Tuesday, March 31, 2009

HW11

A little later than hoped, but here is HW11. Based on what we covered today, and sections 15.1-15.4 in the textbook, you should now enough to do all but 15.63 already. We'll go over some of the first ones in class on Thursday. Text-only version of HW11 below.

Problems due by the end of Thursday, 2 April: 15.26, 15.37

Problems due by the end of Friday, 3 April 15.55, 15.63, 15.106

HW11

Homework 11 will come out some time later today (Tues). They will all relate to Ch. 15 in your textbook. If you want to get a jump, read through Ch. 15 and study the example problems. This week's homework will not be quite as bad as last week's ...

As I mentioned earlier, there are no problems due this week until Thursday, since I'm asking you to give talks later this week.

Monday, March 30, 2009

Talks for Thurs/Fri

Thursday and Friday of this week, I would like each group to give a short oral presentation on their progress with the rocket project so far. A few formal details:

  • It is a group presentation - designate a speaker, or take turns. Either way.
  • Minimum 10 minutes, maximum 15 minutes per group
  • Powerpoint/etc are not required but may be used if you like.
  • We will start on Thurs and finish these on Fri. Pseudo-random numbers will choose your presentation order on Thurs.

As for the content, I would mainly like to see you present your basic approach to the assignment, where you are now, and what you think you still need to do.

  • What is required for the project?
  • What experimental quantities must be determined, and what is your progress?
  • What parameters must be controlled, and what is your approach?
  • What are the relative errors & their sources?
  • How will you determine the trajectory/angle at launch time?
  • What problems are still outstanding?
  • What solutions are you still working on?

I'll discuss this a bit more tomorrow, but hopefully you get the idea. This is really just meant to be a reasonably informal (but not too informal) status report, like an oral version of your memos, updated with all your recent progress.

You can keep the talk fairly general in some respects - I am not asking you to give all your secrets away to the other teams, but I do expect you to give concrete statements about what you have done. For example - quoting your launch velocity and error or discussing the algorithm you are using for your calculations is fairly general. Giving away your trick for highly accurate positioning of the turret or optimizations to basic algorithms is not necessary.

Grade-wise, this will count as a lab.

Tomorrow's class

Tomorrow, we will begin Ch. 15, "Oscillations," and continue with this material for the rest of the week. Start reading :-)

We will defer Ch. 14 on fluids until the end of the semester, for two reasons. First, the material is a bit fluffy at the level presented in the text. It will be nicer to cover this during our last week, when the PH125 Deathwatch begins. Second, I think it is more logically consistent to move on to oscillations now that we have covered all the core topics in Mechanics.

Next week, we'll begin Ch. 16, "Waves I" ... and the week after that is Exam III (April 14).

HW11

HW 11 will come out later today, but there are no problems due until thursday.

That is, I'm giving you the day off from homework tomorrow so you have some time to work on your presentations for Thursday & Friday. I'll post details about the presentation requirements later today to help you get started.

Friday, March 27, 2009

Friday's class

We'll delay harmonic motion until Monday. Tomorrow we will spend some time solving homework problems, and a little bit of time discussing the rocket project and your (informal!) presentations for next week.

For the rockets, I'll sketch out some algorithms and pseudo-code you can use to get the trajectory and then find the launch angle (some of you have already done this). There will be extra points for particularly clever solutions, so it is worth some time to optimize your calculations. I heard talk of pointers today, and it pleased me :-) I should mention that spiffy user interfaces and graphics are not necessary, we would rather focus on cleverness of calculation and efficiency for now.

I will also introduce you to the position sensors you'll be receiving next week (one per team) that you can use to determine the target's range and height.

Homework due at what time?

So I have said that the homework is due at the end of the day formally, i.e., 11:59pm on the day in question. That is still true. There are 'grey areas,' however.

Less formally, it will not be considered "late" if I am still awake after that time. Tonight that means you have about another hour or maybe 90 minutes. In general, 1-2am is a reasonable bet. If you're staying up until 2am doing homework, that is hardcore, and I'll give you some leeway ;-)

Minimal points are taken off for submissions arriving before I begin grading the homework, usually a day or two after the due date. Things happen, you get busy. Don't make it a habit.

Submissions after I have already posted the solutions online are taken on a case-by-case basis, but will receive some credit for at least bothering to copy the solutions. Learning by osmosis can work sometimes.

The message here is again: always turn it it. It is never given zero credit.

