10 5357T x s x s s= = =

1min

5357 89.3 min60

T ss

= =

The moon has a period of 28 days. If the earth's mass is 5.98 x 1024 kg, how far is the moonfrom the earth?

624 360028 2.42 10

1 1

h sT day T x s

day h

= =

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3

2r

TGm

= square both sides:3

2 24r

TGm

=

Solve forr3:

3 2

2

1

4r T Gm

=

2

23

1

4r T Gm

=

( ) ( )2

26 11 24

2 2 23

12.4210 6.67 10 5.98 10

4

kg m mr s x x kg

s kg

=

25 33 5.923 10r x m=

convert 5.923 x 1025 to something times 1024 (we chose 1024 because 24 is divisible by 3)

24 3 83 59.23 10 3.90 10r x m x m= =

183


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Gravity in Orbit: We all know that the astronauts in space orbiting the earth are "weightless".Does this mean that there is no gravity in space? Well, no. Most of our spacecraft are in pretty low

orbits. The distance between the astronauts and the earth is not that much greater than when theyare on the earth. Their weight is only about ten percent less than it is on earth. So why are theyweightless?

The space shuttle and everything in it are falling towards the earth. It is in a state of freefall. Weknow that everything falls at the same rate, so everything in the space shuttle is falling at the samespeed. Because of this there is no relative motion between the space shuttle and everything in it.There is no sense of up or down and the astronauts no longer feel the force of gravity. Its likebeing inside an elevator that is falling down the elevator shaft (the cable broke or something). In anormal elevator, one that isnt falling, gravity exerts a downward force on everything. If you stand

184

Useless Tivia Facts:

15 million blood cells are destroyed in the human body every second. 3,000 teens start smoking every day in the United States. The average person spends two weeks of their life kissing.

Scientists in Australia's Parkes Observatory thought they had positive proof of alien life,

when they began picking up radio-waves from space. However, after investigation, theradio emissions were traced to a microwave in the building.

40 people are sent to the hospital for dog bites every minute. In Japan, 20% of all publications sold are comic books. The average day is actually 23 hours, 56 minutes and 4.09 seconds. We have a leap year

every four years to make up for this shortfall.

In the next seven days, 800 Americans will be injured by their jewelry. Disneyland opened in 1955. It is estimated that at any one time, 0.7% of the world's population are drunk. If you divide the Great Pyramid's perimeter by two times it's height, you get pi to the

fifteenth digit.

The pupil of the eye expands as much as 45 percent when a person looks at somethingpleasing.

50,000 of the cells in your body will die and be replaced with new cells all while youhave been reading this sentence.

67.5% of men wear briefs instead of boxers. 75% of people wash from top to bottom in the shower. The average four year-old child asks over four hundred questions a day. 80% of all body heat escapes through the head. Lightning strikes the Empire States Building more than 50 times a year. 85% of the population can curl their tongue into a tube. A fetus acquires fingerprints at the age of three months. The Future's Museum in Sweden features a scale model of the solar system. The sun is

105 meters in diameter and the planets range from 5mm to 6 km from the sun. Theexhibit also features the nearest star, Proxima Centauri. The distance to the star is also toscale. It is located in the Museum of Victoria. . . . in Australia.


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on a bathroom weight scale, you push down on it and it reads out your weight. But now the cablebreaks. You are still standing on the scale, but the elevator, the scale, and you are all falling downaccelerating at 9.8 meters per second squared. You no longer exert a force on the scale it isfalling at the same speed that you are. It now reads zero.

Any objects in the elevator would appear to be weightless. If you held a ball outward and then

released it, it would not appear to fall down (since it is already falling). It would appear to float inspace in front of you. You would think, Hey, cool, theres no gravity in the elevator. Neat!

It would be pretty neat too. Until the elevator hits thebottom of the elevator shaft.

Okay, lets transfer this idea to the space shuttle. It is,in effect, a falling elevator, one with the advantage ofnot crashing into the floor it never hits the earth!

Thats why the astronauts are weightless.

