(3)+Newton's+Laws,+Circular+Motion,+&+Gravitation

PAGE EDITOR(S): Isabella Norcini, Daniela Reyes-Capo, Stephanie Rice, Carla Pacini, Jason Schwartz


 * Due Date: 11/02/09 (Monday)**

**Natland Note (11/11/09):** Be sure to check the requirements or this assignment on the homepage of this wiki. There should be more pictures/links with related pictures. Additionally, there is still a caption below that has not filled in and there are still broken picture links.

**Natland Note (11/03/09):** Nice job guys. This is really looking good! Remember to write a caption for the formula 1 race car below. Also, make sure you have completed everything (e.g. right number of links, pictures, etc. - excluding the few I posted). Also, ALL MEMBERS of the group need to post something. Part of the grade for the wiki is participation.

Chapter 4: Forces

à Something __capable__ of changing an object’s state of motion, that is changing its velocity
 * Force**: a push or pull on an object due to the object’s interaction w/ __another__ object

When the interaction ceases the objects will no longer experience a force


 * __Newton’s Laws of Motion__**

-An object(s) will stay in its current state of motion (at rest or moving at a constant velocity) unless an external net force is applied to the system/object(s) -If no net force acts on a body (Fnet = 0), the body’s velocity cannot change (it cannot accelerate) (Net force is the same thing as a resultant force) -No net force: 1) No change in magnitude of the velocity (speed) 2)No change in the direction of the velocity vector __In other words:__ **I. Every object in a state of uniform motion tends to remain in that state of motion unless an external force is applied to it.**
 * 1st Law**:

-Quantitative measure of mass -The more quantitative is the more inertia
 * Inertia**: natural tendency for an object to resist change in its motion

-When an external net force (Fnet = ∑F) acts on a system of mass “m”, the acceleration, is directly proportional to the net force ∑ F = ma A =( ∑ F )/ m
 * 2nd Law**

-a is in the same direction as Fnet -Units: kg x m/s2 à N (Newton) __In other words:__ **II. The relationship between an object's mass //m//, its acceleration** a**, and the applied force** //F// **is** //F **= m**a//**. Acceleration and force are vectors (as indicated by their symbols being displayed in slant bold font); in this law the direction of the force vector is the same as the direction of the acceleration vector.**

-Shows the relative magnitude and direction (including the angle of all forces acting on a system in a given situation) -Vector diagram where all forces acting upon a system are drawn “tail-to-tail” -Only show forces applied __to__ the system __not__ by the system
 * Free-Body Diagram** **(FBD)**

1) Identify “the system” and the “environment” -__System__: object(s) whose motion you want to study -__Environment__: everything else (including ropes, springs, ground, etc.) 2) Draw a picture of the situation -List knowns and unknowns -Ask yourself where and how does the environment interact with your chosen system? -Draw F.B.D. (all forces acting on system à MAR) 1. __Contact Forces__: two objects have to be in contact to be considered interacting 2. __Long Range Forces:__ do not require contact (ex: magnet)
 * Identifying Forces**

- **External Force**: a force acting on your chosen system from the environment - **Internal Force**: a force between bodies/objects within your chosen system
 * Types of Forces**

1) __Forces of Gravity (Fg)__: gravitational attraction between to objects w/ mass a. //Equation:// (Gm1m2)/r2 i. **m1m2**- Two masses being attracted to each other (kg) ii. **r**-the radial distance between the centers of the two objects (if they are symmetric) (m) iii. **G**- Universal gravitational constant (6.67 x 10-11 (Nm2/kg2) kg2/ m2) iv. **Ag = GMp/ (Rp +h)2** [h= distance above planet] b. Finding out generally what is the force of gravity for an object near the surface of the earth i. Fg= GMEmobj./ RE2 à Fg = [6.67 x 10-11 (Nm2/kg2) kg2/ m2][5.08 x 1024 kg] [mobj]/ [6.38 x 106m)2 à Fg= mobj (9.8 m/s2) ii. Fg = W (weight) = mg iii. ∑F = ma à Fg = ma à a = Fg/m [[image:GMM.jpg]] 2) __Tension (T or Ft):__ When a cord (or a rope, cable, etc) is attached to a body and is pulled so that it is taut, the cord pulls on the body with a force “T” directed away from the body, along the cord a. __Assumptions:__ i. Cord has negligible mass compared to the body’s mass ii. Assume that the cord is un-strechable (rigid cord) iii. Cord exists __only__ as a connection b/w two bodies [[image:OC0811004_L6Pulley.gif]] 3) __Normal Force (Fn or n):__ When a body presses against a surface the surface (even a seemingly rigid one) deforms and pushes back on the body with a normal force (Fn) that is __perpendicular__ to the surface





4) __Friction (f or Ffr):__ A force directed along a surface, opposite the direction of the impending/intended motion relative to that surface, due to bonding b/w the body and the surface

Mr. Natland recommends reading section on frictional forces on pg. 103!

