Law of Gravitation and Planetary Motion
All Bodies and Objects Attract!
Yes, each body exerts a gravitational force on other surrounding bodies
Distance between 2 bodies and the mass between 2 bodies effects the force exerted between them. This leads to Law of Gravitation.
Law of Gravitation
Definition: A force between 2 bodies is proportional to their masses and distance between them
The proportionality constant G from you reference table front sheet:
F g
= G m
1 m
2
/r
2
where G= 6.67x10
-11
m
3
/Kg·s
2
Even though this information below is not in the curriculum anymore, it is important and you should know it.
Keppler's 3 rd
Law- Cube of a planet‟s average distance from the sun (R
3
) divided by the square of its period (T
2
) is a constant for all planets.
OR R
3
/ T
2
=K
Satellite Motion
Fc= M· s
V
2
/r for satellite in orbit around the earth.
Equating this with Newton‟s Law of Universal Gravitation:
F c
= M s·
V
2
/r= G·M earth ·
M s
/ r
2
= F
G
V
2
/r= G·M earth
/r
2
________
V=√G·M earth
/r
2

Think of spinning an object (usually called a “bob”) on the end of a string. How would you describe it to someone?

The easiest thing to say is that it moves in the shape of a circle. o This means you should be able to measure the radius of the circle, and from that calculate things like circumference.

You might also observe that since it is going in a circle, the motion of the bob as it moves around repeats itself over and over again. o We refer to this as periodic motion , because periodically ( every so often ) it gets back to where it started and keeps on going. o Other objects, like pendulums, can also be described as periodic, as long as they repeat the same motion over and over again.

The bob is also an example of two dimensional motion, just like a projectile. o It is moving in the x and y directions at the same time, not in a perfect straight line.
You might remember that when we originally talked about acceleration we defined it as a change in the speed and/or direction of the movement of an object.

While the bob is going around in a circle, its speed might be constant (and usually will be in the problems you do), but what is happening to the direction it is traveling?

At any particular instant in time, the velocity of the bob spinning in a circle is pointing in the same direction as the tangent line to the circle.

That means that the direction the bob is traveling is constantly changing! o By definition, this means that the bob is accelerating. o If it wasn‟t accelerating, it would be traveling in a straight line. o So how do we calculate the acceleration of the bob?
V
Centripetal force
Ex. stone at end of string, Daytona 500!, other examples etc.
In order to stay/continue movement in circle you must have a force pulling toward center of circle.
You can feel the force pulling on your hand when you swing an object on a string.
F.B.D
V a c
= centripetal acceleration
F
Net
=m·a
F c
=m·a c
V a c, Fc a a c, Fc c, Fc a c, Fc
` F c
= m·a
How do I get V?
V c
pulling into center
V
3 a c,Fc

Eq. a c
=V
2
/r V= 2·π·r/T f f c
=m·a c c
= m·V
2 where T= period for 1 revolution
/r = Centripetal Force circumference of circle= distance travelled
Some textbooks will start showing you all these fancy diagrams and talk about things like
“chords” to explain how they figured out a formula for the acceleration of an object moving in a circle.
 Does this really help you understand the formula any better? I don‟t think so. It seems like students usually get really concerned about trying to figure out this weird math that they might not have seen before, and completely forget about the end result.
 So, instead, here‟s the formula for you… v
2 a c
 r a c
= centripetal acceleration (m/s
2
) v = velocity (m/s) r = radius of the circle (m)
Notice I called the acceleration “centripetal”.
 Centripetal comes from Greek meaning “centre seeking”

The bob is basically being pulled in towards the centre of the circle, which is why it follows a curved path around the centre. It‟s as though it were seeking the centre of the circle, even though it never gets there.
 Do not confuse this with the word “centrifugal”, which doesn‟t even exist! o A person in a car going around in a circle might say that a
“centrifugal force” was pushing them across the seat out of the circle. o In fact, it is just your body obeying Newton‟s first law… it is o trying to move in a straight line.
The car is trying to keep you going in a circle which is centripetal
Example : What is the centripetal acceleration of a person in a car driving at 60km/h in a
. traffic circle that is 120m across?
Change the velocity into metres per second, and since the measurement of the circle is a diameter, divide it by two. v
2
( 17 m / s )
2 a c
 
= 4.6 m/s
2 r 60 m
Velocity in a Circle
If you need to calculate the centripetal acceleration of an object but you don‟t know the velocity, you would have a little bit of a problem.