Thursday, March 26, 2009

HW 10 question 9

Updates to the hint to fix the glaring error. (with the proper link this time)

I might try to revise it a bit more to make it less terse ... I'll post here if I get that done.

Even-numbered problems for this week

13.16 3e-10 N
13.68
a : 92min
b : 7.67e3 m/s
c : 5.77e10 J
d : -1.18e11 J
e : -6.0e10 J
f : 6.64e6 m
g : new period ~89min; difference ~150 sec. Since Picard was initially behind by 90s, this means he is now ahead by about a minute or so.

I should double-check my numbers on #68, since I did it a bit quickly ... but I believe they are correct. Update: they are correct after a double check, except g) is closer to a minute.

Problem 9 / HW10

There is a mistake in the beginning of the hint ... the first expression for velocity v(y) is inverted.

I will fix that and clean it up a bit in the next hour. The proper expression for velocity is

v(y) = (const)*sqrt[y/(R+y)]

where "const" involves G's, M's, and R's. Possibly even a 2.

Keep an eye out here for the revision soon.

Constraints vs. real forces

It is confusing sometimes to realize what is a real, live force, and what is not. For instance, circular motion. You draw a free-body diagram, and sum up the forces. This is one side of the equation. The other side is your constraint on the motion, namely, that for a circular path a net force of mv^2/r is required. Sometimes, we call this "centripetal force," and that is highly misleading. It is not a real force, but a fictitious force, which basically means that if you impose the requirement that the motion is circular, an specific constraint is placed on your force balance.

When you think about it, your force balance always has a constraint, namely, that the net force has to result in mass times the acceleration required to produce the observed path. For a straight line path, zero acceleration, the force balance is zero: equilibrium. For a circular path, the required acceleration is v^2/r, so if circular motion is observed then the net force divided by mass must give v^2/r - if not, then you don't have circular motion. For generic paths, it is more complex: an observed path constrains the force balance, but it depends on the speed along the path and the local radius of curvature.

The confusion is, in my opinion, largely an unfortunate artifact of history and terminology. There is no centripetal force, it is just a boundary condition on your force balance that enforces circular motion.

Anyway: here's what I wrote to one of you earlier. We'll touch on this again in class tomorrow.

In the middle of page 336, in Equation 13-12 (F_n - m*a_g = m(-(omega^2)R)), why do we have a negative sign on the right side. I feel like gravitational acceleration (weight) and centripetal acceleration, both being directed toward the center of the earth, should have the same sign. Here, if you put them on the same side, they have opposite signs.

It is a little bit confusing because the centripetal force is not a true force in its own right, but only the *result* of all other forces. The language in the text is confusing in this regard.

What it is saying is that there are two real forces acting: the normal force acting upward, and the gravitational force acting inward. Since the box follows circular motion by virtue of being on the earth's (rotating) surface, we know that the *constraint* on the force balance in magnitude is that it must sum to mv^2/r. The constraint on direction is that it must be directed radially inward to result in a circular path.

Saying that "centripetal force points toward ..." or "the centripetal acceleration points ..." is misleading. The left side of the force balance contains all the real forces; the right side contains the constraints on the motion from the known path:

(1) straight line: constraint is that forces sum to zero
(2) circle: constraint is that forces give mv^2/r to be consistent with the path, pointing radially inward.
(3) general: what we derived a while back; sum of forces relates to the rate of change of speed and the local radius of curvature of the path.

So it is OK that they have the same sign - it says that the net force balance comes out in favor of gravity, and the net force (which we call the centripetal force, misleadingly) points in the same direction ... we stay on the surface and don't fly off.

Thursday's class (26 Mar09)

Today, we will focus on applying the general ideas I presented on Tuesday. This will start by motivating the full form of Newton's general law of gravitation, and the principle of superposition. This will allow us to handle the gravitational force between arbitrary (non-point-like) objects like spheres and disks and so forth, as well as the gravitational force above the earth's surface. You should have read most of Ch. 13 by now ...

As a side bonus, we'll figure out what really happens if you drill a hole through the earth.

This will lead us to gravitational potential energy, which will reduce to mgh close to the earth's surface, and lead us to the energy required to achieve stable orbits around a planet/star/etc. From there, we already derived Kepler's laws, and that is that.

I hope to spend a good amount of time on the homework problems for this week and some other examples, at the expense of what is normally reserved for lab time. In what time is left, I want to see your progress on rocket calculations and what your next plans are.

Friday, I will announce another small rocket-related project ... next week, each team will have to present a short progress report.

