# Tag Info

0

So you might want to read this wiki article I like to think of a piezo as a voltage source (pressure dependent), with a series capacitance, (and then some leakage resistance.) When you squeeze it you generate some voltage on the cap, but it leaks off. When you release the pressure then you get opposite sign of voltage for a time. (That assumes you are ...

1

I don't think you need to overthink this so much. Mechanical equilibrium in this context basically means that from a macroscopic point of view, all forces are balanced; this usually also means that the system's parts are at rest, though a system in uniform motion could be considered in mechanical equilibrium, I guess. The point that the authors are trying ...

0

Maybe in a perfect world with no thermal or atmospheric disturbance (yes simulation) this could happen. Zeno holds true regarding infinite time, but for all practical purposes the bob would appear to get there in a reasonably short time. In the real world control systems engineers accomplish this feat all the time using feedback to overcome the small ...

3

If you initially give to the bob a velocity $\sqrt{4rg}$, it will actually take an infinite time for the bob to reach the top! A little lesser velocity will cause the bob to stop earlier and come back toward the initial point, while a little greater one will take the bob over the top (the motion will continue, with increasing velocity, to the other side). ...

0

Since it is a rigid rod, you are probably right. If the rigid rod is replaced with a string, then your teacher would be right, as the velocity at the top must be non zero in order for the string to remain tight and not collapse before reaching the top. In real life scenarios, however, it is nearly impossible to maintain an unstable equilibrium.

2

Isn't it just a test of what kind of potential minimum you are sitting in (interpreting $f$ as a force)? So if $d f/d x < 0$, and $f = -d U/d x$, where $U$ is the potential, then your condition reduces to $d^2 U/dx^2 > 0$ at $x=x_0$. If $f$ only depends on $x$ then that clearly marks a local potential minimum (assuming that $f(x_0)=0$). Beware ...

1

One could interpret the update steps as possible discrete steps in a fictitious time and in that case the transitions represent dynamics on the state space of a Markov chain. As an example, there is the relaxational non-conservative Glauber dynamics and the magnetization conserving Kawasaki dynamics which are used to simulate Ising and related systems. The ...

Top 50 recent answers are included