There should be a way to come up with a basic formula that relates velocity in a circle to some of the other concepts we have studied so far.
4

 Let‟s try starting off with a formula that we know from the beginning of the course. v
 d t


Since we are looking at something going around in a circle, the distance it covers each revolution is equal to the circumference of the circle.
C = 2
 r
We will substitute this into the first formula… v

2
 r t
WHICH EQUALS v
 d t

When we are describing something that has periodic motion, we say that the amount of time it takes for it to complete one full cycle (in this case, circle) is the period. o This is just a special term that is used to measure the time of any object as it goes through periodic motions. o It has the symbol “T” in formulas and is still measured in seconds. o This means that the final version of our new formula should look something like this… v

2
 r
T v = velocity (m/s)

= pi, just use 3.14 in your calculations r = radius of the circle (m)
T = period (like time) (s)
Example : What is the length of a student‟s arm if she can swing a pail in a circle at
2.72m/s every 1.5s? v

2
 r
T r = (vT) / 2

= (2.72m/s) (1.5s) / 2
 r = 0.65m
You could also combine this velocity formula with the acceleration formula that we have already looked at…
5

a c
 v r
2 v

2
 r
T


2
T r 

2 a c
 r a c

4

T
2
2 r
 Although you won‟t find this formula on the data sheet, you might find it useful at times.

You need to be able to show how you derive this formula for a long answer question, so don‟t just blindly memorize the final formula.
Example : What is the acceleration of a horse running around a circular race track with a radius of 37m once every 12s? a c

4

T
2
2 r

4
 2
( 37 m )
( 12 s )
2
= 10m/s
2
Centripetal Force
By now you should be saying to yourself that if an object going in a circle is accelerating, there must be a force acting on it.
 This follows from Newton‟s Laws that a force results in acceleration.

If there were no forces involved, the object should be moving at a constant speed in a straight line, which it isn‟t.

We should be able to come up with some formulas for forces that relate directly to circular motion.
If you have already calculated the centripetal acceleration of an object moving in a circle from one of the formulas we have studied so far, then just go ahead and use F = ma to calculate the force.

Because we are referring to a force causing an object to move in a circle, we refer to the force as a centripetal force .

This is written in formulas as F c

A lot of the time you won‟t know the acceleration in advance, so we should have a couple of shortcuts to be able to calculate the centripetal force based on the information you do have.
First Formula
We‟ll start with the basic force formula, then substitute in our centripetal acceleration formula.
F c
= ma c and a c
2
 v r
6

mv
2
F c
 r
This formula is on your data sheet.
Example : What is the centripetal force acting on a
100kg person spun around in a
8.80m circle at… a) 10.0m/s? b) 15.0m/s?
You might have seen movies with an astronaut in training spinning around and around in this big machine to train for their flight. The device is called a centrifuge, like the one pictured at right (click on it to go to the Nasa website for the Center for Gravitational Biology Research ).
Let‟s look at the forces on this person‟s body, and try to relate it back to regular
Earth gravity (one “gee” = 9.81m/s 2
). Keep in mind that just standing on the ground the person in this example would normally weigh ( F g
= mg ) 9.81 x 10
2
N. a) F c
 mv r
2

