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[Physics & Frontiers I] Re: Example "Optional Reading Quiz"

Kyle Sponenberg — Thu, 28 Aug 2008 23:12:22 -0400

Thanks for letting me know, that's really helpful.

[Physics & Frontiers I] Re: Homework 1 Problem 9

cec8 — Thu, 28 Aug 2008 19:25:02 -0400

Hmm. I think tomorrow's lecture will be quite helpful for you,
so if you can, plan to finish the problem after this.

Having said this, remeber that the F in "F=ma" is the [i]net[/i] force.

Remember, the F in F=ma is not a force and it never goes on a FBD.

So here the net force must be zero. But since forces can add in the
positive or negative direction, the individual forces might not be
zero. So when you write F=ma where a=0 what you mean is that
[math]F_1 + F_2 +F_3 +... = 0[/math] where each term might be a positive
or negative term. So for a top book, for example, there are two
forces: a weight force (always there if the object has mass near the
surface of the Earth) and a Normal force which is the table pushing on
the book. You know the table is pushing on the book because if it
were not pushing on the book, the book would accelerate (fall) down.

When my four-year old daughter jumps on my back, I might say in
English that I "feel her weight" so you might say that I should put
a weight force on a FBD for me. But in fact I do not feel her weight.
What I feel is the [i]push[/i] of her feet on my back. A push is a
"normal" force, not a weight.

Now it turns out that the magnitude of that pushing force does in fact
depend upon her weight, but that dependence follows from Newton's Second
Law.

Does this help?

[Physics & Frontiers I] Re: Example "Optional Reading Quiz"

cec8 — Thu, 28 Aug 2008 19:18:09 -0400

No, worries. Students time themselves. You can start as soon as you
arrive. Student self-timing will be done on the "honor system".

When you come in, early or late, grab a quiz, sit down and take it,
and then pass it into the box. If you arrive a little later then
the only drawback is that you must do this whilst I've started the lecture.

[Physics & Frontiers I] Homework 1 Problem 9

Em — Thu, 28 Aug 2008 18:54:38 -0400

I'm not sure how to approach this one. In the hints, it says that there should only be one weight force for each FBD, but wouldn't the table and the book directly on top of that have two weights for each of the books? Also, how are we using Newton's second law if the acceleration is 0? What I have is that F = m*0, therefore F = 0. But there are forces acting, they are just canceled out. So.. I am confused as to how to write it that way.

[Physics & Frontiers I] Re: Example "Optional Reading Quiz"

Kyle Sponenberg — Thu, 28 Aug 2008 17:05:55 -0400

I have to come all the way from the Mandel Center to Rockefeller in the 10 minute gap between classes, and since this is usually close to impossible with the construction going on on Euclid Ave. chances are I'm going to be a couple of minutes late to class. Does this mean that I'm definitely going to miss the reading quiz on a weekly basis?

[Physics & Frontiers I] Example "Optional Reading Quiz"

cec8 — Thu, 28 Aug 2008 14:26:00 -0400

Dear Physics 123 students:

I wrote up two "Optional Reading Quizzes" for the assigned
readings. One of these I have (randomly) chosen to give to
the class tomorrow (Friday) at the start of lecture. The
other is presented below, so that you get an idea of what
I am likely to throw at you:

The following is [i]just[/i] for practice. The quiz
I give Friday morning will have three different questions.

[b]Physics 123 Optional Reading Quiz:{/b]

[b]Note:[/b] Taking this quiz is entirely optional. Quiz is
for Optional Bonus Points only. Students cannot hurt their grade
in Physics 123 by answering incorrectly or by choosing not to
take or submit this quiz.

[b]Name: [/b] [b]Case NetID[/b]:

Rules: HONOR SYSTEM! Please time yourself. Please take
no more than 60 seconds to answer these three questions:
No books, notes, laptops, or consulting with any other students.

Question 1: What do we mean by the word [i]speed[/i]?

(a) The square of the acceleration
(b) Position over time
(c) The absolute value of the velocity
(d) A can of Mountain Dew dropped from the Eiffel Tower

Question 2: Kepler's Second Law tells us that:

(a) A line from planet to sun sweeps out equal areas in equal time.
(b) The sun attracts all planets using gravity
(c) Planets move in perfect circles
(d) Textbooks prices increase as the square of their mass density.

