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91 Cards in this Set

  • Front
  • Back
Dynamics (p. 67-80)
PART 1/8 (10%)
Dynamics
-is the study of the CAUSES of motion.
-covers the "how" and "why".
-FORCES are the lifeblood of dynamics: objects move and change their motion under the influence of different forces.
-Main Emphasis: Newton's Three Laws.
-The Sat II calls upon your knowledge of kinematics and vectors, but these question will probably be simpler than the problems you've encountered in your physics class.
-b/c you won't be asked to do any math that would require a calculator, you should focus on master the concepts that lie behind the math.
What are Forces?
-a force is defined very practically as a push or a pull- essentially it's what makes things move.
-a force is a vector quantity, as it has both a magnitude and a direction.
Newton's First Law (Law Of Inertia)
-an object at rest remains at rest, unless acted upon by a net force. An object in motion remains in motion, unless acted upon by a net force.
-The net force is the sum of all the forces acting on a body.
-Evil Roommate Problem:
F + -F = 0
-Good/Evil Roommate Problem:
F + -F + R = R w/ the velocity of the box, v, in the same direction as the net force.
Inertia
-the tendency of an object to remain at a constant velocity, or its resistance to being accelerated.
-fundamental property of all matter and is important to the definition of MASS.
Mass
-is an intrinsic scalar quantity: it has no direction and is a property of an object, not of the object's location.
-is a measurement of a body's inertia, or its resistance to being accelerated.
-the words "mass" and "matter" are related: a handy way of thinking about mass is as a measure of how much matter there is in an object, how much "stuff" it's made out of.
-"mass" and "weight" refer to two different, but related quantities in physics.
Newton's Second Law
F = MA
-the unit of force is defined, quite appropriately, as a Newton (N).
-b/c acceleration is given in units of m/s(2) and mass is given in units of kg, Newton's Second Law implies that 1N=1 kg. m/s(2).
-in other words, one newton is the force required to accelerate a one-kilogram body, by one meter per second, each second.
Newton's Third Law
-To Every Action, There is an Equal and Opposite Reaction.
-tells us that when object A exerts a force F on object B, object B will exert a force -F on object A.
-When you push a box forward, you also feel the box pushing back on your hand.
-If Newton's Third Law did not exist, your hand would feel nothing as it pushed on the box, b/x there would be no reaction force acting on it.
Weight
-the mass reflects its resistance to being accelerated.
-the weight of an object is a measure of the gravitational force being exerted upon it, and so it varies depending on teh gravitational force acting on the object.
-mass is a scalar quantity measured in kilograms, while weight is a vector quantity measuring force, and is represented in newtons.
-although an object's mass never changes, its weight depends on the force of gravity in the object's environment.
Friction
-which is at work in every medium but a vacuum, and is the bugbear of students pushing boxes across the sticky floors of dorm rooms everywhere.
-there are 2 main types of friction: Static Friction & Kinetic Friction.
Static Friction
-when the box is at rest, it takes some effort to get it to start moving at all. That's b/c the force of static friction is resisting your push and holding the box in place.
-Static friction is only at work when the net force on an object is zero, and hence when F(static) = -F(push).
-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 Friction
-Once you exert a strong enough force, 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 b/c the force of kinetic friction is pushing in the opposite direction of the motion of the box, trying to bring it to a rest.
-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.
-If F(push) were equal to -F(kinetic), the net force acting on teh box would be zero, and the box would move at a steady velocity of 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 F(push) were less than the magnitude of F(kinetic), the net force would be acting against the motion, and the box would slow down until it came to a rest.
3 Reminders about Frictional Forces:
1.) The smaller µ is, the more slipper the surface.
2.) The coefficient of Kinetic Friction is smaller than the coefficient of static friction.
3.) Frictional forces are directly proportional to the normal force.
Tension
-the force exerted on the box from the rope is called the TENSION force, and comes into play whenever a force is transmitted across a rope or a cable.
Kinematics (p. 45-62)
PART 2/8 (6%)
Displacement
-is a vector quantity, commonly denoted by the vector s, that reflects an object's change in spatial position.
-deals only with the separation b/w points A and B, and not with the path the object followed b/w points A and B.
-By contrast, the distance that the object travels is equal to the length of path AB.
Speed & Velocity
-Speed is a measure of teh distance an object travels in a given length of time:

average speed = distance traveled/time elapsed = Δx/Δt

-Velocity is a vector fquantity defined as rate of change of teh displacement vector over time:

average velocity = change in displacement/ time elapsed = Δs/Δt
Instantaneous Speed & Velocity
-When speaking of Instantaneous Velocity speed at a given moment) we don't want to know how many meters an object covered in the past ten seconds; we want to know how fast that object is moving RIGHT NOW.
-Ex: 5 seconds into the launch, the rocket was shooting upward at 5,000 meters per second.
Acceleration

average acceleration = change in velocity/time elapsed = Δv/Δt
-is a vector quantity that measure the rate of change of the velocity vector with time.

Acceleration = Δv/Δt = V(final) x V(initial)/T(final) x T(initial)
Velocity
v = Δy/Δt
-For any position vs. time graph, the velocity at time t is equal to the slope of the line at t.

Average Velocity = (change in displacement)/ (elapsed time)
Displacement
Displacement = velocity x time
-Graphically, this means that the DISPLACEMENT IN A GIVEN TIME INTERVAL IS EQUAL TO THE AREA UNDER THE GRAPH DURING THAT SAME TIME INTERVAL.
Reading Graphs: Rules of Thumb
1.) The slope on a given graph is equivalent to the quantity we get by dividing the y-axis by the x-axis.
2.) The area under a given graph is equivalent to the quantity we get by multiplying the x-axis and the y-axis.
Graphs
Graph-Slope-Area Under the Graph

