Tue, Dec 12, 8:06 a.m. - are we going to be provided information on the test from table 10.2 or do we need to know them

You do not need to have the moments of inertia from table 10.2. The only one that you need is $mr^2$ for a point-like object. (And you also need to know that more than thing spinning together means add the moments of inertia.) We will provide any other moments of inertia that you need.

Mon, Dec 11, 2:53 p.m. - Does a rotating object experience friction with on a no slip surface? If not, what is the difference between no slip and frictionless?

Yes, there is friction between the rotating object and the no-slip surface. In fact, the friction is what causes the rotating object not to slip.

So, when you have an object rolling without slipping, you do need to include a friction force in both your free-body diagram and your torque diagram (unless you are using the point of contact for your observation point in which case the torque due to friction is zero).

Mon, Dec 4, 8:42 a.m. - Addendum to (I'm standing on a wheel with no friction on its axle. There is nothing exerting a force on myself or the wheel. I shake my hand to the right. The wheel and I start moving. The system started with no momentum because it wasn't moving. When I shake my hand, there's momentum. Momentum doesn't seem to be conserved. Please explain!):

In lecture, it was discussed that because I and the wheel have to be moving in the same direction because we are one system. Then, the turntable is NOT moving in the opposite direction as how I moved my hand. I do not understand how ANY (linear or angular) momentum is conserved in this problem. I talked to three other people who attended last lecture who could also not answer my question (I did pay attention).

Okay, the demo from lecture. Let's start with angular momentum, which is the main point of the demo. The key here is that in lecture (and I appreciate you pointing out that you are asking specifically about the lecture demo), I was twirling my hand in a circle over my head. So, my hand (and the weight in it) were going around in a circle over my head, carrying angular momentum in one direction. So since I started completely motionless (with total angular momentum zero), when I started twirling my hand over my head, my body and the turntable had to rotate in the opposite direction to keep the total angular momentum equal to zero. So, before I was doing anything, total angular momentum was zero (nothing moving). While I was twirling my hand over my head, total angular momentum was zero (angular momentum of my hand opposite that of the angular momentum of me/turntable). Zero = zero -- angular momentum is conserved.

Now, about linear momentum, at any moment, my hand and the weight were moving in a particular direction, which means that something else must be moving the opposite way to keep total linear momentum equal to zero. Since the turntable was effectively attached to the floor, it is really the entire Earth that was moving in the opposite direction when I'm moving my hands. But since the entire Earth is so much more massive than my hand (and the weight), its motion is undetectable. It's similar to how when I jump straight up into the air, the downward motion of the Earth is so small as to be undetectable. Again, total linear momentum zero both before and after.

This is probably the best that I can do trying to explain this on-line. Please swing by my office (or any of the pooled office hours) if this is still unclear.

Mon, Dec 4, 2:21 a.m. - I'm standing on a wheel with no friction on its axle. There is nothing exerting a force on myself or the wheel. I shake my hand to the right. The wheel and I start moving. The system started with no momentum because it wasn't moving. When I shake my hand, there's momentum. Momentum doesn't seem to be conserved. Please explain!

It depends on how you shake your hand. If you move your hand straight outward, away from the center of the turntable, then you and the turntable do not move. On the other hand, if you move your hand in an azimuthal direction (e.g., parallel to the edge of the turntable), then your hand has angular momentum relative to the center of the table, so the table must acquire angular momentum in the opposite direction to keep the total angular momentum equal to zero.

From this and the previous question, I get the impression that you are confusing momentum'' and angular momentum'' -- they are two different things. If things are still unclear, pose the question again, but please specify whether you are asking about linear momentum (i.e., $\vec{p} = m\vec{v}$) or angular momentum (i.e., $\vec{L} = \vec{r} \times \vec{p}$ and $\vec{L} = I\vec{\omega}$).

And take a look again at your notes from last Thursday's lecture -- we talked about this in detail. If you question is specifically about something from lecture, please let me know which part of lecture you are referring to so that I can be clear about what I'm responding to.

Mon, Dec 4, 2:18 a.m. - Ch. 11 Wolfson #58. Potential energy of the spring equals the kinetic energy of the mass... and the rotational energy of the turntable or no? <p>How does the turntable have any rotational energy? Let's think about this. There's no friction. The mass can't exert a force on the table, then. How, then, is the table moving? Isn't the mass basically not in contact with the table? Please explain!

