Over, and more!

This course is officially over now!? It feels very abrupt to me, certainly, especially after all the work-load that kept me on the verge of nervous breakdown for the past one week or two, right up until today... I guess I now know a thing or two about teaching a large course like this!!

Thanks to all of you who participated in all the discussions. Many of those discussions were great!

It was a great pleasure to be your instructor for this course! Sorry to have made many of you feel "uncomfortable" in exams and such. Although I did not really mean to do that, I realize that it is something necessary to some extent.

As for the final outcome, that letter grade, I believe you can check it out now, since I posted all of them today.

Also, the solutions to the final exam are uploaded in the sols+ folder.

I have many thoughts about this course. Many things were enjoyable and some weren't. I am sure most of those thoughts will be good for me in the future. If you have some helpful thoughts to share with me, come knock on my office door in the new year.

I hope all of you have happy holiday breaks and come back fresh for more constructive struggles!

In the mean time, when you look up the sky with your loved one(s) at day or night during this holiday season, chew on this question, if you don't have any other thing to think about (or even if!). We had this thing about the circular orbit of a space shuttle. Without doing anything else than just applying very quick forward thrusts it is possible to put this shuttle in another circular orbit with a larger radius. The surprise is that the speed of the shuttle is smaller in the larger circular orbit! Just one thrust will not do, but two exactly measured, very quick, thrusts will do! It may sound a little absurd. How is it possible that each time the thrust is applied the speed increases, but, in the end, the final speed is less than the initial speed? Actually, this is exactly what happens! [Partial answer: the intermediate orbit is not a circle!]

Happy holidays!

The subtle thing

OK, the center of mass moves like a point mass (= total mass of a compound object) under the influence of external forces only.

Now consider a non-center-of-mass point of the compound object. Does the external force have any influence at all on that point? The answer is Yes, Yes, Yes! In general, any point of an object is under the influence of external and internal forces. The center of mass happens to be where all internal forces magically cancel out! It does not mean that external forces do NOT apply to other points.

However, since it often suffices, or it is the most important, to figure out the motion of the center of mass, we are used to reducing a problem involving a complicated compound object to a point mass problem (without forgetting an internal rotation or an internal motion, if applicable), but we should never forget that external forces act on other parts, and the center of mass feels external forces through those other parts around it.

Just a subtle point to make clear, in order to avoid any misunderstanding. Maybe it is already clear to many students?!

Done with homework!

I've downloaded all homework scores and, even though the MP program might be still recording scores, those scores won't be reflected in my grade book any more.

So, we are done with the homework!!!
Kudos to all of you for all the hard work!!!

All the solutions are available at the "sols+" folder at the course web site.
I also plan to upload solutions to the final practice exam "soon"!

Review session schedules

Some students still ask me the question, "is the final going to be cumulative?".  The answer is No, and then Yes.  It is No, since it will cover rotational motions and afterwards, as I announced it before.  However, it is Yes, in the sense that if you do not have cumulative knowledge of what went on before, it would be very hard to do problems!  Practically speaking you should review Homeworks 6-9, PracExam, Bear problem, your note, lecture notes, book.

Fri Dec 5 2008

3:30PM - 5:00PM

Space Assignment(s): Nat Sci Annex 101

(space is limited; only ~ 100 seats;

if you are unable to attend the sunday session, please come early to this one!)

Sun Dec 7 2008

3:00PM to 4:30PM

Space Assignment(s): Thimann 003

That "just for fun" problem

I think the discussions and questions for that just for fun problem of going through the earth were very good.

Some more thoughts here.

(1) If the density of the earth is dependent on radius, becoming larger as radius becomes small, how would it affect the round-trip time?  Ans: It will shorten the time.

(2) In this case, will the shell theorem still applicable?  Ans: Yes.  See my lecture note about the exact statement of the shell theorem.  Newton's shell theorem is applicable when the density is the function of r only.   The density does not need to be uniform.  As long as the density is not orientation dependent (again an assumption here), the shell theorem is valid.

(3) In all these "calculations," simple and "stupid" assumptions like uniform density is made.  Why should I trust the calculation?  Ans: In engineering, the inaccuracy of these results may be viewed as great deficiency.  In physics, the inaccuracy of these results may be viewed as problematic in some cases (if you are trying to build a high precision experimental equipment based on these estimates) but it would not prevent the results from being interpreted as insightful in other cases (if you are after only an order of magnitude estimate, to be compared with other numbers based on other views).  Generally, these order of magnitude estimates are very valuable, and so "stupid" does not really mean that.  Even if you are into doing fancy computerized calculations, these simple calculations give you an idea what the answer should be approximately, and play a great role of preventing mistakes in super duper computer calculations.  Physicists like to simplify things and draw a rough sketch first.  More often than not the sketch already contains all essential physics! 

When I say "stupid" or "crazy" in class, I actually likely to mean "smart" or "cool"!

(4) Just to make sure people are on board on this.  To avoid Coriolis force to mess up with our "travel through the center of mass" (since the earth is spinning), we need to do it from the north pole to the south pole, or the other way around.

Those were great questions!

Comments about the EC (format)?

The Doodle polls about the EC exam are coming in. Nice.

If you like to let me know your reasons why you voted so in the Doodle poll, please do leave comments here or just send me emails.

