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Mikael

  • 3 years ago

The rod is massless. At t = 0 rod is pointing up. It is of length R. rod is rotating without friction around the lower end which is hinged to rotation axis. The mass M is attached at the top end of the rod. The mass is pushed at t=0 with initial velocity . g is gravitational acceleration. All these parameters are given and known.

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  1. Mikael
    • 3 years ago
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  2. Mikael
    • 3 years ago
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    WHEN WILL THE MASS REACH THE LOWEST POINT ?

  3. Mikael
    • 3 years ago
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    @mukushla @UnkleRhaukus @experimentex @estudier

  4. Mikael
    • 3 years ago
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    @ParthKohli @radar @RaphaelFilgueiras

  5. Mikael
    • 3 years ago
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    @sauravshakya

  6. Mikael
    • 3 years ago
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    @Hero @apoorvk @AccessDenied

  7. UnkleRhaukus
    • 3 years ago
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    find the angular velocity

  8. Mikael
    • 3 years ago
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    Well it is not constant and it is not explicit in energy - the linear v is there

  9. Mikael
    • 3 years ago
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    @ganeshie8 @ajprincess

  10. Mikael
    • 3 years ago
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    @ganeshie8 @ajprincess

  11. Mikael
    • 3 years ago
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    @ParthKohli @mathslover

  12. Mikael
    • 3 years ago
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    \[0.5mv^2 + mgh = E_0 = constant\]

  13. Mikael
    • 3 years ago
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    CAN ANYONE SOLVE IT AROUND HERE ?

  14. Mikael
    • 3 years ago
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    @Omniscience ?

  15. Mikael
    • 3 years ago
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    HINTS ?

  16. Mikael
    • 3 years ago
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    Please solve it someone

  17. Mikael
    • 3 years ago
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    At least some SIGNIFICAN part of the way forward ?

  18. Mikael
    • 3 years ago
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    @radar you can help ?

  19. apoorvk
    • 3 years ago
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    Hmm I guess you will just have to find the general term for velocity at any instant, and integrate the term in 'dt' - since energy conservation and angular velocity fundae won't work here.

  20. apoorvk
    • 3 years ago
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    let me try this out...

  21. Mikael
    • 3 years ago
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    You ARE 20% forward

  22. Mikael
    • 3 years ago
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    Hi @ajprincess

  23. Mikael
    • 3 years ago
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    help

  24. Mikael
    • 3 years ago
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    Ok , the prize for the sover will be 25 minute-worth field integral from EM that I am ready to do in return !

  25. Mikael
    • 3 years ago
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    OK the beauty of this situation is that it is VERY common and simple setting. Buuut very unusual answer

  26. Mikael
    • 3 years ago
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  27. Mikael
    • 3 years ago
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  28. Mikael
    • 3 years ago
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    @demitris @Directrix can you help ?

  29. Mikael
    • 3 years ago
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    Hi @sauravshakya

  30. Mikael
    • 3 years ago
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    @demitris @Directrix can you help ?

  31. sauravshakya
    • 3 years ago
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    Hi @Mikael ..

  32. Mikael
    • 3 years ago
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    I AM VERY SORRY I STARTED WITHOUT YOU. Complete please

  33. sauravshakya
    • 3 years ago
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    Wait a minute.... Let me see what I can do.

  34. Mikael
    • 3 years ago
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    @Directrix ?

  35. Mikael
    • 3 years ago
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    @asnaseer Please help

  36. Mikael
    • 3 years ago
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    Hello @Callisto - try to help please

  37. Mikael
    • 3 years ago
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    Well imagine my sour expression when a nice timid student asks me this question in the middle of a nice sunny day ....

  38. Algebraic!
    • 3 years ago
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    Why no rotational energy term?

  39. Mikael
    • 3 years ago
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    Yea , you can use that. Although this is not the only way. Just do it please

  40. sauravshakya
    • 3 years ago
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    This requires physics too

  41. mathslover
    • 3 years ago
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    why not to use law of conservation of energy?

