Recent content by zanick

I was rounding but yes, 1/2Iw^2..... but also 1/2M(wxR)^2 So we then can solve for M which is M=I/R^2 . So it will vary for wheel sizes 13" to 18" for example for values ranging from 1.3 to near 1.5. and going to 2x (as you indicated) when just talking about adding weight to the outer...

i think what was said in more technical terms, is that yes, you have more "force" due to a higher weight (which is a force) but the mass is greater, so it cancels out. 9.8Newtons on a 1kg object will accelerate at 9.8meter/sec/sec (acceleration). since acceleration (g) is constant, if...

Having gone over all possible ways to make my race car perform better, this has come up a number of times with racers, and i have answered them with a much easier description than outlined here . if you think about it, there is a way to equate the rotational mass to what it would equate to if...

the movement for the pendulum is a precession due to a swinging plane deviating from surface alignment. the puck wont suffer Coriolis effects, because it is not moving as noted by the OP. coriolis is the perceived deviation from a path as viewed from the rotating reference frame, so the...

that is NOT due to earth rotation

yes, "straight" but curved in relation to large, massive bodies. (i misspoke there, so if the path is straight, then there would be no precession of a gyro in an orbit around a planet? Quite simply, the other question I had was if the "spaceship" was pointed in the direction of a geodesic...

My original question is whether or not the gyro will be affected when installed inside an object that is following a curved path (the geodesic path) toward a massive object or orbiting a massive object like a planet. In other words, when in freefall, does the gyro always see a straight, inertial...

Einstein said, when describing someone falling off a building, that the Earth accelerating up to meet him/her. Without the Earth getting larger in all directions as the paradox goes, it curvature of space-time which is why you can have the acceleration up without the surface moving up as you...

that makes sense..."viscous damping" and so does it contribution to entropy of both a reversible and Irrev process.

If the entropy is taking place in the gas, in the irrev example, how it is easily calculated? if you can't use S=dQ/T, can you use a boltzman equation instead? S=k.logW ?

that was one of my questions to you. if no friction, I am leaning toward it never coming to rest, but i really don't know. It should act like a spring mass system. i suppose that the entropy calculation would have to assume it comes to rest at some point.

the piston with the weights will move as long there is a net force acting on them. as the pressure reduces the force reduces and the piston will stop moving when equilibrium has been reached (when the pressure on the piston / area (F) = the weight on the piston's weight. the force will be...

That makes more sense, especially since i was probably thinking of an isolated system rather than closed where heat can be exchanged across the border. in the Irrev process, the entropy is due to irreversibility of the process, while in the reversible process, the entropy rise is due to the...

Can you briefly describe in your words why the equation wouldn't /doesnt work for irreversible processes. thx

so the reason that dS=dQ/T only works with a reversible process as that it incorporates teh surroundings and the system, which would be injecting energy control the "reversibility" of the process. if irreversible, this can't be used due to that equation leaving out other factors showing the...