Wednesday, March 25, 2009

Problem set 10, number 9

Having worked it out myself last night, I realized that problem 9 is a bit more ... punishing than it really ought to be. The physics is easy, but the integral you need to solve is somewhat pathalogical.

Thus, the MASSIVE HINT, which sets up the integral for you and gives you the basic result. Note that I said "Use any means necessary to evaluate the integral required." This means you can look it up, perhaps with the Wolfram Integrator.

If you read the hint carefully, there are bonus points for solving the thing the hard way, and further bonus points for proving that it reduces to our usual expression for small heights.

Further hint: don't reinvent the wheel. What are the odds I made this problem up, and what are the odds that it comes from any number of advanced mechanics books?

Turning your homework in late ...

If I haven't started grading it yet, it isn't late. If the solutions are not posted yet, the grade reduction is likely to be minimal.

This is something of a gamble ... but always turn it in. It is highly likely that even disgustingly late and incomplete homework will merit some credit, which is better than nothing.

Tuesday, March 24, 2009

Sigh.

Snipped from an article on wired.com:
Toyota's bland bar graph makes saving gas about as much fun as a physics lecture.
Sigh. Having attended my fair share of physics lectures, they do make a good point ... you can only aspire to be an occasional exception to the rule.

One more hint

11.13
On 11.13, we managed to find alpha, but we can't find how long the ball slides because we don't know omega at that point. We know we can solve for the rest of the problem, but we're stumped right here.
Ah, but you do know omega at the point that slipping stops - at that point, it is pure rolling motion, and v = r * (omega). That's what you found in the first part.

You know that the linear velocity starts out at v_i, and ends up at v at the moment t that the sliding stops. That means

v(t) = v_i + at

where a is the acceleration you already found. For the angular part, you know that the ball starts out *without* rotation, so (omega)_i = 0. At the moment rolling without slipping starts, you know (omega) = (omega)_i + (alpha)t = (alpha)t. You also know what when rolling without slipping starts, v = r(omega). Put that together ...

v(t) = v_i + at = r(omega) = r*(alpha)*t

Now you know everything but t in the equation above ...

compiled HW hints

This is stuff that I sent to various people this afternoon ... just so you all get the same info. Forgive the lack of useful typesetting on the math, this is cut & pasted from plain-text emails.

11.13 (#3)
for number 3. i doesn't give us a mass of the ball, so are we supposed to calculate exact values or just symbolic solutions?
in the end, you shouldn't need it ... the mass should cancel everywhere.

for instance, on part b, you want the acceleration. the only force is friction, f = (mu)mg. acceleration is f/m, or (mu)g.

for part c, you want angular acceleration, which is (torque)/(moment of inertia). The force is the same f as above, acting at a distance r. the mass occurring in the torque will cancel with the factor occurring in the moment of inertia.

the other parts are similar - you only need velocities and so on, and since the accelerations are independent of mass, so are they. Let me know if you get stuck on a specific part. The answers of this one are in the back of the book too.

10.67 (#2)
We've pretty much figured out that our tangential acceleration is not
constant, which is pretty obvious since we're told to find it at 35
degrees, but we have no idea how to derive it. We managed to solve for
omega, but that really only gave us a single number, and there's not
really another way to solve for alpha, and therefore tangential
acceleration, since it isn't constant.
you are on the right track ....

the radial acceleration is just a_r = l (omega)^2, since the radial distance is l. That's half of it.

You can differentiate omega, but you want to use the chain rule when you get to the thetas - theta *is* a function of time.

actually, its easier to differentiate (omega)^2 implicitly and avoid the square root:

d(omega^2)/dt = 2*omega*(d omega/dt) = 2 * omega * alpha

Now if you know (omega)^2 = 3g(1-cos(theta))/l, then you also know

d(omega^2)/dt = (3g/l) * d(-cos(theta))/dt = (3g/l)*(sin(theta))*(d theta/dt) = (3g/l)*sin(theta)*omega

Here you have to use the chain rule when you differentiate sin(theta) with respect to time:

d(sin(theta))/dt = [d(sin(theta))/d theta] * [d theta/dt]

Thus, 2*omega*alpha = (3g/l)*sin(theta)*omega. This gives you alpha, which relates to the tangential acceleration via a_t = r*alpha = l * alpha.

Alternatively, you can find alpha by using torque (even though the problem tells you not to, bah!) ... the total torque is moment of inertia times alpha.