-Pressure = Force / Area __What factors affect friction?__ - Coefficient of friction: Materials in contact with each other - Moving or not? (static and kinetic coefficients)

__Tarzan … on Ice__ Jane pushes off of Tarzan with a force of 100N to the right. Mass of Jane à 60 kg, Mass of Tarzan à 90 kg.

a) What is the magnitude of the force on Tarzan? Jane? - Force on Tarzan would be force that Jane is applying which is 100N. Tarzan would be exerting force of 100N on Jane also.

b) What are Tarzan and Jane’s respective accelerations? - Tarzan: aTarzan= FJane/mTarzan - Jane: aJane = -FTarzan/ mJane

c) If Jane is pushing off for .7s, what are their respective velocities after this time interval? VT = V0 + att, V0 = 0 VJ = V0 + aJt, V0 = 0 VT = 0.78 m/s VJ = -1.167

d) After .7 seconds the velocity stays constant because there is only an acceleration when a force is being applied.

-The tension in a string is the same throughout the string -When you have action/reaction forces within the system they will cancel each other out and cannot cause the system to accelerate, called **internal forces** -Constant velocity = No net force

-All surfaces are rough (even if they appear to be smooth) -Coldwells form between 2 surfaces Ffr = µ Fn -Always in the opposite direction of the movement Examples: 1) Paper falls through the air: 2) Block hanging from ropes: 3) Mass hanging off-center: 4) Hanger, band, and mass: (person squeezes the hanger b/w two fingers) 5) Eraser against the wall: 6) Hold eraser against frictionless surface: 7) Push chair to the right but it doesn't move: 8) Pull the chair to the right across the floor: 9) A book sits on an inclined table: 10) Book slides down an incline: 11) Book slides across a table with a box on top: 12) Box rests on top of the book as it moves across a table: 13) Box rests on a book during initial acceleration: 14) Book with box on top during acceleration phase:
 * Friction:**
 * -Kinetic Friction:** an object slides relative to another
 * -Static Friction:** the two objects are non-moving
 * Steps for Solving Force Problems:**


 * 1. Draw picture of situation **
 * 2. Draw FBD of system **
 * 3. Break any force at an angle to your coordinate system into perpendicular components **
 * 4. Apply component form of Newton's 2nd Law **
 * 5. Use kinematics, if necessary. **

Newton's 3rd Law:

If object 1 exerts a force on object 2, then object 2 will exert a force back onto object 1 that is equal in magnitude and opposite in direction. -These action/reaction pairs are applied to different objects -They must be the same type

__In other words:__ **III. For every action there is an equal and opposite reaction.**



Examples:

You push down with your hand on a desk, and the desk pushes upward with a force equal in magnitude to your push. A brick is in free fall. The brick pulls the Earth upward with the same force that the Earth pulls the brick downward. When you walk, your feet push the Earth backward. In response, the Earth pushes your feet forward, which is the force that moves you on your way.

Example Problems with Solutions:


 * EXAMPLE 1**

The figure above gives us a free-body diagram that shows us the direction in which all forces are acting, but we should be careful to note that vectors in the diagram are not drawn to scale: we cannot estimate the magnitude of //**C**// simply by comparing it to //**M**// and //**L**// .**What is the magnitude of the force Curly is exerting?**Since we know that the motion of the sled is in the direction, the net force, //**M**// + //**L**// +//**C**//, must also be in the  direction. And since the sled is not moving in the direction, the //y//-component of the net force must be zero. Because the //y//-component of Larry’s force is zero, this implies: where is the //y//-component of //**M**// and  is the //y//-component of //**C**//. We also know: If we substitute these two equations for and  into the equation, we have: Now that we have calculated the magnitude of the net force acting on the sled, a simple calculation can give us the sled’s acceleration: We have been told that the sled is moving in the direction, so the acceleration is also in the  direction.This example problem illustrates the importance of vector components. For the SAT II, you will need to break vectors into components on any problem that deals with vectors that are not all parallel or perpendicular. As with this example, however, the SAT II will always provide you with the necessary trigonometric values.**EXAMPLE 2**
 * |||| The Three Stooges are dragging a 10 kg sled across a frozen lake. Moe pulls with force //**M**//, Larry pulls with force //**L**// , and Curly pulls with force <span class="question_inline" style="font-family: 'Times New Roman',Times,serif; font-size: 12px; letter-spacing: 0.05em;">//**C**// . If the sled is moving in the [[image:http://img.sparknotes.com/content/testprep/bookimgs/sat2/physics/0001/phy.total.SMALL114.gif align="absMiddle"]] direction, and both Moe and Larry are exerting a force of 10 N, what is the magnitude of the force Curly is exerting? Assuming that friction is negligible, what is the acceleration of the sled? (Note: sin 30 = cos 60 = 0.500 and sin 60 = cos 30 = 0.866.) ||  ||
 * |||| The Three Stooges are dragging a 10 kg sled across a frozen lake. Moe pulls with force <span class="question_inline" style="font-family: 'Times New Roman',Times,serif; font-size: 12px; letter-spacing: 0.05em;">//**M**//, Larry pulls with force <span class="question_inline" style="font-family: 'Times New Roman',Times,serif; font-size: 12px; letter-spacing: 0.05em;">//**L**// , and Curly pulls with force <span class="question_inline" style="font-family: 'Times New Roman',Times,serif; font-size: 12px; letter-spacing: 0.05em;">//**C**// . If the sled is moving in the [[image:http://img.sparknotes.com/content/testprep/bookimgs/sat2/physics/0001/phy.total.SMALL114.gif align="absMiddle"]] direction, and both Moe and Larry are exerting a force of 10 N, what is the magnitude of the force Curly is exerting? Assuming that friction is negligible, what is the acceleration of the sled? (Note: sin 30 = cos 60 = 0.500 and sin 60 = cos 30 = 0.866.) ||  ||
 * What is the acceleration of the sled?**According to Newton’s Second Law, the acceleration of the sled is <span class="question_inline" style="font-family: 'Times New Roman',Times,serif; font-size: 12px; letter-spacing: 0.05em;">//**a**// = <span class="question_inline" style="font-family: 'Times New Roman',Times,serif; font-size: 12px; letter-spacing: 0.05em;">//**F**/// m. We know the sled has a mass of <span class="question_inline" style="font-family: 'Times New Roman',Times,serif; font-size: 12px; letter-spacing: 0.05em;">10 kg, so we just need to calculate the magnitude of the net force in the [[image:http://img.sparknotes.com/content/testprep/bookimgs/sat2/physics/0013/phy.total123.gif align="absMiddle"]]-direction.
 * |||| Each of the following free-body diagrams shows the instantaneous forces, <span class="question_inline" style="font-family: 'Times New Roman',Times,serif; font-size: 12px; letter-spacing: 0.05em;">//**F**//, acting on a particle and the particle’s instantaneous velocity, <span class="question_inline" style="font-family: 'Times New Roman',Times,serif; font-size: 12px; letter-spacing: 0.05em;">//**v**// . All forces represented in the diagrams are of the same magnitude. ||
 * |||| [[image:http://img.sparknotes.com/content/testprep/bookimgs/sat2/physics/0007/freebodies.gif align="center"]] ||
 * |||| [[image:http://img.sparknotes.com/content/testprep/bookimgs/sat2/physics/0007/freebodies.gif align="center"]] ||


 * 1. |||| In which diagram is neither the speed nor the direction of the particle being changed? ||
 * 1. |||| In which diagram is neither the speed nor the direction of the particle being changed? ||


 * 2. |||| In which diagram is the speed but not the direction of the particle being changed? ||
 * 2. |||| In which diagram is the speed but not the direction of the particle being changed? ||


 * 3. |||| In which diagram is the direction but not the speed of the particle being changed? ||
 * 3. |||| In which diagram is the direction but not the speed of the particle being changed? ||

