Recent content by eddiezhang

I recently came across the concept of 'mean free path', and some similar concepts in thermodynamics (the depth of my understanding is very shallow though, which is why I'm here). This is very much a shower though, so: Suppose I have a box filled with some some metal ball bearings, which I...

Is this an aerodynamic drag problem? This is sort of why I'm asking this question, but are there two distinct forces acting here, i.e. drag / air resistance and then the force induced by the pressure difference (which I will call the 'pressure differential force' from now on)? Or, as I suspect...

I don't think I can edit my original post, so here's a more accurate version of what I'm describing

Pretty much... yeah. I asked this question in this rather... different way because I thought an explanation for this scenario would provide a more 'generalised' explanation, but as I've learned, they're quite different scenarios... I hope I don't sound like a broken record, but I'm curious...

Fair enough - the picture is pretty misleading

Right... I should have phrased that question better. I'm more interested in specifically the air pressure underneath the paper. Knowing, for example, the speed at which I bring it upwards and the air pressure of the air round me, is there some mathematical method of predicting roughly what the...

Another likely very basic question from me Suppose I pinch a piece of A4 paper and pull it up, at a velocity of let’s say 1 or 2m/s (i.e. order of magnitude 10^0). The paper, of course, encounters air resistance / drag as it moves through the air, but there’s also a (larger) force that is...

Yep... kicked myself for not seeing it this entire time

THANK YOU SO MUCH I'd give you a medal but I'm too broke to pay for shipping to Australia

OH OH OH That's just the parallel axis theorem, right...? and so the ##I_n## of the solid is (I'm gonna try finish off the problem):

Thanks for the heads up on the notation I'm in over my head because I don't really understand the second part. Why isn't ##\rho WL\int_{h=0}^H(d^2+\frac 14H^2)dh## the moment of inertia ##I_n## of the solid? Aren't we summing all the MoIs of the individual lamina with width dw, height H and...

I'm just unclear on how to apply this mathematically (i.e. setting up the integral). Here's what I think the computation would look like based on how you were nudging me along, though I'm probably wrong:

My try (probably wrong): (using 'H' as the height and summing infinitesimally short slabs of height dH with width W and length L) dm = ρdV = ρWLdH hence the mass integral becomes ? I don't get how this helps me find the moment of inertia...

I see...

I've only done single-variable calculus, so what does 'integrating over the body' mean...?