( 100 kg
8
)( 10 .
0 m
.
80 m
/ s )
2
= 1.14 x 10
3
N
This is the “weight” that the person will feel themselves weighing.
Remember from the section on mass and weight that weight is based on how much gravity is in the area that you are in. Sticking this person into the centrifuge we are able to make them fell heavier than they really are.
How much “gravity” is his body feeling?
F g
= mg

g = F g
/ m = (1.14 x 10
3
N) / (100kg) = 11.4 m/s
2
This is just a bit more than regular gravity. We would say that he is experiencing 1.16g (read as 1.16
gees ) of gravity. We figure this out by taking the acceleration due to gravity we just calculated (11.4m/s
2
) and divide it by regular gravity (9.81m/s
2
). You might have heard fighter pilots in movies saying that they were “pulling five gees”… that just means they were experiencing five times regular gravity. The person would feel five times heavier than normal. b) F c
 mv r
2

( 100 kg
8
)( 15 .
0 m
.
80 m
/ s )
2
= 2.56 x 10
3
N
Wow! That‟s a pretty big increase in weight for just a small increase in velocity. Keep in mind that velocity is squared in the formula, so even a small increase can make a big difference. How many gees is the person feeling now?
7

F g
= mg

g = F g
/ m = (2.56 x 10
3
N) / (100kg) = 25.6 m/s
2
That would be 2.61g … you would definitely start to feel uncomfortable by this point. This is actually close to the acceleration that astronauts feel going up in the space shuttle. They spend a good part of liftoff at about three gees.
Second Formula
We now have a centripetal force formula to use when we know the velocity of the object.
But what if we didn‟t have the velocity already? Let‟s look at combining these two formulas…
F c
 mv r
2
and v

2
 r
T
F
F c c


2
2 r m

T r
4
 2 mr
T
2
Example : What is the centripetal force acting on a 100kg man in a 8.80m radius centrifuge if he is going at 15 rpm?
 “rpm” stands for “revolutions per minute, so we will be able to figure out a period from that. o If he completes 15 revolutions every minute, that means he makes 15 revolutions every 60 seconds. o To find out how many seconds it takes for one revolution (the period) we take 15 rev / 60s and inverse it!
F c

4
 2 mr

T 2
T = 60s / 15 rev = 4.0s per revolution
4
 2
( 100 kg )( 8 .
80 m )
= 2.2 x 10
3
N
( 4 .
0 s ) 2
Be prepared to solve these two formulas for other variables.

Also, be careful about the squares in the formulas. o You only square the number being squared, not the whole formula. o Also, a lot of people end up with the wrong answer because they‟ll forget to do something like square

in the middle of the formula when they type everything into a calculator.
8

Vertical Circles
All of the formulas and questions that we have been working with up to this point have depended on one characteristic… that the object is moving in a horizontal circle.

By horizontal circle, I mean that the object moves without any vertical motion at all. o Imagine tying a string to a weight. You rest it on the floor, and then start to spin it around in a circle so that the weight is dragging along on the floor. That‟s an example of a horizontal circle.

When something is spinning in a vertical circle, we need to take into account that gravity will have different effects on the object at different times.
Tie something to the end of a string and spin it in a vertical circle.

If you spin it too slowly, then the tension in the string disappears when it is at the top of its swing. The weight might even start to fall.

This never happens at the bottom of the swing.

There must be some sort of relationship between the tension in the string (a force) and the centripetal force.
 Let‟s look at an example problem to see if we can work it out.
Expect to read through this example a few times before it makes sense.
Do not say “Oh yeah, I get it,” and just skim through it!
You will be lost latter on in these questions if you don’t understand the explanation thoroughly!
Example : A ball is spinning in a vertical circle at the end of a string that is 2.0m long. If the ball has a mass of 3.5kg and moves at a constant speed of 8.0m/s… a) what is the tension in the string when the stopper is at the bottom of the circle? b) what is the tension in the string when the stopper is at the top of the circle? a)
First, let‟s define that down is the negative direction. This will mean that g = 9.81m/s
2
.
Next, we will need to focus on the different forces acting on the string when the ball is at the bottom of the circle. We might have to calculate some other things to find all the forces acting on the string.

