Dr. Phil's Home

Lectures in PHYS-2050 (21) / PHYS-2140

Updated: 03 May 2011 Tuesday.

FINAL EXAM RESULTS:

n       73     73
hi     184     200
lo      80     100
ave    150     160

Average Star=        18,000

FINAL GRADES FOR PHYS-2050/2140 CAN BE FOUND HERE.

Week of April 25-29, 2011.

Monday 4/25: Office hours.

Tuesday 4/26: Office hours.

Wednesday 4/27: FINAL EXAM (2 HOURS) 2:45-4:45pm.Office hours.

Thursday 4/28: NOTE: No office hours.

Friday 4/29: LATE FINAL EXAM (2 HOURS) 11am-1pm, come to Dr. Phil's office. See: Office hours.

Week of May 2-6, 2011.

Monday 5/2: Office hours.

Tuesday 5/3: Grades will be done by Noon.


Week of January 10-14, 2011.

Monday 1/10: Class begins. The nature of studying Physics. Science education in the United States. Natural Philosophy. The Circle of Physics. Aristotle and the Greek Philosophers. Observation vs. Experiment - Dropping the book and the piece of paper (2 views).

  • Note that the 1pm PHYS-2050 lecture is a stand-alone class which (a) meets five days a week and (b) does NOT participate in the exams of the other two sections. Do not take quizzes or exams or homework assignments from the other sections.
  • Week 1 Checklist -- These checklists are provided for your benefit only. No information is recorded. You can print them out or click on items and print them out if you wish.
  • Quiz 1 will be in-class on Friday 14 January 2011. It will be for attendance purposes. If you miss class on Friday, you will be able to get some of the points by downloading Quiz 1A from the website and turning it in.
  • Tuesday 1/11: Observation vs. Experiment - Zeno's Paradoxes. Theory and Measurement. "Speed Limit 70" First Equation: Speed = Distance / Time. In terms of variables, a classic three-variable equation: v = d / t . Development of Speed equation for Constant or Average Speed. An Equation is a contract -- the left and right sides must be the same thing and all terms must have the same units. Distribute syllabus.

    Wednesday 1/12: Our equations so far: v = d /t ; x = x0 + v t (for v = constant or average ONLY). Discussion of Formula Cards. delta-x = xfinal - xinitial, so v = (xf - xi) / (tf - ti) = delta-x / delta-t. Taking the limit as delta-t goes to zero, gives the derivative v = dx/dt. A simplified trip to the store -- The S-Shaped Curve. Acceleration. a = dv/dt = d2x/dt2. Integrating to find the set of Kinematic Equations for constant acceleration. Kinematic Equations for Constant Acceleration.

    Thursday 1/13: Speed. 60 m.p.h. = "A Mile A Minute". (1848: The Antelope) SI Metric System. What do we mean by Measurements? What is "1 m/s"? We need a few benchmark values to compare English and SI Metric quantities. 60 m.p.h. = 26.8 m/s. 1.00 m/s = slow walking speed. 10.0 m/s = World Class sprint speed. (The 100 meter dash -- Usain Bolt is the current Olympic (9.59 seconds) and World (9.58 seconds) record holder.) 344 m/s = Speed of sound at room temperature. 8000 m/s = low Earth orbital speed. 11,300 m/s = Earth escape velocity. 300,000,000 m/s = speed of light in vacuum (maximum possible speed).

  • Quiz 1 will be in-class on Friday 14 January 2011. It will be for attendance purposes. If you miss class on Friday, you will be able to get some of the points by downloading Quiz 1A from the website and turning it in.
  • The Physics Help Room will open for business on Tuesday 18 January 2011. (There is no class on Monday -- MLK Day.) The Physics Help Room is located in 0077 Rood Hall, sort of behind and underneath our lecture hall. If you go down the stairs near 1110 Rood, 0077 Rood will be right there when you get to the basement. Hours will be 9am to 4pm Monday through Friday, and Physics faculty and grad students will be on duty for most of those hours. Dr. Phil's hour will be Thursday at Noon, starting Thursday 20 January 2011.
  • Friday 1/14: A simplified trip to the store -- The S-Shaped Curve. Acceleration. a = dv/dt = d2x/dt2. Integrating to find the set of Kinematic Equations for constant acceleration. Kinematic Equations for Constant Acceleration. The Equation Without Time -- Avoiding the Quadradic Formula. The fundamental equations of motion are based on the calculus: x, v and a are related by derivatives (slopes) and integrals (areas under the curve). v = dx/dt, a = dv/dt = d²x/dt². There are higher derivatives. For example, jerk is j = da/dt = d²v/dt² = d³x/dt³. Physics Misconceptions: Things you think you know, are sure you know, or just assume to be true in the back of your mind... but aren't true. Aristotle was sure that heavier objects always fell faster than lighter objects, but we did a demostration on Tuesday which showed that wasn't always true. Example: You're driving a car. To speed up, you need to put your foot on the accelerator (gas pedal), so YES, you are accelerating -- True. To drive at a constant speed, you must still have your foot on the accelerator, so YES, you are accelerating -- Not True because constant v means a = 0. To slow down, you must take your foot off the accelerator and put it on the brake pedal, so NO, you are not accelerating -- Not True because v is changing, so a < 0 (negative). What do we mean by a = 1 meter/sec² ? You cannot accelerate at 1 m/s² for very long. How do we solve kinematic problems? There are six kinematic variables for constant acceleration in 1-D: x0, x, v0, v, a and t. Example: car accelerating from rest to 26.8 m/s (60mph) with a = 3.00 m/s². We know or can assume values for four of the six kinematic variables. Then use the kinematic equations to find t and x. Using another equation to check your answers can be very helpful, especially early on. The Kinematic Equations form a system of equations and all of them must give correct answers for a given problem. Dr. Phil's Reasonable Significant Figures. (Click here for a copy of the handout.) Q1 and your PID number. (If you missed Quiz 1 you will be able to get some of the points by downloading Quiz 1A from the website and turning it in.)

    Week of January 17-21, 2011.

    Monday 1/17: MLK Day -- No Classes.

    Tuesday 1/18: Types of Motion: No Motion (v=0, a=0), Uniform Motion (v=constant, a=0), Constant Acceleration (a=constant). The fundamental equations of motion are based on the calculus: x, v and a are related by derivatives (slopes) and integrals (areas under the curve). v = dx/dt, a = dv/dt = d²x/dt². There are higher derivatives. For example, jerk is j = da/dt = d²v/dt² = d³x/dt³. Note that our simplified trip to the store we had sudden changes in acceleration -- this would create infinite slopes in da/dt. The real world is smoother -- accelerations tend to be turned on and off over a brief time, not instantaneously. However, our approximation that acceleration is strictly constant turns out to work pretty well for many problems, so our simplification remains useful. On the other hand, we could have a seventh derivative of position with respect to time. Comments on Star Problems on our Exams. Back to our constant (or average) acceleration problems. Example: acceleration and time for a bullet fired from a rifle. Q2 in-class.

    Wednesday 1/19: The P-O-R (Press-On-Regardless) road rally problem. "You can't average averages." The only way to find the speed required for the rest of the trip is v = distance remaining / time remaining. Just like part (c) in Quiz 2. Demo: Our class webpages and what you can find on them. Free fall, ignoring air resistance -- all objects near the surface of the Earth will fall at an acceleration g = 9.81 m/s². Re-writing the Kinematic Equations for x, y and y pre-loaded with ay = -g. Note that the g = 9.81 m/s² and is positive. PTPBIP -- Put The Physics Back Into the Problem. For example, compare acceleration to g. We generally cannot accelerate for very long. Example: acceleration and time for a bullet fired from a rifle. The acceleration we calculated yesterday, 80,000 m/s², is more that 8000 g ! Note that NASA defines a hammer blow as 100 to 500 g. The acceleration of a rifle bullet is much larger than this, which is why we can't fire humans in a 1.00 meter rifle barrel from rest to over Mach 1.

