Course Announcements
« PHYS521 Fall 2013
Class cancelled for this Tuesday, Nov. 26
Sun 24Nov2013 9:14AMAs the remainder of the compensation for the 2-hour midterm exam, the class scheduled for this Tuesday, Nov. 26 has been cancelled (as most of you had requested). Have a nice Thanksgiving!
Comments on Homework 10
Mon 11Nov2013 12:48PMFirst, I made some edits of my earlier followup on Homework 9. There were some misleading wording as well as a typo for the distance d travelled by charges for maximum separation in Problem 6 - it's \pi p cos (\theta) / B, that is, "cos" not "sin" as was written before. (The answer to part (a) does have "sin" in contrast.)
Now for Homework 10:
Problem 2: It may be simplest to think first of one conducting line segment of length a oriented radially. The ends of such a segment develops a potential difference as it cuts across the region of B. Then, you can consider the region of a by a by t of the conducting disk as made up of parallel such segments.
Problem 4: Again, think about the EMF induced in a conducting segment as it cuts through B. The sense (which side is higher than which) of the induced EMF) depends on the orientation and motion of the segment relative to B and some will tend to develop current while others will not. You will get a differential equation that can be put in the form of (d/dt)\theta as a function of \theta, a form very useful to answer (b).
Problem 5: Only the z-component (B_z) is given in the problem as that's all you need to calculate the terminal velocity. However, the terminal velocity is reached only because there is a retarding force. Would there be a retarding force at all if there is no B_x or B_y? Think about this.
Problem 6: As given in the Hint, calculate the angular momentum. If it's zero as the charge leaves the region of B, then the velocity has no tangential component.
Problem 10: We will discuss the general entropy of mixing in class on Tuesday.
Good luck!
Followups on Problems 1 and 6 from Homework 9
Fri 08Nov2013 11:28AMWe left some ends loose in Problem 1 of Homework 9 in class yesterday. Here is a followup.
First, since the equation of motion with or without viscous drag turns into a linear, first-order differential equation for y" and z", you can apply the same method as for coupled oscillators. Namely, construct a 2x2 matrix equation and identify normal coordinates. In this case, y"+iz" and its complex conjugate y"-iz" are the normal coordinates. From this, using the given initial conditions, you can solve (y",z"), then (y',z'), and finally (y,z), exactly in closed form. For example,
y' (horizontal) = D [ e^{-at}cos(bt)-1 ] -E e^{-at}sin(bt)
z' (vertical) = -E [ e^{-at}cos(bt)-1 ] - D e^{-at}sin(bt)
where a=k/m, b=qB/m, D=bg/(a^2+b^2), E=-ag/(a^2+b^2). This solution is good for both a=0 and a>0. So you can come up with the condition for v_{max} even for the a>0 case, though you can't get the v_{max} in closed form since the condition is a transcendental equation. Indeed, v_{max} for a>0 occurs during the first oscillation just before hitting the lowest point in that oscillation and its value is greater than the terminal velocity (which can be gotten from the above exact solution by letting t go to infinity as well as from the condition of zero net force as we saw in class). At the point of v_{max}, the net force is indeed perpendicular to the velocity.
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There was also a request for me to work out Problem 6 for which we didn't have time in class. Let me just give a crude sketch. +e and -e both undergo helical motions of the same radius (given by p sin \theta/(qB) in SI units) but in the opposite rotational sense. So the maximum separation of these charges occur when each undergoes a phase \pi rotation (semi-circle). At that point, the separation is 4r where r is the radius. So s_{max} = 4 p sin(\theta)/(qB) again in SI. This maximum separation occurs every time the charges travel a longitudinal distance of v|| times (\pi/\omega). This is \pi p cos(\theta)/(qB) in SI. You can then convert these into the units given by using the value of "a" in the problem.