Since the ball has mass, there must be a force due to gravity ( F g
) acting on it downwards. This will cause some pull on the string down . Even without spinning the
F g
 ball, this force would be there.
F g
= mg = (3.5kg)(-9.81m/s
2
) = -34N

The force is negative because the string is being pulled down by the weight.
9

F c

Since the ball is moving in a circle, there must be a centripetal force acting on the ball to keep it moving in a circle.

That force must be pointing upwards (centripetal = to the centre) Let‟s calculate it…

F c
 mv
2

( 3 .
5 kg )(
2 .
8
0
.
0 m m / s )
2
= 112N = 1.1 x 10 r

Remember, this is the centripetal force acting on the ball! By
2
N
Newton‟s 3 rd
Law, there is an equal and opposite force acting on the string.

That means there is a force acting on the string of -1.1 x 10
2
N. It is negative because it is opposite to the centripetal force on the ball.
So the total net force acting on the ball is…
F
NET
= -34N + -1.1 x 10
2
N = -146N = - 1.5 x 10
2
N
This force is bigger than the weight of the ball alone, so maybe the string will snap.
b) What happens at the top of the circle? The same sort of thing, except now the centripetal force points in the opposite direction.
 At the top of the circle, gravity is still pointing down… it has to! It‟s gravity.

So we would still have the same value of -34 N for the force due to gravity.

The centripetal force acting on the ball still points towards the centre of the circle (like it always does).

It will still have the same magnitude of 1.1 x 10
2
N, but now it is negative because it is pointing down. So the centripetal force acting on the ball is - 1.1 x 10
2
N.
 Again, by Newton‟s 3 rd
Law, this means there is an equal but opposite force acting on the string of + 1.1 x 10
2
N.
Now the net force acting on the string will be…
F
NET
= -34N + (+ 1.1 x 10
2
N) = + 78N
The force is less than we calculated in part (a), so if the string is strong enough to swing the ball through the bottom of the swing, it should also be strong enough at the top of the swing. This is why the string doesn‟t feel as taut ( tight ) at the top of the swing. There isn‟t as much tension in it.
Example : Using the numbers from above, what is the minimum speed at which you could swing the ball so that the string won‟t go slack ( limp ) at the top of the swing?

The minimum speed would happen when the centripetal force and the force due to gravity are exactly equal at the top of the swing.
10
F c

o When this happens, the net force on the ball would be zero!

Since F g
in the question was 34 N, then F c
= 34 N
F c
 mv
2 r
 v

F c r m

( 34 N )( 2 .
0 m )
3 .
5 kg
= 4.4 m/s
If I told you the string can withstand up to 400 N of force, how would you calculate the maximum speed the ball could move at before the string would snap?
Pendulums
Pendulums are a great example of SHM because they are simple to make and follow a pretty well established pattern.

It is true that if you hold a weight on the end of a string and let it swing like a pendulum it will eventually slow to a stop (because of friction). Still, by making pendulums with big masses, they can actually swing for a long time.

SHM motion needs a force that causes an object to return to its equilibrium point. In the case of a pendulum this force is supplied by gravity. o When the bob is pulled away from the vertical, it has a large force trying to move it back. As it gets closer to its equilibrium this force decreases. o This follows the model we already established for SHM with springs.
One of the first people to realize how important pendulums are was Galileo.

According to Galileo, he was visiting a local church one day when he happened to look at a person lighting the lamps that were hanging in the church. o These lanterns were hanging from the roof by long chains. The lanterns themselves were quite heavy.

The person lighting the lanterns was using a very long pole with a taper on the end to light them… each time he reached up with the pole, he would nudge the lantern a bit which started it swinging.

Galileo noticed that although all the lanterns were the same mass, the length of the chain they were dangling from seemed to change how fast they were swinging.