  • Quiz Solutions are posted on the class web page.
  • Discovery Channel and YouTube come through! I've been telling the story of North American Aviation company test pilot Scott Crossfield and the static test stand explosion of the X-15 for years, and just now discovered that there's video online of this! You can see more successful flights of the three X-15 aircraft in this video and the first segment of this newsreel video. For more detailed shots of training, flights, landing, and the damage from a white covered X-15 flight that reached Mach 6.7, one last video. The X-15 lands as a glider, like the Space Shuttle, but it's a lousy glider. The only way the F-104 chase planes can follow it easily to the dry lakebed runway is to drop flaps and landing gear, and the X-15 still falls out of the sky like a brick. (grin)
  • Thursday 1/20: Motion in Two-Dimensions: x and y directions are perpendicular to each other and are independent of each other. You may be able to break a two-dimensional problem down into two one-dimensional problems, connected by time, which you can already solve. Example: The guy with the fedora and the cigar. There are 6 variables from the first dimension (x0, x, v0x, vx, ax, t), but only 5 from the second (y0, y, v0y, vy, ay), because time is the same. We need to find v0x , but we don't know the time. So we can find the time it takes to fall from the top of the building in the y-problem, then use that in the x-problem. Another problem with two motions linked by time: Classic Simple Pursuit (Cop and the Speeder). Same Place at the Same Time. x1 = x2 . Note that in our algebra for car 1 and car 2, we cancelled a factor of t at one point -- this represents t = 0, which by definition is also a solution. But... because our problem is Simple Pursuit, when the car 2 reaches car 1, the cop is traveling at twice the speed of the speeder. Demo: Reaction time. Back to our Simple Pursuit problem. If car 2 doesn't start at the same time car 1 passes, perhaps becuase of reaction time, then instead of x10 = 0 for car one, we can set x10 = v10 × Reaction Time.

    Friday 1/21: Quiz 3 in class.

    Week of January 24-28, 2011.

    Monday 1/24: Motion in the y-direction. The consequences of Falling Down... ...and Falling Up. The Turning Point ( v=0 but a = -g during whole flight). The illusion of "hanging up there in the air" at the turning point.

    Tuesday 1/25: More about Exams. Two kinds of numbers: Scalars (magnitude and units) and Vectors (magnitude, units and direction). Adding and subtracting vectors: Graphical method. To generate an analytical method, we first need to look at some Trigonometry. Right Triangles: Sum of the interior angles of any triangle is 180°, Pythagorean Theorem (a² + b² = c²). Standard Angle (start at positive x-axis and go counterclockwise). Standard Form: 5.00m @ 30°. Practical Trigonometry. SOHCAHTOA. First Sample Exam 1 (Click here for a copy.) Quiz 4 take-home quiz on Falling Up and Falling Down plus Simple Pursuit, due Friday 28 January 2011, in class or by 5pm.

    Wednesday 1/26: Adding and subtracting vectors: Graphical method. To generate an analytical method, we first need to look at some Trigonometry. Right Triangles: Sum of the interior angles of any triangle is 180°, Pythagorean Theorem (a² + b² = c²). Standard Angle (start at positive x-axis and go counterclockwise). Standard Form: 5.00m @ 30°. Practical Trigonometry. SOHCAHTOA. Adding and subtracting vectors: Analytical method and Sketch of the problem. (Check to make sure your calculator is set for Degrees mode. Try cos 45° = sin 45° = 0.7071) Why arctangent is a stupid function on your calculator. Finding the final vector velocity of The guy with the fedora and the cigar problem.

    Thursday 1/26: Return Q2, Q3. Unit Vectors: i-hat, j-hat, k-hat (point in x, y, z directions, respectively), have unitary length (length of 1). Allow you to describe a vector with x- and y-components times the i- and j-hat unit vectors. Using Vector Addition for Velocities: Upstream, downstream (rivers), Headwind, tailwind, crosswind (airplanes). Ballistic (or Projectile) Motion -- applies equally to a thrown football and a cannonball. Still working with ax = 0 and ay = -g. History of early cannons.

    Friday 1/27: Ballistic (or Projectile) Motion. Two Dangerous Equations. You can only use the Range Equation if the Launch Height = Landing Height. But the sin (2*theta) term in the Range Equation means that (1) 45° gives the maximum range for a given initial velocity and (2) that all other angles have a complementary angle (90° - theta) that gives the same range (but a different time and height). High and low trajectories for Range Equation. Topic 1 assigned. (Click here for a copy of the handout.) Second Sample Exam 1 (Click here, here and here for a copy.) Quiz 5 take-home quiz on Vectors, Standard Form and Vector Addition, due Tuesday 1 February 2011, in class or by 5pm.

  • Note: The actual Topic 1 Handout is 27 pages long. You've been given pages 1,2,11 and 27, and the link to the Topic 1 Assignment webpage. There you can see the whole handout as a PDF or as a Searchable HTML page with links to jump to the main topics.
  • If this all seems like Too Much Information, come see Dr. Phil and he'll help you find something of interest.
  • Please remember that you are advised NOT to choose a book you have already read.
  • Week of January 30-February 4, 2011.

    Monday 1/31: Remember, the Range Equation can only be used if the landing height is the same as the launching height, y = y0 . At the maximum height, y = h, this is the turning point. vy = 0 , but vx = v0x , since ax = 0 . If the landing height is different from the launching height, you might want to break the problem into two free-fall problems on either side of the turning point, y = h. If the y > y0 , there are two positive times when you reach the landing height, while going up and coming down. If y < y0 , one time is positive and one is negative (before the problem started). Monkey Hunter problem: Not sure the demo equipment is working right, so just did this on the board. If the gun is pointed right at the monkey AND the monkey lets go from his branch exactly when he sees the muzzle flash, then the monkey gets hit every time. The fall of the bullet from a straightline trajectory and the fall of the monkey from rest are both equal to -½gt² . Demo: Toss a piece of chalk horizontally and drop a second piece -- they both hit the table at the same time, because they have the same y-problem of free-fall in the vertical. Cannot aim directly at an object, have to allow for the drop. Uniform Circular Motion (UCM): speed is constant, but vector velocity is not; magnitude of the acceleration is constant, but the vector acceleration is not. Velocity is tangent to circle, Centripetal Acceleration is perpendicular to velocity and points radial INWARD. For UCM, ac = v²/r.

    Tuesday 2/1: A final note on ballistic motion: You have to have some positive v0y if you want to jump a gap, because otherwise you start falling immediately once you are no longer supported. Uniform Circular Motion (UCM): speed is constant, but vector velocity is not; magnitude of the acceleration is constant, but the vector acceleration is not. Velocity is tangent to circle, Centripetal Acceleration is perpendicular to velocity and points radial INWARD. For UCM, ac = v²/r. Space Shuttle in Low-Earth Orbit. (There's still gravity up there!) Comments on Free Fall vs. "zero gravity" in space. Though we think of Low-Earth Orbit as being pure vacuum, there is still a tiny bit of atmosphere, which is why Skylab fell back to Earth. My sketch of the Space Shuttle in orbit has 3 flaws. The first being no human has ever orbited the Earth in polar orbit (going strictly north and south), though the US Air Force planned to launch Space Shuttles into polar orbits from Vandenberg AFB before the Challenger disasater.Quiz 6 take-home quiz, due Thursday 3 February 2011, in class or by 5pm.

    Wednesday 2/2: WMU closed due to blizzard. No classes held.

    Thursday 2/3: Recap: Our studies so far have described "How" things move, and allow to say "When" and "Where" things move, but not "Why" things move. For that we have to start talking about Forces -- and that means Newton. Some stories about Sir Isaac Newton. Newton's Three Laws of Motion: Zeroeth Law - There is such a thing as mass. Mass is a measure of how much "stuff" an object made of matter contains. SI unit of mass = kilogram (kg). First Law - An object in motion tends to stay in motion, or an object at rest tends to stay at rest, unless acted upon by a net external force. The Normal Force is a contact force perpendicular to a contact surface. Example: A book laying on a table experiences a Normal Force from the table pushing up. Second Law - F=ma. Third Law - For every action, there is an equal and opposite reaction, acting on the other body. (Forces come in pairs, not apples.) "If I were to punch the wall, then the wall punches back." The Normal Force and the weight may be equal-and-opposite forces, but if they both apply to the same object, this is First Law, not Third Law. Force is a vector. Comments about Exam 1 tomorrow. Questions? Uh, no one had any questions.

    Friday 2/4: Exam 1.

    Week of February 7-11, 2011.