185

F r e e F a l l i n g i n a ne l e v a t o r


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186

Dear Cecil:

Tycho Brahe, the astronomer, is buried here in Prague in a church. That's not so

unusual. What is rather strange is the persistent rumor that he had a silver nose,

something to do with a duel. Did it tie on around his face with silk strings? Did it have

little hooks that went around his ears, like eyeglasses? Was it surgically attached? I'm

truly curious. I also wonder if it tarnished, and whether he had to polish it. Did it make

a note like a whistle when he blew his nose? --Raymond Johnston, chief copy editor,

Prague Post

Cecil replies:Glad to be of help, Ray. I know how tough it is for journalists to come by this kind ofinformation. What you heard was no rumor. Tycho Brahe (1546-1601), the father (or at leastthe godfather) of modern astronomy, really did wear an artificial nose, owing to the fact thatthe real one had been sliced off in a duel. You may think: this does not sound like thescientists I know. Tycho Brahe, however, was no ordinary stargazer.

It happened in 1566 while the 20-year-old Tycho was studying at the University ofRostock in Germany. Attending a dance at a professor's house, he got into a quarrel

with one Manderup Parsbjerg, like himself a member of the Danish gentry. Over awoman? Nah--tradition has it that the two were fighting over some fine point ofmathematics. (My guess: Fermat's Next-to-Last Theorem, which posits that 2 + 2 = 5for very large values of 2.) Friends separated them, but they got into it again at aChristmas party a couple weeks later and decided to take it outside in the form of aduel. Unfortunately for Tycho the duel was conducted in pitch darkness with swords.Parsbjerg, a little quicker off the dime, succeeded in slicing off the bridge(apparently) of Tycho's nose. Reconstructive surgery then being in a primitive state,Tycho concealed the damage as best he could with an artificial bridge made ofprecious metals. He carried some nose goop with him always, either to polish thenose or to glue it more firmly in place. But no hooks or string, and probably nowhistling either.

High-handed and irascible, Tycho Brahe was the kind of guy who got into duels. Luckily hewas also a genius. Fascinated by the stars since his youth, he discovered that existingastronomical tables were grossly inaccurate and set about making his own meticulousobservations of the heavens, a project that occupied him for most of his life. To keep himfrom going abroad, the king of Denmark and Norway gave Tycho a prodigious quantity ofcash ($5 billion in today's money, by one estimate) and his own island. There Tychoconstructed an observatory where for 20 years he compiled the impressive body ofastronomical data that his assistant Johannes Kepler subsequently used to deduce the laws ofplanetary motion. All this, mind you, with the naked eye; the telescope hadn't yet beeninvented.

To give you a further indication of the type of guy we're dealing with there, Tychodidn't marry the mother of his eight children, employed a dwarf as a jester, kept a petelk (which died after breaking a leg while going downstairs drunk), dabbled inalchemy, and tyrannized the local peasantry. After his royal patron died of excessivedrink he managed to tick off everyone in Denmark, had his subsidies revoked, andeventually found it wise to leave the country. Having relocated to Prague, he diedafter drinking heavily at dinner, obviously a pretty common fate in those days.


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187

Dear Doctor Science,IF YOUR SO SMART TELL ME HOW THE INERTIAL DAMPENERS WORK ON

THE USS ENTERPRISE-NCC-1701-D? BET I STUMPED YOU ON THAT ONE.

-- JON STIGLER from EADS, TN

Dr. Science responds:They didn't work, and that was the big problem with the series. The ship slipped into

hyperdrive whenever it felt like it, sending the crew zipping across the galaxy in a

nanosecond. This made it especially hard on the navigators, who were constantly comingup with novel ways to get their bearings, including monitoring talk radio from nearby

planets in the hope of learning their approximate location. Eventually, captain and crew

developed a pervasive sense of chronic apathy, and instead of piloting their craft, spenttheir time watching reruns of the first Star Trek, where Jim and Spock traded profundities

on the planet of scantily clad women.

Tycho's tomb was reopened in 1901 and his remains were examined by medical experts.The nasal opening of the skull was rimmed with green, a sign of exposure to copper.Presumably this came from the artificial nose, which supposedly had been made of silveror gold. The experts put the best face on this, as it were, saying that Tycho was an expert

in metallurgy and probably wanted an alloy that was durable and skin colored. Sure, guys.I say Mr. Astronomy got nicked in the nose department twice.