Update: The overall poll results so far (Dec 05, 2008) are in.

Poll: Was the EC better LEARNING experience?

Yes, it was better learning experience.	No, I did not learn much.	Don't know -- probably no difference.
53	34	27

Poll: Was the EC better experience for being EVALUATED?

Yes, I felt better about being evaluated this way.	No, I think students can't get evaluated fairly using this type of exam.	Don't know -- probably no difference.
29	52	27

All these votes, except couple votes, were collected before my actual evaluation went out. I do not know whether there is any difference between "before" and "after."

Thanks to those who voted!

HW #9: the last homework!

NOTE (for students who do not know the meaning of these answers)

--------------------------------------------------

these are not answers to your questions.  i am sorry for the misunderstanding.  i have been posting answers to the even numbered problems which are randomized for each student.  this is the helping device to student.  it is not meant to give the answer itself.  it was explained in one of the previous blogs.  i started doing it from requests of couple of students.  these are answers to textbook version questions.  the idea is that you figure out the procedure to do the problem using the textbook numbers, cross-check your answers with these, and then move on to calculate the answer for the MP questions.

--------------------------------------------------

Hint on 9.31 (how to do it without moving your hand)

--------------------------------------------------

In today's (Dec. 3) class, I told you that the 1D elastic collision with m much less than M corresponds to 9.31.  Here m is the mass of the ball.  M is the mass of the car.

However, note that M is not at rest initially, in 9.31, as opposed to what we did in lecture.  How can we still make use of what we did in class?

(1) Convert the velocity of the ball, as reference to the moving frame of the car.  (2) In the moving frame of the car (i.e. from the driver point of view), the car is always at rest, so what we did in class is applicable.  So, figure out what is the final velocity of the ball within that frame.  That was a very simple result!  (3) Convert the velocity back to the child's rest frame velocity.

These calculations are mouth-fuls to describe, but they are simple enough so you could do them in your head, in principle.

Or, you could just set up the momentum conservation equation (where you would neglect certain terms based on masses), and that "grand result equation" for 1D elastic collision: v1i - v2i = v2f - v1f, and then just solve them.

--------------------------------------------------

9.12: 75 cm from the center

9.76: 0.33 J
13.64: (a) 6.5 cm, (b) 0.51 s

Good luck!!

Please note -- this homework is due on Thursday, the 4th, and no late homework will be accepted.  This is so that we can move on to the final exam, as smoothly as we can.

QR: Bear on High Wire

Due to a request, the demo today became an instant quiz question to discuss. I like you to think about this carefully. This is a great problem.

NOTE: If you submitted a late quiz report, you have gotten a 50 % credit for this one by default, without your doing anything. In order to get the full credit, you need to write a report about this problem. I strongly recommend you do it!!

We had a little bear riding a small unicycle (pulley). The pulley was put on a "high wire". The bear happily went back and forth on it! It was thanks to the balance due to the heavy weights at the end of the bear's extremely long plastic arms, (one of which I, not any student, helped break!!).

A student also pushed the bear from the side gently and then let go. The bear was seen to oscillate back and forth, and did not fall down, as long as the angular displacement of the bear was small.

Based on these observations,
(1) where was the center of mass of this bear (including the long arms and weights of course) -- was it above the wire, on the wire, or below the wire?
(2) how can you explain the (in)stability of the system in all three above cases of the position of the center of mass, in terms of the torque that the center of mass motion of the bear experiences?
(3) how can you explain the (in)stability of the system in all three above cases of the position of the center of mass, in terms of the potential energy function U(theta), where theta is the small angular displacement?

Additionally, how can you verify where the center of mass of the bear is, by using a string? Ans: hold the bear twice at two different orientations, and each time draw a vertical line using the string. The crossing point of the two lines is the center of mass.

Lastly, how can you measure the rotational inertia of this bear? Ans: Measure the center of mass position for this bear, and measure the frequency of the small angle oscillation. The angular frequency should be sqrt (g / (gamma L)) from our physical pendulum discussion, where L is the distance of the CM to the center of the rotation. So, from the measured values of L and omega, one can extract gamma. Finally, if one measures the mass of the bear using a scale, then we are done. The rotational inertia is then given by I = gamma ML² (from our definition during our discussion of the physical pendulum).

Other comments on extra credit problems

(1) Question 49:  The equation v2 = Agr was meant to be v2 = Agr*cos(theta_d).  In either case, A is just a constant, but I much prefer the latter equation.  However, you can give the answer assuming either equation, and I will grade it with two possible correct answers.

(2) Questions 90, 91, 92:  Please consider the monkey at the same height of the center of the rotor (the origin of the coordinate system).  I.e., the monkey is hugging the blade just so that his center of mass is at the same height as the center of the rotor.   So, no potential energy (due to gravity) to consider for these two particular questions, since it is zero.

(3) Question 37:  This question asks whether or not the direction of force n that you derive from solving Newton's 2nd law equation (radial components only) is physically correct.  I.e., OK means "physically plausible."   It is a choice problem between "is" and "is not".

(4) Misconception Warning!!!:  I heard that somehow there is a notion floating around that, in part 1 of problem 1, the bug can stay on the sphere up to 90 degrees of rotation.  This is incorrect!  The whole point of this problem is to show that this is not the case, and the bug leaves the surface much much earlier on.  I might point out to you that, if you are still confused about the answer of this problem, you shoul read problem 7.62 of the textbook carefully.  Also, you may have taken notes when I went over this problem in class.  Please master those notes.