    Monday 2/7: Newton's Three Laws of Motion: Zeroeth Law - There is such a thing as mass. First Law - An object in motion tends to stay in motion, or an object at rest tends to stay at rest, unless acted upon by a net external force. Second Law - F=ma. Third Law - For every action, there is an equal and opposite reaction, acting on the other body. (Forces come in pairs, not apples.) SI unit of mass = kilogram (kg). SI unit of force = Newton (N) = (kg·m/s²). English unit of force = pound (lb.). English unit of mass = slug (Divide pounds by 32. For English units, g = 32 ft/sec².). Force is a vector. Free Body Diagrams. Normal Force (Normal = Perpendicular to plane of contact). Sum of forces in x or y equations -- either will be equal to 0 (Newton's 1st Law) or ma (Newton's 2nd Law). The sum of forces equations are the analytical version of the graphical Free Body Diagram -- we generally need both to solve Force problems. SI unit of mass = kilogram (kg). SI unit of force = Newton (N). Examples: Pushing a 125 kg crate around. (Near the surface of the Earth, you can use the relationship that 1 kg of mass corresponds [not "equals"] to 2.2 lbs. of weight. So multiple 125 by 2 and add 10%... 250 + 25 = 275... so a 125 kg crate has a weight of mg = 1226 N or 275 lbs.). "The Normal Force is NOT automatically present -- you have to be in contact with a surface. The Normal Force does NOT automatically point up -- FN is perpendicular to the surface. The Normal Force is NOT automatically equal to the weight. FN = mg only if there are no other forces in the y-direction." Continue with the 125 kg crate. Variations as we allow for an applied force that it at an angle.Quiz 7 take-home quiz on F.B.D. and Sum of Forces equations, and due Thursday 10 February 2011, in class or by 5pm.

    Tuesday 2/8: Continue with the 125 kg crate. Variations as we allow for an applied force that it at an angle. "The Normal Force is NOT automatically present -- you have to be in contact with a surface. The Normal Force does NOT automatically point up -- FN is perpendicular to the surface. The Normal Force is NOT automatically equal to the weight. FN = mg only if there are no other forces in the y-direction." Push down and Normal Force increases; pull up and Normal Force decreases -- though it cannot go negative. "You can't push on a rope." Since the force from a wire/string/rope/chain/thread/etc. can only be in one direction, Dr. Phil prefers to call such forces T for Tensions rather than F for Forces. Simple pulleys (Massless, frictionless, dimensionless, only redirect the forces). "There is no free lunch." The bracket for the pulley will have to support a force greater than the weight of the hanging object. Mechanical advantage: multiple pulleys allow us to distribute the net force across multiple cables or the same cable loop around multiple times. Tension in the cable is reduced, but you have to pull more cable to move the crate.

    Wednesday 2/9: Elevator Problems. The Normal Force represents the "apparent weight" of the person in the elevator. For the elevator at rest or moving at constant speed, the Normal Force = weight, and the tension of the cable = weight of loaded elevator. But if there is an acceleration vector pointing up, the apparent weight and the tension of the cable increase; if the vector points down, the apparent weight and the cable tension decrease. In true Free Fall, without any air resistance, the Normal Force = 0 and you are floating. Atwood's Machine: Two blocks whose motion is link via a common cable and a pulley. Note that though they have a common magnitude of both speed and acceleration, the velocity vector has no bearing on either the F.B.D. or the solution to the Tension and acceleration. Inclined Planes: Change the co-ordinate system, change the rules. In the tilted x'-y' coordinates, this is a one-dimensional problem, not two-dimensional.

    Thursday 2/10: Inclined Planes: Change the co-ordinate system, change the rules. In the tilted x'-y' coordinates, this is a one-dimensional problem, not two-dimensional. Friction Force: Two kinds of Friction: Static (stationary) and Kinetic (sliding). Friction is a contact force, but whereas the Normal Force is perpendicular to the contact plane, Friction is parallel to the contact plane. For any given contact surface, there are two coefficients of friction, µ, one for static and one for kinetic. Static is always greater than kinetic. Static Friction is "magic", varying between zero and its maximum value of µ times the Normal Force. Kinetic Friction is always µ times the Normal Force. Kinetic Friction always opposes the motion, Static Friction opposes the direction of impending motion (since the object is not moving yet). If object is at rest, need to "test" to see if an applied external force exceeds the maximum static friction force ("breaks the static friction barrier"). Static Friction can vary from zero to its max value in either direction. Demonstration of block sliding down inclined plane with friction. Finding the coefficient of static friction by tilting. µs = tan(thetamax). Similar for kinetic friction, except one has to tap the board to "break the static friction barrier". Rubber on concrete. Tires rolling with friction on good roads -- this is static friction not kinetic friction because the tires aren't sliding on the pavement. Friction while driving. Rubber on dry concrete, coefficients are 1.00 and 0.800 . Tires rolling with friction on good roads -- this is static friction not kinetic friction because the tires aren't sliding on the pavement. Quiz 8 take-home quiz on Tensions & Simple Pulleys and Forces & Friction, and due on Tuesday 15 February 2011, in class or by 5pm.

    Friday 2/11: Two kinds of Friction: Static (stationary) and Kinetic (sliding). For any given contact surface, there are two coefficients of friction, µ, one for static and one for kinetic. Static is always greater than kinetic. Static Friction is "magic", varying between zero and its maximum value of µ times the Normal Force. Kinetic Friction is always µ times the Normal Force. Examples using our 125 kg crate sliding on the floor. If object is at rest, need to "test" to see if an applied external force exceeds the maximum static friction force ("breaks the static friction barrier"). Static Friction can vary from zero to its max value in either direction. Friction while driving. Rubber on dry concrete, coefficients are 1.00 and 0.800 . Tires rolling with friction on good roads -- this is static friction not kinetic friction because the tires aren't sliding on the pavement. If you panic and lock up the wheels (stop their rotation) and switch from static to kinetic friction, you will take longer to stop because µk < µs and so there is less available friction force. Anti-Lock Brakes and Traction Control. ABS works by monitoring the rotation of all four wheels. If one wheel begins to "lose it" and slip on the road while braking, it will slow its rotation faster than the other tires, so the computer releases the brake on that wheel only until it is rolling without slipping again. This can be done many times a second, much faster than the good old "pump your brakes to stop on ice" trick older drivers are familiar with. Traction control uses the ABS sensors to monitor the wheel slip during acceleration -- keeps the wheels from spinning. First Sample Exam 2. (Click here for a copy.)

    Week of February 14-18, 2011.

    Monday 2/14: Return X1. Discussion of "The Great Reality Check". Reminder that your X1 grade isn't the end of the world.

    Tuesday 2/15: More with Forces and Tensions: Hanging a sign with angled wires -- still the same procedure: Sketch of the problem, Free Body Diagram, Sum of Forces equations in the x- and y-directions, solve for unknowns. Note that when the two wires have different angles, 30° and 45°, that T1x and T2x still have to cancel each other. Also the two tensions, T1 and T2, are each supporting more than half the weight of the sign, but less than all the weight of the sign, because they are pulling against each other. Work: A Physics Definition (Work = Force times distance in the same direction). Work = Energy. Pay particular attention to Units. Dot products: one of two methods of multiplying two vectors -- this method generates a scalar, which is a good thing because Work happens to be a scalar, which is Work's virtue (i.e. why we care). Dot products: run through two 3-dimensional vector case. W = F d can only be applied when (1) the Force is constant and (2) the Force is parallel to the displacement vector, i.e. the angle between the two vectors is 0°.

  • Linked for your amusement: Climbing a 1700 foot television tower... via The Register in the U.K.
  • NOTE: If you are following the textbook chapters, you may notice that Dr. Phil is not taking the same order as Serway. We are not yet done with Chapter 5, even though we are doing some topics in Chapter 6.
  • With the definition of Work and the Work-Energy Theorem, it is possible to start here and work backwards, developing the concepts of Force, Acceleration, Velocity, Position and Displacement, along with Newton's Laws of Motions, all the way back to our first lecture.
  • Wednesday 2/16: We can talk about: the Work done BY something, the Work done ON something and the net Work (total work) done ON something. Kinetic Energy -- an energy of motion, always positive, scalar, no direction information. Work-Energy Theorem (net Work = Change in K.E.). Re-derive using algebra: Wnet = Fnetd = mad and the equation without time. Get K.E. and Work-Energy Theorem, but can only claim that it is a general result -- calculus derivation earlier didn't have that restriction. Lose angle and directional information because energy is a scalar, not a vector. Example: Revisit The guy with the fedora and the cigar. Initial speed is v = v0x, find final speed. Only force is gravity. Get exactly the same answer as before, but without x- and y-components. Hooke's Law (Spring force) is an example of a Force which is not constant. Coil springs can be open coil (stretched or compressed) or closed coil (stretched only). Quiz 9 take-home quiz on Friction, Work and the Work-Energy Theorem, and due on Friday 18 February 2011, in class or by 5pm.