1. Position Vs. Time-Velocity-(none)

2. Velocity vs. Time- Acceleration-Displacement

3. Acceleration vs. Time-(none)-Change in Velocity

4. Force vs. Time-(none)-Impulse (p or J)
Kinematic Equations (MEMORIZE):
•x = x(o) + (1/2) (v + v(o))t
•v = v(o) + at
•x = x(o) + v(o)t + (1/2)at(sq)
•x = x(o) + vt - (1/2)at(sq)
•v(sq) = v(o)(sq) + 2a(x-x(o))
•√v(x)(sq) + v(y)(sq)
Work, Energy, & Power (p. 85-98)
PART 3/8 (6%)
Work Problems w/ Graphs
-There's a good chance the SAT II may test your understanding of work by asking you to interpret a graph. This graph will most likely be a force vs. position graph, though there's a chance it may be a graph of Fcosθ vs. position.
-In the latter case, you'll be dealing with a graphic representation of a force that isn't acting parallel to the displacement, but the graph will have already taken this into account.
-Bottom line: all graphs dealing with work will operate according to the same easy principles.
-THE WORK DONE IN A FORCE VS. DISPLACEMENT GRAPH IS EQUAL TO THE AREA B/W THE GRAPH AND THE X-AXIS DURING THE SAME INTERVAL.
Conservation of Energy
-tells us that the energy in the universe is constant.
-Energy cannot be made or destroyed, only changed from one form to another form.
-Energy can also be transferred via a force, or as heat.
-This law applies to any closed system. A closed system is a system where no energy leaves the system and goes into the outside world, and no energy from the outside world enters the system.
Kinetic Energy
KE = (1/2) m v(sq)
-is the energy a body in motion has by virtue of its motion.
-Ex: when teh cue ball strikes the eight ball, the cue ball comes to a stop and teh eight ball starts moving. This occurs b/c the cue ball's kinetic energy has been transferred to the eight ball.
-There are two main types of kinetic energy: translational and rotational. TRANSLATIONAL KINETIC ENERGY is the energy of a particle moving in space and is defined in terms of the particle's mass, m, and velocity, v.
Hockey Puck Example:
•1 kg •v=10 (m/s) •F(frict.)=1N
-Find v after 32 m gliding.
•KE = (1/2)mv(sq) = 50J
•W = F.s = (-1N) 32m = 32J
•Work decreases kinetic energy
= 50-32= 18J
•Plug back into KE eq. and the answer is v=6 m/s
Potential Energy (U)
Net Work: W=-ΔU
-is a measure of an object's unrealized potential to have work done on it, and is associated with that object's position in space, or its configuration in relation to other objects.
-It may be tempting to thing that the work done on an object increases its potential energy, but the opposite is true. Work converts potential energy into other forms of energy, usually kinetic energy.
Gravitational Potential Energy
W = -mgh
-registers the potential for work done on an object by the force of gravity.
-the work done by the force of gravity as you lift the water balloon is the force of gravity, -mg, times the water balloon's displacement, h. So the work done by the force of gravity is W=-mgh.
-THE HIGHER AN OBJECT IS OFF THE GROUND, THE GREATER ITS GRAVITATIONAL POTENTIAL ENERGY.
•Gravitational Potential Energy when dropped: U(g)=mgh
Example:
-A student drops an object:
•m= 10kg
•h= 5m
•Assume g=-10 (m/s2)
•Find the velocity when it hits the ground.
•When the object is dropped, it has a gravitational potential energy of:
U(g)=mgh= 10x-10x-5= 500J
•(1/2)mv2=500J
v2 = 2(500J)/10kg
√100
V= 10 (m/s)
Thermal Energy