In order to get some sort of measurement of the time it took them to swing, he actually used his own pulse to time them!
Although we have come a long way from the work that Galileo started, we still have a pretty simple equation that we can use to figure out how long it takes for a pendulum to swing…
T

2
  g
T = period of the swing (s)
11

ℓ = length of wire (m)
2 g = gravity (m/s )
Notice that the mass of the bob (the weight at the end of a pendulum) is not in the formula.
 This agrees with Galileo‟s observation that different masses on the end of the pendulum will not affect the period of the swing.

The mass does affect how many times the pendulum swings. o Think of it terms of inertia… a heavy bob has lots of inertia, so the pendulum swings back and forth for a long time before it stops. A lighter bob has little inertia, so it only swings for a little while before coming to rest at its equilibrium position.

What really matters in the case of the pendulum is the length of the pendulum.
If you‟ve ever looked carefully at the pendulum on a Grandfather clock, you might have noticed that the bob can be slid up and down a little. This is to adjust the period of the pendulum so that the clock runs at the right speed.
Be careful when you are using this formula. Remember three things:
1.
The period is the time it takes to complete one full swing… that means if you let go of the bob (the weight on the end), it will swing away from your hand and back to your hand. That‟s one complete swing. Back to where it started and ready to
2.
move in the original direction again.
This formula only works well if the pendulum is held at an angle of less than 15
 from rest. As the angle gets further past 15

errors start to creep in. (You‟ll read why in the text book)
3.
Be careful if you need to solve it for “ℓ” or “g”. You‟ll see in the following examples how to do this.
Example : What is the period of a pendulum that is 12.5 m long?
I will assume that it is in regular earth gravity…
T

2
  g

2
 12 .
5 m
9 .
81 m / s
2
= 7.09 s
I used 3.14 for pi , which gave me three sig digs, and the “2” is an exact digit, so it has an infinite number of sig digs.
12

Example : We decide to measure gravity in a particular location on Earth. I use a 2.75m long pendulum and find that it has a period of 3.33 s. What is the acceleration due to gravity in this area. g

4
 2 

Do you get this same formula? Double check!
T 2 g

4
 2
( 2 .
75 m )
= 9.78m/s
2
( 3 .
33 s )
2
Example : What length of pendulum would give me a period of one minute?
We can already make a pretty good guess that this is going to be a pretty long pendulum to take that long to swing.
  gT
2
4
 2

( 9 .
81 m / s
2
)( 60 s )
2
4
 2
= 895m
13

Name_______________________________
Regents Physics
Date_________
Mr. Morgante
Circular Motion, Gravitation Notesheet
Basic Data:
Circumference:
Units:
_________________________________ ____________
Period:
Frequency:
Uniform circular
Motion:
Instantaneous
Velocity:
_________________________________ ____________
_________________________________ ____________
_________________________________ ____________
_________________________________ ____________
Centripetal
Acceleration:
Centripetal
Force:
Algebra:
If a c
= v
2
/r,
If F c
= m a c
,
_______________________________
_______________________________
____________
____________ v = ______________ , r = _________________ v = ______________ , r = _____________, m = _________
If F c
= mv
2
/r , and r is doubled, F = ______________________________________
14

1. Solve, and label the diagram
A 1.1 kg ball on a string is rotated clockwise in a horizontal circle of radius 0.75 m..
If the ball completes a rotation every 0.5 seconds,
Calculate the circumference of the circle:________________
Calculate the average speed of the ball: ________________
Calculate the centripetal acceleration on the ball:______________
Calculate the centripetal force on the ball:____________________
At point A on the circle draw the instantaneous velocity vector,
The acceleration vector, the force vector.
2. Graph practice:
a c
r (v=const.)
Fc
v (r=const.) a c
v (r=const.)
Fc
v (m&r const.) a c
A
Fc r (a c
=const.)
r (v&m const.)
15