10.54 (#1)
For 10.54, is part (b) talking about directions in terms of x- and y-
coordinates or in terms of clockwise/counterclockwise? I'm a little
confused about that because the forces are represented as vectors in the figure.
Just clockwise or counter-clockwise is enough. It wants the direction of the angular acceleration for (b). The angular acceleration is an axial vector, so it has a direction, but also specifies the direction of rotation about an axis. You can specify either one and that is enough, but it is just simpler I think to say clockwise or counter-clockwise ...

So basically, write down which way should it rotate, that's enough.

Tonight's even-numbered question

10.54: 9.72 rad/s^2

The other two problems due tonight are odd-numbered, and therefore have answers in the back of the book.

Special lecture TONIGHT

Something I forgot to mention in class today that I just found out about:

At 7pm tonight Dr. Nathan Smith, UAB School of Medicine's assistant dean for admissions (and professor of Psychology) will be speaking at the AED meeting in 30 ten Hoor.

If you are interested in medicine, and contemplating medical school, this is a great chance to hear from someone who actually evaluates medical school applications for a living ...

More action photos

Rocket launchers in action.

Coding for fun and profit

By the way: if you enjoy coding, I can probably find you an independent study in the department doing it for a real, live research project. Seriously - there are really not many students around who 1) enjoy coding, and 2) are any good at it. If you can answer yes to even one of those criteria, there is a job for you ...

In particular, the astronomers are always looking for someone, as are the condensed-matter theorists. The particle physicists also do a serious amount of numerical work, and could probably use your skills.

If you think you would be interested, let me know. You have the possibility of getting 400-level physics or astronomy credit during the semester, and actual cash during the summer. There is also no reason why it couldn't be a CBH project, that is not hard to arrange.

N.B. - Fortran is still in common use in physics, as are things like C/C++, python, etc.

Tuesday's class / Gravitation

For better or worse, there will be much mathness tomorrow. Two things that can help either before or after the fact:
The idea is that we're going to "derive" Kepler's laws of planetary motion and Newton's law of universal gravitation from some basic principles and one astronomical observation (viz., the existence of orbits described by conic sections).

In the end, you will not be responsible for the derivations, only the main results. You will see the derivations again in PH301 or PH302 (and possibly MA227). The hope is to show you that the main points of Ch. 13 can be derived with a bit more work, which will ideally help you appreciate them a bit better. Also, all that scary math at the beginning of the semester will pay off again, which is nice.

Once that is out of the way, we'll work on some of the homework problems. Thursday, we will make use of our shiny new results, and be able to show that the remainder of Ch. 13 is a bunch of special cases.

Anyway: tomorrow will be a lot of 'I derive stuff not in the book and you watch' than usual, but not without good reason.

Sunday, March 22, 2009

HW 10

Yes, the spring break is over. Sigh.

Homework 10 is out, first problems are due Tuesday following the break. The first problems are more angular momentum and torque, after that you will need to read Ch. 13 in your text.

Final exam

Wouldn't this be fun?

Kidding. Good practice though, they are not all easy questions ... some of them do require Physics II stuff you haven't seen yet. Note 80% is considered a 'pass' by the Diesel Technician Society.

By the way: Moodle is now up to date, I spent quite a while grading over the break catching up. One thing I have noticed: the rate of missed (or partially missed) homework seems to be increasing ever so slightly lately. This is not shocking, being near the end of the term and all ... but be careful. It is not a large percentage of your grade, but the practice is crucial.

Finally: next week is the last week of advising. I am the undergraduate adviser for physics and astronomy. If you want/need to take more physics and astronomy courses, I'd be happy to talk with you about your options and give some friendly advice (that is, in addition to your regularly-scheduled advising sessions if you are not a physics major).

Wednesday, March 11, 2009

Homework 9

If you are stuck so far ... don't fret too much yet. Tomorrow we will spend most of the first hour of class on problem solving, specifically setting up the homework problems.

What we did Tuesday was very general and high level, what we missed was putting it all into practice ... and the homework problems are not easy ones. I think by the end of class tomorrow you'll be able to burn through the homework problems pretty quickly.

Photos

Here are a few cell-phone-camera photos of you sciencing today ...

(If for any reason you object to having your photo online, let me know and I will remove it.)

(Probably should have warned you that I was taking photos ... and I should bring a real camera next time.)

Tuesday, March 10, 2009

Homework 8 solutions / Tues class

Here you go.

They are a bit more terse than I would like, in the interest of getting them out quickly (and as a result of it being late). Hopefully they are still somewhat readable ...

For Today' class, we'll begin to discuss angular momentum and torque, which we will finish up on Thursday. Please read/skim Ch. 11 before class if you haven't done so already. We will devote about an hour to the rocket launcher project.