<span style="border-collapse: separate; color: #333333; font-family: georgia,times,fantasy; font-size: 14px; line-height: normal; webkit-border-horizontal-spacing: 0px; webkit-border-vertical-spacing: 0px;">The answer to question 1 is **B**. The two forces in that diagram cancel each other out, so the net force on the particle is zero. The velocity of a particle only changes under the influence of a net force. The answer to question 2 is **C**. The net force is in the same direction as the particle’s motion, so the particle continues to accelerate in the same direction. The answer to question 3 is **A**. Because the force is acting perpendicular to the particle’s velocity, it does not affect the particle’s speed, but rather acts to pull the particle in a circular orbit. Note, however, that the speed of the particle only remains constant if the force acting on the particle remains perpendicular to it. As the direction of the particle changes, the direction of the force must also change to remain perpendicular to the velocity. This rule is the essence of circular motion, which we will examine in more detail later in this book. The answer to question 4 is **D**. The net force on the particle is in the opposite direction of the particle’s motion, so the particle slows down, stops, and then starts accelerating in the opposite direction.
 * 4. |||| In which diagram are both the speed and direction of the particle being changed? ||  ||
 * 4. |||| In which diagram are both the speed and direction of the particle being changed? ||  ||


 * Atwood's Machine**:

system: m system: M ∑F = ma ∑F = Ma T-mg = ma Mg-T = Ma

system: m and M ∑F = (m+M)a Mg-mg = (m+M)a


 * How do you determine the normal force?**


 * *MISCONCEPTION: the normal force is NOT always equal to the weight of an object. **


 * Six Basic Cases of the Normal Force: **



When the elevator moves down your inertia cause lighter. When the elevator moves upward your inertia cause you to push more into the floor and feel heavier. Friction

Static: -not moving/impending motion

not moving: **f****s****= F(applied)**

object just about to move: **f****s** **= µ** **s x** **F(normal)**

Imagine, once more, that you are pushing a box along a floor. When the box is at rest, it takes some effort to get it to start moving at all. That’s because the force of static friction is resisting your push and holding the box in place. In the diagram above, the weight and the normal force are represented as <span class="question_inline" style="font-family: 'Times New Roman',Times,serif; font-size: 12px; letter-spacing: 0.05em;">//**W**// and <span class="question_inline" style="font-family: 'Times New Roman',Times,serif; font-size: 12px; letter-spacing: 0.05em;">//**N**// respectively, and the force applied to the box is denoted by. The force of static friction is represented by, where. The net force on the box is zero, and so the box does not move. This is what happens when you are pushing on the box, but not hard enough to make it budge. <span style="color: #333333; font-family: georgia,times,fantasy; font-size: 14px; font-weight: normal; line-height: normal;">Static friction is only at work when the net force on an object is zero, and hence when. If there is a net force on the object, then that object will be in motion, and kinetic rather than static friction will oppose its motion.

Kinetic: -moving relative to other surface in contact


 * f** **k** **= µk** **x** **F(normal)**


 * µ** **s** greater than **µk** because harder to get something moving than to keep it moving

The force of static friction will only oppose a push up to a point. Once you exert a strong enough force, the box will begin to move. However, you still have to keep pushing with a strong, steady force to keep it moving along, and the box will quickly slide to a stop if you quit pushing. That’s because the force of kinetic friction is pushing in the opposite direction of the motion of the box, trying to bring it to rest.

<span style="color: #333333; font-family: georgia,times,fantasy; font-size: 14px; line-height: normal;">Though the force of kinetic friction will always act in the opposite direction of the force of the push, it need not be equal in magnitude to the force of the push. In the diagram above, the magnitude of is less than the magnitude of. That means that the box has a net force in the direction of the push, and the box accelerates forward. The box is moving at velocity <span class="question_inline" style="font-family: 'Times New Roman',Times,serif; font-size: 12px; letter-spacing: 0.05em;">//**v**// in the diagram, and will speed up if the same force is steadily applied to it. If were equal to, the net force acting on the box would be zero, and the box would move at a steady velocity of <span class="question_inline" style="font-family: 'Times New Roman',Times,serif; font-size: 12px; letter-spacing: 0.05em;">//**v**// , since Newton’s First Law tells us that an object in motion will remain in motion if there is no net force acting on it. If the magnitude of were less than the magnitude of, the net force would be acting against the motion, and the box would slow down until it came to a rest.