Regents Physics
Name_______________________________
Gravitation Notesheet
1.Newton‟s Universal Law of Gravitation:
Date_________
Mr. Morgante
2. Regents Reference Table Resource Review
List of Physical Constants
Name
Universal
Symbol
Gravitational Constant
Acceleration due to gravity on Earth‟s surface
Mass of Earth
Mean radius of Earth
Mean distance-
Earth to the Sun
Mechanics
Equation
F g
= Gm
1 m
2
r
2
Variables/constants g = F g m
Label the sketch below using F g
, G , m
1
, m
2
, r
Units
Value
Can be used to find…
Why are there two F g vector arrows ?
16

Compare: little g = _______ and BIG G = ___________
Are they the same values? ______ Do they have the same units? ______
If you use BIG G instead of little g in an equation, will you be incorrect?_____
3. Practice problems: a. Using the given constants, calculate your weight on Earth using both methods:
G = 6.67 x 10
-11
N kg
2
/m
2
Radius earth
= 6.4 x 10
6
m
1) F g
= (GMm)/r
2
M earth
= 6 x 10
24
kg m you
2) F g
=______________
= mg
_____________ _____________
Comment on the results:______________________________________________ b. Calculate the gravitational force between the Earth and the Sun:
Sketch
1) This gravitational force can also be described as _________ Force
4.
Set (Gm
1 m
2
)/r
2
= m
2 g. Solve for g: substitute values for Earth Algebraically g = ___________________
The above result is useful because: g = ___________________
5.
If masses remain constant, sketch a rough graph of F g
(ord) vs. r (abs) below.
What type of relation is shown? ________________
Using F g
= (Gm
1 m
2
)/r
2
; what happens to F g
if:
1) r is doubled 2) mass1 is tripled 3) both masses doubled; r is tripled
17

Name_______________________________
Regents Physics
Date_______
Mr. Morgante
Gravitational Field Strength + Weight Notesheet
Define:
1. Gravitational field:
A
2. Gravitational field strength:_______________________________________________
3. Test mass:
Diagram sketch:
4.Sketch the direction of gravitational force vectors above the Earth at the various points:
K
M
E
G
B
F
D
S
L
C
P
H a) Which point would have the greatest gravitational force due to the Earth? ____ b) Which point would have the least gravitational force due to the Earth? _______ c) What happens to the concentration of gravitational force vector lines as you approach the surface of the Earth?_________________________ d) The gravitational field strength _________ as the distance from Earth _______
5.
Calculate and sketch the value of “g” (gravitational field strength) versus Earth radii: g
----------+----------+----------+----------+----------
1R
E
2 R
E
3 R
E
4 R
E
5 R
E
18

Weight
1. Explain the differences between mass and weight using the following terms: vector, scalar, kilogram, Newton, matter inya, force of gravity onya
2. Describe the slope of a Weight versus mass graph:
F g
Earth slope = _____________________ (units)
Eetnorb
Is the value of „g‟ on Eetnorb greater or less than Earth?______________
Mass a) On object falls from rest and drops 6.75 m to the planet surface in 3 seconds.
What is the acceleration of gravity at the surface of this planet? b)What is special about g = 9.8 m/s
2 a)__________
? Is this value of g the same value on all planets?
3. Apparent weight : Use the spring scale and mass to simulate an elevator and passenger.
Mass value: _____g = ______kg Weight of mass at rest: ______
Case 1
Elevator accelerates quickly
Upward
Case 2
Elevator travels upwards Elevator accelerates or downwards with constant velocity
Case 3 quickly Downward
What happens to the apparent weight (spring scale reading) in each case?
19