No, it is not correct that potential energy of the spring equal kinetic energy of the mass. Please, please, please, please ... don't ever start a problem with potential energy of (something) equals kinetic energy of (whatever)'' -- I can promise you that we will ALWAYS deduct at least 2 or 3 points if you start a problem like that. The starting principle is that since $W_{NC} = 0$, that means that mechanical energy is conserved, i.e., $E_{before} = E_{after}$.

It is true that $E_{before} = K_b+U_b = U_{spring}$ since everything is motionless in the before picture (which I know that you have drawn because it is the first step in the step-by-step approach and I trust that you now know never to leave out any steps). $E_{after} = K_a+U_a = K_{trans} + K_{rot} + 0$, so, yes, the turntable does have rotational energy.

By the way, don't forget conservation of angular momentum as well -- that plays an important role.

Answering your question -- it is correct that the mass is not in contact with the table. It is flying off in one direction, and the table is then rotating. Total angular momentum is clearly zero in the before picture that you have sketched, so with the mass flying off and clearly carrying angular momentum (calculate it using $\vec{L} = \vec{r} \times \vec{p}$), the turntable must be carrying the opposite angular momentum (it must be spinning) to keep the total at zero.

By the way, conceptually, when the spring is pushing the mass, it is also pushing back on the table (by Newton's 3rd Law), so the table gets a torque while the spring is pushing that causes it to rotate.

Sun, Dec 3, 4:23 p.m. - For Question A44c, why in the world does the yoyo move towards the string, let alone accelrate so quickly? My brain is litterally on the verge of melting; this is trully the first time in this course where I cannot fathom a reason why something is happening. But I'm sure it will be a simple answer.

Yeah, it's a great puzzle, isn't it? I love this particular question, because almost no one ever expects it to roll the way that it does.

The answer, of course, can be found in the standard step-by-step approach that we have discussed. Draw a free-body diagram for the yo-yo, and write down Newton's Second Law in component form (you only need the component in the direction of acceleration). Draw a torque diagram and apply $\tau_{net} = I\alpha$ (warning: you'll need two different radii, one -- let's call it $r_1$ -- for the small radius of the spool around which the string is wrapped, and a second -- let's call it $r_2$ -- for the outer radius of the yo-yo. And then $a = r_1\alpha$. Using those three equations and solving for $a$, you'll find that it will accelerate in the direction of the tension in any case where $r_2 \gt r_1$.

By the way, the result is quite interesting, because if you use the center of the yo-yo as the origin for torque calculations, the torque from the tension causes the yo-yo to twist one way, and the torque from the friction causes the yo-yo to twist the other way. The friction torque wins this battle, resulting in the yo-yo accelerating the way you found; however the tension force has to be stronger than the friction force, resulting in the yo-yo accelerating in the direction of the tension. These statements are consistent because of the fact that the radius for the friction force is larger than the radius for the tension force, so the friction torque can be (and is) larger than the tension torque, even though the friction force is smaller than the tension force.

By the way, there is a quicker way to argue that the yo-yo has to accelerate in the direction that you found: draw the torque diagram, but use the contact point with the table as the observation point when determining the net torque. When you do that, there will only be one force that causes a torque, and it should be easy to figure out the direction of that torque. And then consider that your observation point is the contact point, and then you'll be able to see which way the yo-yo will rotate. This is a quick approach, but a bit subtle.

Mon, Nov 27, 9:34 a.m. - Suppose you have a rope around a pulley with a nonzero mass. Both ends of the rope have masses attached to them. Is the tension across this rope the same across every point in the rope?

Absolutely NOT! (i.e., the tension is NOT the same everywhere in the rope, as long as masses hanging from both ends aren't the same.)

This was the basis for ConcepTest Number 3 for Lecture 22 (Thursday, November 16). If the tension in the rope was the same on both sides of the pulley, then there would be no net torque on the pulley, so nothing would accelerate.

(Note that if the pulley is massless and frictionless, then you can assume that the tension is the same on both sides, because a massless, frictionless pulley requires infinitesimally small net torque for non-zero angular acceleration. But if $I \ne 0$, then you need a non-zero torque for non-zero $\alpha$ since $\tau_{net} = I\alpha$, so the tensions are different on the different sides of the pullely.)

So, if there is a pulley that has nonzero mass, do not assume that the tension is the same on both sides of the pulley.