Recent content by Matuku

Ah, thank you; I understood where the sin(\theta) came from but didn't really phrase my enquiry very well. What I really wanted to ask was do you not have to consider the fact that the disks aren't vertically sided (as you do for the rings) which you have already answered; thanks! ^^

Whilst following my textbooks advice and "proving to myself that the inertia listed are true" I considered the Moment of Inertia of a hollow sphere by adding up infinitesimally thin rings: dI = y^2 dm = y^2 \sigma dS = y^2 \sigma 2\pi y dz = 2\pi y^3 \sigma dz This didn't work...

Ah, that brings back memories from A level actually; let me see if I've got this right. Let, \] Acos(\theta) + Bsin(\theta) = Rsin(\theta + \phi)\\ \therefore Acos(\theta) + Bsin(\theta)=Rcos(\theta)sin(\phi) + Rsin(\theta)cos(\phi)\\ \therefore A = Rsin(\phi), B= Rcos(\phi)\\ \therefore R =...

I've been going through and proving to myself that the solution to the SHM equation is correct; at A level we were just told what it was but never really shown why. I've gone through the problem as a standard 2nd Order ODE (with complex roots of course) and ended up at: x = A_{real}...

If you have a fraction, for example, \frac{1}{{1.0091532\times10^{-12}} + 1} Is there a simple way to convert it to a more easily calculated form, specifically, 1-x (where x is a very small number)

You're right, it is -cot(u) rather than tan(u); don't know where I got that from. But you don't believe there's anyway to get it into a form x(t) [or even t(x)]?

\frac{dx}{dt} = \cos(x+t) I'm having real troubles with this; I tried a substitution of u=x+t but it just ends up as, t=\tan{u} + \csc{u} + C And I can't see where to go from there. (the middle function is cosec; it didn't come out very clearly on my screen).

Homework Statement Find, \int \left( \frac{1}{x} - \frac{1}{x^2} \right)e^x ~dxHomework Equations NoneThe Attempt at a Solution I tried integrating by parts, \]\int \left( \frac{1}{x} - \frac{1}{x^2} \right)e^x ~dx\\ Let ~\frac{dv}{dx}=\left( \frac{1}{x} - \frac{1}{x^2} \right), and...

Oh of course it is! We're told what x is!

I'm sorry I'm still sightly confused; I now have: a=-A\omega^{2}cos(\omega t+\phi)=\frac{-kx}{m} Which implies that x=Acos(\omega t+\phi) but doesn't really show why? Is this what you intended or am I missing something?

So differentiate the solution given to us to get it in terms of acceleration and then just compare that with the a=-kx/m?

Homework Statement A particle of mass m moves in one dimension under the action of a force given by -kx where x is the displacement of the body at time t, and k is a positive constant. Using F=ma write down a differential equation for x, and verify that its solution is x=Acos(\omegat+\phi)...