What NOT to do for extra credit problems...

First what do to: There is only one (simple) principle governing the correct format of your answer sheet. "Do not modify any of the lines starting from, and including, 'YOUR ANSWERS START HERE.' except for writing down your answers."

Practically speaking, this is what you can do: Take my email to you with the subject line "Answer sheet for extra credit problems for midterm2," hit the "reply" button, fill in your answers, and then hit the "send" button. You don't need to, and better not, delete any text of that email!

Please remember that a computer program will grade your exam.
The following "what not to do list" may grow as I notice problems in the format of the solutions returned.

What NOT to do:
(1) Do not merge lines, even if answers repeat.
This is OK.
3: G (write one key, chosen from A to AC, next to the colon)
4: G (write one key, chosen from A to AC, next to the colon)
5: G (write one key, chosen from A to AC, next to the colon)
This is NOT OK.
3: 4: 5: G (write one key, chosen from A to AC, next to the colon)

QR: Three Basic Problems on SHM

It's been a while since I posted Quiz questions to re-visit. Feel free to discuss these three basic questions.

Q1:

Q2:
Q3:

Extra Credit Problems etc.

Hi, I just thought that it may help some people if I included some rough drawings like those below.

Your solutions can be submitted by email to me by 9 am, Dec. 2. [However, don't let these problems spoil your thanksgiving events. Do these problems sooner than later!]

The final exam will cover rotational motions, angular momentum conservation, SHM, and whatever we cover for the rest of the quarter. I will let you know more about it after thanksgiving.

Also, there WILL be a Wednesday class.
There is no lab this week.

Sig-fig reminder!

Sig-fig is an important matter, and here is one reminder to everyone (from the early lecture).

Keep more (at least one more) sig-figs in intermediate steps than what you need at the end. This is so that rounding errors do not accumulate. If you are storing intermediate results in variables (in your calculator or on your computer), then you would not need to worry about this.

What's in a name? (A and x)

There was one question after yesterday's class. I like to summarize my answer one more time here, since it touches upon the trivial but important and confusing concept. I may call the concept "all names are dummies."

For all names (A, x, y, z, v, omega, alpha, theta, ...) their meaning can change (sometimes very dramatically) depending on context.

The example here is the meaning of A and x in x = A cos (omega t + phi), in the context of "a person stretches a mass on a spring and then releases it to observe the consequent motion."

In this context, the initial stretch is A and the position of the mass at any arbitrary time thereafter is x.

However, the student noted that in previous lectures, x was used to mean the initial stretch (or compression) in many problems. Indeed this was the case. Actually, Hooke's law was introduced by using name x for the initial stretch or the initial compression.

The use of name x in apparently two different ways is not just me but also our textbook and other textbooks.

What you need to understand here is that in the SHM context we do need two symbols since we are considering the entire oscillation of a mass on a spring while in previous lectures we did not really need two symbols since we were not concerned with such a motion but we were just considering other things (like how much energy is stored in spring). So, in the context of the SHM, we have two things to describe -- the initial stretch and the time-dependent position of the mass -- which one should we call x? When given such a choice, x is usually used as a variable, namely in this case x is used to mean the latter.

Of course, it is extremely important to note that Hooke's law F = -kx holds at any time, in this problem, and so we are not comletely messing up with names, really.

I hope this is clear.

Vector/cross product, once and for all

It is neither surprising nor satisfactory to see that there is a fair amount of struggle about the vector product, i.e. the cross product. It remains absolutely true that the first two pages of Lec_11-10.pdf contain all there is to know about the vector product at this point of your life. However, I like to go over the vector product one more time to help those of you who are struggling with this.

Given two vectors A and B how do we obtain the vector/cross prodcut, A x B?
(1) Geometric method
(i) Align the tails of the two vectors, by parallel-shifting one of them.
(ii) Imagine rotating A to B with the tail fixed.
(iii) The right-hand rule is then applied to figure out the direction of A x B.
(iv) The magnitude is given by ABsin(theta), where theta is the angle between A and B.
(2) Numerical mothod
A = (Ax, Ay, Az), B = (Bx, By, Bz) => A x B = (AyBz - AzBy, AzBx - AxBz, AxBy - AyBx)

These two methods are equivalent to each other, of course. In my course, you are not asked to use (2). In later courses of Phys 6, you might have a chance to do that.

Here is a simple example, in the context of the torque.

EMAT Analysis (Mid-Quarter Student Feedback)

Hi class,

thanks to everyone who filled out the scantron form a little over a week ago to let me know how this course is going for you.  The majority of you participated in this EMAT (Electronic Mid-Quarter Analysis of Teaching) survey.  I just got results back from the CTE of UCSC.

I am happy to say that the vast majority of students view this course positively.   Please see the results summarized in the image below.

I am very happy that "respect, fairness, and communication" were rated extremely highly.  Indeed, I would regard the course as a failure if I did not establish trust with you, no matter how the course is going in other aspects, and this is the reason why your trust makes me very happy.

On the other hand, it seems that I should spend more time introducing new concepts, doing examples, and picking relevant homework problems.  Here, I have my own opinions, but also do respect what you think and will accordingly make some adjustments.