--CECIL ADAMS


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AP Physics Gravity WrapupYou get three equations to work with, and only three. Everything else you have to develop fromthese three equations (also have to use some other ones). Here they are:

1 2

2GGm mF

r= Newtons law of gravity.

2

c

va

r= Centripetal acceleration equation.

NetF F ma = = Second law.

Heres what you have to be able to do:

1. You should understand the uniform circular motion of a particle so you can:

a. Relate the radius of the circle and the speed or rate of revolution of the particle to themagnitude of the centripetal acceleration.

You have this equation to work with:

2

c

va

r=

In this equation, v is the linear speed, so its fairly easy to relate it to the radius (r) or to

the centripetal acceleration (ac). You just use the old equation as it is given. If r gets

bigger, then the centripetal acceleration gets smaller, etc.

The rate of revolution is the rate that the thing rotates. This would be the number of

rotations divided by the time it took to do them. This is called the angular velocity, .Unfortunately, you dont got no equation for this. Just remember that any rate is simply

a quantity divided by time, so the angular velocity is simply:

t

=

Where is the angular displacement, which will most likely be the number of revolutionsthat have been made?

What happens to the centripetal acceleration if the velocity is increased and you want to

keep the same radius? If the radius increases, but the velocity stays the same, what

happens to the centripetal acceleration?

188


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You should be able to derive an equation that relates linear speed to angular velocity. We

did that in your gravity handout. The Physics Kahuna showed you how to develop the

equation:

2v r =

Basically, the velocity is proportional to the angular velocity. So whatever happens when thelinear speed increases also happens if the angular velocity increases.

You might be able to get away without developing the equation at all. Using this concept,

it is a simple matter to relate the rate of revolution of the particle to the magnitude of the

centripetal acceleration.

b. Describe the direction of the particles velocity and acceleration at any instant duringthe motion.

The acceleration is always towards the center.

The velocity is always tangent to the objects path.

For any type of circular motion, such as the moon orbiting a planet, the direction of the

centripetal acceleration is towards the center and the velocity direction is a tangent to the

path.

c. Determine the components of the velocity and acceleration vectors at any instant andsketch or identify graphs of these quantities.

This is a simple matter use trigonometry to resolve the acceleration and or velocity

vector into its components. Again, graphing these quantities is quite simple.

189

P l a n e t

M o o n

P a t h o f O r b i t

v

a c


35/44

Above is a drawing of an object that is following a circular path. The velocity and centripetal

acceleration vectors are shown at five different positions. To the right of that is a graph of

velocity Vs time for the same motion. Please now figure out how the graph works.

The velocity is either only up and down or only right or left at the four cardinal positions, 1,

3, 4, and 5. So the speed is either all vx or all vy. For all other positions on the path, there

will be both a horizontal and vertical component for the velocity. Can you see how the speed

v is the sum of the squares of the components? Think about till you can.

Here is a graph of the centripetal acceleration Vs time for the same set of points.

2. Students should be able to analyze situations in which a body moves with specifiedacceleration under the influence of one or more forces so they can determine themagnitude and direction of the net force, or of one of the forces that makes up the netforce, in situations such as the following:

190

v

a c

1

2

3

4

5t

v

5 2

3

14

v x

v y

2

yt

5

2

3

1

4

a x

y

a ca c

a


36/44

a. Motion in a horizontal circle (e.g., mass on a rotating merry-go-round, or carrounding a banked curve).

The main thing here is using the second law and the centripetal acceleration to find the

centripetal force.

2 2 2

c c cv v mvF ma a F m Fr r r

= = = =

The centripetal force can then be used to calculate the frictional force acting on a car

traveling in a circle or some such thing.

In a previous unit we looked at several problems along this line.

The centripetal force will be used in several other sections. One that really requires it is

when we study a charged particle moving through a uniform magnetic field. Youll see howthis works later on.

b. Motion in a vertical circle (e.g., mass swinging on the end of a string, cart rollingdown a curved track, rider on a Ferris wheel).

Draw a free body diagram. Analyze the forces. The centripetal force will always end up

being the net force acting on the body and will always be directed to the center of the

circular path. There are two forces working on the vertical circle the weight of the

object and the centripetal force. At the very top for the instant of time that the body is up

there, the sum of the vertical forces must be zero since the body is moving sideways and

not up and down. So you can solve for the tension which will turn out to be The idea here

is that the centripetal force must equal the weight of the object. From this you can

calculate an equation for the minimum velocity a body needs to travel along a vertical

circular path. The equation is: v gr=

3. You should know Newtons Law of Universal Gravitation so you can:

a. Determine the force that one spherically symmetrical mass exerts on another.