Coefficient of friction is normally greater than 0 but less than 1



The amount of force needed to overcome the force of static friction on an object, and the magnitude of the force of kinetic friction on an object, are both proportional to the normal force acting on the object in question. We can express this proportionality mathematically as follows: where is the **coefficient of kinetic friction**,  is the **coefficient of static friction**, and <span class="question_inline" style="font-family: 'Times New Roman',Times,serif; font-size: 12px; letter-spacing: 0.05em;">//N// is the magnitude of the normal force. The coefficients of kinetic and static friction are constants of proportionality that vary from object to object.Note that the equation for static friction is for the //maximum// value of the static friction. This is because the force of static friction is never greater than the force pushing on an object. If a box has a mass of <span class="question_inline" style="font-family: 'Times New Roman',Times,serif; font-size: 12px; letter-spacing: 0.05em;">10 kg and = <span class="question_inline" style="font-family: 'Times New Roman',Times,serif; font-size: 12px; letter-spacing: 0.05em;">0.5, then: If you push this box with a force less than <span class="question_inline" style="font-family: 'Times New Roman',Times,serif; font-size: 12px; letter-spacing: 0.05em;">49 newtons, the box will not move, and consequently the net force on the box must be zero. If an applied force is less than, then  = –.

click link to view picture. [] (Static Friction Car Picture) W: Weight N: Normal Force F: Force of Static Friction The force of static friction between the car and the table keeps the car from sliding down off of the table

click link to view picture. [] (Kinetic Friction Box Picture)

F k : Force of Kinetic Friction N: Normal Force F: Force applied W: Weight Once the force applied surpasses the force of static friction, the object begins to move to the right. However, the surface still exerts a force on the object in motion. This is the force of Kinetic Friction, symbolized by the symbol F k

__**Uniform Circular Motion**__ - objects always move at constant speed - for an object to move in a circe, there must be a net force (∑Fnet) pointing towards the center of the circle at all times - if there is a net force pointing towards center of circle, there is an acceleration in the same direction - known as CENTRIPETAL ACCELERATION - **ac = v2/r** - applying Newton's 2nd law to UCM 1. sum up forces directed towards the center of the circle ∑Fc = mv2/r
 * Question to ask yourself: What forces in your free body diagram is (are) causing the centripetal motion?



<span style="font-family: Verdana,Arial,Helvetica,sans-serif; font-size: 12px; line-height: 14px;"><span style="font-family: arial,helvetica,sans-serif; font-size: 13px; line-height: 19px;"> __**Question**__ <span style="font-family: Verdana,Arial,Helvetica,sans-serif; font-size: 12px; line-height: 14px;">1. A tube is been placed upon the table and shaped into a three-quarters circle. A golf ball is pushed into the tube at one end at high speed. The ball rolls through the tube and exits at the opposite end. Describe the path of the golf ball as it exits the tube.

<span style="font-family: Verdana,Arial,Helvetica,sans-serif; font-size: 12px; line-height: 14px;">__**Answer**__**__:__** The ball will move along a path which is tangent to the spiral at the point where it exits the tube. At that point, the ball will no longer curve or spiral, but rather travel in a straight line in the tangential direction.


 * [[image:http://www.physicsclassroom.com/Class/circles/u6l1c10.gif height="62" align="center"]] || [[image:http://www.physicsclassroom.com/Class/circles/u6l1c3.gif height="71" align="center"]] || [[image:http://www.physicsclassroom.com/Class/circles/u6l1c4.gif height="108" align="center"]] ||
 * As a car makes a turn, the force of friction acting upon the turned wheels of the car provides centripetal force required for circular motion. || As a bucket of water is tied to a string and spun in a circle, the tension force acting upon the bucket provides the centripetal force required for circular motion. || As the moon orbits the Earth, the force of gravity acting upon the moon provides the centripetal force required for circular motion. ||

__Free Body Diagram Examples__

1) rounding a flat curve in your car



2) dice in your car while rounding a flat curve

3) plane

4) satellite

5) gravitron

6) swinging stone in vertical circle

7) banked curve

8) artificial gravity



Physics can be seen in every day life, like the Olympics!!

__Other Important Equations__

V = 2πr/T //F//c = //mv//2///r//

//F//c = 4π2//mr/////T//2

__Sample Problems__

1) A 2 kg ball on a string is rotated about a circle of radius 10 m. The maximum tension allowed in the string is 50 N. What is the maximum speed of the ball?

