Recent content by dalterego

Ok when I use L'Hopital's rule it goes like: ln([x+ln2]/x)/(1/x) = (x/[x+ln2])/(-1/x^2) which simplifies to -x^3/(x+ln2) = -3x^2 which is still infinity

Homework Statement limit x --> infinity, (1+ [(ln2)/x])^x Homework Equations The Attempt at a Solution I tried raising the whole thing to e ln [function] so that the x power would get eliminated. What I've got right now is this: e lim(x->infinity) ln([x+ln2]/2)^x e...

Alright, so is the answer inverse of the function = (ln x)^1/3 ?

The inverse of the exponential function... Homework Statement Find the inverse of the function = e ^ (x^3) Homework Equations The inverse of the exponential function = the natural logarithm of that same function The Attempt at a Solution inverse of f(x) = ln(x^3) ? This...

Integration Issues...a bit urgent Hey having problems with five of my assignment questions, any help would be appreciated. Homework Statement (problems attached) Homework Equations The Attempt at a Solution For the first one, I tried to solve it by using the sin-1 x as 'u'...

Thanks very much!

By the way, thank you Doc Al for your help, right now I'm kind of assuming that the c.o.m. is equal to the length and I managed to get an answer for the angular speed: 6.28. I'm not sure if it's correct I hope it is, but I kind of have to go now and I'll check back in the morning. I understand...

Right...forgot to square the r s, sorry! Just wondering, since the rotation in question is from its initial state to when AB is perpendicular to the axis at A (or vertical), wouldn't the distance between the initial centre of mass and the final centre of mass be equal to the length of each...

Erm, I found the rotational inertia using the formula I = Sum of m r(squared) where r is the perpendicular distance from the axis, and I got: (0.2)[(0)+(0.5)+(0.5)+(0.707)] = 0.3414 And since rotational K.E. is = (0.5)I(omega)squared, it will be (0.5)(0.3414)(omega)squared... When you said...

Well I don't quite understand what "through the vertical position" means, is it when it passes through the line perpendicular to A? And yes it is released from rest; mechanical energy is the sum of the kinetic and potential energy right? So this would turn out to be something like: Initial...

Lol, no the other particles are not labelled anything. And it doesn't tell you 'when'. I'm assuming, kinetic energy will be conserved? (er..I think)

Oh the exact phrase is "What is the angular speed of the body about axis A when rod AB swings through the vertical position?"

Yeah, here's the question: Four particles, each of mass 0.20 kg, are placed at the vertices of a square with sides of length 0.50 m. The particles are connected by rods of negligible mass. This rigid body can rotate in a vertical plane about a horizontal axis A that passes through one of the...

How can you find angular speed (average), if its a question of rotation about a fixed axis, and all you know really is the rotational inertia of the system? You have no values of the angles (with regards to the reference axis) and no values for time either.