Question 3: A Free Body Diagram does what?

(a) It includes a "point like representation" of one body in a problem.
(b) It includes all forces on the body.
(c) It often includes a coordinate system.
(d) It precedes the application of Newton's Second Law.
(e) All of the above
(f) None of the above.

[General Physics II., E & M] Recitation times

gab23 — Thu, 28 Aug 2008 14:17:30 -0400

Its hard to find a recitation time that suit everyone, but given the lack of attendence I assume 2pm on Thursday doesn't work for anyone?

Open to suggestions.... let me know which times on Thursday do work and I'll try to find an empty room.

[Physics & Frontiers I] Anonymous posting allowed here!

cec8 — Thu, 28 Aug 2008 14:06:59 -0400

Dear Physics 123 Students:

This Phorum bulletin board is one of several "safe places" where
students in Physics 123 are allowed -- nay, even [i]encouraged[/i] -- to be
wrong. Other "safe places" are office hours, during lectures and
recitations. In fact the only place where it is "unsafe" to be "boldly
incorrect" is during exams, and even then, if you can justify your answer
in terms of physics principles you will find me lenient with partial credit.

Many of the best students are by training or temperment highly adverse to
finding themselves shown to be wrong about anything of an academic nature.
Students find it painfully embarrassing to be shown to be wrong. This is
compounded if this happens in the presence of an instructor, for example.

That's a concern in a class like this. I do not expect students to
"know" the material already. I expect you to learn it. And for this kind
of material, unless you are Einstein himself, chances are pretty good that
you are going to flounder a little bit before you "get it". It takes
practice and you have to let yourself fall down and get up a few times.

So forget about it. I am not interested in being impressed by any student's
background or knowledge. I have equal affinity for all student, and perhaps
if I am honest I will feel slightly more affection for the student who is
willing to put him or herself out there with a flawed or confused
question or approach, since this is a student who is
willing to take a personal risk for the cause of learning.

So please, go ahead and post your "stupid questions". There's a really good
chance that others are just as stuck as you are.

And if it helps, please feel free to post "anonymously" -- simply "logout"
of the Phorum (or just don't log in in the first place) and in this case
you can fill in any made-up name you want for your question.

-Corbin Covault

[Physics & Frontiers I] Re: HW#01: Problem 7?

cec8 — Thu, 28 Aug 2008 11:42:53 -0400

You've made a "very small mistake". Hint: one of your values is off
by exactly a factor of two...Another hints: your final answer is precisely
four times too small...

[Physics & Frontiers I] Hints on Homework #01, part 2

cec8 — Thu, 28 Aug 2008 11:38:16 -0400

Here are more hints on the current homework:

[b]Problem 4[/b] This is a "standard problem" that you see in high school.
You need to use the words "Free fall kinematics" and/or "Constant acceleration"
before you charge ahead with equations. The Fundamental Equations are these:

[math]a_y(t) = -g [/math] (a constant) where [math]g \equiv[/math]+9.81 m/s[math]^2[/math] ([math]g[/math] is a [b]positive[/b] number.)
[math]v_y(t) = v_{y0} -gt [/math]
[math]y(t) = y_0 + v_{y0} -\frac{1}{2}gt^2[/math]

Please solve the problem [i]symbolically[/i] before you plug in any number!

[b]Problem 5:[/b]

This is two or three lines of algebra. Draw a small sketch to indicate
clearly what is going on. You must start with the fundamental equations
of [b]Constant Acceleration[/b]. The whole idea here is that kinematic
variables can have positive or negative values. So make sure that the time
that you calculate is a positive number.

[b]Problem 6:[/b]

I already posted some fairly comprehensive hints [url=http://www.phys.cwru.edu/forums/read.php?43,6434]here[/url] and [url=http://www.phys.cwru.edu/forums/read.php?43,6446]here[/url].
This trip corresponds to two (symmetric) trips each taking the same
period of time, so don't mix up the total time and/or distance with
the "half-way point" time and distance.

[b]Problem 7:[/b] The purpose of this problem is to see if you can make
the connection between Universal Gravity and Weight:

[math] F_{UG} = \frac{GM_{Merc}m}{r^2} [/math]

while

[math] W = mg_{Merc} [/math]

where [math]M_{Merc}[/math] corresponds to the mass of the planet Mercury
and [math]g_{Merc}[/math] corresponds to the "acceleration due to local
gravity on the surface of Mercury (which is [i]not[/i] 9.81 meters per second).