This is pie. You just use Newtons law of gravity.

b. Determine the strength of the gravitational field at a specified point outside aspherically symmetrical mass.

This is another slice of really good pumpkin pie with whipped cream on top (yummy).

Again, merely apply good old Newtons law of gravity.

4. You should understand the motion of a body in orbit under the influence of gravitationalforces so you can, for a circular orbit:

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a. Recognize that the motion does not depend on the bodys mass, describequalitatively how the velocity, period of revolution, and centripetal accelerationdepend upon the radius of the orbit, and derive expressions for the velocity andperiod of revolution in such an orbit.

This is a derivation special. The Physics Kahuna demonstrated how to do all of this in aprevious unit.

You have to derive the equation for centripetal force and set it equal to the force of gravity

from Newtons law of gravity. Then you solve for the velocity. This gives you the orbital

velocity.

1Gm

vr

=

To find the period, you use the equation for velocity,x

vt

= and the

equation for the orbital velocity we just found. The distance x is the circumference of the

circle. Which is 2 r . Put them together and you end up with:

3

2r

tGm

=

We did a number of problems that required developing and using these equations.

192


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AP Test Examples: From 1997:

To study circular motion, two students use the hand-helddevice shown, which consists of a rod on which a spring scaleis attached. A polished glass tube attached at the top serves

as a guide for a light cord attached the spring scale. A ball ofmass 0.200 kgis attached to the other end of the cord. Onestudent swings the ball around at constant speed in ahorizontal circle with a radius of 0.500 m. Assume frictionand air resistance are negligible.

a. Explain how the students, by using a timer and theinformation given above, can determine the speed of the ball as it is revolving.

Measure radius and calculate circumference. Use stop watch to measure the period

(time of one revolution). Usex

vt

= to find the speed of the ball.

b. How much work is done by the cord in one revolution? Explain how you arrived at youranswer.

No work. The motion is perpendicular to the force.

c. The speed of the ball is determined to be 3.7 m/s. Assuming that the cord is horizontal asit swings, calculate the expected tension in the cord.

2 2

T C c C

v mvF F F ma a F

r r= = = =

( ) ( ) 20.200 3.75.5

0.500C

kg m sF N

m= =

d. The actual tension in the cord as measured by the spring scale is 5.8N. What is thepercent difference between this measured value of the tension and the value calculated inpart (c)?

( )5.8 5.5

% 100% 5.5%5.5

N Ndiff

N

= =

e. The students find that, despite their best efforts, they cannot swing the ball so that the cord

remains exactly horizontal. On the picture of the ball below, draw vectors to represent theforces acting on the ball and identify the force that each vector represents.Explain why it is not possible for the ball to swing so that the cord remains exactlyhorizontal.

The tension in the string, t, provides the centripetal force. mg is

the weight of the ball.

There is always a downward force acting on the ball, its weight,

mg. So there must be a component of the tension in the opposite

193

m g

t


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direction since the ball is not moving in a vertical direction. The sum of the vertical

forces must be zero.

f. Calculate the angle that the cord makes with the horizontal.

( ) ( )21 1 o

2

0.200 9.8sin sin 19.8

5.8

kg m smg

kg mTs

= = =

From 2001:

A ball of massMis attached to a string of lengthR and negligible mass.The ball moves clockwise in a vertical circle, as shown to the right.When the ball is at pointP, the string is horizontal. Point Q is at thebottom of the circle and pointZis at the top of the circle. Air resistance isnegligible. Express all algebraic answers in terms of the given quantitiesand fundamental constants.

(a) On the figures below, draw and label all the forces exerted on the ballwhen it is at pointsPand Q, respectively.

(b) Derive an expression forvmin, the minimum speed the ball can have at pointZwithout leavingthe circular path.

C

F t mg= but the tension is zero CF mg=2

c c c

mvF ma F ma F

r= = = Derive the equation for centripetal force.

