Recent content by joshd

Yes, this seems to be the answer. Someone else said the same thing to me.

Had a quick search, didn't find anything. So, we have been told on our course, that a spinning extended object will "re-align" itself such that it is spinning along a principle axis with the highest moment of inertia, such as to conserve angular momentum, but minimise rotational kinetic...

Here: What the question is aiming at is this: what if you switch your wheels for 24" dia. instead of 26"? Assume mass of wheel stays the same. The axle of the wheel is at the same position, so the wheel's mass is centred at the same point. But, the "back edge" of the wheel is 1" closer to...

I was discussing this with a friend: Say you are doing a tailwhip on a bike (bike spins around pivot at headtube) Simpify this by looking at it as a bar pivoted at one end, with a wheel mounted to the bar with the radius in the direction of the bar. We want to spin this around the pivot in...

Is that simply because they coalesce? We take collisions to be elastic a lot in mechanics questions in maths, but I can't remember whether that was only when they don't coalesce... I guess it can only be elastic if you are using Newton's law of impacts, and e=1? :-\

Plot y=f(x) on a graph, and pick an integer starting point near to the root. The method will then usually converge, but it doesn't always work, eg, when it crosses asymptotes etc.

I'd have "generic superpower" that can do anything. :D

I thought x \rightarrow 0 means "x tends to 0", which means x is infinitly small, it doesn't mean that x IS 0? I know I miseed a lot of stuff out, hence why I just said the line was just part of it... We started with a curve, equation y=u.v, with two points, P and Q. The gradient...

darn, yea, swap all \partial for \delta etc... sorry. :blushing: Umm, not sure. After re-reading my notes, it says "Take limits as x -> 0".

We covered the product rule in maths last lecture, and as part of the derivation of it, we got this line: \frac{\partial y}{\partial x}=\frac{u \partial v}{\partial x} + \frac{v \partial u}{\partial x} + \frac{\partial u \partial v}{\partial x} And were told that as x \rightarrow 0...

It can take off just fine. The plane moves forward as usual (the wheel bearings are the magicaly frictionless ones, OK?) and takes off, since the plane's engines are pushing against the air, not the groud, the moving ground has no effect apart from to make the wheels spin twice as fast as it...

I would be interested to watch this also. :D

If it is slowing the Earth's rotation gradually, wouldn't that change the length of the day rather than the length of the year? EDIT: ^^^ exactly. We must have posted at the same time.

spin? Isn't that the reason that the charge is different in the first place? (spin = helicity?)

That is interesting, I've not heard that before. Yea, I am attending an open day at cambridge soon, and also a maths "intro" in manchester. I suppose I will see how those go. I think physics would make for a more interesting career maybe... My physics lecturer lent me a book called "Black...