Comments on HW#9
Mon 04Nov2013 1:08PMProblem 1: A very tricky problem in my opinion. Note that all forces involved are in the vertical plane perpendicular to B (i.e., the y-z plane). Note that you need a viscous retarding force in order to get to points where the net force is zero - without it, there is no point where the net force is zero - think about why. So the trajectory is qualitatively different for k=0 vs k not equal to zero.
Problem 3: Recall that both particles and fields can have angular momentum.
Problem 4: Method of images and then solve for dynamics.
Problem 6: In this day and age, it is unfortunate that some still use CGS units. I suggest you do everything in SI units and only convert them back to CGS at the end. Please remove the "1/e" that I added to the end of "a = ..." line in the problem.
Problem 7: Interpret "just after t_3" in part (d) as the order of 10^{-12} sec after t_3, i.e., as having felt the full blunt of the acceleration of the charge at the origin.
Problem 8: Draw a diagram and make sure to indicate your sign conventions for the currents and induced EMF. That is label the direction of I_1 and I_2 which is considered "positive" by arrows, as well as label by "+" and "-" the poles of the primary and secondary coils which will be considered positive EMF. As you might surmize, the signs are very important in this problem.
Problem 9: Use the relativistic transformation formula given with the midterm exam (or one uploaded to the Supplmements area of this web site).
Problem 10: This problem discusses adiabatic demagnitization. Assume that B is removed slowly and adiabatically.
Comments on HW#7
Tue 22Oct2013 4:48PMSome comments on HW#7, especially for those who missed today's class, are given below (though there are some additional comments for everyone as well):
Problem 2: If you use both the Gauss's Law in free space (that is, the microscopic law with E that takes into account all charges) and the corresponding macroscopic Gauss's Law with D judiciously, this problem should be straight-forward. If needed, you can also use \rho_b = - (div) P (for bound charge density) and \sigma_b = -(P_2 - P_1) dot n (for the surface bound charge density where the righthand side uses the discontinuity in polarization), though you can do without them and just sticking to the Gauss's Laws directly.
Problem 5: One of the boundary conditions that you might miss is the one that the entire plane that bisects the metal sphere and perpendicular to E_0 is at the same potential (which can be taken to be zero). Why?
Problem 6: The pressure that must be applied on the hemispheres to keep them from flying apart is equal to \sigma^2/(2 \epsilon) just as in the case of attraction between the two plates of a capacitor. I explained why in class, but do make sure that you understand this.
Problem 7: Apply both forms of the Gauss's Law as appropriate. For uniformity, let's say the "dielectric constants" (\epsilon_1 and \epsilon_2) given in this problem to be the dimensionless ones (like the dielectric constant \kappa for a capacitor with dielectric material between the plates).
Problem 8: Don't underestimate this problem. Basic, but very instructive. Assume that the spheres are conducting and that the separation d >> radius r as stated in the problem.
Problem 9: First consider a single dipole and focus on the angular average. (The positional and momentum degrees of freedom are irrelevant in this problem.) Then you can define an orientational partition function and the desired average from it. Finally, you use the given concentration to arrive at the polarization.
Problem 10: As I discussed last Thursday in class, the classical partition function can be obtained by integrating the Boltzmann factor (e^{-\beta E}) over space and momenta (and normalizing by h^3). The space integral just gives the overall volume V.
Good luck!
Midterm Exam: 7 - 9 pm on Friday, October 11 in Rm.338
Tue 01Oct2013 4:47PMAs in the title. The exam has 5 actually used Qualifier problems from some previous years. The topics should mirror what we will have covered in the first 6 homework sets. That is, mechanics (including relativity) and some thermal (mostly thermodynamics and also combination of thermal and mechanical) problems.
The class on Thursday, Oct. 10 will be an optional review session.
Since there is a chance that you may not have a graded HW#6 back prior to the Exam (especially if you do not attend the optional review session), you might want to make a photocopy of your HW#6 prior to submitting it this Thursday (if you want to make sure that you have a copy to study for the Exam). This is entirely up to you, though.