One thing that I am a little surprised by is the relatively low rating on the course reader (# 21).  Possibly there has been some misunderstanding here (or not?), since usually a course reader means a separate booklet or something like it(?!).  I meant to ask you about my lecture notes.

Another surprise is the time that you say you spend on this course.  I am sorry to say this, but you have to spend more time...  Minimum 15 hours per week seems to be the consensus in the physics department (that wouldn't be even half of what I spend! :>)  Looking at how much time you spend on homework, I was under the impression that many of you spend close to that, but maybe I was mistaken...

Well, all in all, I am quite happy about your trustful responses.  Now, we are well over half of this course.  I'd say that Newton's laws, energy conservation, angular momentum conservation, and momentum conservation are the main conceptual framework of this course, and so I feel that in a way we are almost done.  On the other hand, we do have several big concepts left (simple harmonic oscillator, collisions, gravity, statics).  So, let us not slow down.  I will change the way the course is going a little bit to reflect your opinions, so that your finish is easier.  Also, feel free to make any constructive suggestions.

PS: Do not worry too much about midterm 2.  If the results show that the majority of you struggled with this one, then I may come up with one or two "extra credit" problems for you to do at your own leisure.  This may not be too much of a burden, since this week's homework is going to be very thin.   However, I have not decided on anything yet, since I do not know how well you guys did on midterm 2, but I will keep you updated.

EX: Where is Nat Sci II Annex?

EX: Discuss your [repeat] questions here...

If you have some remaining questions on problems that I already showed how to do in class or in homework solutions, I like to ask you to discuss your remaining questions between your classmates, first. If you do not have any classmate nearby to talk to, then please leave your question here for your classmates to help you. This would be a more efficient and more meaningful way to help you learn and to help me evaluate your (and your classmate's) progress. If no solution emerges, then I may intervene.

[General Note] Your email inquiries about problems in homework or practice exam (or anything) will continue to be OK with me, but if the nature of your question is a simple "please do it again" type, then I will ask you to do this first: post your question on this blog site as comment and ask for help from your classmates. Also, although I try to respond to your email questions as best as I can, it is sometimes not possible to do so due to my time constraint.

[More Notes] In preparing for the 2nd midterm, please know that your efforts will pay off also in the final exam (since the final will cover many of the materials you are studying). For general advice about this type of course, I just found a yet another, potentially helpful list: http://physics.ucsc.edu/~josh/6A/studying_physics.html. One last thing: you should make sure that you get enough sleep tonight.

HW #7

You should not worry about this homework too much, until the midterm2 is over.
You should go over the practice exam first, and do all other preps for the midterm2 first.
So, come back to this posting later.

10.40: (a) 30 J, (b) 12 J
11.12: 61 rad/sec (You should figure the direction out! Discuss with your friends before submitting your solution!)
11.38: (a) 14 J s, (b) 57 N m

NOTE: L = I omega for almost all problems in this course, as in many problems of this homework set. Next week, we will see an example, where we have to use L = r x p (the more fundamental formula), but such an example is more an exception than a rule.

Keep up your good work in doing the homework!!!

HW: Math, math, math ...

Don't get me wrong. I love math, but this course is not about math, so I like you to have the easiest time with regards to math. You are supposed to think physics rather than math.

So, let us deal with some math issues that seem to be bothering you in connection with the current homework.

Trigonometry: For the Tarzan problem, you need to do the following.
From Tarzan's final position (but just before he drops from the vine), draw a horizontal line back to the vertical line, which represents the vine at the starting point. Then, consider the triangle generated by that horizontal line segment and the line that represents the vine in the final state.

Vector Product: For that mouse on the clock problem, figure out carefully what is the r vector and what is the F vector. Then, figure out the angle between them. Hints: (1) It will help you if you parallel-transport one vector so that their origins coincide. (2) For some strange reason, many of you think that the angle is 30 degrees. It is not. Neither the r vector nor the F vector is horizontal!?

Integration: For the last problem, you are not supposed to use the raw integration method! This is not a calculus class. I mentioned in class that I will NOT ask you to do any integration to obtain the rotational inertia. Instead, this problem is a "parallel axis theorem" problem. No integration required.

Good luck. Please post some questions that bother you. If it bothers you, it is most likely bothering other fellow students as well. Do not struggle alone. Keep bothering other people. However, do remember that in the end you have to be the one who puts everything together. Read my previous blog postings where I posted hints. If you are spending more than 30 minutes on a problem, it is time to stop and do something else, like posting your question on this site or have your favorite snack.

HW: What is the deal with those units?

A statement like this is quite common in the scientific literature.
"V = 3.8 h where h is in meters and V is in jules."
An equivalent sentence would be
"V = a h, where a = 3.8 jules/meters."
or
"V = 3.8 jules/meters h"
or
"V = 3.8 J/m h."
The last two are not recommended in your writing, since units and variables can be confused.

Why am I saying this? It is in relation to Problem 7.28. That coefficient in front of x^2 is NOT dimensionless/unitless. Rather its dimension is [U(x)] / [x^2].