Solution for Problem 1 >> The centripetal force in this case is provided entirely by the tension in the string. If the maximum value of the tension is 50 N, and the radius is set at 10 m we only need to plug these two values into the equation for centripetal force:

T = F c = implies that v = thus

v = = 15.8 m/s

2) A satellite is said to be in geosynchronous orbit if it rotates around the earth once every day. For the earth, all satellites in geosynchronous orbit must rotate at a distance of 4.23×107 meters from the earth's center. What is the magnitude of the acceleration felt by a geosynchronous satellite?

Solution for Problem 2 >> The acceleration felt by any object in uniform circular motion is given by a =. We are given the radius but must find the velocity of the satellite. We know that in one day, or 86400 seconds, the satellite travels around the earth once. Thus:

v = = = = 3076 m/s thus

a = = = .224 m/s2 Close

Orbits:

The centripetal force acting on the satellite is the gravitational force of the Earth. Equating the formulas for gravitational force and centripetal force we can solve for <span class="question_inline" style="font-family: 'Times New Roman',Times,serif; font-size: 12px; letter-spacing: 0.05em;">//v// :
 * How Do Orbits Work?**Imagine a baseball pitcher with a very strong arm. If he just tosses the ball lightly, it will fall to the ground right in front of him. If he pitches the ball at <span class="question_inline" style="font-family: 'Times New Roman',Times,serif; font-size: 12px; letter-spacing: 0.05em;">100 miles per hour in a line horizontal with the Earth, it will fly somewhere in the neighborhood of <span class="question_inline" style="font-family: 'Times New Roman',Times,serif; font-size: 12px; letter-spacing: 0.05em;">80 feet before it hits the ground. By the same token, if he were to pitch the ball at <span class="question_inline" style="font-family: 'Times New Roman',Times,serif; font-size: 12px; letter-spacing: 0.05em;">100,000 miles per hour in a line horizontal with the Earth, it will fly somewhere in the neighborhood of <span class="question_inline" style="font-family: 'Times New Roman',Times,serif; font-size: 12px; letter-spacing: 0.05em;">16 miles before it hits the ground. Now remember: the Earth is round, so if the ball flies far enough, the ball’s downward trajectory will simply follow the curvature of the Earth until it makes a full circle of the Earth and hits the pitcher in the back of the head. A satellite in orbit is an object in free fall moving at a high enough velocity that it falls around the Earth rather than back down to the Earth.
 * Gravitational Force and Velocity of an Orbiting Satellite**Let’s take the example of a satellite of mass [[image:http://img.sparknotes.com/content/testprep/bookimgs/sat2/physics/0021/phy.total481.gif align="absMiddle"]] orbiting the Earth with a velocity //v//. The satellite is a distance <span class="question_inline" style="font-family: 'Times New Roman',Times,serif; font-size: 12px; letter-spacing: 0.05em;">//R// from the center of the Earth, and the Earth has a mass of [[image:http://img.sparknotes.com/content/testprep/bookimgs/sat2/physics/0021/phy.total482.gif align="absMiddle"]].

<span style="color: #333333; font-family: georgia,times,fantasy; font-size: 14px; line-height: normal;">As you can see, for a planet of a given mass, each radius of orbit corresponds with a certain velocity. That is, any object orbiting at radius <span class="question_inline" style="font-family: 'Times New Roman',Times,serif; font-size: 12px; letter-spacing: 0.05em;">//R// must be orbiting with a velocity of. If the satellite’s speed is too slow, then the satellite will fall back down to Earth. If the satellite’s speed is too fast, then the satellite will fly out into space.

-The wing on the front of the car adds a greater force downward which, in turn, increases the normal force. Since the normal force is increased, the static frictional force increases, which causes the car to skid less and travel faster.
 * FUN STUFF**

<span style="font-family: Verdana,Arial,Helvetica,sans-serif; font-size: 12px; line-height: normal;"> - In a roller coaster, what keeps the car moving along the curved path is the centipetal force. The centripetal force prevents the car from exiting the curve by continuously making it change its direction toward the center of the circle. In this example, for the most part the centripetal force is the normal force. Gravity also pulls down on the car with a constant force, whether it moves uphill, downhill, or through the loop. At the very bottom of the loop, gravity and the centripetal force act in different directions. On the other hand, at the very top of the loop, both the normal force and the force of gravity act in the same direction to produce the centripetal force. No matter what, the centripetal force always points towards the center of the circle. - Note: If you are asked how fast the car may be going just before losing contact with the tracks, then the normal force between the car and tracks is zero.