NAME________________________________
Regents Physics
Gravitation Exercises
Show all work. Sketch all problems!
1.Calculate the gravitational force between the Earth and the Moon:
Date________
Mr. Morgante
2. If F g
= (Gm
1 m
2
)/r
2
, sketch the graphs of the following:
1._____________ a) F g
vs. m
1
; m
2
& r const. b) F g
vs. m
2
; m
1
& r const. c) F g
vs. r, m‟s const.
3. Draw the force of gravity (weight) vectors on the objects.
Note: Earth unless otherwise note; bottom of page is towards surface of Earth a) Object on inclined plane: b) Object hanging on wire c) Object in free-fall d) Object in a parabolic trajectory e) Object on table f) Helium balloon on ceiling
4. What is your mass on the Moon?_________ Earth?________ Jupiter?_______
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5.American astronauts have just landed on a newly-discovered planet.
The astronauts experiment and find that an object falling from rest free-falls
10 meters in 2 seconds. a) Calculate the acceleration of gravity (g) for this planet:
Sketch Knowns Equation(s) a)_______________ b) The astronauts know that the diameter of the planet is 2.4 X 10
7 meters.
What is the mass of the planet?
Sketch Knowns Equation(s) b)________________ c) What is the weight of a 90 kg astronaut on this planet?
Sketch Knowns Equation(s) c)________________
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Name:______________________ Date:___________
Regents Physics Mr. Morgante
Show all work including FBD‟s
1) Elevator Problem: a
DOORS OPEN CASE:
The doors of an elevator are open so you can see the direction of the person as they go by as shown below.
ELEVATOR
P = push of the floor on the student m= mass of the person a=acceleration of the elevator
P-mg=ma
OR
P=m(g+a)
Note: If you see an increase in the persons weight then P>mg
SCALE
P
mg
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a
DOORS CLOSED CASE:
The doors of an elevator are closed so you cannot see the direction of the person as they go by as shown below.
ELEVATOR
P = push of the floor on the student m= mass of the person a=acceleration of the elevator
P-mg=ma
OR
Note:
P=m(g+a)
If you see an increase in the persons
weight then P>mg
SCALE
`
ma P
mg
If the student does not know the elevator is accelerating he/she considers themselves to be in equilibrium under the action of the two forces P and a “gravitational force” the equation of which is –m(g+a).
He/she can say that either “someone accelerated the elevator upward” or someone turned on an extra downward gravitational force.
So what happens when a is going down? You sketch the diagram below.
1.
What is the force reading on the scale for a student whose mass is 60 kg and the elevator is accelerating upward at 4 m/s?
2.
What is the force reading on the scale for a student whose mass is 5000 dg and the elevator is accelerating downward at 5 m/s?
3.
Million dollar question…..what will the scale read if the elevator moves up or down with constant velocity and the student has a mass of 100 kg?
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Name:___________________________
Regents Physics
Date:_______
Mr. Morgante
Gravitation & Fc Worksheet – Show All Work
1) Two equal masses of 3.8 x 10
8
kg are 6.0 x 10
2
m apart. A) What is the force that attracts them? B) If the distance is made twice as great, what will be the new force of attraction?
2) A) How much does an 80 kg man weigh on the surface of the earth? B) How much would the same man weigh when he is 1 x 10
7
m from the center of the earth.
3) A car with a mass of 1000 kg is moving around a curve at 10 m/s. The radius of the curve is 20 m. A) How far will the car travel in 15 seconds? B) What frictional force is needed for him to go around the curve without slipping off the road? C) How much does the car weigh?
4) The blades of an electric fan rotate at 120 revolutions per minute (rpm). The weight of the blades is centered at 0.3 m from the center of rotation and their mass is 0.3 kg. A) What is the period of revolution of the blades? B) What is the velocity of the blades? C) What is the centripetal force acting on the blades?
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Name:_____________
Regents Physics
Date:_________
Mr. Morgante
Gravitation Free Response