Monday, March 9, 2009

Homework 9

All problems are due by the end of Friday (meaning, realistically, before you leave for break). With one exception, they are not too bad. Here you go.

In the end there 6 questions on angular momentum and torque, and one odd math problem that somehow found its way to your problem set. You will need to know what a geometric series is ...

The even-numbered H&R problems have the following numerical results, for reference:
  • 11.14: 1.34 m/s
  • 11.16: 0.25
  • 11.66: 32 degrees (this is the bastard problem, FWIW)

Homework this week

For the record: there are no homework problems due tomorrow. There will be problems due on Thursday, which will come out this evening.

Physics Advising

This week begins the advising season for the F09 semester, which will continue through the week following spring break. If you are a physics major, thinking about a physics major, or plan on taking more physics in the near future, I would like to meet with you.

Individual meetings only take 15-30 minutes, please let me know if you are interested, and when you would have some time either this week or the week following spring break.

One new thing I should mention is our new seminar course, PH111. It is a 1-credit seminar course with no prerequisites, meeting once per week, and it is devoted entirely to discussing neat topics in physics at a completely introductory level. In the fall, it will meet on Mondays from 4-5:30. Anyone interested in physics is encouraged to take the course, it should be a lot of fun. A tentative syllabus can be found here.

Finally, I would like to point out that all physics majors are eligible for E. Scott Barr scholarships, and we could use some more applicants ... it is not too late to be considered for a fall scholarship! Details here.

Friday, March 6, 2009

Project memos

By the way: you can submit your project memos in whatever electronic format is convenient. PDF, Word, Pages, Tex, etc., or something more exotic. The odds are very good I can read it.

Trajectory calculations

Here's a quick snippet of C code that illustrates the algorithm I was talking about today. Some key bits are missing, but it should give you an idea of what to do.

You start with the initial velocity and parameters, and from that calculate everything a time increment dt later. Repeat for many (many) such dt ...

If you don't read C, I can translate it to some other language upon request. If you request something odd like postscript or the like it may take longer ...

Note also that this only calculates the trajectory given some initial inputs. What you want is the launch angle that gives a certain range ... calculating the trajectory is only a portion of that problem.
while (y>=0) {
v = pow((vx*vx)+(vy*vy),0.5);
theta = atan(vy/vx);
ax = ... ; // get these from your force equation
ay = ... ;
vx = vx + ax*dt;
vy = vy + ay*dt;
x = x + vx*dt+0.5*ax*dt*dt;
y = y + vy*dt+0.5*ay*dt*dt;
if (y<0)
y=0; // negative y means we hit the ground
// store various values here
t+=dt;
}

Error propagation

A nice little post on error propagation, which may be useful for your trajectory calculations ...

Next Friday

Next Friday - the day before spring break - we will not have a formal recitation. One must be realistic about these things ...

If you are still around, you can optionally show up and we will do some more work on characterizing the rockets and launchers. That is, you can use the time to make some measurements or work on your trajectory calculations and I'll be there to help.

Theory

This loosely sums up how I feel about theory & calculations. At the end of the day, someone still has to go and look. A reality check, if nothing else.

Also, it is a thin excuse to link to XKCD. Be sure to read the mouse-over text.

Commentary.

More succinctly
, and in t-shirt form.

Thursday, March 5, 2009

Rocket launcher progress

Before the end of tomorrow, I want you to write me a memo regarding your plans. About 2 pages (max), one per group, due by the end of tomorrow.

The details will become more clear after today's class, where we will spend a good amount of time on further characterizing the rockets.

Exam II results in more detail

Here are a few more details on the exam results. First, a plot how how many of you chose to answer each question, and a second showing what the average score was among that group.


Basically, this tells me you perceived questions 2 and 4 as far more difficult than the others. This was borne out by the average score for number 2. It is a difficult problem, mainly because it seems like you should be able to solve it by energy conservation, but you can't. Number 4 on the other hand was not so difficult as it seemed - those who tried it did quite well. Compare this to the poll at right; most of you thought question 3 was the most difficult, but you did well.

The second set of plots is just a histogram, with the second plot having more narrow bins. Broadly, it is very good - the average was much higher than I expected, and there are not a lot of people on the left-hand 'tail' of the distribution.

Looking at the finer-grained plot doesn't tell me much more, except that the large majority of you have no cause for worry - you should consider 75 or above on this exam to be a good score, it was very tough. Those of you that were at 90 and above did truly outstanding work, and I was quite impressed.

