Recent content by mr bob

Just ran through it trying splitting the triangle and i got OB = kmg(4a\cos\theta -4a\cos\frac{\theta}{2}) As this is equal to the GPE of the rod. mga\cos\theta = kmg(4a\cos\theta -4a\cos\frac{\theta}{2}) V = mga[(4k - 1)\cos\theta - 4k\cos\frac{\theta}{2}] Which is the correct...

Hmm. Since the triangle OAB is isosceles, could i split it into 2 right angled triangles to get the length of the extension in terms of \cos\frac{\theta}{2}? I'm not too clear why the sine of the angle theta can't just be used. Is there a fundamental flaw in my understanding of mathematics...

Just come across a question and I'm at a point where i see no further. A uniform rod AB, of mass m and length 2a, is free to rotate in a vertical plane, about the end A. A light elastic string of modulus kmg and natural length a, has one and attached to B and the other end to a fixed point O...

Thank you both so much! I finally got the answer.

How would i do this, impulse is applied. Would it be:-0 0.5 = 0.2v, where v is velocity?

A particle P of mass 0.2kg is attatched by an elastic spring of modulus 15N and natural length 1m to a point A of the smooth horizontal surface on which P rests. P receives an impulse of magnitude 0.5Ns in the direction AP. Show that while the string is taught, the motion of P is simple harmonic...

This is probably a really easy question. But alas the answer has eluded me thus far. Anyway, here is the question:- Points O, A and B lie in that order on a straight line. A particle P is moving on the line with S.H.M period of 4s, amplitude 0.5m and centre O. OA is 0.1m and OB is 0.3m. When...

Hmm, I've canceled it all down to get \frac{3mgl}{4} - \frac{2mgl}{3} + \frac{mgl}{3} = \frac{3mg(y - l)^2}{4l} - (Y - l)mg Which goes to:- 3(y - l)^2 - 4l(y- l) - \frac{20l^2}{12} Which after applying the quaratic formula, equates y - l = \frac{10l}{6}. (y - l) is the distance i...

Not too far off the reaction i was expecting:biggrin:

Hey guys, Have a listen to what i do when I am not doing maths and physics (music wise). www.myspace.com/thevolatilegentlemen I hope you enjoy. It might make for some interesting late night (Britain) conversation. Or at the very least mildly annoy you.

EPE = \frac{3mg(\frac{2}{3}l)^2}{4l} The string is initially stretched by 2/3L as natural length is L and stretched length is 5/3L. I really appreciate your help Doc Al, this question has been giving me trouble for the last couple of days. Thank's once again.

Hmm. Still cannot get the correct answer. Please help, this is turning me mad. Thank you for all the help i have received so far.

I'm calling the entire length of the string (at maximum stretch after projection) Y. The energies are measured from the equilibrium point. Before projection:- KE = \frax{3mgl}{4} EPE = \frac{3mg(\frac{2}{3}l)^2}{4l} PE = 0 After projection:- KE = 0 EPE = \frac{3mg(y -...

In my book, Lambda is defined as \lambda = \frac{mgl}{x} where x is the compression/extension. As the equilibrium length is \frac{5l}{3} the extension is \frac{2l}{3}. Subbing that into the above gives: \lambda = \frac{3mg}{2} Why is this wrong?

Ok, We've only been taught the elastic modulus, so i assume the question should be able to be done using that. Do i use K and i would use lambda in the EPE equation? Thanks once again, Bob