Recent content by Tkennan

Yeah, I probably will. I went ahead and answered with cos* rather than /cos even though I don't understand it yet (I haven't put a lot of thought into it though, now that I can work backwards from the answer it will be a hell of a lot easier). For now though, my brain demands I take a break...

I just can't grasp this. Isn't your analogy equivalent to the last part of my previous post, and therefore worked ala 5/cos(30)? And I see no difference between mv^2/r being the full force or the horizontal component, because either way, isn't all of the force pointing towards the center of...

Then it would be 5cos(30). But if you only knew the x component, which is the case here, and IT was 5 m/s, to find the speed it travels at 30 degrees above the horizontal, it would be 5/cos(30), right? I'm sure that you're right and I'm missing something (or I wouldn't have come here for...

Parallel to the ramp, as I understand it, would be cos(theta)=((mv^2)/r)/F, where F is the force parallel to the ramp. Thus, F=(mv^2)/r*cos(theta). Obviously I'm missing something, but what?

Just a few more hours before I need to have it answered, so here's a pic of the problem and all the answer choices to see if I left something out. I'm personally out of ideas. http://img124.imageshack.us/img124/8600/q14nj8.jpg http://img133.imageshack.us/img133/2864/q14alc3.jpg I tried...

Well that was one of the choices and it counted it wrong, so that's pretty intimidating. Not sure what you mean by taking full advantage of the angle. Anyhow, I'm going to be out for 3 hours or so. Thanks for all your help, and if you have any more hints or ideas on that question that could...

So F=ma, and a=v^2/r (centripetal acceleration), correct? Now I know I need to use trig in some fashion, I assume to get this force, which is horizontal, in the direction parallel to the angle. That seems really weird though. I initially tried (mv^2)/r*cos(theta), which made the most sense to...

Okay, gotcha. F=mg *slaps self*. Muchos gracias. So, onto number two...my first inclination was indeed to find the net force, but one thing that bothered me was that none of the answers had any variables other than theta, r, v, and m. I couldn't picture how to find the netforce without...

I suppose it either points to the center of the circle or in the direction of gravity...I'm not sure which, friction tends to confuse me. Assuming then that it's just downward, would µmg be all that is at work for the frictional force? That would make sense. Thanks :) Edit: nevermind, that...

The net force in that direction would simply be the centripetal force, no? I've tried (mv^2)/r and that's not the answer. :frown:

I've drawn a diagram, but that only helps in understanding the problem, not the question. For both of these I don't really know what they're asking. Is frictional force the coefficient of friction, or some component of the maximum frictional force? On number two, is the force parallel to the...

Well, that's the problem. I can do these problems when the circular rotation and the object are both on the horizontal plane, i.e. a car on pavement, but wouldn't the formula need to be altered to accomidate a horiztonal rotation and vertical contact/friction? There's nothing whatsoever in my...

And to start off the discussion, I could have sworn the answer to number 1 was F=(µmv^2)/r, but that was wrong. That was the first choice too, just to tempt me into an easy answer. >_<

Overall, I've done pretty well on this physics homework (which is due 10-13 @ 11PM Central), but these two questions are bugging the hell out of me. 1. An amusement park ride consists of a large vertical cylinder that spins about its axis fast enough that any person inside is held up...