Those of you on the lower side have absolutely no cause for worry yet. No one has below a C average in the course now. This exam is only 10% of your grade; homework is worth 12.5%, the final is worth 25% of your grade. If you scored (say) 20 points lower than you would have liked to, this is only a (roughly) 2 point change in your grade - barely enough to change a + to a -. The message is: keep at the homework diligently, and work hard at the 3rd exam and the final.

(By the way: there is an 80% correlation between homework grades and overall grades ... the highest of any segment of your grade. Meaning, I like to think, there is some sense in slogging through the homework each week.)

(Also: the last poll option seems to roughly correlate with the tail of the distribution, which makes sense to me. At least one of you should have picked the first option though.)

Wednesday, March 4, 2009

Problem 10.42

Right .... so I screwed up a unit conversion myself.

The numerical answer to 10.42a should be 8.32e-6, not 8.32e-3.

(I missed that the mass was in mg, not g ... my bad.)

Exam II grades are on Moodle

Check it out.

Average: 82% / std. dev.: 13%

Details to follow.

(UPDATE: details in the next post. Homework 6 and Lab 5 are also on Moodle now. The only outstanding grades right now are those from HW7.)

Tomorrow's class

Thursday, we will devote about an hour to more measurements of the rockets and characterizing the launchers. This, as opposed to the rotational dynamics lab previously scheduled.

Remember your end goal: predicting, for a given angle of launch, where the thing is going to land. Keep in mind the target may not be at the same level as the launcher. Tomorrow or Friday we'll also discuss in more detail how to use your data to model the drag force empirically and use that in your trajectory calculations.

Exam II solutions

Here you go. All problems except the first have solutions; I'll post an update on that one later today I hope, as well as adding some better double-checks to the other problems.

Tuesday, March 3, 2009

Exam II problem 2

The boy sliding on the hemisphere is trickier than it seems.

An energy balance will not tell you when he leaves the sphere ... finding the normal force as a function of position and finding when it is zero will.

As I grade more of the problems I'll put little comments here on what seemed to trip you.

PS - this was a graduate school qualifying exam (i.e., a comprehensive test of undergrad physics) a few years back.

Problem 1, exam II

Read the question text very carefully.
A block of mass m is released from rest at a height d=40cm and slides down a frictionless ramp and onto a first plateau, which has length d and where the coefficient of kinetic friction is mu_k=0.5. If the block is still moving, it then slides down a second frictionless ramp through height d/2 and onto a lower plateau, which has length d/2 and where the coefficient of kinetic friction is again mu_k=0.5. If the block is still moving, it then slides up a frictionless ramp.

Where is the final stopping point of the block? If it is on a plateau, state which one and give the distance L from the left edge of that plateau.
There is one crucial difference between this problem and the practice problem you did on HW7. One key word, in fact.

Put another way: can the final stopping point be on a frictionless ramp?

Tuesday's class

We'll begin rotational motion today, starting with angular equations of motion and working our way up to rotational kinetic energy. I'm presuming you've skimmed Ch. 10 a bit; if not, be sure you have read through it by Thursday certainly.

We will also go over a few of the exam problems. Probably not all of them, some will have to wait until Thursday in the interests of time. Solutions to the exam will come out tomorrow after class.

After that, we'll work on the ballistics project for 45-60min. If the parts come in, we can start with that already ... if not, we will work on ballistics calculations and determining drag forces. The order for our rocket launchers has been placed, it is only to be seen if they show up tomorrow or not ...

There will be a project-related assignment for Thursday which I'll discuss tomorrow. Most likely, it will be to determine a reasonably-accurate form of the drag force on our rockets, but this also depends on whether said rockets are delivered or not ...

Even-numbered problems for this week

Numerical answers for the even-numbered problems ... just so you can see if you're on the right track.

10.32
  • a: 2.3e-9 rad/s^2
  • b: 2600 yrs
  • c: 0.024 s
10.42
  • a: 8.35e-3 kg m^2
  • b: 0.22%
10.66
  • 1.42 m/s

Monday, March 2, 2009

HW 8

I was going to finish it tomorrow, but somehow I was inspired. So ... Here is homework 8.

You still only have one problem due for Tuesday, viz., one of the exam problems.

Next HW

The next homework set in full will be out tomorrow. My struggle with FAIL continued through a good portion of Saturday, and I am a bit behind ...

For now, you have only one problem to do for Tuesday: solve one of the exam problems you did not choose to solve on the exam itself.

You can find the exam here.