2 2mv v

mg g v rg v Rg r r

= = = =

(c) The maximum tension the string can have without breaking is Tmax. Derive an expression forvmax, the maximum speed the ball can have at point Q without breaking the string.

The Tension in the string is the upward force and the weight is the downward force, the sum of

these two forces is ma so:

Maxma T mg =

But the acceleration acting on the system is the centripetal acceleration.

194

Q

P

Z

M

S i d e V i e w

P

m g

t

Qm g

t


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( )2

2

Max Max

v rm T mg v T mg

r m= =

( ) ( )ax axM Mr R

v T mg T Mg m M

= =

(d) Suppose that the string breaks at the instant the ball is at pointP. Describe the motion of the ballimmediately after the string breaks.

The ball would go straight up, slowing down as it is accelerated downward by gravity, it will stop,

then fall down, accelerating as it goes. Its motion is tangential to the path, which is straight up.

From 1999:

5. A coin Cof mass 0.0050 kgis placed on a horizontal disk at a

distance of 0.14 m from the center, as shown below. The diskrotates at a constant rate in a counterclockwise direction asseen from above. The coin does not slip, and the time it takesfor the coin to make a complete revolution is 1.5s.

a. The figure to the right shows the disk and coin as viewed fromabove. Draw and label vectors on the figure below to showthe instantaneous acceleration and linear velocity vectors forthe coin when it is at the position shown.

b. Determine the linear speed of the coin.

The coin makes one revolution is a time of 1.5 s. So the distance

it travels in that time is the circumference of the circle it makes,

which is 2 r .

( )

( )

2 0.1420.586

1.5

mx r mv

t t s s

= = = =

c. The rate of rotation of the disk is gradually increased. The coefficient of static frictionbetween the coin and the disk is 0.50. Determine the linear speed of the coin when it justbegins to slip.

cF f=

2vm mg

r=

( ) ( )2

0.14 0.5 9.8 0.828m m

v r g ms s

= = =

d. If the experiment in part (c) were repeated with a second, identical coin glued to the top ofthe first coin, how would this affect the answer to part (c)? Explain your reasoning.

There would be no change. The mass cancels out.

195

v

a


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From 1995:

Part of the track of an amusement park roller coaster is shaped as shown below. A safety bar isoriented length-wise along the top of each car. In one roller coaster car, a small 0.10 kilogramball is suspended from this bar by a short length of light, inextensible string.

a. Initially, the car is at rest at pointA.

i. On the diagram to the right, draw and label all theforces acting on the 0.10-kilogram ball.

ii. Calculate the tension in the string.

20 0.10 9.8 0.98

mt mg t mg kg N

s

= = = =

The car is then accelerated horizontally, goes up a 30 incline, goes down a 30 incline, andthen goes around a vertical circular loop of radius 25 meters. For each of the four situationsdescribed in parts (B) to (E), do all three of the following. In each situation, assume that theball has stopped swinging back and forth. 1) Determine the horizontal component Th of thetension in the string in newtons and record your answer in the space provided. 2) Determinethe vertical component Tvof the tension in the string in newtons and record your answer in thespace provided. 3) Show on the adjacent diagram the approximate direction of the string withrespect to the vertical. The dashed line shows the vertical in each situation.

b. The car is at pointB moving horizontally to the right with an acceleration of 5.0 m/s .

2 2

0.10 5.0 0.49h

m m

T ma kg s s

= = =

2 20.10 9.8 0.98

v

m mT mg kg

s s

= = =

c. The car is at point Cand is being pulled up the 30 inclinewith a constant speed of 30 m/s.

0hT =

2 20.10 9.8 0.98v

m m

T mg kg s s

= = = d. The car is at pointD moving down the incline with an

acceleration of 5.0 m/s2.

o

2cos 0.10 5.0 cos 30 0.43

h

mT ma kg N

s

= = =

196

t

m g


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The sum of the forces in the y direction must equal Vma . Therefore:

v v v vmg T ma T mg ma = =

The acceleration in the vertical direction is sinma .

o

2 2sin 0.10 9.8 0.10 5.0 sin 30 0.73

V

m mT mg ma kg kg N

s s

= = =

e. The car is at pointE moving upside down with an instantaneous speed of 25 m/s and notangential acceleration at the top of the vertical loop of radius 25 m.