-is the most common form of energy produced in energy transformations.
-It's hard to think of an energy transformation where no heat is produced. Take these examples:
•Friction is at work everywhere, and friction produces heat.
•Electric energy produces heat: a light bulb produces far more heat than it does light.
•When people talk about burning calories, they mean it quite literally: exercise is a way of converting food energy into heat.
•Sounds fade to silence b/c the sound energy is gradually converted into the heat of teh vibrating air molecules. In other words, if you shout very loudly, you make the air around you warmer!
Power
Average Power =
ΔW/Δt or ΔE/Δt
-is an important physical quantity that frequently, though not always, appears on the SAT II.
-Is a measure of the amount of work done in a given time period.
-is measured in units of watts (W), where 1W=1 J/s.
Power Example:
-A piano mover pushes on a piano w/ a force of:
•F=100N •Moving it 9m in 12s
•With how much power does the piano mover push?
*First we need to calculate how much work the piano mover does, and then we divide that quantity by the amount of time the work takes.
•W = F(s) = 100N x 9m= 900J
•P = ΔW/Δt = 900J/12s = 75W
Instantaneous Power
P = F.v
-the amount of power output by that person at any given instant.
-In such cases, there is no value Δt to draw upon. However, when a steady force is applied to an object, the a change in the amount of work done on the object is the product of the force and the change in that object's displacement.
Special Problems in Mechanics (p. 103-129)
PART 4/8 (5%)
The Three-Step Approach to Problem Solving
1.) Ask yourself how the system will move.
2.) Choose a coordinate system.
3.) Draw free-body diagrams.
*Test writers don't want to test your ability to recall a formula or do some simple math but TO SEE WHETHER YOU UNDERSTAND THE FORMULAS YOU'VE MEMORIZED.
Pulleys
-are simple machines that consist of a rope that slides around a disk, called a block.
-main function is to change the direction of the tension force in a rope.
-almost always consist of idealized, massless and firctionless pulleys, and idealized ropes that are massless and that don't stretch.
These Unrealistic Pulley System Parameters Mean That:
1.) The rope slides w/o any resistance over the pulley, so that the pulley changes the direction of the tension force w/o changing its magnitude.
2.) You can apply the law of conservation of energy to the system w/o worrying about the energy of the rope and pulley.
3.) You don't ahve to factor in the mass of the pulley or rope when calculating the effect of a force exerted on an object attached to a pulley system.
-Exception: problem involving to take the pulley's torque applied to a pulley block-take the pulley's mass into account.
Understanding a Standard Pulley Problem: M & m attached.
1.) B/c the masses are connected, we know that the velocity of mass m is equal in magnitude to the velocity of mass M, but opposite in direction. Likewise, the acceleration of mass m is equal in magnitude to the acceleration of mass M, but opposite in direction.
2.) Up is the positive y direction and down is the negative y direction.
3.) T (pointing up) and mg (pointing down) on each block.
Pulley Problem:
1.) Find acceleration of M?
-If mass m is a, then the acceleration of mass M is -a.
Newton's 2nd Law:
for mass M: T - Mg= -Ma
for mass m: T - mg= ma
_______________________

