Recent content by Nathanael

Yes, that is correct. In your OP you mixed the mass and momentum terms. I guess you can say that... In my eyes those 3 variables represent one unknown with zero equations. 7 and 17 are just identities which combine to give more identities. You did assume that (I believe you have to). When I...

Equation (17) is wrong. Equation (12) is wrong. I don’t see the point of (7) because if you think combining 7 and (the corrected) 17 will lead to anything but a trivial identity... then I disagree. This is what happens when you work with every variable: you get lost in blind alleys. You...

Yes (as opposed to giving just the absolute value). This is the number you give for part c. So the change in internal energy is Q?

Is this reasonable? A man who weighs 70.1 grams? Must be a world record... (Posted at the same time as haruspex, oops)

“The attempt at a solution” means your steps and thoughts. Please show effort.

You can write a specific equation for v in terms of w based on what has been said. I do think you would get more out of this problem if you retried in the morning.

No. I’m not sure how to hint at it, so I will just say it. They are all the same. If you imagine an arbitrary body purely rotating about some point, you can see it rotates about its CoM at the same rate as it goes about the CoR. Well now you’re almost done, just a few steps left outlined in...

Right. And this frame is moving horizontally relative the original frame, so the vertical component of velocity is unchanged, and so the v =what? (dθ/dt is unchanged as well if that wasn’t clear) If you find v correctly in terms of w, then you can find w from the energy equation and then you...

Yes... but you can be more precise. You should know the exact CoR from that reasoning. Btw what I said at the end of post#12 is used in the solution that avoids finding the CoR. If we do it your way, then we don’t need to worry about that equation. Also it’s good to sleep on things :wink:

The two points I’m speaking of are the center of mass and the contact point. Are you implying you don’t know the exact inclination of these velocities...?

We don’t know the velocity of any point yet, but we know the direction of those two points. Isn't that enough to find the instantaneous CoR? It’s not “about that point on the ground” in the sense of pure rotation; its just the rate of change of orientation. Anyway I thought you defined “w”...

Yes you have the right idea for finding the center of rotation. We can also choose another point we know the motion of, then the CoR is the intersection of the lines normal to their motion. (The contact point is a nice choice, as it only moves horizontally.) About my question about w... did...

You are familiar with p = ϒmv right? And also with ϒ = 1/√(1-(v/c)2) right? Eliminate v from the two equations and what do you get?

Mg√3L/4 is not the energy released by gravity. That’s not the correct center. The correct center has every point on the rod moving perpendicularly. But in the absence of horizontal forces, the center moves straight down, which is not perpendicular to the point you chose. I think we could...

Yes that is correct and yes we can infer it, quantitatively. It is a law of its own, independent of conservation of energy.