  • As pointed out in class the other day, we are doing chapters out of order -- and I have finally gotten some sample problems from the end of Chapters 7-9 for your studying purposes.
  • Sample Book Problems (not to be handed in): Chapter 7: 1, 5, 9,11, 13, 15, 17, 23, 27, 31, 39, 41, 43, 49, 57. NOTE: these are from the WMU 8th edition.
  • Sample Book Problems (not to be handed in): Chapter 8: 3, 5, 7, 15, 22, 43. NOTE: these are from the WMU 8th edition.
  • Sample Book Problems (not to be handed in): Chapter 9 (Set 1): 1, 7, 18, 19, 23. NOTE: these are from the WMU 8th edition.
  • Thursday 2/17: Work through some examples of Work, the Work-Energy Theorem, Conservation of Energy. For a conservative force, U = -W done by that force. Gravitational P.E. Ug = mgh = mgy , Spring Force P.E. Us = ½kx² . For Conservation of TME, K1 + U1 = K2 + U2   yields: U1 - U2= K2 -K1 or delta-K = - delta-U. Total energy limits maximum height. Lose angle and directional information because energy is a scalar, not a vector. Example: Roller-Coaster. If speed at top of the first hill is about zero, then this P.E. is all we have. Cannot get higher, but we can change height for speed.

    Friday 2/18: For a conservative force, U = -W done by that force. Gravitational P.E. Ug = mgh = mgy , Spring Force P.E. Us = ½kx² . For Conservation of TME, K1 + U1 = K2 + U2   yields: U1 - U2= K2 -K1 or delta-K = - delta-U. For non-conservative forces, like friction, Wnc < 0. Non-conservative forces "waste" energy, they do not store it and can not restore the energy to the system. K1 + U1 = K2 + U2 - Wnc . Power = Work / time. You should know that 1 h.p. = 746 W. We need another Physics quantity, one which describes the "relentless quality" of motion, one that includes mass. Inertia or Linear momentum: p = m v. This is a vector. Newton's form of 2nd Law: F = dp/dt, not F = ma. More Conservation Laws in Physics. Two extremes in collisions: Totally Elastic Collision (perfect rebound, no damage) and Totally Inelastic Collision (stick together, take damage). Linear momentum is conserved in all types of collisions. Example: The Yugo and the Cement Truck. Third Sample Exam 2 (Click here for a copy.) Quiz 10 take-home quiz on Conservation of Energy and the Work-Energy Theorem, and due on Tuesday 22 February 2011, in class or by 5pm.

    Week of February 21-25, 2011.

    Monday 2/21: Dr. Phil has canceled his classes due to treacherous roads.

    Tuesday 2/22: Linear momentum is conserved in all types of collisions. Three example collisions: head-on, rear-end, 2-D. (The Non-Collision -- if the car following is going slower, it isn't going to run into the car ahead. PTPBIP.) Seat belts, shoulder belts, steel beams in doors and crumple zones. What happens in a wreck. The myth of "better to be thrown from the wreck." How airbags work. Momentum, p, is conserved in all collisions, but K.E. is not conserved, except for T.E.C. We can see this in the case of the head-on collision with identical momentums -- the final speed is zero and so all K.E. is lost during the collision. Where does all the energy go? Into the non-conservative work to break, bend and stop things. Fifth Sample Exam 2. (Click here for a copy.)

    Wednesday 2/23: Review for Exam 2. Quiz 11 on Totally Inelastic Collisions, not due until Tuesday 8 March 2011, in class or by 5pm.

    Thursday 2/24: Exam 2.

    Friday 2/25: Spirit Day. (No Classes)

    Week of February 28-March 4, 2011.

    WMU SPRING BREAK

    Week of March 7-11, 2011.

    Monday 3/7: Momentum, p, is conserved in all collisions, but K.E. is not conserved, except for T.E.C. We can see this in the case of the head-on collision with identical momentums -- the final speed is zero and so all K.E. is lost during the collision. Totally Elastic Collisions -- perfect rebound, no damage, conserve both momentum and K.E. The equations get messy because each object has both an vi and a vf. Worse, momentum is a vector and can have components, while K.E. is a scalar and a square (½mv²). Two special cases: (1) m1 = m2 , v2i = 0, so v2f = v1i and v1f = 0. All the momentum and K.E. transfer from object 1 to object 2. (2) m1 = m2 , v1i = - v2i , so they just bounce off each other and go the other way. Close approximations: The Executive Time Waster. Demo: a suspended bowling ball shows conservation of T.M.E. All P.E. when swings up to a stop on either side, all K.E. at bottom of swing. (There must be non-conservative forces, such as air resistance and friction in the pivot point on the ceiling -- because the bowling ball never quite gets up as high as it starts.) Since a Totally Elastic Collision takes no damage, there must be a contact phase of compression and rebound -- this can be modeled with a Hooke's Law force and we can use conservation of Total Mechanical Energy. Why you want inelastic collisions in a wreck. 5 mph versus 3 mph impact bumpers.

  • What if... you made a car with soft, deformable body parts? So after a wreck you could just mold it back into shape? From Saturday Night Live: "Adobe: The Little Car Made of Clay". Alas, I cannot find the video itself online.
  • Tuesday 3/8: What's the opposite of a collision? An explosion. Or recoil. Example: When a gun is fired, the bullet goes one way and the gun barrel goes the other way. Example: A pitcher on ice skates at rest -- when he hurls a fastball to the right, he goes to the left. Total momentum of the system remains constant (in this case, zero). The Rocket Equation -- use conservation of momentum. NOTE: Serway's derivation is similar, but he does not make two points totally clear: (1) dm = -dM , the differential change of the mass M of the rocket is negative; (2) since Mf < Mi, then ln (Mf / Mi) is negative, but we already have a minus sign in the equation, so using ln (Mi / Mf) = -ln (Mf / Mi) puts in a second minus sign to cancel the first. Quiz 12 on Collisions-vs-Explosions, due on Thursday 10 March 2011, in class or by 5pm.

    Wednesday 3/9: The Ballistic Pendulum -- Old School Physics, in the days before all our modern electronics: We can find the speed of a projectile through an Inelastic Collision followed by Conservation of TME. Resistive Forces: Air Resistance. Low speed ( Fdrag = -bv ) and high speed air resistance ( Fdrag = -cv² ). If allowed to drop from rest, then a real object may not free fall continuously, but may reach a Terminal Velocity (Force of gravity down canceled by Drag force up) and doesn't accelerate any more. Ping-pong balls versus turkeys or pennies. Terminal velocity of a human depends on clothes and orientation.

    Thursday 3/10: Final thoughts on Terminal Velocity: World's Record Free-Fall. And 27 May 2008 Failed Attempt. UCM Revisited: Centripetal Force, Fc = mac = mv²/r. No such thing as Centrifugal Force. Only the Centrepital Force, which points radial inward, just like the centripetal acceleration. Note that the Centripetal Force is an ANSWER to the sum of forces equation -- it does not show up in the F.B.D. directly -- something has to CAUSE the Centripetal Force, such as a Normal Force (or component), tension, friction, or a combination of forces, etc. For a car on a flat road making a turn, the maximum safe speed for a turn comes from the maximum static friction. You aren't forced to the outside in a turn by some outward pointing "centrifugal force), you merely move in a straight line unless there is a force to keep you on the circle. Making "artificial gravity" for long-duration space flight by living in a rotating object. Test tube example. The story of the 50,000 rpm Ultra-Centrifuge and the Fresh Rat's Liver.

