Recent content by ozmac

Ahh ok, fair enough (and awesome btw!). I presume for your force balance you can ignore the forces on the cylindrical part of the wall as they all balance each other, and just look at forces along the direction of the tube. So you have the force on one end p(r)*A and on the other end p(r+Δr)*A...

Density is not constant though, the gas is compressible, I'm not sure how that helps me It seems most literature is based on assumptions of incompressible fluids and constant acceleration/gravity, none of which come close to a suitable approximation in this case.

I have what seems like a straight forward question, but am unable to find a formula. The question originates from my previous topic which was trolled: https://www.physicsforums.com/threads/blowing-up-a-balloon-using-centrifugal-force.848892/ Basically I have a 1.5m hollow length of 10mm tube...

Hilarious

Stop it you're not helping anymore. I looked at it. The comprehensibility of the fluid and increasing rate of acceleration make it difficult for me to calculate, hence why I'm asking here. If you don't know yourself please stop trolling.

No offence but it's no help to continually tell me to create a differential equation for the scenario. If I knew how I wouldn't be asking here. Everything is straight forward when you know how to do it. We shall wait!

It's not quite that simple for two reasons, one because the effective gravity is dependent on the position in the tube, and secondly because of the compressible nature of the fluid. So the pressure gradient would actually increase at an increasing rate because these two things working together...

I presume you had the spinning end sealed off with a balloon also. If you had both ends open you will end up the generation of a vacuum due to Bernoullis venturi effect. But yeah the logic of increased density at the end of the sealed tube does still hold, it's just at a much smaller magnitude...

Oh wait I think I see now. Atmospheric pressure is related to all of the mass of the air pressing down on you. And all that air results in a pressure (similar to hydro static pressure) of 101.3 KPa. In this case, even if all of the air in the tube was at a density of 24 times atmosphere, the...

Yes which would mean the balloon feels air of that full weight, and as the air outside the balloon is atmospheric, the balloon would inflate until it stretches to equilibrium (or pops) right? Davenn, simple calculation or online calculator: Fc = m v2 / r = m (n 2 π r / 60)2/r = 0.01097 m r...

Hey guys, So I had what I thought was a cool simple idea to show the effects of the apparent centrifical force. I glued the mouth of a balloon onto the end of a 1.5m length of a 10mm flexible tube. The idea in my head was that by spinning it around the inertial forces would result in effective...

Please guys I'm really stuck here. Am I suppose to work with a Cx and Cy instead? I tried that, and added a few equations, for example solving the x forces and y forces to be zero on ABC, and then I took a moment about D considering both bodies, (so forces E, Bx, By, T acting on the relevant...

Hmmm ok, so if instead of Fc, I should designate a Cx and a Cy? then solve the two equations simultaneously? For arguments sake, let's say that B and C are not horizontal, i.e. C is a little below B, meaning you can't ignore Cx when calculating moment about B.

Hey guys, So it's been a while since I've done F.B.D's, and what seems like a simple problem is causing me grief. I have this scenario below, where there are two rigid bodies, first one is ABC which pivots at fixed point B. This is connected with a pin to CDE, where D permits horizontal motion...

Hmm ok but it doesn't really help me, I'm having trouble relating it to any existing examples. Most examples of statically indeterminate beams in torsion have one or both ends fixed, whereas this is not the case. Also, I think I agree I need to consider the properties of all arms (even though...