a = g(M-m)/M+m
2.) Find Velocity of mass m after it travels a distance h?
-Kinematics would be a bit unwieldy.
-take this in terms of energy.
-b/c mechanical energy, E, is conserved, we know that any change in the potential energy, U, of the system will be accompanied by an equal but opposite change in the kinetic energy, KE, of the system.

-Applying the law of conservation of mechanical energy:
•ΔKE = -ΔU
•(1/2)M(v2) + (1/2)m(v2) = -(mgh-Mgh)
•1/2 (M+m)v2 = (M-m)gh
________________
• v =√ 2gh x (M-m)/M+m
3.) What is the work done by the force of tension in lifting mass m a distance h?
-Since the tension force, T, is in the same direction as the displacement, h, we know that the work done is equal to hT.
-BUT WHAT IS THE MAGNITUDE OF THE TENSION FORCE? We know that the sum of forces acting on m is T-mg which is equal to ma.
-Therefore, T=m(g-a). From the solution to question 1, we know that a=g(M-m)/(M+m), so substituting in for a, we get:

•W=hT=m(g-a)h=m x (g- g(M-m)/M+m (h))
•=mgh(1-(M-m)/M+m
Complex Formulas Question 1
1.) WHICH OF THE FOLLOWING FIVE FORMULAS REPRESENTS THE ACCELERATION OF THE PULLEY SYSTEM?
-you would then be given five different math formulas, one of which is the correct formula. The test writers would not expect you to have memorized the correct formula, but they would expect you to be able to derive it.
Complex Formulas Question 2
2.) WHICH OF THE FOLLOWING IS A WAY OF MAXIMIZING THE SYSTEM'S ACCELERATION?
-Don't even need to know the correct formula, but do need to understand how the pulley system works.
-We would maximize the downward acceleration by maximizing m and minimizing M and µ.
Complex Formulas Question 3
3.) IF THE SYSTEM DOES NOT MOVE, WHICH OF THE FOLLOWING MUST BE TRUE?
-You would then be given a number of formulas relating M, m, and µ. The idea behind such a question is that the system does not move if the downward force on m is less than or equal to the force of friction on M, so mg≤µMg.
Frictionless Inclined Plane Problem (p. 111)
•10 kg box
•Frictionless 30° inclined plane
•horizontal distance of d meters and a vertical distance of h meters.
1.) HOW WILL THE SYSTEM MOVE: Acceleration and velocity of the box to be affected by the angle of the plane.
2.) CHOOSE A COORDINATE SYSTEM: The x-axis runs parallel to the plane, where downhill is the positive x direction, and the y-axis runs perpendicular to the plane, where up is the positive y direction.
3.) DRAW A FREE-BODY DIAGRAM:
•Force of Gravity (mg)-straight downward)
•Normal Force (Perpendicular to the inclined plane)
•mg x sine 30 (+X): Downward Force on Slope
•mg x Cos 30 (-Y)
Frictionless Inclined Plane Problem (p. 111)
Question 1/4
•g = 10 m/s (sq)
•cos30 = 0.866
•sin30 = 0.5
1.) WHAT IS THE MAGNITUDE OF THE NORMAL FORCE? [Same as the opposite of (-y)]
•N = mgcos30
• = (10kg)(10 m/s)(0.866)
• = 86.6 N
Frictionless Inclined Plane Problem (p. 111)
Question 2/4
•g = 10 m/s (sq)
•cos30 = 0.866
•sin30 = 0.5
2.) WHAT IS THE ACCELERATION OF THE BOX? (Downward) [-y]
•MA = F (From F=MA): 2nd Law
•ma = mgsin30
•a = gsin30
• = (10)(.5)
• = 5 (m/s (sq))
Frictionless Inclined Plane Problem (p. 111)
Question 3/4
•g = 10 m/s (sq)
•cos30 = 0.866
•sin30 = 0.5
3.) WHAT IS THE VELOCITY OF THE BOX WHEN IT REACHES THE BOTTOM OF THE SLOPE?
-Use Law of Conservation of Mechanical Energy:
• Top: KE=0
• At bottom: PE is KE/ KE (1/2mv(2)) = PE (mgh)
• E = KE + U = 0 + MGH = MGH
• V = √2gh
• = 4.47 √h
Frictionless Inclined Plane Problem (p. 111)
Question 4/4
•g = 10 m/s (sq)
•cos30 = 0.866
•sin30 = 0.5
4.) WHAT IS THE WORK DONE ON THE BOX BY THE FORCE OF GRAVITY IN BRINGING IT TO THE BOTTOM OF THE INCLINED PLANE?
-Use Work-Energy Theorem: •W=KE
•Top: KE=0
• BOTTOM: KE=PE (at top) [mgh]
• W = mgh
• 10 x 10 x h
• 100h J
*Work done is independent of how steep the inclined plane is, and is only dependent on the object's change in height when it slides down the plane.
Frictionless Inclined Planes w/ Pulleys
•m on plane & M on Pulley
•M > m sinθ
1.) Acceleration of masses?
• a = g(msinθ - M)/ (m+M)
*b/c M>msinθ, the acceleration is negative, which, as we defined it, is down for mass M and uphill for mass M.
2.) WHAT IS THE VELOCITY OF MASS m AFTER MASS M HAS FALLEN A DISTANCE H?
• Law of Conservation of Mechanical Energy: since KE=0
• Mass m's PE increases by
mgh x sinθ
• Solve for V:
KE + KE - PE + mghSinθ