    Friday 3/11: UCM Revisited: Centripetal Force, Fc = mac = mv²/r. No such thing as Centrifugal Force. Only the Centrepital Force, which points radial inward, just like the centripetal acceleration. Note that the Centripetal Force is an ANSWER to the sum of forces equation -- it does not show up in the F.B.D. directly -- something has to CAUSE the Centripetal Force, such as a Normal Force (or component), friction, etc. For a car on a flat road making a turn, the maximum safe speed for a turn comes from the maximum static friction. For a car on a banked (angled) curve, there is a "natural" design speed where you can go around a curve with that radius without friction. Demo: Rodney Reindeer on a string. Example of a 14" and a 2" hard disk drive spinning at 3600 rpm ( f = 60.0 Hz). Frequency (Hz) = (1/sec), f = 1 / T. The guard around a circular saw blade takes the sawdust and broken bits which shoot out tangentially from the blade and redirects them to a bucket -- improves safety and makes less of a mess. With spinning objects is very easy to come up with enormous centripetal accelerations. Quiz 13 on Rockets and UCM, due on Tuesday 15 March 2011, in class or by 5pm.

    Week of March 14-18, 2011.

  • Week 9 Checklist.
  • Did you remember to reset your clocks an hour ahead for Daylight Saving Time?
  • Mid-Term Grades are now available via GoWMU.
  • Monday 3/14: Return X2. Short discussion of events in Japan over the weekend -- 9.0 magnitude quake, safety systems, tsunami damage, what's happening in the ten damaged nuclear power plants.

    Tuesday 3/15: General Rotational Motion: Translating Linear physics to Rotational physics (as "easy" as changing Roman/English variables to Greek). The radian is a "quasi-unit" -- it's not really a unit, but represents a fraction of a circle. (We can "wish" it away when we need to.) Angular position, angular velocity, angular acceleration, and the Kinematic Equations for constant angular acceleration. Continuing to fill in our Linear vs. Rotational table: Newton's 3 Laws of Motion applied to rotations, Moment of Inertia = "rotational mass", angular force = torque.

  • Regarding the situation with Fukushima I Nuclear Plant after tsunami damage in Japan, historical context with Three Mile Island in Pennsylvania and Chernobyl in the former Soviet Union:
  • Posted video of tsunami damage on my LiveJournal blog here.
  • Japanese Reactors Fukushima I (Units 1-4) (ongoing 2011).
  • Three Mile Island (1979).
  • Chernobyl (1986).
  • Article on Michigan and Midwest nuclear reactors.
  • Wednesday 3/16: Angular Kinematic Equations -- Example: A car traveling at 27.0 m/s in the +x direction comes to a stop in 5.00 seconds. The tires have a diameter D = 0.760 m and a radius r = 0.380 m. Assuming the tires are in good contact with the road (static friction), then we can use the linear information to find the rotational problem. Note that the tire rotation is clockwise, which is in the NEGATIVE direction. The angular acceleration alpha will therefore be positive.Find omega0, alpha, theta -- and then convert theta to the linear distance traveled during the stop. Although the variable names have changed, we solve the rotational kinematic problem exactly as we solve linear kinematic problems. We can even use the concept of an average angular velocity to solve problems. Continuing to fill in our Linear vs. Rotational table: Newton's 3 Laws of Motion applied to rotations, Moment of Inertia = "rotational mass", angular force = torque, rotational Work, rotational K.E., angular momentum, rotational collisions. Since Work is in Joules, we can find that torque must have units N·m, and the moment of inertia has units kg·m². Quiz 14 is two-page quiz on Rotational Motion, due on Monday 21 March 2011, in class or by 5pm.

  • Sample Book Problems (not to be handed in): Chapter 9 (Set 2): 25, 26, 33, 34, 35, 37, 41, 49, 51, 55, 70. NOTE: these are from the WMU 8th edition.
  • Sample Book Problems (not to be handed in): Chapter 10: 1, 3, 5, 11, 13, 15, 25, 31, 33, 35, 39, 41, 45, 49, 70. NOTE: these are from the WMU 8th edition.
  • In the class example -- a car traveling at 27.0 m/s comes to a stop in 5.00 seconds -- we were able to use linear kinematics to find the stopping distance -- and we got the same answer.
  • Note that if you do Q14 correctly: (b) = (e) and (c) = (d).
  • Fears vs Reality Over Radiation in West Michigan.
  • Chalk River reactor in Canada (1952) and then Navy Lt. Jimmy Carter.
  • Thursday 3/17: Extended Objects: We have been treating our objects really as dimensionless dots, that have been allowed to have mass. Now we want to start considering how that mass is distributed. An airplane with mass unevenly concentrated in front, back or to one side, may not be flyable. Center of mass is a "weighted average", meaning it combines a position with how much mass is involved. Center of mass in the x-direction: discrete case and 1-D uniformly distributed mass (Example: A meter stick balances at the 50 cm mark.) We have been calculating the motion of the center of mass all this time -- it's the dot in the Free Body Diagram. Mass per unit length (lamda = M/L), mass per unit area (sigma = M/A), mass per unit volume (rho = M/V). 2-D uniformly distributed mass -- Center of mass in x-direction and in y-direction. Rectangular plate. Note that the center of mass value depends on the coordinate system, but the center of mass point remains in the same place. Triangular plate -- parameterizing y = y(x) (y as a function of x) and x = x(y) (x as a function of y).

    Friday 3/18: Center of Mass (con't.): Demo: Suspending real objects from different points to find the center of mass -- hung from the center of mass, the object is perfectly balanced. Include: irregular plate, rectangular plate, triangular plate, Michigan (Lower Pennisula), Florida. The center of mass does NOT have to be located ON the object -- the obvious example is a ring or hoop, where the center is empty. Demo: The toy that "rolls uphill" -- actually, whether with the cylinder or the double-cone, the center of mass is going downhill. Start discussion of Moment of Inertia. We will be reproducing the results from Table 10-2 in your textbook, but you can use these results on your Formula Card and on this next quiz. Moment of Inertia, discrete case. Parallel Axis Theorem. Moment of Inertia by Integration, 1-D uniformly distributed mass. Axis through center of mass: I = 1/12 ML² . Axis from end: I = 1/3 ML² -- note that this is four times larger than through the center of mass, we had predicted that the moment of inertia would be higher when rotated from the end. Checked results with parallel axis theorem! First Sample Exam 3. (Click here for a copy.) Quiz 15 on Discrete Cases of Center-of-Mass and Moment of Inertia, due Tuesday 22 March 2011, in class or by 5pm. And yes, that means there are TWO quizzes out right now.

    Week of March 21-25, 2011.

    Monday 3/21: Review of 2-D and 3-D Integration (as PDF handout). Rectangular (area, volume), Polar (circumference, area), Cylindrical (volume, surface area). Spherical Coordinates (volume, surface area, hollow volume). Moment of Inertia of Ring, I = MR² by inspection, by integration.

    Tuesday 3/22: Moment of Inertia by Integration, Double- and Triple-Integrals in Spherical Co-ords. Moment of Inertia of Ring, Solid Disk. Moment of Inertia of Solid Cylinder, Hollow Sphere, Solid Sphere. Note that Ithin ring > Isolid-disk and Ihollow-sphere > Isolid-sphere, which makes sense because in the solid, more mass is near the axis of rotation, while in the hollow, more mass is far away from the axis of rotation. Rotational K.E.: Rolling objects down an incline (rolling without slipping). mgh = ½ mv² + ½ Iw², the energy available from the P.E. is split between linear K.E. and rotational K.E. Because I is a multiple of MR² and omega = v / r, the rotational K.E. term ends up as some multiple of mv². So we get cases, based on what type of round object we have rolling, and what its moment of inertia I is. Start with thin ring / hoop / thin-walled cylinder, with I = MR². Final speed v = sqrt(gh), which is less that v = sqrt(2gh), the speed of a block sliding down the same incline with no friction. With friction, however, the block may not even move!

    Wednesday 3/23: Rotational K.E.: Rolling objects down an incline (rolling without slipping). mgh = ½ mv² + ½ Iw², the energy available from the P.E. is split between linear K.E. and rotational K.E. Because I is a multiple of MR² and omega = v / r, the rotational K.E. term ends up as some multiple of mv². Depending on the moment of inertia of the rolling object, the final speed at the bottom of the incline varies, but does not depend on the mass m or the radius R. All are slower than a block sliding down an incline without friction. (With friction, we don't know what the block may do without more information -- it may not move at all.) Demo: A "race" down an incline between two steel balls of different sizes is pretty much a dead heat (mass M and radius R is not a factor), and finally between a metal ring, a solid disk, a hollow ball and a solid ball (here the finish order depends on the I used). Rolling objects need both linear K.E., because it takes work to move the center-of-mass, and rotational K.E., because it takes work to rotate the moment of inertia.

