Recent content by Fluxthroughme

So the consensus appears to be I just need to attach it to a conductor that is, roughly speaking, a fair bit larger than my apparatus. Do I have to do this in any fancy way, or could I do something as simple as taping a wire from the casing to a large metal object?

I'm making a small capacitor for a project at uni, where I'll be encasing the whole thing in a metal box for shielding, which needs to be grounded. However, I don't know how to actually go about doing this? The device will be indoors, so I can't just throw a steak in the ground. Any advice is...

So I am given B=\begin{array}{cc} 3 & 5 \\ 5 & 3 \end{array}. I find the eigenvalues and eigenvectors: 8, -2, and (1, 1), (1, -1), respectively. I am then told to form the matrix of normalised eigenvectors, S, and I do, then to find S^{-1}BS, which, with S = \frac{1}{\sqrt{2}}\begin{array}{cc} 1...

My own small bit of input: There are people who can't do math, but it's not quite that simple. I am pretty sure it's down, hugely, to your upbringing. I have seen a LOT of people who can't do basic maths, and if I try and help them, it's not directly because they don't get the material, but...

Ok, I got the correct answer this time, by using 1.5g - T = 1.5a, a=r\alpha, and \tau = rF :) Thank you!

Image included to show I've made an attempt, but my attempt left me realising I just don't know where to go. If someone could point me in the right direction, it'd help a lot.

Ahh! Converting to rad/s first and converting to rev/min at the end does work. So it WAS something simple :P Thank you!

Haven't done much work with gyroscopes, so I think I'm just missing something obvious, but since I'm using very basic formulas, I cannot see my error. The stated answer is 1.8*10^3 rev/min, so I am clearly off.

Ok, a=r\alpha Solving from there is just basic algebra. Prior to googling, I had not seen this, nor thought about a concrete relationship between them, which must've been why I could not see how to do the problem. Thank you for putting up with me :P

I don't know how it relates it general, but at that instant, they are the same? 5g - T = 5a \tau = rF=r(5g-5a) = 61.3 - 6.25a = Ia, a(I+6.25) = 61.3 \frac{61.3}{\frac{1}{3}*15*1.75^2 + 6.25} = a a = 2.84

Which doing will give me 2.84 as the final answer, which is still incorrect? =/

Yeah, that's what I meant - LaTeX troubles :P Well, regardless, I still get the wrong answer: And I checked, in case I mixed my signs up, adding 6.25 rather than subtracting, which gives 2.84, still not the correct answer.

If you mean rF=I/alpha, then isn't that a different acceleration to the acceleration for the block?

I'm not sure how I can be any more direct: I am 100% lost at the post. It's not for lack of thinking, I just don't know what the equation ought to be?

I'm not even guessing; I just don't know how to set up the equations. T - 5g = 5a, I think? Then I am 100% lost at the post.