Also, note the following definition of the turning points, given in 7.26: The "turning points" for this type of motion can be defined as the minimum and maximum values of x for the motion. What does this mean? Let me give you an artificial everyday example. Say, you are hiking by walking forward or backward only. Assume that the altitude changes on your path, so you are also going up and down as you go forward and backward. Say, you go a certain distance and at point A, you turn around. You then go some distances and at point B, you turn around. You repeat this... (because here "you" are not a real person but a roller coaster car or a pendulum or something of that sort). If one uses an x-y coordinate system, where x measures the horizontal position and y measures the vertical position, then the x values of A and B define the turning points of the motion. A and B are collectively minimum and maximum values of the x values covered by your motion, although it is not possible to say which is minimum and which is maximum, since that depends on the choice of the axis.

The Frictional Force Paradox (and the presidential election campaign)

Here is a summary of what I tried to get across in class, today, in regards to the frictional force. It is a confusing subject, and so I summarize it again. You can think of this as a more advanced topic, since it is not commonly discussed in these courses, but I tell you, if you are curious and bothered about this, you are quite normal!

Here goes the frictional force "paradox." Note that a paradox usually means that some assumption is invalid, and we will see that in the two solutions to this paradox as presented below.

Assume that there is a wheel and it is a perfect circle, and it is rolling on a horizontal plane. As explained in class, the friction between the wheel and the plane, if it exists, is a static one, since the circle and the plane touches only at one point, and the velocity of that point is zero. Other than this possible frictional force, we do not consider any other force. From this, the following three observations can be made, which may seem like a crazy situation -- i.e., it is a paradox.

(1) The work-energy theorem says that the kinetic energy change of this wheel is only due to work done by external forces. In this case, that work is zero, since at the point of contact the mass elements of the wheel is not moving. Since power, P=dW/dt, is F dot v, it means the power delivered from the frictional force to the wheel is zero. This means that there is no work done by the frictional force, and thus, the wheel cannot change its momentum or velocity.
(2) From Newton's law, the frictional force gotta be included in the equation of motion, F_net = ma. As we will see later, this equation describes the center of mass motion for a compound object. In fact, since the frictional force is the only force along the horizontal direction, it must be that the wheel slows down (assuming the frictional force is pointed opposite to the motion).
(3) As I mentioned in class, Newton's law in the rotational form tau = I alpha is applicable witin the center of mass frame. If the frictional force is applied opposite to the motion, it is easy to see that the torque due to the frictional force with respect to the center of mass is in the direction to accelerate the wheel!

It is quite a bothersome situation -- (1),(2),(3) are claiming completely different things! What is the solution? There are two solutions. (i) On a level surface there is no friction between a non-deformable wheel and the surface, or (ii) If there is a frictional force (whether it is a small rolling friction, the frictional force that will stop the car very slowly if gas runs out and we don't bother to use the brake, or a large (near-)static friction, the frictional force that kicks in when we brake) between the surface and the wheel while the wheel is rolling on a level surface, it is due to the deformation of the wheel (i.e. the contact is not just a point).

The paradox also disappears if one considers the motion of a wheel on an incline. See the following note, which by the way also explains today's quiz (not from the velocity point of view, but from the acceleration point of view). Note that, as theta goes to 0, f_s goes to zero as well, consistent with the possible solution (i) above. Of course, on an incline, there can be additional frictions due to deformation as well.

Side remark: In this presidential election campaign, there was an episode regarding one candidate saying "keep your tires properly inflated to save energy." This is a valid argument for the reasons discussed here.

EX: PracMid2

The practice exam for midterm 2 is uploaded as a pdf file, complete with solutions.

However, I advise you not to download and read the pdf file just yet, but simply to just know that that file is there.

Tomorrow morning, I will distribute hard copies of the practice exam (without solutions). You should work on the practice exam without any help first. Then you can come back to this message, click the following link to read solutions to compare with your answers.
UPDATE: The first file is without solutions for those of you who may not be able to make it to class tomorrow. The second file contains solutions.
http://physics.ucsc.edu/~gweon/teaching/current/PracMid2.pdf
http://physics.ucsc.edu/~gweon/teaching/current/PracMid2-w-Sols.pdf

QR: Rolling Revisited and Gyroscope

Here are tomorrow's quiz questions. For the first one, consider carefully the consequence of the energy conservation, including the rotational kinetic energy, if applicable.


This question is related to the Newton's equation which says tau = dL/dt. The point here is that the gyroscope is spinning very fast and the hand is exerting a small torque to change L = I omega. The answer is quite non-intuitive, due to the nature of the angular motion.

QR: Tension, Rotation, Rolling and Sliding

Here is the quiz question that we did in the last class, and the quiz question that we did NOT have time to do in that class but we saw the demo for anyway. Answers are indicated. For the first one, one needs to consider the tau= I*alpha form of Newton's equation, that can be worked out using the free body diagram of the pulley. For the second one, we need to consider the energy conservation including not only the translational kinetic energy but also the rotational kinetic energy. We have not covered the rotational kinetic energy, but we will do that on 11-10.

HW: New and Old

New homework is posted, as you may have noticed.
Here are the answers to the textbook version of the even-numbered "random-#" problems.
7.26: -2.0,2.0 m
7.28: (a) -6.7 N, (b) 0 N, (c) 4.5 N
10.22: 1.2 m

Here are some hints.

