Thursday, November 6, 2008

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.

7 comments:

cpoday said...

Very helpful. Thank you for posting this!

Alex said...
This comment has been removed by the author.
Alex said...

A particle slides back and forth on a frictionless track whose height as a function of horizontal position x is given by y = ax^2, where a = 0.90 m^-1. The "turning points" for this type of motion can be defined as the minimum and maximum values of x for the motion.

If the particles max speed is 8.8 m/s then what is the x value at the turning points?

Im treating this as a conservation of energy problem as the
motion is simple harmonic. so in my thinking

1/2mv^2 = mgh is reasonable to find the height of the particle
at max speed and zero speed. When the particle is at the
turning points on the parabola, the height is defined h = v^2 / gh. So v at the max is v = 8.8 and v at min = 0 (turning points ~ extrema)

This should give the height which is equal to f(x) or y in
the parabolic path? So then solve for x?

Im still thinking about my theory here, any insights?

J. Quinn said...

I think that you have it exactly right. Use the equation 1/2v^2=g(.90)x^2. This is no longer an equation defining the motion of the particle, but rather it tells you what the potential energy (mgy) has to equal at the turning point (x). So if you plug in v=8.8 and solve for x I would think you'd get the right answer.

[I am assuming that this is problem 7.60 in the book.]

Sam said...

Comment from an anonymous student
Topic: Problem 7.26 (mbS)
-------------------------------------
Make sure people are inputting these in the form x1,x2. No spaces would be best. Technical problems can be a pain. I say this as it sounded like a guy in the office hours had trouble with that today.
-------------------------------------

Sam said...

A collection of comments from students
(please help out by replying to some of them!!)
----------------------------------
Item Title: Problem 10.24

Comment:
I still can't figure out how to plug in the numbers to get this answer...Any help, please?

Comment: (another student)
I didn't realize that the mouse's mass was in grams and not kilograms until after asking to see the answer.
----------------------------------
Item Title: Problem 7.66 (mbS)

Comment:
I had no idea how to even start this problem.
----------------------------------
Item Title: The Parallel-Axis Theorem

Comment:
Even though the problem was a bit difficult, I actually learned a lot from it. I feel that it was useful in teaching me exactly what it was trying to teach. I'm not too bummed about not getting a lot of it right because through all the tries I ended up taking a lot away from this problem.
----------------------------------
My help Problem 10.65

(please check your note of my lecture; i did an even more difficult example in class -- disk with a square punched out).
I = I(big-full-disk) - I(small-missing-disk)
I(small-missing-disk) = I(small-missing-disk-around-center-of-mass)
+ m(small-missing-disk) * d^2
d = separation between the actual axis and the center of mass axis
Hope this helps.
----------------------------------
Again, fellow students, please help out here by sharing your wisdom!

J. Quinn said...

I'm not sure if the person who asked about 7.66 needs help, but here is how I did it: Work = integral P dt. This integrates to W=Pt (plus a constant but that goes away if you consider when t=0). Work is also the change in kinetic energy so it also = (mv^2)/2. Then you equate Pt to (mv^2)/2, solve for t and you get (mv^2)/(2P). Remember to convert everything to meters/second and multiply P by whatever percentage they gave you.