Walter's Corner
Sun 1 May 2005
Problem Set #12
Hello 8.01-ers!
Special Relativity is confusing as it is counter our daily
experiences, but it is fascinating as that is the way the "fast" world
ticks (no pun implied). It dramatically changed the way we look at
the world, the universe and at our selves. There will be no more
Walter's Corners as it was not part of PIVoT and not part of my Fall
1999 8.01 lectures either. In the old (pre-web) days (late 60s and
early 70s) it was part of 8.02. There is countless - useful and not so
useful - information on the web. This is your chance to make a trip
and return to Earth in the prime of your life in the year 3000. Try "Twin Paradox" on Google - you'll find more than a thousand entries!
Professor Guth is offering you a chance of a lifetime (pun implied)
that you may not want to miss :-).
Greetings!
\\/\/////@lter
Tues 19 April 2005
Problem Set #11
There is alottttttttttttttt to be covered! However, we have 9 days!
=====>>>>>>>>>>> !WATCH THE VIDEOS - START TODAY! <<<<<<<<<<========
PIVoT Keywords: conservation of angular momentum, orbital motion, elliptical orbits, black hole, gravitational potential energy, torsional pendulum.
MUSTS
* Watch ALLLLLLLLLL of Lect #22
Kepler's Laws, Elliptical Orbits, Satellites, Change of Orbits,
Ham Sandwich [& Romance!]
* L29d Conservation Laws for a Satellite's Orbit (05:57)
* L23e Black Holes (10:23)
* Watch ALLLLLLLLLLL of Lect #10
Springs, Simple Harmonic Motion, Pendulum, Small angle Approximation
* L21c Physical Pendulum (with demos) (18:36)
* L29c SHO of a Suspended Rod (08:39)
* L30a SHO of a Physical Pendulum (20:48)
* L13a Potential Energy U(x) due to Gravity (06:55)
* L13d Parabolic U(x) and Conservation of ME => SHO (05:05)
* L13e1 Circular U(x) and Conservation of ME => SHO (15:29)
* L13e2 Circular Track Demonstrations of SHO (11:07)
HELPFUL
* L18e Newton's Law of Universal Gravitation (06:27)
* SG Satellite Approaching the Sun (03:48)
1 YF12.32 piece of cake.
2 YF12.34 piece of cake.
3 YF12.37 a) piece of cake
b) show that the total energy (PE + KE)<0 that's the criterion for bound orbits
4 SG9B2 (S) Important problem.
The angular momentum relative to the sun (called "orbital angular momentum") is conserved as the torque relative to the sun, S, is ZERO; the gravitational force due to the sun, acting on the planet, goes through S, no matter where a planet is in its orbit. NOTICE that angular momentum is ONLY conserved with the sun as origin.
5 SG9B3 (S) Apply conservation of angular momentum relative to the sun.
6 SG9B4 Disposing of nuclear waste.
OCW look at Problem 8.3 (8.01 Fall 1999) and the solutions. This is nearly identical to SG9B4.
7 YF12.43 Astrophysics - Black holes!
The ring, as mentioned by YF is probably introduced to make things easy. I am not sure, I have not worked this problem yet. However, I'd like to suggest that you watch
* L23d X-ray Binary Systems (12:39)
* L23e Black Holes (10:23)
In addition, on OCW (8.01 Fall 1999) you also find problem 8.4 with solutions regarding the determination of the mass of a black hole in the binary system Cyg X-1.
8 YF12.87 Tidal forces - Cool!! This is not a difficult problem.
I don't think I covered tidal forces in my 1999 lectures, and I do not find it on PIVoT - sorry.
9 YF13.27 Spring (see "Musts" above).
10 SG2D2 (S) - Important concept - Small angle approximation.
11 SG2D3 * SG Mass Sandwiched Between Two Springs (19:14)
Watch also
* SG Period of Oscillation of Mass Sandwiched Vertically Between a Pair of Springs. (08:47)
12-13 SG4E2 (S) Watch on PIVoT
* L13e1 Circular U(x) and Conservation of ME => SHO (15:29)
* L13e2 Circular Track Demonstrations of SHO (11:07)
14 SG4E5 Important problem - not hard
Gravity has no effect on the period of oscillations. However, gravity changes the equilibrium position.
Solved on PIVoT
* SG Mechanics of a Mass Hanging on a Spring (11:03)
15 YF13.36 Torsional pendulum
* L30c Torsional Pendulum (17:14)
16 YF13.47 Relation between a simple pendulum and a physical pendulum.
Piece of cake.