7.41: Simple energy conservation problem.
7.56: How much mechanical energy does the bug lose, each time it passes the sticky region? Think the work-energy theorem.
7.59: This is a GREAT problem. Energy conservation + Example 5.7. I.e., is the speed (and thus K) 0 or not, at top of the loop?
7.64: Simple energy conservation problem.
7.66: The power, the unit, and the work-energy theorem.
7.26: Energy conservation. What is the total mechanical energy, given v_max? What is the value of K at turning points?
Potential Energy Graphs and Motion: Even if you do not do this problem, please do note the following point. As we learned in class, an equilibrium point for a stable object in nature is described as the minimum point of a potential energy curve. However, this is not the only equilibrium point that can happen in nature. There can be a maximum point of a potential energy curve. This is called an unstable equilibrium point, as opposed to the former kind, which is a stable equilibrium point. Why are both of these called "equilibrium" points? It is because F(x)=-dU/dx=0, whether it is a minimum point, or a maximum point (or else). So, an object at that value of x, for which F(x)=0, with zero initial velocity will stay there forever. Thus, an equilibrium. Now, ask the question, what happens if the initial velocity is very very small, but still finite. [Strictly a side remark (outside the scope of this course): Indeed, you might ask "how can any object in nature maintain a truly zero velocity?" This is a deep question. Answering that question from any point of view you might have, the answer comes down to "not possible at all."] For a stable equilibrium point, the object oscillates back and forth around the initial point (e.g. a mass on a string, a child on a swing, etc.). For an unstable equilibrium point, the object never comes back to the equilibrium point again (e.g. that bug who starts from the top of that bald man's head; the top is an equilibrium point). I will say a few words about this in Wednesday's lecture.
7.28: How do we get F(x) from U(x)? (See the previous hint, if you don't remember.)
7.51: What is the equilibrium point? (Read that hint again, if you don't remember, interpreting r as x.)
Graphs of Linear and Rotational Quantities Conceptual Question: Remember that homework problem where you derived the centripetal acceleration (again) from scratch using your differential calculus technique alone?
10.19,46: 1D kinematics in the "angle language." (like Example 10.2)
10.22: Definition of torque. (Assume angle = 90 degrees, i.e. the most efficient angle.)
10.24: Definition of torque.
10.26: Definition of I.
The Parallel Axis Theorem: Another simple example for this cool theorem.
10.65: Identical to what I did in class, except the hole punched is a circle this time. This is one of those "must-do" problems in the first course in physics.

I have also updated files in the "sols+" folder of the course website. Solutions to HW #4 are uploaded. Also, please note that there are files, Sols-HW01-MP.pdf etc. (They were there before, too, up to HW03. Now they are up to HW04, and some of the printing problems of all previous files are corrected.) These "Sols...MP" files are the print-outs of the MP version of the solutions. Why do I have them there? It is because you can review those files to review past homework in preparation for exams. Let me know if these "Sols...MP" files serve this purpose for you.

QR: Rotational Inertia

Hi, I like to have this quiz question revisited! Please leave comments as to why the answer is B. In particular, could you please explain why the HS has a greater rotational inertia than the SC in this question? You should discuss qualitatively, but rigorously, without relying on the calculated values of rotational inertia.


OK, here is something. In the following proof, the parallel axis theorem is used implicitly when we conclude that IA > IB for each pair of pieces A, B means IHS > ISC. Here, A is an arbitrary piece out of all sliced pieces of a hollow sphere, and B is the corresponding sliced piece for the cylinder. That is r goes from 0 to R.

QR: Bug sliding on bald head

Unfortunately, we could do only one quiz question today, but that is alright. I think there is a lot to discuss for this one [classic] question. Please feel free to work out and discuss your answer to this question as much as you like.

This question is what I would call an "inverse roller coaster loop" question, but the idea remains the same -- the normal force and the energy conservation. So, you can discuss the roles of these two to determine when the bug will leave the head. [The answer is when the bug is down by r/3 in terms of height.] I went kind of too fast at the end of today's lecture and it may be that a lot of people did not get the details of the solution.

If you are really curious, you can also set up Newton's equation for this problem, and appreciate why that equation would be impossible to solve on paper.

QR: The force that gives back...

It appears that not all students were able to hear the good explanation that was given by the student in class on today's (only) quiz question. Sorry about that. So, let us have one or two students explain the answer to this question here on-line. Please volunteer and leave comments (especially if you need some extra credits).

LN: Potential function is well-defined for a conservative force

Up to a constant, that is.

We went over this in today's lecture, but I re-summarize it below in a slightly different way. I think it is better. In particular, the substitution of dr with -dr, as done in class, was not necessary, and that step was not particular clear in the lecture, I think.

HW: Set #5

6.62: You need to look up the last table ("energy contents of fuels") of Appendix C of the textbook.
6.65: Review Example 6.9 first.
7.21,7.24: Review Examples 7.4,7.5 first.

Good luck!
Questions can be posted as comments here.

Someone suggested that I give the answers to even numbered questions of the book that I use for homework (with numbers modified by a random number generator). I think it is a fair request, since then you can practice from the book to see if you get the answer right, before working on the actual MP question. So, here are the answers to the selected textbook questions (not the MP questions):
6.16: 1.9E8 N
6.80: 870
6.69: 42 kJ
7.16: 7.0 MJ and 1.0 MJ
7.18: 5.2 J
7.24: 2.3 kN/m

HW: Set #4, RPM and Chandelier

Perhaps this is a cultural thing. Vinyl record listeners probably knew "RPM" better.
RPM = revolutions per minute
If you express the period T, in minutes, then that is "minutes per revolution," i.e. the inverse of RPM.
Hope that helps.