0HT =

In the vertical direction, we have a centripetal FC , the

tension, and the weight of the ball. The centripetal force

must equal the sum of the other two forces.

2

C V V C

mvF T mg T F mg mg

r= = =

( )( )

2

2

0.10 25 0.10 9.8 1.525

V V

m

kg msT kg T N m s

= =

197

m g

TTV

a v a


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198

More Foolish Trivia:

A jogger's heel strikes the ground 1,500 times per mile. A lifetime supply of all the vitamins you need has a mass of only about

225 g (eight ounces). 3 out of 4 optometrists wear eyeglasses. Computer users, on average, blink seven times per minute. Dark circles under the eye are an inherited trait. The iris membrane controls the amount of light that enters your eye. In 1992, when EuroDisney first opened in France, park visitors beat up

some of the costumed characters (oh please not Snow White!) becauseat the time most people had been against the park being built.

Until the 1960's men with long hair were not allowed to enterDisneyland.

Eiffel Tower was the tallest structure in the world before theconstruction of the Empire State Building in 1930. A B-25 bomber airplane crashed into the 79th floor of the Empire State

Building on July 28, 1945.

Sunbeams that shine down through clouds are called crepuscular rays. The Apollo 11 Lunar Module had only 20 seconds of fuel left when it

landed on the moon in 1969.

Kite flying is a professional sport in Thailand. A normal raindrop falls at about 7 miles per hour. A penny whistle has six finger holes.

The human eye sees everything upside-down, but the brain turns it rightside up. A shell constitutes 12 percent of an egg's weight. A silicon chip a quarter-inch square has the data processing capacity of

the original 1949 ENIAC computer, which occupied a city block.

All the dirt from the foundation to build the former World Trade Centerin NYC was dumped into the Hudson River to form the communitynow known as Battery City Park.

Rice paper does not contain rice or any rice products. The height of the Eiffel Tower varies as much as six inches depending

on the temperature.

At April 2000, Hong Kong had 392,000 faxlines - one of the highestrates of business fax use in the world.

The only part of the human body that has no blood supply is the cornea.


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(CNN) --On Saturday the Earth reaches its farthest point from the sun during its orbit, butthe average global temperature actually goes up this time of year. What gives? Like the other planets in the solar system, the Earth moves in a lopsided orbit around the sun. InJuly, our planet reaches the most distant point of its annual revolution, called aphelion. In January,it dips to its nearest position, or perihelion.Compared to other planets, the Earth's orbit is only slightly askew. It has an eccentricity of 1.7

percent, while Mars' is more than 9 percent and Mercury's more than 20 percent.Looking at a page-sized map, an observer might mistake the Earth's path around the sun for an exactcircle. Scientists know better."When we're closest the sun, the distance is 147.8 million kilometers [91.8 million miles]. Thisweekend we will be 152.6 million kilometers [94.8 million miles] away, a 5 million kilometer (3.1million mile) difference," University of Florida astronomer George Lebo said.The greater the distance, the less intense the rays of the sun."Averaged over the globe, sunlight falling on Earth at aphelion is about 7 percent less intense than itis at perihelion," NASA climate researcher Roy Spencer said.

Then why is it so warm now in the Northern Hemisphere? The reason is that the Earth tilts 23.5degrees on its axis, and because of where it is in its orbit, the northern half is the portion mostexposed to the sun during the summer months.The conditions are reversed in the Southern Hemisphere, which basks in the solar rays half a year

later, when the South Pole tips more toward the sun.Overall, the entire globe averages higher temperatures when the Earth is farther from the sun. Theplanet is about 4 degrees F (2.3 degrees C) warmer during aphelion than perihelion, NASAscientists said.The reason? Continents and oceans are distributed unevenly. The north has more land and the southmore water. When sunlight strikes the former, it heats up more quickly than the latter, which canabsorb more heat before rising in temperature.In other words, Earth is slightly warmer in July because the sun is shining on all the northerncontinents, which heat up rather easily. In January, the sun focuses its rays on the southern oceans,which have higher heat capacity.Another quirk of physics contributes to even more planetary heat this time of year. During aphelion,

planets move in their orbits more slowly than during perihelion.As a consequence, the summer in the north is several days longer than in the south, affording thesun more time to cook the landmasses in the Northern Hemisphere.

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