 * __Videos & Interesting Links__**

[] ( <span style="font-family: arial,sans-serif; font-size: 15px; line-height: normal;">//Aboard NASA's// '//Vomit Comet//' - <span style="font-family: arial,sans-serif; line-height: normal;">RIT students experience zero gravity while conducting reduced-gravity scientific experiments //**aboard NASA's**// "//**Vomit Comet**//" aircraft; reported by Kelly Downs of RIT University News.)

[] (An accurate, not-so-useful comic discussing the centripetal vs. centrifugal force)

[] (An article about the 5 points of Lagrange, where gravitation forces balance so something placed at one of these points will remain there - useful for potentially building a space station?) A diagram showing the five Lagrangian points in a two-body system with one body far more massive than the other (e.g. the Earth and the Moon). In such a system [|L] [|3] –[|L] [|5] will appear to share the secondary's orbit, although in fact they are situated slightly outside it. An object placed in orbit around [|L5] (or [|L4]) will remain there indefinitely without having to expend fuel to keep its position, whereas an object placed at [|L1], [|L2] or [|L3] (all points of unstable equilibrium) may have to expend fuel if it drifts off the point.

http://science.howstuffworks.com/space-station.htm/printable (How Stuff Works article about space stations)
 * [[image:http://static.howstuffworks.com/gif/space-station-space-settlement.gif width="400" height="527" align="center" caption="This is an artist's conception of what the inside of a future space colony could look like. What do you think?"]] ||

__Apparent Weight__

code <object wdth="560" height="340"><param name="movie" value="http://www.ytube.com/v/25CmOdcKWN8&hl=en&fs=1&"> <param name="allowFullScreen" value="true"> <param name="allowscriptaccess" value="always"> <embed src="http://www.youtube.com/v/25CmOdcKWN8&hl=en&fs=1&" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" w code


 * Condensed Guide to :media type="youtube" key="kr-fJHA8fyQ" height="344" width="425"**
 * (3 1/2 minutes to understanding forces)**

http://www.youtube.com/watch?v=kr-fJHA8fyQ

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 10px; line-height: normal; white-space: pre;"><object width="560" height="340"><param name="movie" value="http://www.youtube.com/v/S0tMhlhnga0&hl=en_US&fs=1&"> <param name="allowFullScreen" value="true"> <param name="allowscriptaccess" value="always"> <embed src="http://www.youtube.com/v/S0tMhlhnga0&hl=en_US&fs=1&" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="560" height="340">

Way better Video than the one above... Centripetal forces and friction media type="youtube" key="S0tMhlhnga0" height="340" width="560"

This video shows how a banked curve allows cyclists to achieve greater speed without sliding off the course. However, once the friction between the tires and the track is overcome by the crash, the bikers slide towards the outside of the the track due to the lack of a significant centripetal force. You can tell that the curve is banked because all the bikers slide back into the middle after the crash.

http://www.youtube.com/watch?v=S0tMhlhnga0

__**SOURCES**__

__http://science.howstuffworks.com/space-station.htm/printable__ (artist's conception of space station)

http://en.wikipedia.org/wiki/L5_Society (picture showing the 5 points of Lagrange)

http://www.physicsclassroom.com/mmedia/circmot/rcd.cfm (animation of the forces varying during a roller coaster loop-de-loop)

[] (picture of formula 1 race car)

[] (friction picture)

http://library.thinkquest.org/2745/data/newton.htm (picture of Newton's third law)

http://www.staff.amu.edu.pl/~romangoc/graphics/M3/7-atwood-machine/M3-7-fig1.gif (Atwood's machine)

http://blog.dotphys.net/2009/06/the-physics-of-michael-jacksons-moonwalk/ (graph of friction) ***really cool article about the physics of MJ's moonwalk

<span style="font-family: Arial,sans-serif; font-size: 10px; line-height: normal; white-space: pre;">http://www.youtube.com/watch?v=kr-fJHA8fyQ <span style="font-family: Arial,sans-serif; font-size: 10px; line-height: normal; white-space: pre;">(Quick Guide to Forces Video)

http://www.youtube.com/watch?v=S0tMhlhnga0 (biking video about banked curves with centripedal accel.)

<span style="font-family: 'Calibri','sans-serif'; font-size: 11pt; line-height: 115%;">[] (Static Friction Car Picture)

[] (Kinetic Friction Box Picture)