* L30a SHO of a Physical Pendulum (20:48)
Greetings,
\\/\/////////@lter
Mon 11 April 2005
Problem Set #10
PIVoT keywords: angular acceleration, torque, angular momentum, static equilibrium
MUSTS on PIVoT
* Watch ALLLLLLLL of Lect #24 - Gyroscopes - great demos!
* Watch ALLLLLLLL of Lect #25 - Static Equilibrium
1. PIVoT[ed] disk with attached weight -- See solutions to Q#8
NOTICE a small change though in the new text.
2. Slightly revised from Q#9. Cylinder on a string. This is a CLASSIC!
"WL Discusses Unrolling a Rope Around a Disk [12:05]"
NOTICE: solve (i) using as origin the center of the cylinder => find omega about the center. (ii) origin at point P where the rope "meets" the cylinder. REMEMBER that in the case of pure roll (no slip), at any moment in time, the cylinder also rotates about P with angular velocity omega. Do not overlook that I_P is different from I_C! HINT part (d) If a person accelerates the string upwards with acceleration "a", as far as the tension in the string is concerned, the problem is equivalent to NO acceleration upwards but an increased gravitational acceleration (g+a). Suppose the person accelerated the string DOWNWARDS with acceleration g. WHAT WOULD THE TENSION BE? GET IT??
3. A CLASSIC!!! This problem is similar to the Baseball problem of last week.
"WL Discusses a Hit Ball on Frictionless Surface [25:19]"
The following two videos contain all necessary ingredients. Make sure you see the connection between the three videos.
"WL Discusses a Hit Rod on Frictionless Surface - Part 2 [10:45]"
"L21b Dynamics of a Spinning Rod [18:05] with demos."
4. SG9C2 (S) L_vect=r_vect x p_vector
take the time derivative on both sides and you got it! Keep in mind that (dr/dt) x p = v x p = zero.
5. SG9D1 (S) Solved on PIVoT! Important problem -- NOT EASY! Omega and L not in the same direction!
"SG Tilted Rod on Rotating Table [07:12]"
I realize that this problem is solved in the SG. Yet, you will greatly benefit from also watching the solution on PIVoT!
6. SG9D9 rotating meteoroid. Important problem -- NOT EASY!
Omega and L not in the same direction! It's NOT on PIVoT.
7. SG9D4 (S) Gyroscope - Lect #24
NOTICE on page 345 (top line): "we use again our assumption that the only relevant contribution to the angular momentum is a vector of magnitude I*omega". This is one way of saying that the angular momentum due to precession is negligibly small. Also see my note under Problem 8.
8. YF10.48 Gyroscope - Lect #24
Keep in mind (see the bottom of page 388 in YF) that eq. 10.36 only holds if the angular momentum of the spinning disk (I*omega) is MUCH larger than the angular momentum due to precession. When the angular velocity of the spinning disk goes to zero (thus the disk no longer rotates), the precession angular velocity is NOT infinitely high as eq. 10.36 might imply. Please do check whether this condition is met.
9. YF11.12 When will the beam tip?
Required for static equilibrium: the sum of all forces = zero, and the sum of all torques (relative to ANY point) = zero.
10. YF11.46 Balancing pieces of steel.
Required for static equilibrium: the sum of all forces = zero, and the sum of all torques (relative to ANY point) = zero.
11. SG9A6 Torque on non-uniform rod
Extend the two ropes till they intersect. The center of mass of the rod MUST lie vertically above the point of intersection. WHY?
12. SG9A7 Will the ladder slip? Solved on PIVoT.
"SG Ladder Stability [7:05]"
\\/\//////////@lter
Mon 4 April 2005
Revised PS#9
The new version is highly reduced -- it's still difficult but
no longer "murder"! Now that I have worked most (not all)
problems, I can be a bit more specific (helpful?) than I was
Saturday.
PIVoT
keywords: angular momentum, rolling motion, torque, yo-yo
ABSOLUTE MUSTS!!!!!!!
* Watch ALL of Lect #20 (nice demos)
* L21b Dynamics of a Spinning Rod (with demo) [18:05] !!!!!!!!!
* SG Conical Pendulum and Angular Momentum.
Torque (tau) and angular momentum (L) are RELATIVE to a chosen point
of origin. In principle you are FREE to CHOOSE that point. Though
more often than not, for a given problem, there are "obvious"choices. However, there can be more than one point that leads to the
desired solutions. Also keep in mind that angular momentum in ONE and the SAME problem, may be conserved relative to one point (or more
points) but not relative to others! It is sometimes preferred to
choose points for which L is conserved, but that is not always an
absolute must (it depends on the problem). In the case of the conical
pendulum, for instance (this is NOT on PS#9, however I recommend it,
see above), you can get solutions by choosing as the origin the center
of the circle (angular momentum is conserved about this point and ONLY
about this point), but also by choosing as the origin the suspension
point of the pendulum about which angular momentum is NOT conserved.