    Thursday 3/24: The Cross Product and Right-Hand Rule (R.H.R.). Using Right Hand Rule to assign directions to x,y,z coordinates and the sense of rotations for theta, omega (angular velocity), alpha (angular acceleration) and tau (torque) -- the vectors for these variables ends up pointing up or down the axis of rotation. Torque = r F, when applying a linear force perpendicular to a radius line to the axis of rotation. (Most torques we apply are done this way.) A "breaker bar" is a pipe used to extend the handle of a wrench -- this increases the torque for a given applied force, but the use of a breaker bar may damage the thing you are trying to torque. Angular momentum L = r p, when the linear momentum vector is perpendicular to the radius vector. The Cross Product (or Vector Product) is the exact opposite of the Dot Product (or Scalar Product). Multiplying two vectors together by a cross product gives us another vector (instead of a scalar). And the cross product is not commutative, vector-A × vector-B = - (vector-B × vector-A), so the order is paramount. How we solve force problems: (1) Free Body Diagram, (2) Sum of Forces equations, (3) Newton's Laws. How we solve torque problems: (1) Free Rotation Diagram, (1A) Choose an axis of rotation. (2) Sum of Torques equations, (3) Newton's Laws. Real pulleys vs. Perfect Massless Pulleys. Atwood's Machine with a real pulley. Get 3 equations with 3 unknowns -- the two tensions and the common acceleration. Note that two tensions, T1 and T2, are no longer equal, because they have to supply the net torque to rotate the pulley. Each of the tensions, T1 and T2, attach tangent to the pulley and therefore, by definition, are already perpendicular to the radius line. The equations depend on the moment of inertia, I, of the pulley, which depends on its mass and how that mass is distributed. Note that in connecting the rotational problem with the linear problems, the radius R of the pulley cancels. The acceleration of the two masses is less than the acceleration with a simple pulley, because it takes work and energy to rotate the real pulley. Still, the tensions and the common acceleration only change a little. When we simplified the Physics to do Atwood's Machine with a perfect pulley, the answer isn't too far off from using a real pulley. The more we add to the Physics, sometimes we don't change the answer much. So sometimes taking a simplified Physics approach is a useful approximation. Second Sample Exam 3. (Click here for a copy.)

    Friday 3/25: The "Free Rotation Diagram". Statics: objects not translating in any direction and objects not rotating in any direction. The teeter-totter or seesaw -- (1) if the pivot is located at the center of mass (c.m.), then two children of equal mass will balance the teeter-totter when located equal radius arms from the pivot; (2) if the pivot is located at the c.m., then an adult of mass M can balance against a child of mass m when the adult radius R = (m/M) r, or we can say that the adult has "cheated" or "scooted" forward on the board; (3) if the adult of mass M and the child of mass m are both located at the ends of the board, then they can balance if we move the pivot point closer to the mass M, but then the weight of the board at the c.m. of the board now provides a torque which must be accounted for in the F.R.D. Free Body Diagrams, Free Rotation Diagrams (sum of forces, sum of torques). Helpful hint: You can choose to put your pivot point (axis of rotation) anywhere you like, because if an object is not rotating, it is not rotating around any axis. So put the pivot point where one of your unknown forces is attached, and the algebra is easier. Unloaded bridge supported at ends by two support pier forces F1 and F2. Easy to show that F1 = F2. Loaded bridge, with truck located closer to pier 2, then F1 < F2. Diving board is a bridge where pier 2 is located to the left of the board's c.m. When we treat F1 and F2 as both pointing up, as in the bridge problems, it isn't apparent at first that F1 will be negative and therefore points down. This is not a problem -- a minus sign merely tells us that the vector force points the other way. Third Sample Exam 3. (Click here and here for a copy.) Quiz 16 on Torque Problems, due Tuesday 29 March 2011, in class or by 5pm.

  • Sample Book Problems (not to be handed in): Chapter 11: 1, 3, 5, 7, 11, 21, 23, 30, 31, 41, 45. NOTE: these are from the WMU 8th edition.
  • Sample Book Problems (not to be handed in): Chapter 12 (Set 1): 1, 3, 5, 8, 11, 19, 23, 25. NOTE: these are from the WMU 8th edition.
  • You can work out the Diving board problem with numbers at home and show that if F1 and F2 both point up in your diagrams, then F1 will be negative and therefore really points down: L = 5.00 m, distance between supports is d = 2.00 m, mboard = 655.0 kg, mdiver = 55.0 kg.
  • Week of March 28-April 1, 2011.

    Monday 3/28: Statics problems (no translation, no rotation about any axis) con't. The ladder leaning on the wall. Choose pivot point at floor to eliminate two of the three unknown forces from the sum of torques equation. Figure out whether the perpendicular components of the weight and the wall (normal) force use sin or cos of theta, the angle the ladder makes with the floor. Is there enough static friction to hold it? (theta = 70°, m = 15.0 kg, L = 3.00 m, mu's of 0.600 and 0.700 for floor only, no friction with wall.) If you're keeping score at home, Statics is in Chapter 12 of your book, and teeter-totters and the leaning ladder are both examples in the book. Stability of objects -- demo with heavy lead brick. Falls over when center-of-mass is unsupported. Tall & skinny objects much easier to tip over, than low & wide.

    Tuesday 3/29: Stability of objects. Objects fall over when center-of-mass is unsupported. Tall & skinny objects much easier to tip over, than low & wide. Stability around a curve also connected with previous discussion of stability and center-of-mass. Rollovers, "J-Turns" (a U-turn with a rollover), Jeep CJ (narrow width, high c.m.) vs. Jeep YJ (wider width gave more stability). Ground clearance and the HUMVEE. (Very wide width, relatively low c.m., and rim wheel drive means no drive shaft or differential housing hanging underneath vehicle.) Extended Objects -- Allowing for Deformation. Tension, Compression, Shear, Bulk. Stress = Force / Cross-sectional Area. Pressure also is Force = F/A. (SI units = N/m² = Pascal = Pa) Strain = Delta-L / L0 , the amount of deformation divided by the original length. Stress versus Strain graph. Linear region (no damage), elastic limit, plastic deformation, brittle and ductile failures -- failure occurs before the curve turns down in brittle failure, after the curve turns down in ductile failure. Tensile Strength, necking, voids, failure. Young's Modulus. Tension, Compression. Bulk Modulus. Table 12.1, p. 359. Steel has Young's Modulus Y = 20×10¹° N/m², Shear Modulus = 8.4×10¹° N/m², Bulk Modulus 14×10¹° N/m². Compare to Bulk Modulus of "nearly incompressible" water, 0.21×10¹° N/m². Example: Find the change in length, delta-L, of the steel cable that we used to hang the bowling ball for the conservation of energy demonstration. (L0 = 2.00 m, 16 lbs. bowling ball m = 7.27 kg, diameter of steel cable D = 4.00 mm.) Quiz 17 on Elastic Deformation by Tension, due Thursday 31 March 2011, in class or by 5pm.

    Wednesday 3/30: Example: Find the change in length, delta-L, of the steel cable that we used to hang the bowling ball for the conservation of energy demonstration. (L0 = 2.00 m, 16 lbs. bowling ball m = 7.27 kg, diameter of steel cable D = 4.00 mm.) We found delta-L to be 5.68×10-5 m = 0.0568 mm -- essentially no detectable stretching of the cable. (A slightly smaller delta-L was found by those who calculated using a square cross-sectional area of sides d = 4.00 mm, which makes sense because the square has a larger area than the inscribed circle.) For many materials, the Young's Modulus for tension and the compression modulus are the same. But some are different. Wood, for example, has different values depending on which way the grain of the wood points. Knotholes, where branches meet a larger part of the tree, have weak spots (voids). One way around these problems is to make plywood -- alternating thin sheets of wood with the grain in different directions held together by layers of glue. Some boards in construction are now made out of plywood, rather than single pieces of wood. On the other hand, particle board -- small pieces of wood bound together in a glue matrix -- is very weak because there are no long pieces of wood together, so it fractures easily. Pre-Stressed Concrete: Concrete is strong in compression, weak in tension. But you can cast large concrete beams with wires inside and tighten the wires to put the concrete into net compression, even if you put it in a tension situation. The four fundamental forces in nature, from weakest to strongest: Gravity, Electromagnetism, Weak Nuclear Force, Strong Nuclear Force. Gravity may be the weakest, but it holds us onto this planet and "binds the galaxies together". Fourth Sample Exam 3. (Click here for a copy.)

