Recent content by mr bob

  1. M

    Stability physics problem

    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...
  2. M

    Stability physics problem

    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...
  3. M

    Stability physics problem

    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...
  4. M

    Simple Harmonic Motion

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

    Simple Harmonic Motion

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

    Simple Harmonic Motion

    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 recieves an impulse of magnitude 0.5Ns in the direction AP. Show that while the string is taught, the motion of P is simple harmonic...
  7. M

    Simple Harmonic Motion - Alevel M3

    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...
  8. M

    Mechanics - Elastic springs and strings

    Hmm, Ive cancelled 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...
  9. M

    Breakbeat techno metal thingy

    Not too far off the reaction i was expecting:biggrin:
  10. M

    Breakbeat techno metal thingy

    Hey guys, Have a listen to what i do when im 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.
  11. M

    Mechanics - Elastic springs and strings

    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.
  12. M

    Mechanics - Elastic springs and strings

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

    Mechanics - Elastic springs and strings

    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 -...
  14. M

    Mechanics - Elastic springs and strings

    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?
  15. M

    Mechanics - Elastic springs and strings

    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
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