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ParthKohli

  • one year ago

Physics (continued)

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  1. Empty
    • one year ago
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    nice art m8

  2. ParthKohli
    • one year ago
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    |dw:1443169743112:dw|

  3. ganeshie8
    • one year ago
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    in that image, let \(d = L/2\) and \(\theta = 90^{\circ}\) now if I release it that way, find the maximum height reached by the pendulum bob before the string becomes slack

  4. ParthKohli
    • one year ago
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    \[g(d) = g\left(1 - \frac{d}{R}\right)\]If \(d = R\), \(g=0\) so \(F=0\).

  5. ParthKohli
    • one year ago
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    Firstly, you can see that if the initial height of the pendulum is less than that of the peg, then it'd reach the same height.

  6. ganeshie8
    • one year ago
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    why is that so ?

  7. ganeshie8
    • one year ago
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    "string becomes slack" is that same as "velocity of bob =0" ?

  8. ParthKohli
    • one year ago
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    Initial energy is \(mgh_1\) Final energy is \(mgh_2\) \(h_1 = h_2\) The reason why this doesn't apply when the bob is at a higher height than that of the peg is that the string becomes slack before reaching its initial height. Slack meaning that \(T = 0\). When the string becomes slack, then the path carried out by the bob is a projectile as you would imagine.

  9. ganeshie8
    • one year ago
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    Ohk, it is in circular motion till the tension becoems 0, afterwards the string is not pulling the bob

  10. ParthKohli
    • one year ago
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    If you want to delve deeper into the analysis of vertical circular motion, here it is: The minimum velocity at the bottom to complete the whole circle is \(\sqrt{5gR}\). The maximum velocity at the bottom to do a pendulum-kind motion like on a clock is \(\sqrt{2gR}\) (use conservation of energy for this too) But if the velocity is between those two, then the string would become slack at one point in the first-quadrant of the circle.

  11. ParthKohli
    • one year ago
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    Yes. And the maximum height reached by the bob is not when the string becomes slack. If the velocity points upwards at that moment, then it'd go further up in parabolic motion.

  12. ganeshie8
    • one year ago
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    Interesting, I think I have understood the question and why the heights must be same when the bob is released below that peg. Let me try working what happens when the bob is released from 90 degrees..

  13. ParthKohli
    • one year ago
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  14. ParthKohli
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  15. ganeshie8
    • one year ago
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    Awesome! so the string wont become slack at all if the bob is released from a height below the peg

  16. ParthKohli
    • one year ago
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    ya

  17. ganeshie8
    • one year ago
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    Since the forces involved here are conservative, I think conservation of energy can be applied : PE + KE = constant Let us define \(0\) at the bottom most of pendulum. Then total mechanical energy at \(\theta=90\) is \(mgL +0\) Total mechanical energy at the position where the string becomes slack is \(mgL' + \frac{1}{2}m{v_{L'}}^2\) These muse be equal : \[mgL = mgL' + \frac{1}{2}m{v_{L'}}^2\tag{1}\]

  18. ganeshie8
    • one year ago
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    two unknowns in that equation, \(L'\) and \(v_{L'}\) so need to cookup another equation hmm..

  19. ParthKohli
    • one year ago
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    yeah, one with the forces

  20. ganeshie8
    • one year ago
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    I'm not good with these tensions and stuff, so im not so sure what happens when the string loses tension and how to take advantage of that..

  21. ParthKohli
    • one year ago
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    |dw:1443173019236:dw|

  22. ganeshie8
    • one year ago
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    Oh we can consider that bit anchored to peg as another circular motion is it |dw:1443173202385:dw|

  23. ParthKohli
    • one year ago
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    Let's draw the free-body diagram of the bob in the frame of the bob itself.|dw:1443173207901:dw|

  24. ParthKohli
    • one year ago
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    Since we're in the bob's frame, it's not accelerating towards or away from the string at this moment, so \(mg \sin \theta = 2mv^2/L\)

  25. DanJS
    • one year ago
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    is this like tha tprob where a bullet strikes a block on a string, and the block moves a certain angle to a height depending on the mass and velocity of the bullet

  26. ParthKohli
    • one year ago
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    no, we can't use conservation of momentum here.

  27. DanJS
    • one year ago
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    ok

  28. ParthKohli
    • one year ago
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    The height at this moment can be expressed as \(L' = L + L\sin\theta \)

  29. ParthKohli
    • one year ago
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    so now we have two equations that we can solve! yay!

  30. ParthKohli
    • one year ago
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    \[2mgL = mgL(1 + \sin \theta) + \frac{mv^2}{2}\]\[mg\sin \theta = \frac{2mv^2}{L}\]

  31. ganeshie8
    • one year ago
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    I may be wrong, but don't we also have another new variable, \(\theta\) ?

  32. ParthKohli
    • one year ago
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    Two variables: v and theta.

  33. ganeshie8
    • one year ago
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    nvm, i see..

  34. ParthKohli
    • one year ago
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    Using that, we can find that the height at which the string becomes slack. To find the maximum height in the path, we know that the angle is \(90 - \theta \) to the horizontal.