    Thursday 3/31: Review for Exam 3.

    April 4/1: Exam 3. (NOT an April's Fool Joke!)

    Week of April 4-8, 2011.

    Monday 4/4: Newton's Universal Law of Gravity (or Newton's Law of Universal Gravity). Gravity is attractive between any two masses. Same magnitude, opposite direction by Newton's Third Law. Use Universal Gravity to check "g". The value we calculate is close, 9.83m/s², which turns out to be only off by 0.2%. Why is it off? Because using Univeral Gravity in this manner makes the assumption that the entire Earth is uniform and homogenous from the surface to the core -- which it is not. We would need to integrate over layers to get the observed value of 9.81m/s². On the other hand, an error of only 0.2% suggests that treating a roughly spherical Earth as sphere of radius r with its mass at the center of mass, works pretty well. The Shuttle in Low Earth Orbit (Revisited). Calculating g(r) for r = 6,770,000 m (the radius of the Earth plus the height of 400 km for Low Earth Orbit), we get a value somewhat different than we found for the centripetal acceleration. Working backwards, we discover for this radius that the period T = 5542 sec and NOT the estimated 5400 sec (90 minutes) we had started with before. Each radius of circular orbit has a different value of g(r). As r increases, v decreases and T increases. Orbital mechanics: Speed up and radius decreases, slow down and radius increases. For the Moon, the period is around 28 days at a quarter of a million miles away.

    Tuesday 4/5: Orbital mechanics: Speed up and radius decreases, slow down and radius increases. For the Moon, the period is around 28 days at a quarter of a million miles away. Geosynchronous orbits occur T = 1 day exactly, and for geosynchronous communications sattelites, the orbit must be directly over the equator -- hence all sattelite dishes in the U.S. face south. Tides (high/low, spring/neap). Planetary Orbits. Ptolemy to Copernicus to Johannes Kepler. Models of the Universe eventually became Models of the Solar System as we began to understand just how big the Universe is. Early Man would certainly have felt as if the Earth was the Center of the Universe (geocentric). But the Sun is very bright and powerful, and it is also possible to construct models where the Sun is the Center of the Universe (heliocentric). One could also suggest which objects in the sky were nearer and which were farther. The BIG stumbling block was that some of the planets, most especially Mars, exhibited this bizarre retrograde motion, whereby the seem over the course of some nights to slow down, stop, turn around, go backwards against the fixed stars, then resume the normal progress through the stars. The problem of Mars in retrograde. Epicycles, elliptical orbits and Occam's Razor. Tycho Brahe's observatory and his data. Quiz 18 Take-Home on Universal Gravity and Kepler, due Friday 8 April 2011, in class or by 5pm.

    Wednesday 4/6: Kepler's First Law -- All orbits are ellipses, with the larger mass at one focus. Circular orbits are a special case where the semi-major axis is the same as the semi-minor asix: a = b = R. Kepler's Second Law -- The Equal Area Law is equivalent to a statement of Conservation of Angular Momentum. Kepler's Third Law -- T² = C R³, where R is the radius of a circular orbit, or the semi-major axis a in an elliptical orbit. There is one value for the constant C for every orbital system, i.e. one C for objects orbiting the Earth, another C for objects orbiting the Sun. More on Kepler's Laws: 1st Law -- Elliptical orbits, where a is the semi-major axis, b is the semi-minor axis and c is the offset between a focus and the center. For an ellipse, a² = b² + c² and the eccentricity is e = c / a . The circle is a special case of an ellipse with a = b = r and c = 0, making e = 0. 3rd Law -- Again, starting with the universal the graviational force FG , and using a circular orbit, we can set F = ma = mac = mv² / r . Using v = d / t = C / T = 2 pi / T, we can derive the relationship that T² = C r³ = (4 pi² / G M) r³ . Or T² = (4 pi² / G M) a³, since all elliptical orbits with the same semi-major axis a have the same period T. The constant (4 pi² / G M) holds for all orbits about the mass M. There is one constant for orbits about our Sun and a different constant for orbits about the Earth.

    Thursday 4/7: Three Classical States of Matter: Solid, Liquid, Gas. If chemical reactions do not occur, then there is a progression from Solid to Liquid to Gas as temperature increases, and vice versa. Combinations: Condensed Matter (covers both Solids and Liquids) and Fluids (covers both Liquids and Gasses). Two Extreme States of Matter: Plasma (electrons stripped off, high temperature), Cryogenics (extreme cold, odd behavior). Mass-to-Volume Ratio (Density). NOTE: Do not confuse the Density of the Materials with the Mass-to-Volume Ratio of the OBJECT.

    Friday 4/8: Mass-to-Volume Ratio Density). NOTE: Do not confuse the Density of the Materials with the Mass-to-Volume Ratio of the OBJECT. Density of Water built into the SI metric system (1 gram/cm³ = 1000 kg/m³). Note that a 1 m³ fish tank would have 1000 L of water, or over 250 gallons, and weigh over one ton. Pressure = Force / Area. SI unit: Pascal (Pa). Example: Squeezing a thumbtack between thumb and forefinger. One Atmosphere standard air pressure = 1 atm. = 14.7 psi = 101,300 Pa. Note that 1 Pa is a very small unit. Pressure at a depth due to supporting the column of liquid above, P = (rho) g h . The difference between Gauge Pressure (pressure difference inside and out, can be positive, negative or zero) and Absolute Pressure (total pressure, always positive or zero only for vacuum).

    Week of April 11-15, 2011.

    Monday 4/11: Return X3. Pressure at a depth due to supporting the column of liquid above. Water pressure = 101,300 Pa at depth h = 10.33 m. Why you need a qualified SCUBA instructor.

    Tuesday 4/12: Water pressure = 101,300 Pa at depth h = 10.33 m. How a straw works in a cup of liquid -- Physics Does Not Suck. Using a column of liquid to make a barometer to measure air pressure. Switch from water to mercury (rho = 13,600 kg/m³) changes h at 1 atm. from 10.33 m to 0.759m. The aneroid barometer. Pressure from a column of liquid looks like P.E. Create a Kinetic Pressure term which looks like K.E. to create Bernoulli's Equation. Water Tower and the Faucet Problem.

    Wednesday 4/13: Fluid flow = volume/time = Av. Bernoulli's Equation and the Continuity Equation. When speed goes up, pressure goes down. Example: The aspirator -- a vacuum pump with no moving parts. Example: Air flow around a wing. (Faster air over top means lower pressure on top, so net force is up -- generating Lift.) Classic F.B.D. for straight level flight: Lift vs. Weight and Thrust vs. Drag. The spoiler is a vent door in a wing designed to allow air to flow from bottom to top and thus "spoiling" the pressure difference and "spoiling" the lift. Why the Mackinac Bridge has grates on the inside north- and soundbound lanes -- inside lanes are open metal grates and cannot support a pressure difference. Quiz 19 Take-Home, on Fluids, due Friday 15 April 2011, in class or by 5pm.

    Thursday 4/14: Why Boats Float. Example: Front lab table as a 250 kg boat with 4.00 m³ volume. Not only floats, but floats very high. Calculating how much load can be (safely) added to our "boat". Buoyant Force = Weight of the Boat = Weight of the Water Displaced by the Submerged Part of the Boat. Calulating the amount of the boat submerged, by using the fact that the mass of the boat and the displaced water are the same. Why many ships leave seacoast harbors during high tide. First Day to turn in your Topic 1 Paper. First Sample Final Exam. (Click here and here for a copy.)