[b]Problem 8:[/b] Okay this is the most challenging problem in this
homework set. Unfortunately I did not (quite) get as far during Wednesday's
lecture. I've already given some important advice in [url=http://www.phys.cwru.edu/forums/read.php?43,6438]this post[/url].
I will talk more about applying Newton's Laws in class on Friday. However,
everything you need to solve this problem is spelled out for you in
remarkable detail in the [url=http://www.phys.cwru.edu/courses/p123/notes/Ch5Aug21.pdf]Online Course Notes Chapter 5[/url]. Note that I will
tend to use a slightly different convention for labeling forces:

When Bob Brown says "[math]F_{\mbox{on 1 by 2}}[/math]" Corbin Covault says "[math]F_{12}[/math]".

[b]Problem 9:[/b] Again, just like Problem 8: Draw a diagram to
show what is going on. Be sure to define everything in terms of
symbolic variables, (not numbers!) and then write out a [b]Free Body Diagram[/b]
for each relevant body in the problem. Remember that there is only one
weight force on each body. If you have more than one weight on any FBD
you are not doing it correctly. When you are sure that your FBD's are
correct, then [b]write down and solve Newton's Second Law[/b] for [i]each[/i]
of the FBDs. Take your time and do not skip steps. The good news here
is that you know the acceleration is zero (right)?

[b]Problem 10:[/b] Now the acceleration is not zero, but it is known.
This has no impact on your FBDs which are used [i]only[/i] to deal with
the left-hand side of Newton's Second Law, (right?) Being accelerated
does not change the weight of the body. Ask yourself this: if you were
in the same elevator, what would you experience? If the elevator had no
windows, would you still know that it was accelerating downward at 8 m/s[math]^2[/math]?
How would this feel to you?

[Physics & Frontiers I] HW#01: Problem 7?

wrk7 — Thu, 28 Aug 2008 11:20:25 -0400

I am having trouble with problem 7. I looked up on the internet that the gravity on Mercury is aproximately 38% of that on earth, so it should be
(9.81m*s^2 * .38 ), or 3.72 m*s^2.

I used the equation a = (G*M) / (R^2) from notes chapter 4
with the values
G = 6.67 X 10^-11 m^3/(s^2*kg)
M = 3.3 X 10^23 kg
R = 4.878 km * (1000 m / km ) = 4,878,000 m

Putting these values into the equation, I am getting a value of a = .925, which does not agree with published values of Mercury's gravity. Where did I make a mistake?

[Physics & Frontiers I] Re: 1 AU

jpm63 — Thu, 28 Aug 2008 00:37:29 -0400

Ah, 1 AU is 150 million kilometers, I see that and immediately got it crossed up with the 50 AU given by the problem. Will go back and check through it again tomorrow. Thanks for the help.

[Quantum Mechanics I] Re: HW1, Problem 2.i

arh17 — Wed, 27 Aug 2008 21:51:03 -0400

I found that the math is easier if you make a substitution at the beginning. Instead of [math]x(t)=A\cos\sqrt{\frac{k}{m}}t[/math], make the substitution [math]\tau=\sqrt{\frac{k}{m}}t[/math]. So [math]x(\tau)=A\cos\tau[/math], and [math]\tau[/math] ranges from 0 to [math]\pi[/math]. Since [math]\tau[/math] is proportional to [math]t[/math] (and with the assumption of a uniform probability distribution in [math]t[/math]), you can say that there is a uniform distribution in [math]\tau[/math].

Therefore, [math]\rho(\tau)=\frac{1}{\pi-0}=\frac{1}{\pi}[/math], and the math has now become a whole lot easier.

[Quantum Mechanics I] Re: HW1, Problem 2.i

kxs213 — Wed, 27 Aug 2008 20:34:30 -0400

I'm having trouble showing that the probability density is the equation given. When I do it, I get dt/T (or p(x)) to be
p(x) = sqrt(m)/[2t sqrt(kA(A-x))]
which doesn't seem right. I'm not sure where pi comes in, and how A and x become squred (to become A^2-x^2)....any suggestions?