Comments on HW#6
Mon 30Sep2013 12:30PMProblem 6: If you start off with a correct guess on the presence/absence of relative motion (and if so, the directions) among the 3 masses, then the problem is very simple. However, since the problem doesn't tell you about this, you will have to justify it. With the need for this justification, the problem becomes much more difficult. Hint: If it were not for friction, then both 4M and M blocks would move to the right relative to the platform. Thus, if instead both were to be stuck to the platform by (static) friction, you know which direction the friction should act on 4M and M blocks. You should then be able to argue that such a situation is impossible. In other words, at least one of the blocks must move relative to the platform. Which one(s) and in which relative direction(s)?
Problem 7: The 4-momentum (E/c, p) transforms the same way as the 4-coordinate (ct, r) (where p and r are the usual 3-vectors). The "Center of Mass frame" should be interpreted as "Zero Mometum frame" as the former is a bit confusing at least in the intial state where there is also momentum in the incoming radiation.
Problem 8: I hope to discuss related issues in class on Tuesday. Just in case I don't get to that discussion, here are some thoughts. The problem is asking about the light that is emitted when the bulb just passes the point shown in the picture. The observer would then measure the (apparent) transverse velocity of the bulb by obtaining the (apparent) transverse displacement of the bulb in small time duration (Delta t) and the corresponding time duration between the receipt of the light at the end and beginning of the (Delta t) period at his/her location. Assume that the observer knows the angle theta. This situation can lead to so-called super-luminal (apparent) velocity of an ordinary object. Justify all of this.
Problem 9: You may assume that both the super-conducting and normal metal forms of the entropy would have a zero off-set at T=0. Otherwise the problem is ill posed.
Problem 10: This is called "free" expansion, where a gas expands into vacuum without any resistance, thereby not having to do any work to displace whatever might have been there to begin with.
Comments on HW#5
Tue 24Sep2013 3:56PMProblem 3: As discussed in class, please use AU/yr as the unit for velocities and acceleration in AU/yr^2. This way, answers to parts (a) and (b) can be given only in terms of the given masses (m_1, m_2, and M=m_1+m_2) (plus some simple numbers and pis). For part (c), the M/M_s comes out to be a pure number with no reference to m_1 or m_2.
Problem 4: If you are stuck in (c), one way to proceed is convert the time derivatives that appear in the Euler-Lagrange equation to derivatives with respect to theta (noting, e.g., dr/dt=L_0/(mr^2)(dr/dtheta) where L_0 is the angular momentum). This should gets you a differential equation of u=1/r with respect to theta. In part (d), you can use the geometry I mentioned in class to write down u(theta) immediately.
Problem 5: Don't forget both repulsive and attractive cases are possible physically. But do you need different expressions?
Problem 8: Here, the Carnot engine is operated in reverse as a refrigerator. Then, that is used to remove heat that comes in to the house so that the house is kept at a constant temperature.
Problem 9: Refrigerators that use a "flame" for power source do exist such as those powered by propane or natural gas, especially for recreational uses, e.g. Here, you combine a heat engine that creates power, which is used in turn by a refrigerator, in effect using two heat cycles in series. The equivalent one-device "cycle" would involve heat from/to 3 reservoirs (with no work by or to the cycle). You can then use the so-called generalized Carnot inequality, where
(sum over heat reservoirs i) Q_i/T_i <= 0
where Q_i is the heat taken from the reservoir i. The equality is for the reversible case that leads to highest performance.
Another way is to write the overall performance eta as the product of the engine efficiency and refrigerator performance and then consider maximizing each separately. Note that the highest coefficient of performance for a refrigerator occurs for a reversible refrigertor and it is then equal to T_L/(T_H-T_L) [think of a Carnot engine operated in reverse].
Problem 10: This problem actually describes a reversible Carnot cycle operated on a photon gas rather than a classical ideal gas. As long as it is reversible and operates between two heat reservoirs (with 2 isothermal and 2 adiabatic legs), all such engines should have the same (maximum) efficiency, right? Check this with your answer.