PURE ROLL = NO SKIDDING: For objects with constant radii (such as
spheres and cylinders) the tangential speed at the circumference
equals the speed of the center of mass of the rolling object:
V(tan)=V(cm). V(tan) is ALWAYS omega*r, but ONLY in pure roll is
V(cm)= omega*r. When such an object is in pure roll, the point of
contact with the surface (P) stands still and the object is rotating
about P with angular velocity omega which is the same angular velocity
about the center of mass (see YF page 371). This can be used to get a
very quick and very elegant solution in problem 4.
1. SG8D.2 & 8E.4 L21b Not very difficult.
2. YF10.39 A classic! Not too tough. The tension in the rope is the centripetal force. If you pull on the rope, the radius gets smaller, and the angular velocity will go up because angular momentum about the hole in the table is conserved, but ONLY about that hole because the torque (= r x tension) relative to that point is zero. There are two ways to calculate the work that you have to do. The easiest is to calculate the difference in rotational KE between the two situations. Alternatively, this is harder, you do an integral of (F dot dl) as you pull the rope downwards over a distance dl.
3. SG10.1 (S) KEY Problem!!!
"L21b Dynamics of a Spinning Rod [18:05]"
L_P = not zero; L_Q = zero. Yet, angular momentum relative to the PIVoT P is conserved and angular momentum relative to Q is NOT conserved. WHO SAID THIS WAS EASY? Notice that it is assumed here that there is no friction in the PIVoT at P.
4. SG8E8 (S) VERY important problem!!
"L24a Pure Roll of Hollow & Solid Cylinders with demos" As the origin, you may choose the center of the rolling object, or the point (P) where the object is in contact with the incline. In the latter case you do NOT need Newton's 2nd law, but you now have to calculate (using the parallel axis theorem) the moment of inertia for rotation about point P. I did both methods in class today. The second method works because in the case of PURE ROLL (NO SKIDDING), the point of contact with the surface STANDS STILL and the rolling object rotates about this point with the same angular velocity with which it rotates about the center of mass (see above). As the object rolls down, the friction is NOT doing any negative work. NO HEAT is produced as the point of contact with the incline STANDS STILL!!
5. SG9D6 Another KEY problem! Here is the scenario that may give you some insight:
* As the wheel touches the ground it has a speed but omega about its axle of rotation is zero - the wheel is not rotating.
* The wheel skids at first - heat is produced, you are burning rubber (blue smoke!) - the friction will be the maximum possible.
* The friction will torque the wheel up omega(t)= alpha*t.
* The friction will decelerate the wheel's axle according to Newton's second law: F(fric)= Ma, V(t) = V(o) - at. NOTICE the "-".
* Thus, omega increases, and the plane's speed decreases UNTIL the NO slip (pure roll) situation is reached. At that time the friction will go to ZERO, and the speed of the plane will NOT change any more!!!
Notice that in Problem 4, there was ALWAYS pure roll, yet the friction
is NOT zero. The reason is that, because of the incline, the object
continues to accelerate down hill; thus the friction continues to
torque up the rolling object.
NONE OF THIS IS EASY!!!!!!!!!!!!
NOTICE!! It would be MORE REALISTIC to do this problem with a bowling
ball. The idea that the pilot does not brake, makes no sense to me.
6. YF10.24 This is a classic! Cute problem, I have not worked it out in detail yet. But it seems that it requires (i) a good understanding of pure roll (see above) (ii) the role that friction plays in changing the angular velocity (see problems 4 and 5).
7. SG10.11 (S) This is a famous Brain teaser!
WL Discusses a Yo-Yo Brain Teaser (with demo) [01:33]
It's a GREAT problem, and easy ONLY if you realize a KEY property of "pure roll" as mentioned above!! See also YF page 371 Fig. 10.14. Pure roll (no slip) means that the angular velocity about the center of mass is ALSO the angular velocity ABOUT the point of contact (P) with the table. Thus, by evaluating the torque about P, and NOT about the center of mass, the problem becomes a one-liner and very TRANSPARENT! I strongly advise you to do this experiment at home!!!
8. YF10.98 This is a classic. This is a KEY problem and it is NOT easy! The concepts behind it are discussed in WL Discusses a Hit Rod on Frictionless Surface - Part 2 [10:45] and L21b Dynamics of a Spinning Rod (with demo) [18:05]. I plan to discuss this one in class (with demo).
9. YF10.27 Not too difficult.
10 YF10.55 Very useful problem - Not very difficult but very fundamental.
Good Luck!
\\/\////////@lter
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