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  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
  16. M

    Mechanics - Elastic springs and strings

    I get the Modulus of elasicity to be mg = \frac{\lambda\frac{2}{3}L}{L} Therefore:- \lambda = \frac{3mg}{2} Am i correct? Thanks.
  17. M

    Mechanics - Elastic springs and strings

    Argh, I've tried the question again and again and cant get the answer. Are the above values for energy correct? Thank you.
  18. M

    Mechanics - Elastic springs and strings

    Sorry, i dont understand.
  19. M

    Mechanics - Elastic springs and strings

    A particle P of mass m is attached to one end of a light elastic string of natural length L whose other end is attached to a point A on a ceiling. When P hangs in equilibrium AP has length \frac{5l}{3}. Show that if P is projected vertically downwards from A with speed \sqrt(\frac{3gl}{2}), P...
  20. M

    Mechanics - Elastic springs and strings

    A particle P of mass m is attached to one end of a light elastic string of natural length L whose other end is attached to a point A on a ceiling. When P hangs in equilibrium AP has length \frac{5l}{3}. Show that if P is projected vertically downwards from A with speed \sqrt(\frac{3gl}{2}), P...
  21. M

    Messed up somewhere in my integration

    Thank you very much. My maths has been pretty bad over the last couple of days (you can tell by the number of posts).
  22. M

    Messed up somewhere in my integration

    But isn't C just a constant. Which wouldnt matter if i multiplied it by 2 or not.
  23. M

    Messed up somewhere in my integration

    \frac{dv}{dt}= -x^{-3} \frac{1}{2}v^2 = \frac{x^-2}{2} + C As at rest when x = 1, v = 0 then C = -1 v = \sqrt(x^-2 - 1) Then i continue to integrate as above.
  24. M

    Messed up somewhere in my integration

    \frac{dv}{dt}= -x^{-3} when t=0, the particle is at rest with x=1 Therefore by integrating i get v = \sqrt(x^-2 - 1) \frac{dx}{dt}= \sqrt(\frac{1 - x^2}{x^2}) dx\frac{x}{(/sqrt(1 - x^2))} = dt -\sqrt(1 - x^2) = t + C C=-1 Therefore:- t = 1 - \sqrt(1 - x^2) However i cant...
  25. M

    Imaginary multiplication with answer to be in polar (variables)

    Multiply the numbers out first. Then you will have just one complex number. Then just find the modulus and argument using pythagoras and arctan like you mentioned.
  26. M

    Yet another FODE question

    When i integrate i get t = 1 - (1 - x^2). I'm getting confused by this now, i think my brain's just died, ah well. Sorry for keeping asking questions, i can see how annoying it would be if the answer was so obvious yet i dont grasp it. Thanks, Bob
  27. M

    Yet another FODE question

    I've integrated to t = \sqrt(x^2 - 1) but cant get the answer correct. Find t when x = 1/4, which gives the root of a negative number, not real. But the book's answer is \sqrt(15) which i notice is \sqrt(x^-2 - 1). How is this possible? Thank you, Bob
  28. M

    Yet another FODE question

    Thank you very much. I really appreciate the help. Some of these mechanics questions are a bit hard. Need to improve my integration skills abit.
  29. M

    Yet another FODE question

    Hey. I've been doing more mechanics recently - further kinematics in M3. I've come across another question i'm confused by. A particle moves on the positive x-axis. When its displacement from O is x meters, is acceleration is of magnitude x^-3 m/s^2 and directed towards O. Given that...
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