Good Luck!!
Some comments on HW#4
Wed 18Sep2013 12:24PMIt's kind of late, but I thought I might add some comments on the current homework set (#4).
Problem 2: The reason that the drag force is typically proportional to v^2 (and not v as in Stokes Law) is because, at the velocities involved in a problem at a macroscopic scale like this one, the dominant retarding effect is due to pushing air (or surrounding fluid) out of the way as the object moves (rather than viscosity or friction). The work that needs to be done to give the air a velocity of the order of v during time dt is on the order of F_drag x v dt = (mass of displaced air) x v^2. Since mass of displaced air during dt is proportional to v dt, it works out that F_drag is proportional to v^2.
Problem 4: As I alluded to in class, the principal moments and axes calculated for the moment of inertia with the origin not at the center of mass of the 3 mass points are only meaningful if the origin is a stationary point, forming the 4 the point of a rigid body. This is not relevant to the problem itself, but I think physically important for you to know.
Problem 9: If you haven't thought about phase transitions before, you might want to take a look at Schroeder (or similar text) - it's p.180-185 or so on Schoeder. For ideal gas, the P-V curve is strictly monotone and there cannot be a phase transition. You need a kink in P-V curve (as for the Van der Waals gas).
Problem 10: The pressure difference between the inside and outside of the bubble being equal to 2 sigma / R is called Young-Laplace equation. (Everything has a name, it seems.)
Supplement on Homework 3 Discussions
Fri 13Sep2013 11:30AMProblem 1:
One of you had an interesting comment on the problem during class that I didn't completely understand. We had another discussion later and I also thought some more about this. Essentially, the idea was that the initial speed with which the grains leave the upper portion of the hourglass would not always be zero. Even though the very first grain to fall starts with a zero velocity, those that come later have increasing velocities as they leave the top part. This is true as a grain falls from above toward the hole at bottom, its potential energy is converted to kinetic energy; in other words, they had already started "falling" prior to hitting the hole at bottom. (Consider water leaking from the bottom of a tank.) If one wishes to take this into account, then the solution would be much different from what we discussed, in fact, depending on the geometry of the hourglass. (Consider the difference between a very narrow and tall hourglass vs very fat and short one, e.g.)
Now, however, the rate of mass leaving the top part per unit time is the product of the cross sectional area of the hole, the density (# of grains/unit volume), and the velocity with which the grains are leaving the top part. Since the problem specifies that this rate is kept constant, if the velocity varies, then the density must also vary. This is not likely even with grains (and not possible with an incompressible fluid), however.
So what gives? Ultimately, the premise of the constant rate of mass exit from the top part is unphysical. However, (1) we are given this premise and must assume it - in which case, the exit velocity should then be constant too and since that is zero for the first grain, we need to assume that remains zero for later grains, too, and (2) the assumption is an approximation that the exit velocity remains close to zero fro all grains. The latter approximation is not at all unreasonable for typical dimensions of an hourglass. So, in the end, this is another of the ambiguities associated with "idealizations" that are often not explicitly stated (even if not unreasonable and considered "standard" by those in this line of work).
Problem 4:
We discussed briefly on what if there were forces between blocks A and B and ended up saying that it would still not affect the velocity of block A immediately after B and C collide. This, I now consider, was incorrect. In this problem, there are no horizontal forces between A and B, and thus this is irrelevant. However, if there were forces, the velocity of A right after B and C collide would depend on the nature of the forces and of the collision itself. This is evident by considering an extreme limit where A and B are bound together with strong forces - in such a case, A and B (also C) together will have speed equal to 2/3 of v_i. With different forces, it would depend on its details and duration of the B-C collision. Whatever the impulse B provides to A during the collision with C would determine the momentum change of A during the collision.