Essay Questions – Show all of your work and draw a FBD with all vectors and associated information or else POINTS WILL BE DEDUCTED!!!!!!!!!
1.
What is the force of attraction between the earth and the moon? (Hint: use the reference table).
2.
Calculate the force of attraction between a 50 kg mass and a 3000 kg mass that are 200 cm apart.
3.
An object weighs 1 N on the surface of the earth. How much will it weigh at a distance of one earth radius above the earth? How much will it weigh at two earth radii?
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Name:___________________________
Regents Physics
Uniform Circular Motion Worksheet
Date:_______
Mr. Morgante
61.What is the magnitude of the centripetal force acting on the object?
( 1 ) 2.5N
( 2 ) 10.N
( 3 ) 25N
( 4 ) 50N
62. While the object is undergoing uniform circular motion, its acceleration
( 1 ) Has a magnitude of 0
( 2 ) Increases in magnitude
( 3 ) Is directed towards the center of the circle
( 4 ) Is directed away from the center of the circle
63. If the string is cut when the object is at the position shown, the path the object will travel from this position will be
( 1 ) Toward the center of the circle
( 2 ) A curve away from the circle
( 3 ) A straight line tangent to the circle
64. If the magnitude of the force applied to the string by the student's hand is increased, the magnitude of the acceleration of the ball in its circular path will
( 1 ) decrease
( 2 ) increase
( 3 ) remain the same
Base your answers to questions 65-67 on the information and diagram below.
A 4.0-kilogram model airplane travels in a horizontal circular path of radius 12 meters at a constant speed of 6.0 meters per second.
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65. What is the direction of the centripetal force on the airplane?
( 1 ) north
( 2 ) south
( 3 ) east
( 4 ) west
66. What is the magnitude of the centripetal acceleration of the airplane?
( 1 ) 0.50 m/s
2
( 2 ) 2.0 m/s
( 3 ) 3.0 m/s
2
2
( 4 ) 12 m/s
2
67. If the speed of the airplane is doubled and the radius of the path remains unchanged, the magnitude of the centripetal force acting on the airplane will be
( 1 ) half as much
( 2 ) twice as much
( 3 ) one-fourth as much
( 4 ) four times as much
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Name:___________________ Date:________
Regents Physics
Score:_________
Mr. Morgante
Fc Fc Fc Fc Fc Fc Fc Fc Fc Fc…………………….
1. Use the pendulum diagram below to answer the questions that follow:
Pt. A
Length of pendulum string = 2 m
Mass of pendulum ball = 2 kg
Pt.C
Pt. B
The pendulum ball is raised from Pt. C to Pt. A and then released. It takes 2 seconds for the ball to return to Pt. C. a.
What is the a c
of the pendulum ball when it returns to Pt. C (4 pts)? b.
What is the Fc of the ball at Pt. C (2 pts)?
2. You are an automotive engineer, blah blah blah. You are trying not to be responsible for a racecar driver‟s death so you decide to take a trial run around the track to test your car design. It is a circular asphalt track. There is a slight drizzle on the morning of your test run. The track diameter is 200 meters, and it takes you 20 seconds to complete a lap around the track. The car has a mass of 2,000 kg. a.
What is the a c
of the car as it rounds the track? (3 pts) b.
What is the Fc of the car as it round the track?(3 pts) c.
Does the car stay on the track? SHOW ALL OF YOUR WORK to prove your answer (4 pts).
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d.
If the embankment of the track was 45° and the car was stationery (a.k.a. not moving) and the track is dry, what is the “actual” F f
between the tires and the road? (3 pts)
4. Use the diagram below to answer the questions that follow:
Pt. B Sad drawing of car on track.
*
*
Pt. A
There is a car on a vertical track that travels from Pt. A to Pt. B as shown above. The following information applies to the problem:
Diameter of track = 3 m
Velocity around track = 10 m/s
Minimum F net
required to stay on track is 266 N.
Mass of car = 4 kg a.
Sketch all of the vector components on the above diagram that are associated with the problem.(4 pts) b.
Show all calculations below to prove that the car stays on or falls off the track. (6 pts)
Z:\Physics\Regents Physics\Class Material\Unit 2F GRAVITATION & Fc 1-11-10.doc
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