  35. ParthKohli
    • one year ago
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    ohh!!

  36. ParthKohli
    • one year ago
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    height is \(L/2 ( 1 + \sin \theta)\)

  37. ganeshie8
    • one year ago
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    \(v= 2\sqrt{L}\) plugging this in first equation gives \(L' = \dfrac{4L}{5}\)

  38. ParthKohli
    • one year ago
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    heyyy the first equation is \(2mgL = mgL/2 (1 + \sin \theta) +1/2 mv^2 \)

  39. ganeshie8
    • one year ago
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    Nice problem, I think the key thing was observing that the "centripetal acceleration" and the component of "acceleration due to gravity" along string are equal just at the instant of string losing tension.

  40. ganeshie8
    • one year ago
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    Oh was there some mistake ?

  41. ParthKohli
    • one year ago
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    yes, I forgot that the radius was L/2 in the second part of the motion so the height is \(L/2 + L/2 \sin \theta\)

  42. ParthKohli
    • one year ago
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    I have a huge collection of mechanics problems so if circular motion bores you

  43. ganeshie8
    • one year ago
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    after fixing : \(v = \sqrt{\dfrac{10L}{3}}\implies L' = \dfrac{5L}{6}\)

  44. ParthKohli
    • one year ago
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    That is correct!`

  45. ParthKohli
    • one year ago
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    Ganeshie, have you ever studied gaseous state/thermodynamics/calorimetry and the like?

  46. DanJS
    • one year ago
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    the initial energy is mgL right, sorry

  47. DanJS
    • one year ago
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    \[2mgL = mgL/2 (1 + \sin \theta) +1/2 mv^2\] what is that 2

  48. ParthKohli
    • one year ago
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    yes true the last time I did this question I just multiplied the equation by \(2\) so it became \(2mgL = mgL(1 + \sin \theta ) + mv^2\) which is why I also made the mistake earlier...

  49. ParthKohli
    • one year ago
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    whenever you do a question the second time you don't treat it the same way you do when you do it the first time ... your memory sorta reminds you of the solution and then you do wrong things :|

  50. DanJS
    • one year ago
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    how would you do that without using conservation of energy, with angular velocities? will that double as soon as you shorten the radius by 2?

  51. ParthKohli
    • one year ago
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    yes it does... you can try that too

  52. ganeshie8
    • one year ago
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    @ParthKohli physics and me don't get along that well, Never had to study thermodynamics before... I'm just a dumb engineer..

  53. ParthKohli
    • one year ago
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    wait so you didn't study that in high school?

  54. ganeshie8
    • one year ago
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    It was more than 10 years ago, can't say I remember much...

  55. ParthKohli
    • one year ago
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    CBSE? Or state board?

  56. ganeshie8
    • one year ago
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    AP state board, it is tougher than CBSE maybe, lets work a question together..

  57. DanJS
    • one year ago
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    So th ething will have an angular accel up to some tangential velocity at the bottom, then double that velocity in the smaller circle, and with that will reach a certain height until the pull of the rope, or the radial accel is zero?

  58. DanJS
    • one year ago
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    I took thermo about 8 yrs ago,

  59. DanJS
    • one year ago
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    i can still picture those bell curve looking state diagrams in my mind

  60. Jhannybean
    • one year ago
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    thermo was really difficult for me :\

  61. Jhannybean
    • one year ago
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    I need to review that stuff all over again sometime haha

  62. DanJS
    • one year ago
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    lots of tables, change of state of something

  63. DanJS
    • one year ago
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    internal energies and entropy at a state, blah blah, The 3 space PVT survace is nice to remember, the other curves are slices of it

  64. DanJS
    • one year ago
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    I would go through that again sometime if you want, so we both are to ask questions

  65. DanJS
    • one year ago
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    i still have all my practice and hw probs and stuff

  66. ganeshie8
    • one year ago
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    I doubt if I ever get hands on thermodynamics... I seriously wish sometimes that it is possible to learn without putting any effort by directly downloading to brain, like in matrix..

  67. ParthKohli
    • one year ago
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    Yes. Mechanics is wayyy better.

  68. ganeshie8
    • one year ago
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    classical mechanics is really enjoyable.. with professors like walter lewin / leonord suskind, they almost spoonfeed you.. there is absolutely no chance of finding it tough by following their lectures

  69. DanJS
    • one year ago
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    hah right, but with feynman i can follow along great, then think what the heck did i learn again a few days later..

  70. DanJS
    • one year ago
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    You already know half of it, the first law is a more general energy conservation equation that includes more things that is like 1/3 the class

  71. DanJS
    • one year ago
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    given state 1 of a system at these conditions, and state 2 at some other conditions, find some value that is missing

  72. DanJS
    • one year ago
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    incompressible fluids, or, ideal gasses

  73. DanJS
    • one year ago
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    Energy transfers through heat and work and mass

  74. DanJS
    • one year ago
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    usually ignore the mechanical energy in those systems, mostly internal energy for closed systems

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