[Physics & Frontiers I] Re: 1 AU

cec8 — Wed, 27 Aug 2008 19:34:32 -0400

Well, okay. I think you make some calculator boo-boos here. Half of
50 AU is not "75 AU". You just need to be more careful with your
unit conversions, I think. I get a time that is somewhat longer.
But not very much longer. Also, I do not see how you calculated
your maximum speed here. How did you do this?

You say that 2.9 days seems awfully short. Yes this is unrealistic that
a "current technology rocket" can get to Pluto this fast. In particular,
a current era rocket can only carry fuel to thrust at "1 G" for a rather
short total period of time (minutes at most). In this problem the
fact that we are able to sustain the thrust is what gets us there quite
quickly.

In particular, the speed of the fastest current rockets is something
less than about a few times 10 miles a second. That's rather lower than the
speed achieved by our "1 G" rocket here.

As a "cross check" can you determine the "average velocity" of the
spacecraft for this whole trip. If a ship travels at constant
acceleration then the velocity is increasing linearly (right?) and so
the average velocity is exact have the maximum velocity (right?).
Again if you work this out, you will convince yourself that the
average velocity is pretty fast compared to our fastest rockets.

In fact, for a fun diversion, it is interesting to calculate the maximum
speed of such a spacecraft against the "fastest possible physical speed".
In Newtonian Mechanics, there are no constraints on the velocities that
any body can achieve. But as we will see later in the course, Einstein's
theory of Special Relativity puts a limit on the velocity of any object.
You can also use relativity theory to get a rough estimate of how much
energy getting to these kinds of speed requires: Roughly [math]K = (1-\gamma) mc^2[/math] where
[math](1-\gamma) \approx v^2/c^2[/math] for [math]v< is the mass of the rocket times the speed of light squared. If you work this
out you will realize that getting a "normal sized" rocket up to even a
relatively small fraction of a percent of the speed of light requires an
enormous amount of energy. ("[i]I need full warp drive, now, Scotty"[/i])

[Physics & Frontiers I] Re: 1 AU

jpm63 — Wed, 27 Aug 2008 18:01:07 -0400

The time I calculated is 247,000 seconds, or aprx. 2.9 days. The max. velocity I calculated is 8.73 x 10^5 km/hr. I expected a time value much greater (upwards of a month). I mean, Pluto is about 150 million km away...

not sure though, maybe I did do it correctly

[General Physics I. Mechanics] Re: Problem 2-2

wxl2 — Wed, 27 Aug 2008 16:31:31 -0400

The boxed equations on p.2-4.

[General Physics I. Mechanics] Re: Chapter One 1-3 HW

alv16 — Wed, 27 Aug 2008 16:28:27 -0400

Just above it we had proved that (lowercase are subscript) Va/b = -Vb/a.

You can then take your velocity relative to the ground and subtract the same so that you have Va/b - Vb/a. In this case you have one set of coefficients cancelling out and can get either Va/a or Vb/b. I would assume that we could cancel these as well, but since the problem asked for Va/a, we can end here.

I hope that is helpful (and that it is right :P)

[General Physics I. Mechanics] Problem 2-2

alv16 — Wed, 27 Aug 2008 16:23:48 -0400

What is the original equation that we are supposed to start with?

[Physics & Frontiers I] Re: 1 AU

cec8 — Wed, 27 Aug 2008 16:21:54 -0400

Alright. Let's keep going here...

I'm especially interested in the fact that you say the time "seems way too low."

What time is that, exactly? And how/why does this time strike you as
"way too low?" What sort of number would seem more in line with your
expectations?

-CC

[Physics & Frontiers I] Re: 1 AU

jpm63 — Wed, 27 Aug 2008 16:02:43 -0400

I went with the estimate. Math works out well, but now I'm getting an answer I feel is far too low.

My time works out to be 2*sqrt(2x/a). Does that seem reasonable?

It is two times the sqrt because I used 75 AU for x, and then doubled the time. (I said it takes exactly long to get to the halfway point from earth as it does to get from the halfway point to pluto).

[Physics & Frontiers I] Re: 1 AU

cec8 — Wed, 27 Aug 2008 15:55:44 -0400

Hmmm. What do you think? What would you guess is the answer to your
own question?

Note: I often (usually) start answering student questions with a question
so if you answer my questions, I promise I will respond fully to your original
question.