    Friday 4/15: Steel canyons and plywood towers -- pressure difference popping out the windows in a Boston skyscraper (see above). Note that the building passed its wind tunnel tests by itself, the problem came from not being the only tall building in Boston. (grin) More on Water Displacement: Archimedes and the Crown -- essentially a non-destructive test to find the volume displaced by the crown and comparing to the volume displace by an equivalent mass of gold. Floating in Air: The mass-to-volume ratio for air is 1.29 kg/m³. Balloons, blimps, zeppelins, etc. are all Lighter Than Air aircraft and can float or rise in air by either using a gas like hydrogen or helium with a density less than air, or by heating the air (hot air balloon) and expanding it to lower the total mass of the balloon. (As opposed to Heavier Than Air aircraft, like airplanes, which depend on Lift to stay in the air.) Periodic Motion, Waves and Resonance. Recall that for any periodic motion, such as U.C.M., there is a Repeat Time T (Period). Frequency f = 1/T . SI units (1/sec) = (Hertz) = (Hz). Revisit Mass on a spring, but this time in motion. F = -kx = ma. For an open coil spring, at x=0 there is no force. Stretch the spring to x = +d and let the mass go from rest and will oscillate back and forth. At x = ±d, v = 0 and |a| = amax. At x = 0, |v| = vmax and a = 0. (Think of conservation of energy, without friction energy goes back and forth between K = ½ mv² and Us = ½kx² .) Next, we need a set of equations that do this, as Fs is not a constant and we cannot use kinematic equations. Revisit Mass on a spring. F = -kx = ma. Generates a 2nd Order Differential Equation - sine & cosine solutions. Any time you have a conservative linear restoring force that can act as periodic motion you have a Simple Harmonic Oscillator that undergoes Simple Harmonic Motion. S.H.O. & S.H.M. Mass on a spring has an angular frequency omega = 2pi f= sqrt(k/m). Second Day to turn in your Topic 1 Paper. Second Sample Final Exam. (Click here and here for a copy.) Quiz 20 Take-Home on Floating and Sinking, due Tuesday 19 April 2011, in class or by 5pm.

    Week of April 18-22, 2011.

    Monday 4/18: Mass-on-a-Spring -- S.H.M. of a S.H.O. Angular frequency omega = sqrt (k/m). f = (1/2pi)(sqrt (k/m)). T = 1/f = (2pi)(sqrt(m/k)). Simple Pendulum -- All the mass m is in the bob, a distance L from the pivot point. The string or rod is considered massless. The Small Angle Approximation is that if thetamax is kept "small", then sin(theta) is approximately equal to theta (in radians ONLY). Although it has omega instead of x, and g/L instead of k/m, this is exactly the same 2nd order differential equation as the mass-on-a-spring, so the simple pendulum is also an S.H.O. undergoing S.H.M. The Simple Pendulum has an angular frequency omega = sqrt(g/L). For a Grandfather clock with a simple pendulum, period T = 2.00 sec gives L = 0.9940 m. For a mantlepiece clock with a simple pendulum, period T = 1.00 sec, L = 0.2485 m. Grandfather clocks made on Earth won't work right in space or on the Moon. (But a tortional pendulum mantlepiece clock will -- see Serway.) Physical Pendulum -- omega = sqrt (mgd / I). At first it looks like the mass factors into the angular frequency, which is not the case in the simple pendulum. But it is not the mass, rather how it is distributed, because the moment of inertia I also contains the mass m. Last Day to turn in your Topic 1 Paper. (Unless you had a Draft paper looked at by Dr. Phil.) Quiz 21 Take-Home, on Simple Harmonic Oscillators, announced in class on Monday 18 April 2011, hardcopies in class on Tuesday 19 April 2011, and due Wednesday 20 April 2011, in class or by 5pm.

  • Note what DOESN'T affect the angular frequency omega -- for a mass on a spring, the initial amplitude, x(0) = +d , does not matter, as long as we don't overstretch the spring beyond its elastic limit ; for a simple pendulum, the mass m does not matter, only that all (most) of the mass is in the bob and that the initial displacement is kept to a small angle.
  • Watch out for calculations requiring RADIANS versus DEGREES on your calculators!
  • Comments on the Final Exam: 200,000 points (twice the time, twice the exam), 4 problems, 50,000 points/problem, each problem has 5 parts, 10,000 points/part (same as regular exams). Star Problems are still 2(a), 2(b), 2(c) and 2(e), but there's 40,000 Star points (nearly half the total Star points) so 10,000 Star points per Star Problem. That's a total of 240,000 points on the Final, or about ¼ of your course grade.
  • It's Week 14 and I'm Still Seeing: (1) an entire quiz turned in without units on any answer or any regard to significant figures; (2) Star Problems written down without any calculus expressions, (3) confusing integration with differentiation, (4) failure to draw a proper F.B.D. or F.R.D., (5) guessing at units instead of using dimensional analysis.
  • We're covering a wide range of material this week -- why? (1) Because having exposed you to wave motion and the wave equation, when you see these things in PHYS-2070 next semester, you will have some background. (2) There are some simple problems -- even "simple" Star Problems -- that we can do with waves and wave motion.
  • Remember: If in a Star Problem you are asked to "show if this equation is a solution", then plug the solution into the Physics equation and see if left side equals the right side of the equation. If Yes, then it is a solution, if No, then it isn't.
  • Tuesday 4/19: Rewriting our solution: Since the sine curve and the cosine curve are the same shape, just offset by 90°, we can write x(t) = A cos ((omega) t + phi), where phi is called a phase angle to shift between cos and sin functions and linear combinations of the two. Quick mentions: Damping with Drag Force = - bv. Adds an exponential term to our S.H.O. solution, because our differential equation now has both a first and a second derivative in it -- see Serway. Three cases: Underdamped, overdamped, critically damped. Need a tuned suspension with shock absorbers to drive a car safely on the road. Waves: Single Pulse vs. Repeating Waves. The motion of the material vs. the apparent motion of the wave. Demonstration: the Slinky shows both longintudinal (string type) and transverse waves (sound type). Hand out Q21.

    Wednesday 4/20: For Repeating Waves, we have a Repeat Length (wavelength) and a Repeat Time (Period). Frequency = 1/Period. Angular frequence omega = 2 pi f. Wave Number k = 2 pi / wavelength .Wave speed = frequency × wavelength. Using our knowledge of the differential equation of the S.H.O. and its solution, we can generate the Wave Equation for a traveling wave moving through space (x) and time (t). Superposition Addition of Waves: Constructive and Destructive Interference. Resonance allows us to see the wave confined to the geometry of the problem. Standing Waves on a string. Fundamental, First Overtone, Second Overtone, etc. Demonstration: First and higher overtones on a string driven by a saber saw. Vary the wave speed by changing the tension -- v = sqrt(T / µ), where µ = mass/length = m / L . Standing Waves in a tube. Demo: Variable length organ pipe (closed at one end), plastic tube (open at both ends). Musical instruments: Accoustic string instruments have a resonance box and can have many strings (piano) or few (guitar, violin). Brass instruments start from the "natural trumpet", which can only play the fundamental and overtones for the pipe. It is a mixture of overtones in various proportions to the fundamental which allows us to tell instruments apart. Quiz 22 Take-Home on Resonance and Standing Waves, due Friday 22 April 2011. NOTE: This is the Last Quiz which requires Physics. Remember, of Q1-23, the lowest three scores (including zeroes) are automatically dropped.

    Thursday 4/21: Musical instruments: Accoustic string instruments have a resonance box. Brass instruments start from the "natural trumpet", which can only play the fundamental and overtones for the pipe. Woodwind instruments get more complicated. Demo: Tuning forks require both tines to work -- the "sound of a tuning fork with one tine" is that of silence. 256 Hz tuning fork attached to a resonance box optimized for the wavelength of sound from f = 256 Hz at room temperature. A second indentical box will sympathetically resonate in response. "Normal" human hearing is frequencies from 20 Hz to 20,000 Hz. Artilleryman's ear -- mid-range hearing loss. Beat frequencies occur when two sounds have almost the same frequency -- get a distinctive wah-wah-wah sound, whose beat frequency = | f1 - f2 | .Constructive and Destructive Interference. Acoustics of concert halls. If you exceed the wave speed in a material, you get a Shock Wave -- distinctive V-shaped pattern from front and back of moving object. Sonic booms in air (actually get a double-boom, because of the two V's.), wake from a boat in water.

    Friday 4/22: Last Day of Class. Review.