V = √2gh(M-msinθ)/m+M
Inclined Planes w/ Friction (p. 115)
-There are two significant differences b/w frictionless inclined plane problems and inclined plane problems where friction is a factor.
1.) THERE'S NO EXTRA FORCE TO DEAL WITH: The force of friction will oppose the downhill component of the gravitational force.
2.) WE CAN NO LONGER RELY ON THE LAW OF CONSERVATION MECHANICAL ENERGY: b/c energy is being lost through the friction b/w the mass and the inclined plane, we are no longer dealing with a closed system. Mechanical energy is not conserved.
Inclined Planes w/ Friction (p. 115)
1.) WHAT IS THE FORCE OF KINETIC FRICTION ACTING ON THE BOX?
• F(f) = µN = (0.5) 86.6N
• = -43.3 N
*Force of friction is exerted in the negative x direction, so that's why the answer is -43.3N.
Inclined Planes w/ Friction (p. 115)
2.) WHAT IS THE ACCELERATION OF THE BOX?
• 2nd: F=mgsin30-F(f)
• ma = mgsin30-F(f)
• 10 a = 10 x 10 x .5 - 43.3
• a = 50N-43.3N/10kg
• 0.67 m/s(sq)
Inclined Planes w/ Friction (p. 115)
3.) WHAT IS THE WORK DONE ON THE BOX BY THE FORCE OF KINETIC FRICTION?
• W = F(f) . d
•F(f)=-43.3N •Disp.=√d(sq)+h(sq)
•W=-43.3 √d(sq)+h(sq) J
SPECIAL PROBLEMS: SPRINGS (P.116)
-SAT II will not test you on the motion of springs involving friction, so for the purposes of the test, the mechanical energy of a spring is conserved quantity.
-Will ask qualitatively about the energy or velocity of a vertically oscillating spring.
-Example May Include:
-A DIAGRAM CAPTURING ONE MOMENT IN A SPRING'S TRAJECTORY AND ASKED ABOUT THE RELATIVE MAGNITUDES OF THE GRAVITATIONAL AND ELASTIC POTENTIAL ENERGIES AND KINETIC ENERGY.
-AT WHAT POINT IN A SPRING'S TRAJECTORY THE VELOCITY IS MAXIMIZED. (ANS: IT IS MAXIMIZED AT THE EQUILIBRIUM POSITION).
-Far less likely that you will be asked a question that involves any sort of calculation.
Oscillation
-is the natural world's way of returning a system to its EQUILIBRIUM POSITION, the stable position of the system where the net force acting on it is zero.
-oscillation is a way of giving off energy (thermal energy caused by friction).
Harmonic Motion
-the movement of an oscillating body
-if you were to graph the position, velocity, or acceleration of an oscillating body against time, the result would be a sinusoidal wave; that is, some variation of a y=a sin bx or a y= a cos bx graph.
-applies to anything that moves in a cycle.
Amplitude
-when you release the mass, the spring will exert a force, pushing the mass back until it reaches position x=x(max), which is called the AMPLITUDE of the spring's motion, or the maximum displacement of the oscillator.
-Note that x(min)=-x(max)
Hooke's Law: F=-kx
•x is displacement
•k (spring constant) is a constant of proportionality
-The force, F, that the spring exerts on the mass is defined by this law.
-it gives us the manner in which the spring exerts a force on the mass attached to it and is needed in order to understand velocity, the energy, or anything else about the mass's motion.
-the spring constant is a measure of "springiness": a greater value for k signifies a "tighter" spring, one that is more resistant to being stretched.
-THE GREATER K IS, THE TIGHTER THE SPRING IS.
Simple Harmonic Oscillation
-is any object that moves about a stable equilibrium point and experience a restoring force proportional to the oscillator's displacement.
-ex: a mass oscillating on a spring.
-For an oscillating spring, the restoring force, and consequintly the acceleration, are greatest and positive at x(min). These quantities decrease as x approaches the equilibrium positive and are zero at x=0. The restoring force and acceleration-which are now negative-increase in magnitude as x approaches x(max) and are maximally negative at x(max).
Period of Oscillation
____
T = 2(pi) x √m/k
-is the amount of time it takes for a spring to complete a round-trip or cycle.
-equation tells us that as the mass of the block, m, increases and the spring constant, k, decreases, the period increases.
-Ex: a heavy mass attached to an easily stretched spring will oscillate back and forth very slowly, while a light mass attached to a resistant spring will oscillate back and forth very quickly.
Frequency
f = 1/T
-tells us how quickly the object is oscillating, or how many cycles it completes in a given timeframe.
-is inversely proportional to period.
-given in units of cycles per second, or hertz (Hz).
Potential Energy (elastic energy)
U(s) = 1/2 (kx(sq))
-results from the spring being stretched or compressed.
-greatest when the coil is maximally compressed or stretch, and is zero at the equilibrium position.
Kinetic Energy
• Maximum Compression:
-the velocity, and hence the kinetic energy, is zero
-Mechanical Energy is equal to the potential energy
U(s)=1/2(kx(sq))