[Physics & Frontiers I] HW#01: Problem 6: 1 AU?

jpm63 — Wed, 27 Aug 2008 15:41:02 -0400

For problem 6, many websites said that 1 AU is approximately (key word approximately) 150 million kilometers. Is it acceptable to use this approximate value for my calculations, or did you have something else in mind.

[Physics & Frontiers I] Re: Problem 1 parts b and c

cec8 — Wed, 27 Aug 2008 13:34:49 -0400

Yes that is correct.

Alternatively, if a function [math]f(t)[/math] is a constant as a function of
time, then it must be true that [math]\frac{df}{dt}=0[/math]. In
other words, if you take the derivative and show that the expression is
exactly zero for all values of time, then that function must be constant
as a function of time.

Either way works...

[Physics & Frontiers I] HW#01: Problem 1 parts b and c

ces58 — Wed, 27 Aug 2008 13:28:31 -0400

Just a quick question. We are supposed to find out if the velocity and acceleration are constant as functions of time. That simply means the velocity or acceleration would be a straight line if it were graphed (with t as the x-axis and v/a as the y-axis), correct?

[Physics & Frontiers I] Re: problem 8 normal force

cec8 — Wed, 27 Aug 2008 13:08:51 -0400

Yes, you can talk about Normal for in many ways.

For me, the simplest idea is that a normal force is a "push", and
it has an "unknown" value, that is to say, you usually cannot figure
out how larger a normal force is unless you apply and solve Newton's
Laws to work it out.

In contrast the Weight force is "known". It always has value like [math]W_m=mg[/math]
and points down.

In this problem there are four different normal forces:

(1) [math] N_{ms} [/math] which is the Normal force on the small block due to the surface. This points up.
(2) [math] N_{Ms} [/math] which is the Normal force on the large block due to the surface. This points up.
(3) [math] N_{mM} [/math] which is the Normal force on the small block due to the large block. This points in the same direction as the applied force [math]F_{app}[/math]
(4) [math] N_{Mm} [/math] which is the Normal force on the large block due to the small block. This points in the opposite direction as the applied force [math]F_{app}[/math]

The other forces in this problem include [math]W_M[/math] and [math]W_m[/math],
(the weight forces on each block, respectively) and [math]F_{app}[/math],
the applied external force (given). Ideally, your solution should have
two Free Body Diagrams where all of these forces above are shown.
Admittedly, the Weight forces and the forces normal due to the surface
do not actually come into the calculation of quantities asked for, however.

Note that Newton's Third Law tells you that [math] N_{Mm}= N_{mN} [/math]

Note that in order to determine the Normal force [math] N_{Mm} [/math] you
must should down a [b]Free Body Diagram[/b] for each individual block and you
[i]must[/i] apply [b]Newton's Second Law[/b]. I do not see that there is
any way to solve this problem without explicitly applying Newton's Second Law.

Does this help?

[Physics & Frontiers I] Re: problem 8 normal force

Chelsea Spengler — Wed, 27 Aug 2008 12:43:10 -0400

My high school physics teacher explained the normal force as the force caused by one object sitting on another. For example, the weight of one of the blocks exerts a force on the table, and the table exerts a normal force in response (and that's why the block doesn't go through the table but instead sits on top of it). Not sure if that's completely accurate though, since it has no way of explaining the force between the two blocks.

However, my guess of explaining it would be this: the normal force is defined as a force that acts perpendicular to the surface of the object...so a force between the two blocks is exerted to the side, perpendicular to the vertical surfaces of the sides of the blocks.

[Physics & Frontiers I] Re: problem 8 normal force

cec8 — Wed, 27 Aug 2008 09:28:02 -0400

40N force is an "applied force". I do not specify what kind of force it is,
but it does not matter.

What does the word "normal" mean, exactly? Can you tell me? It's important.

There are at least two normal forces on each block: One due to the
frictionless surface and one due to the other block.

Does this help?

[General Physics I. Mechanics] Re: Chapter 1-2 Homework

wxl2 — Wed, 27 Aug 2008 09:25:15 -0400

I suppose you mean for problem 1-2? Use a graph that helps you to solve the problem.
it asks to show the intersection of the two paths. Can you do that on a velocity graph?

[General Physics I. Mechanics] Chapter 1-2 Homework

dpp18 — Wed, 27 Aug 2008 08:27:53 -0400

Is the graph for (a) supposed to be a position-time graph or a velocity-time graph?