It appears that the "Chandelier" extra credit problem causes lots of grief, due to the computer formatting problem. I tried to give all due credits back when students email me explaining [calmly, in an ideal world] how dumb the computer program was not to recognize their correct answer. Fair enough. Did I miss giving you the credit that you deserve? Did you have a similar problem that you did not email me about? In either case, let me know by email, right away!

You can ask for your due credit for any problem, when you are sure that your answer was correct. For instance, if you can verify that your answer is just another form of expressing the answer that the computer gives you, then you know that your answer must be correct. Also, please note that parenthesis and other formatting matters get in the way of computer understanding your answer.

When you do send me an email to get your due credit back, you should say that explicitly. That is, don't just say what headaches that you are getting from the MP program, but do say that you want your credit back if that is what you want.

HW: Beware of Red Herrings

Please know that some problems have too much information. If you use the information that is unnecessary for solving the problem, you may find that you are required to know strange things... Typically, if there is something that we really did not discuss in class, then it is likely to be a red herring.

In homework #4, the "rolling friction and bicycle tires" problem, is the pressure information a red-herring? Why don't you try solving the problem without using the pressure information first?

An example of a red herring in the past was the terminal velocity of a sky jumper. That terminal velocity was not essential for figuring out the answer.

A red herring does not necessarily mean a bad taste, in my opinion.

One more note: The rolling friction is similar to the static friction in its origin, but is much weaker. For the purpose of the above problem, just treat it as "another friction," like we did in class for static or skidding (thus kinetic) friction of automobile tires.

HW: Hints for 5.70

This problem is a "mechanical advantage" problem. Here is a hint question to help you, if you are unsure about this problem. How many ropes are pulling the load, effectively? In other words, if the tension of the rope that the worker is pulling is T, what is the effective force that the rope is applying to the pulley? If you do not know the answer to this one for sure, consider the free body diagram for the pulley. You may ignore the mass of the rope (implicit assumption here) and may draw a small rectangle that tightly fits the pulley. If you do so, you may consider all things included in that rectangle as a single compound object. Or, even more simply, you may draw a bigger rectangle that encloses the pulley and the load, and consider all objects inside that rectangle as a single compound object as heavy as the load (again, assuming other things do not weigh any at all).

HW: Hints and Grading in MP

The grading scheme in the case when hints are involved can be confusing. The grading policy says you "get credit for correctly answering a question in a Hint," but it also says you lose credit if you exhaust your answers or request the answer to the Hint. But, how does this work exactly?

Here are some texts from the MP help file, for you to read through and understand, in case this is still a mystery. Some points worth knowing are: (1) if you do not open hints at all you get 2 % bonus, (2) if you open hints but simply read them without submitting any answers, there is no penalty and no bonus (as long as your answer to the main part is good, I think), (3) if you open a hint and submit a solution to that hint, then that hint becomes part of the problem, weighed equally as any other part, (4) if you request an answer to, or exhaust the allowed number of tries for, a hint or the main problem part, then you lose points.

I hope this gives you some ideas, while it may seem (to me also) at times that it is impossible to understand the scoring system.

MP help texts below this line.

--------------------------------------------------------------------------------------------

For example, suppose a problem part has:

▪	3 question hints, each with 1 question to answer

▪	1 final answer

Case 1:

Student answers all hint questions correctly and also gets the final answer. Each hint question counts for 1/4 of the possible grade. Student has answered 3 hints and the final answer correctly, for a total of 4 out of 4. Student gets 4/4, or full credit.

Hint 1	100%
Hint 2	100%
Hint 3	100%
Main Part	100%
Part Score	100%

Case 2:

Student opens hint 1, answers it, understands how to work the problem, doesn't bother with the other 2 hints, goes right on and answers final question correctly. The student has correctly answered 1 hint and the final answer, and gets credit for having answered the other 2 hints correctly because the final answer is correct. 4/4, or full credit.

Hint 1	100%
Hint 2	100%	not answered; 100% because main part is correct
Hint 3	100%	not answered; 100% because main part is correct
Main Part	100%
Part Score	100%

Case 3:

Student opens hint 1, answers it correctly, goes directly on to final answer but gets that wrong (without exceeding the number of allowed answers). Student then goes back, opens other hints, and works them before succeeding in answering the part. Student eventually gets 4 out of 4 answers correct, for full credit.

Case 4:

Student answers all hints correctly but exceeds the number of allowed answers for the main part without answering it correctly. Student gets 3 out of 4 answers correct, for 3/4 credit. (This is a rare case; most students who answer all the hints correctly will get the final answer as well.)

Hint 1	100%
Hint 2	100%
Hint 3	100%
Main Part	0%
Part Score	75%

Case 5: Deferred credit

Student does not answer the main part correctly, does not exceed the number of allowed answers, but works 1 or more hints correctly. When the assignment is due, student does not get any credit because the main part has not been answered. (Remember: the main part must be answered.)

▪	However, if the professor gives partial credit for late work, the student can go back anytime until the end of the course and continue working for additional credit.