• Equilibrium Position:
-Potential Energy is zero, and the
-Velocity and Kinetic Energy are maximized.
-The kinetic Energy is equal to the mechanical energy:

• KE(max)= (1/2)mv(sq) = (1/2)kx(sq)(min)
• V(max)= x√(k/m)
SPECIAL PROBLEMS: VERTICAL OSCILLATION (P.123)
-Most questions will be qualitative (DON'T NEED TO MEMORIZE EQUATIONS)
-Questions May Include:
-CONSIDER A MASS ATTACHED TO A SPRING THAT IS SUSPENDED FROM THE CEILING
-AT WHAT POINT IS THE TENSION IN THE ROPE THE GREATEST (ANS: AT THE EQUILIBRIUM POSITION)
-WHERE IS THE BOB'S POTENTIAL ENERGY MAXIMIZED (AT θ=(+/-)θ(max)).
-It's highly unlikely that you'll be asked to give a specific number.
Equilibrium Position
-the equilibrium position is the point where the net force acting on the mass is zero (the point where the upward restoring force of the spring is equal to the downward gravitational force of the mass.
Vertical Oscillation Equation Derivation
•F=-kh & F=mg
•-kh = mg
•-h = mg/k
VS in Motion
-If the spring is then stretched a distance d, where d<h, it will oscillate b/w x(max)=-h-d and x(min)=-h+d
Mechanical Energy:
E = KE + U(g) + U(s)
-U(g) = gravitational potential energy
-U(s) = spring's (elastic) potential energy.
-Velocity: is zero at x=m(min) and x=x(max), and maximized at the equilibrium position, x=-h.
-Kinetic Energy: zero for x=x(max) and x=x(min) and is greatest at x=-h.
-Gravitation Potential Energy: increases with the height of the mass.
-Elastic Potential Energy: greatest when the spring is maximally extended at x(max) and decreases with the extension of the spring.
SPECIAL PROBLEMS: PENDULUM (p.125)
-It is highly unlikely that the formulas discussed in this section will appear on the test; instead you will be asked conceptual questions such as:
-AT WHAT POINT IN A SPRING'S OSCILLATION IS THE KINETIC OR POTENTIAL ENERGY MAXIMIZED OR MINIMIZED.
-You qualitative understanding of the relationship b/w force, velocity, and Kinetic and potential energy in a spring system is far more likely to be tested than your knowledge of the formulas discussed above.
Pendulum
-defined as a mass, or bob, connected to a rod or rope, that experiences simple harmonic motion as it swings back and forth w/out friction.
-Equilibrium Position: hanging directly downward.
Pendulum Free-Body Diagram:
-Rope: y -Perpendicular to rope: x
•Y: F(T) •-Y=mgcosθ •down: mg
•X: mgsinθ
Period (Pendulum) (T)

•T = 2(pi)√L/g
•T=period •g=gravity •L=length of the pendulum
-The longer the pendulum rope, the longer it will take for the pendulum to oscillate back and forth.
-Mass of the pendulum bob and the angle of displacement play no role in determining the period of oscillation.
Energy
-Mechanical Energy: is a conserved quantity.
-Potential Energy (mgh): increases w/ the height of the bob
-Potential Energy: minimized at the equilibrium point and is maximized at θ=(+/-)θ(max).
Velocity
V=√2gL(1-cosθ(max))
-Calculating the Velocity needs you to:
•Make h=0

•Equilibrium:
-ME=KE
-U=0

•(+/-)θ(max):
-ME(Tot)=PE(Tot)
-KE=0
•All together:
E = (1/2)mv(sq) = mgh

-Deriving V
• h = L-Lcos(-θ(max))
• (1/2)mv(sq) = mgL(1-cosθ(max))
•v = √2gL(1-cosθ(max))
*THE LONGER THE STRING AND THE GREATER THE ANGLE, THE FASTER THE PENDULUM BOB WILL MOVE.
Circular Motion & Gravitation (p.185-197)
PART 5/8 (4%)
Uniform Circular Motion
-occurs when a body moves in a circular path w/ constant speed.
-Ex: say you swing a tethered ball overhead in a circle: Gravity and Tension of the string are the only forces acting on the ball.
-The force acting on a tetherball traveling in a circular path is always directed toward the center of that circle.
-Although the direction of the ball's velocity changes, the ball's velocity is constant in magnitude and is always tangent to the circle.
Centripetal "Center-Seeking" Acceleration

A(c) = V(sq)/r
-the acceleration of a body experiencing uniform circular motion is always directed toward the center of the circle, so we call this acceleration Centripetal Acceleration A(c).
-Even though the direction of the centripetal acceleration vector is changing, the vector always points toward the center of the circle.
-ACCELERATION VECTOR IS CONSTANT IN MAGNITUDE, HAS A DIRECTION THAT ALWAYS POINTS TOWARD THE CENTER OF THE CIRCLE, AND IS ALWAYS PERPENDICULAR TO THE VELOCITY VECTOR.
Centripetal Force

F = ma(c) = mv(sq)/r
*F may also be T (Tension)
-the vector for this force is similar to the acceleration vector: it is of constant magnitude, and always points radially inward to the center of the circle, perpendicular to the velocity vector.
Gravitational Force