Problem 10:
The problem asked us to "estimate" the effusion flux. I gave a solution to calculate this flux "exactly" (under ideal gas assumption, of course). As I alluded to in class, you can estimate it without doing a single integral but totally physically, relying on the equi-partition theorem and kinetic theory of gases (as one of you offered to me privately). This goes roughly as follows:
- Without the hole, the force the gas applies on an area A of the wall is PA. This is equal and opposite to the force the wall applies on the gas particles.
- During time dt, this force provides impulse to change the momenta of the gas particles that hit the wall and get reflected. This impulse (PA dt) is equal to the number of particles that hit the area A of the wall per unit time (J) times the momentum change of one particle.
- The momentum change of one partilce is (2 m <v_z>) where m is the particle mass and <v_z> is the component of the mean velocity normal to the wall.
- <v_z> can be estimated from the equi-partition theorem as:
(3/2) k_B T = 3 x (1/2) m <v_z^2> ~ (3/2) m (v_z)^2
where root-mean-square v_z is used in place of <v_z>.
- Finally, the number of particles that would hit the area A of the wall per unit time (J) is in fact the exit flux the problem asks for when the hole actually exists there to let them out.
When you put all of the above together, you get a correct order of magnitude estimate for the flux: PA/[2 (m k_B T)^(1/2)]
This would be a perfectly good answer/solution.
Comments on Homework 2
Thu 29Aug2013 12:39PMHW2 collects problems having to do with rotation (even the thermal problem #10), but they do involve different kinds of "rotation".
Problem 1. This is not really a problem of angular momentum conservation in the normal sense. Why? If nothing is supporting the axles of the cylinders, they will not generally keep the same relative positions when the cylinders touch; so clearly, there is external torque that is acting on the axles relative to some selected statrionary axis.
However, you can discuss the problem as if it were one of angular momentum conservation but only if you define the positive direction of the angular momentum in alternately reversing directions for the adjacent cylinders. Why? Think about what (friction) forces (thus torque) act on them if two adjacent cylinders touch. The forces involved are of equal magnitude and in opposite directions. This gives rise to the above pseudo-conservation "theorem".
Problem 2. This problem needs to specify the mass of the wedge. Let's say it is "m" (lower case).
Problem 3. Tension in the string is the key quantity. Consider the force the beads exert on the hoop, the force the hoop exerts on the beads, and the tenstion T of the string. Think about the physics when the hoop is actually rising. What's causing this to happen?
Problem 4. Just to make sure, this is not a full 3-body problem. The 3 bodies are constrained to keep the same relative positions.
Problem 5. In all these Yo-yo problems, the key is the mechanical constraints that relate the anglular motion and both the linear motion and the length of the string that extends. The Part II-Problem 2 of Qualifier Fall 2013 was one of these Yo-yo ones.
Problem 6. This problem involves rotation and collision. When bouncing off of a surface, both linear and angular momenta change. These changes can be described by using impulse and torque (which are related to each other). In the direction normal to the surface, the impulse on the spining ball can be assumed the same as if the ball were not spinning. More collision and similar problems will be collected in another HW set.
Problem 9. Relate the changes in momentum and angular momentum of the cylinder to the impulse due to static friction.
Problem 10. To be complete, you should find the coefficient that appears in the solution to n(r) using the conservation of the number of particles. You may be surprised.
Good luck!!
Class cancelled on Tuesday, Sep. 3
Tue 27Aug2013 2:39PMAs I wrote in the syllabus, the class scheduled for Tuesday, Sep. 3 has been cancelled as a compensation for an evening mid-term exam (whose date/time is yet to be determined). Since the exam will be for 2 hours, we might have another class cancellation in addition later (or we might not, depending on how we progress).
Unfortunately, this means we will not have a dedicated time to discuss rotation-related material prior to Thursday, Sep. 5 when Homework 2 (mostly rotation-related problems) will be due. I will try to be available for individual questions (by Email preferred) on Tuesday/Wednesday next week. (I will be hard to contact between this Friday and next Monday, however.)
Hisao Nakanishi