Case 6: The only case in which student is guaranteed to lose credit:

Student requests answer for a main part or a hint.

▪	If student requests answer to a hint: The student gets no credit for the hint even if the main part is correct.

▪	If student gets some hints right, but requests answer to the main part: The student gets credit for those hints, but not for the main part. The student gets no credit for any unanswered hints.

Student is not allowed to request the answer to the main part before opening at least one hint.

After the entire item has been completed, the student's total points are divided by the maximum points for the item equal to:

Max_points = number of main parts

This ratio is their score for the item, and can be greater than 1 if the student got mostly correct answers without using any hints and the hint bonus is on.

QR: The log and the friction

Hi, I like to re-visit this quiz question as well, since it appears that not everyone is 100 % comfortable with it.

Feel free to make comments.

QR: The Sun doing any work on the Earth?

Hi, this is "quiz revisited" for today's quiz. The question that I select is the last one, because I want to make sure that everyone understands the solution to this.
Please make comments about what the correct choice to this question is, and explain why. Also, change "the period of one year" to "one month," and then answer the same question.

NC: Scott Adams Quote (Dlibert)

Hi, my email mentioned something about the book "The Dilbert principle." The following quote is what I meant. This quote should totally lift you up (it definitely does me), not the other way around.

http://www.quotationspage.com/quote/32021.html

QR: Quiz Questions Revisited

Hi, class, as I receive late reports on the quizzes, I notice a pattern of some remaining misconceptions. Also, we simply do not have time to discuss everything in class. So, I suggest the following solution. I will upload Quiz questions as images here, so that you guys can revisit the quiz, at your own convenience.

I'd encourage you to leave your explanations to quiz questions in the form of comments to this message (and similar other future messages). I would like to see the simplest scientific explanation within the principles that we covered in class. Explanations based on everyday experience are OK, but only as far as the connections between them and the physics principles of our lectures are clearly made.

If a good final solution emerges from someone's comment, that person will be inducted into "the hall of fame of the best commentators," and she/he will be given the same bonus as a person who spoke out in class with a good explanation. As in class, I will also consider giving reasonable partial bonus points, for reasonable but incomplete comments. [Note: if you already got bonus points for the same quiz question or the same quiz session, please refrain from speaking out here.]

At this time, I have 5 quiz questions for discussions.

Here is Q1:

Here is Q2:

Here is Q3:

Here is Q4:

Here is Q5:

HW: Homework Grading Policy etc.

Some of you still ask me about homework grading policy and late homework policy. The answer is "check your homework." The following images should be self-explanatory. One thing that it does not say, though, is how late homework is accepted. That is one week, by which time you will get ~ 80 % of credit for the problem that you submit (as per the policy stated in the syllabus). Please note that the MasteringPhysics system is more lenient about late homework, than a traditional system, since it is per day and per problem basis. However, please do not make it a habit of submitting homework late, since it will be detrimental to your progress.

EX: That Tarzan problem

The solution to that Tarzan problem may be mathematically sound, but perhaps psychologically difficult to accept. In that case, try this "other solution," which tracks what goes on in Tarzan's mind as he pushes water and scans the shore line at each stroke.

LN: Weight and apparent weight ...

Weight can be a persistent, nagging, problem in physics, as we are finding in this class. Let me say a few words about it.

We defined weight to be mass times gravitational acceleration. One good (or bad?) thing about definitions is that we do not have to think about them!

Now, what is this concept about "apparent weight"? This is the weight that you actually measure. Right here is the confusion... Why do we not say "measure apparent wight" then? We could, and perhaps should, say that, but we usually don't. So, when questions ask "what is the measured weight" that means "apparent weight" not (the true) "weight"!! With this in mind, please review the last problem of Quiz_10-20 ("measured weight") and the first problem of Quiz_10-22 (weight vs. apparent weight).

Here is a related question for you. Suppose you are within a reference frame A (say an elevator accelerating down), which is accelerating relative to an inertial reference frame B (say your laboratory reference frame on a planet), in which the true weight is equal to the apparent weight. Let us say that unfortunately you do not know the gravitational acceleration value (taken to be constant; we are near the surface of the planet) in the frame B (i.e., you do not know which planet you are on!), nor do you know what the acceleration of frame A is relative to frame B. You have lived all your life in frame A. Can you ever figure out your true weight on that planet (i.e., in frame B), not the apparent weight in frame A? If so, how? If not, why?

Leave some comments, if you have brilliant/clever/mundane-but-sincere/just-so-darn-curious ideas, comments, or questions.

HW: New homework, new experiment

Hi, students and TAs of 6A, I created this blog to facilitate useful discussions pertaining to our on-going class. I have been receiving emails from students regarding homework problems or technical issues regarding the homework site. I've been thinking that an open forum like this may benefit more students.

I have never created any blog. So, this is a new experiment for me.

Please leave comments on this blog site if you have questions about homework or any topics worth discussing in relation to our class. Any students or TAs can leave comments and reply to other comments. Comments may be moderated. Any academic or technical topics related to class are welcome.

I would like very much to see interactions between students. If students interact within themselves and come up with solutions, that would be quite ideal and would benefit you the most in the long run. I will facilitate discussions, of course, and TAs are also welcomed to make comments as well.

OK, so much for the new experiment... The new homework (#4) is open for you. There are some extra credit problems. Good luck!