F(g) = G x (m1m2)/r(sq)
-states that the force of gravity, F(g), b/w two particles of mass m(1) and m(2) has a magnitude of:
F(g) = G x (m1m2)/r(sq)
Gravitational Force & Velocity of an Orbiting Satellite

v = √G m(earth) /R
•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 v:
•F(g)-Gravitation = F(c)-Centripetal
G x m(s)m(e)/R(sq) = m(s)v(sq)/R
•v = √G m(e)/R
-Any object orbiting at radius R must be orbiting w/ a velocity of this equation. If the satellite's speed is too small, then the satellite will fall back down to Earth. If the satellite's speed is too great, then the satellite will fly out into space.
Gravitational Potential Energy
-is equal to the amount of work that must be done to arrange the system in that particular configuration.
-depends on how high an object is off the ground: the higher an object is, the more work needs to be done to get it there.
-is not an absolute measure. It tells us the amount of work needed to move an object from some arbitrarily chosen reference point to the position it is presented in.
-For instance, when dealing w/ bodies near the surface of the Earth, we choose the ground as our reference point, b/c it makes our calculations easier. If the ground is h=0, then for a height h above the ground an object has a potential energy of mgh.
Gravitational Potential in Outer Space

U = -G (m1m2)/r
-Off the surface of the Earth, there's no obvious reference point from which to measure gravitational potential energy. Conventionally, we say that an object that is an infinite distance away from the Earth has zero gravitational potential energy w/ respect to the Earth.
-Because a negative amount of work is down to bring an object closer to the Earth, gravitational potential energy is always a negative number when using this reference point.
Gravitational Potential Energy Problem:
-A satellite of mass m(s) is launched from the surface of the Earth into an orbit of radius 2r(e), where r(e) is the radius of the Earth. How much work is done to get it into orbit?
-U(1): PE on Earth
-U(2): PE in orbit
•W=U2-U1
•-G x m(s)m(e)/2r(e) - (-G x m(s)m(e)/r(e))
•G x m(s)m(e)/2r(e)
Energy Of an Orbiting Satellite

E = -G X [m(s) x m(e)]/ 2R
•F(g) = KE
•E (TOT) = PE (U) + KE

•E = -G [m(s)m(e)/R] + G [m(s)m(e)/2R
• = -G [m(s)m(e)/2R]
Weightlessness
•Use Bathroom Scale in Elevator to measure Weightlessness:
-Elevator Stopped = a:0 = W:mg
-Elevator Up = a:g = W:2mg
-Elevator Down (Free Fall) = a:-g = W:0
•Popular Misconception: Astronauts in satellites experience weightless b/c they are beyond the reach of the Earth's gravitational pull. REAL REASON: This is not b/c they are free from the Earth's gravitational pull. Rather, their space shutle is in orbit about the Earth, meaning that it is in a perpetual free fall. Because they are in free fall, the astronauts, like you in your falling elevator, experience weightlessness.
Kepler's Three Laws
•KEPLER'S FIRST LAW: states that the path of each planet around the sun is an ellipse with the sun at one focus.

•KEPLER'S SECOND LAW:
-states that if a line is drawn from the sun to the orbiting planet, then the area swept out by this line in a given time interval is constant.
-relates a planet's speed to its distance from the sun. Because the planets' orbits elliptical, the distance from the sun varies. This means that when the planet is farthest from the sun it moves much more slowly than when it is closer to the sun.

•KEPLER'S THIRD LAW (T(sq)/a(cubed)):
-states that given the period, T, and semimajor axis, a, of a planet's elliptical orbit, the ratio T(sq)/a(cubed) is the same for every planet.
-The semimajor axis is the longer one, along which the two foci are located.
Kepler's Law Problem
•Every 76 years, Halley's comet passes quite close by the Earth. At the most distant point in its orbit, it is much farther from the sun even than Pluto. Is the comet moving faster when it is closer to Earth or closer to Pluto?

-According to Kepler's Second Law, objects that are closer to the sun orbit faster than objects that are far away. Therefore, Halley's comet must be traveling much faster when it near the Earth than when it is off near Pluto.