My intent with this question was to see how you integrate acceleration in polar coordinates. Whether it's as easy as in Cartesian coordinates because it constantly changes orientation with the polar unit vectors which adds difficulty. To do that, I defined an acceleration for which I already...
Okay, I tried integrating in polar coordinates and in Cartesian, but nothing seems to work.
I feel like I'm missing some condition. This can't be so hard to solve.
I assume there are other trajectories, yes, the set of all values of ## \ddot r, r,\dot \theta^2## that give k.
If this is what you're referring to, then yes I can see I need the initial conditions.Ok, so I did the position to acceleration for circular motion in Cartesian. Then I tried doing...
I give a particle a circular trajectory at 2 units of distance from the origin, and an angular velocity of 4 rad/s, both constant. Then I derive the velocity to obtain the radial acceleration, -32. I then say, "Given -32 radial acceleration, can I obtain the original position vector with the...
I made this exercise up to acquire more skill with polar coordinates. The idea is you're given the acceleration vector and have to find the position vector corresponding to it, working in reverse of the image.
My attempts are the following, I proceed using 3 "independent" methods just as you...
Okay, I see that now. It seems to me that for all common cross-sections this is true, at least the ones I can think of, even compound ones such as an H-beam.
But what about any cross-section? By any I mean, an area enclosed by a loop that doesn't cross itself such as a horseshoe, a...
Hi, I have a problem with inclined planes. The idea is to calculate the stress in an inclined plane of a bar under tension for which you need the surface. I have no idea how this surface is derived, though. In the attached file, you can see what I mean. For a rectangular cross-section, it's...
Sorry I'm posting so late, but I couldn't do this earlier. Now I tried 2 different websites that compute integrals, and also tried it myself but got nowhere. I computed 3-4 different expressions of that integral and substituted r for R, but I either got -358 I think or 3307, which is about 55...
So, I tried that, I put those two and got C=-7911 and then tried to integrate it in dt=dr/v(r) between upper limit R and lower 2R, to the final point from the starting one, but I got a problem, how do I integrate that\int^{R}_{2R}\frac{dr}{\sqrt{\frac{2gR^2}{r}}-7911} so I first tried searching...
It is a definite integral when I do dt=dr/V(r), but in order to get V(r), what do I have to do? Because the first velocity I got was a definite integral to the final point from the starting point \int^{R}_{2R}a(r)dr, and I can't integrate the result of a definite integral which is a value, so I...