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visits member for 2 years, 7 months
seen Jul 15 at 21:02

Jul
15
awarded  Scholar
Jul
15
accepted Interpretation of dispersion relation
Jul
14
comment Why plane stress condition is taken for thin plates
As I wrote in my answer, this is not a proof, nor a consistent derivation. Plane-stress equations are an APPROXIMATION, which is not exactly valid. Specifically for the profile you suggested, it is not enough in order to determine whether it is a possible solution. You need the other stress components, and if they satisfy the force-balance equation $$\sum_j \frac{\partial \sigma_{ij}}{\partial j}=0$$ then you're good.
Jul
14
answered Why plane stress condition is taken for thin plates
Jul
14
asked Interpretation of dispersion relation
Dec
15
awarded  Yearling
May
17
comment How can I determine whether the mass of an object is evenly distributed?
This is true, but the OP only wants to know if it's evenly distributed or not, which is an easier prblem than "find the distribution". For example, you could measure the moment of inertia around an axis, and compare that to what you'd get if it were homgenously distributed. If there's a discrepancy - you can catch it. I wonder whether one can find a counterexample of a body that has the full tensorial moment of inertia of a homogeneously distributed one, but is actually not.
May
2
comment Will a hole cut into a metal disk expand or shrink when the disc is heated?
@jamesdlin I agree that it's heuristic and as such one could argue differently. The real solution is sketched by David Z. If you want more justification, you can say that the thermo-elastic equations with $T=const$ and stress-free BC will result in a stress free configuration (easily verified). Therefore, having the disc in place or cut out has no effect on the surrounding - the "interaction" is the stress, and therefore stress free boundary conditions are equivalent to not having a disc at all. Again, I stress that this can be solved analytically and then there's no ambiguity.
Mar
18
revised From momentum to solid angle
edited tags
Feb
7
comment When driving uphill why can't I reach a velocity that I would have been able to maintain if I started with it?
@Gugg In my experience, the phenomena is pretty common. Also, there are surely many other factors involved (the computer regulating fuel injection, exauhst, wind/friction dissipation...) so this simplified model does not capture the difference between "hard" and "impossible".
Feb
7
comment When driving uphill why can't I reach a velocity that I would have been able to maintain if I started with it?
@dmckee I agree, but I think the car is not yet at this regime.
Feb
7
comment When driving uphill why can't I reach a velocity that I would have been able to maintain if I started with it?
I'd say that the newtonian-gravity is not relevant.
Feb
7
answered When driving uphill why can't I reach a velocity that I would have been able to maintain if I started with it?
Feb
6
comment How does the correlation length of weather emerge?
I agree, but that's a totally different effect. It would be awefully wrong to say that the typical length-scale of the waves in lakes stems from the lake size, scaling-wise.
Feb
6
comment How does the correlation length of weather emerge?
The wave length in lakes is determined by the lake size? This sounds weird to me. The wavelength is typically a few orders of magnitude smaller.
Feb
6
asked How does the correlation length of weather emerge?
Jan
27
comment Scattering from a box potential of width $L$ doesn't reproduce a step potential in the limit $L \rightarrow \infty$
@Joe Well, unless your function grows exponentially for large $x$, you have no freedom in choosing the sign of $k$.
Jan
27
answered Scattering from a box potential of width $L$ doesn't reproduce a step potential in the limit $L \rightarrow \infty$
Dec
15
awarded  Yearling
Nov
3
awarded  Revival