  42. Mikael
    • 3 years ago
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    Calculus and mechanics are essentially one united universal soul

  43. Mikael
    • 3 years ago
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    Good @mathslover USE it and do the soln. please

  44. mathslover
    • 3 years ago
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    potential energy + kinetic energy = constant. I cn help after 1 n half hours

  45. Mikael
    • 3 years ago
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    Actually I have already pointed that in a hint above. But go forward - when is the bottom reached ?

  46. Mikael
    • 3 years ago
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    @ganeshie8 @UnkleRhaukus help pls

  47. Algebraic!
    • 3 years ago
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    \[\frac{ t ^{2} }{2 } + \frac{ V _{o}}{ R } t =\int\limits_{ }^{ }\int\limits_{ }^{ } \frac {-R }{ g \cos \theta } \] 0..pi

  48. Mikael
    • 3 years ago
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    A) What is the integration variable ? B) Why (suddenly ) double int ???? c) What is the reason you write this so categorically ?

  49. Algebraic!
    • 3 years ago
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    A) g, of course B) equation of motion C) integral is tough, not bothering

  50. Algebraic!
    • 3 years ago
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    theta is the angle the rod makes with the vertical 0...pi

  51. Mikael
    • 3 years ago
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    g CANNOT CHANGE - it is CONSTANT !

  52. Algebraic!
    • 3 years ago
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    it's theta, why would you ask?

  53. Algebraic!
    • 3 years ago
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    would it be g or R?

  54. Mikael
    • 3 years ago
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    Please Show the FULL argument. Double integral seems, on the face of it, out of place in this problem

  55. Mikael
    • 3 years ago
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    PLEASE SOLVE @mukushla @TuringTest

  56. Algebraic!
    • 3 years ago
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    argument from torque = -mgRcos(theta) I=mR^2 d^2/ dt^2 (theta) =- gRcos(theta)

  57. Mikael
    • 3 years ago
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    \[E_0 = 2mgR +0.5 mv^2\]

  58. Mikael
    • 3 years ago
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    Giving the solution soon in 3-4 minutes

  59. Mikael
    • 3 years ago
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    E = CONSTANT

  60. Coolsector
    • 3 years ago
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    h for you is like the y axis ?

  61. Mikael
    • 3 years ago
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    yes

  62. Mikael
    • 3 years ago
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    OK here goes the solution

  63. radar
    • 3 years ago
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    I have concluded this problem may be above my pay-grade.

  64. Mikael
    • 3 years ago
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    \[t(h) = \int\limits_{h=2R}^{h=0} \frac{dl}{v(h)} dh\]

  65. Mikael
    • 3 years ago
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    Now \[dl = \frac{dh}{\cos \theta}\]

  66. Mikael
    • 3 years ago
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    Forgot to say - thta is the angle with the horizon of the rod.The relation dh to dl is a geometric fact that follows from\[h = Sin \theta ====> dh = dl * Cos \theta\]

  67. Mikael
    • 3 years ago
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    sorry h = R*Sin theta

  68. Mikael
    • 3 years ago
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    Thus gathering all the above observations until now we obtain the following integral

  69. Mikael
    • 3 years ago
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    \[\int\limits_{h=2R}^{h=0} dh \frac{1}{\sqrt{1-\frac{h^2}{R^2}}*\sqrt{\frac{2E_0}{m} -2gh}}\]

  70. Mikael
    • 3 years ago
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    Here the 1-st square root in the denominator is simply 1/Cos theta

  71. Mikael
    • 3 years ago
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    The second square root is the velocity as a function of h - of course derived from the law of conservation of energy stated above

  72. Coolsector
    • 3 years ago
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    can you please explain how h = rsintheta means dh = dl costheta

  73. Mikael
    • 3 years ago
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    Now by rescaling the distance by R and taking one convenient constant (from the denominator roots) out of the integral we simplify it all to

  74. mathslover
    • 3 years ago
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    I did this

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  75. Mikael
    • 3 years ago
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    |dw:1347813111455:dw|

  76. mathslover
    • 3 years ago
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    \[\large{t = \frac{-1+\sqrt{1+8\pi R}}{2v_0}}\]

  77. Coolsector
    • 3 years ago
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    thanks

  78. mathslover
    • 3 years ago
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    did that make sense? @Coolsector and @Mikael

  79. Mikael
    • 3 years ago
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    @mathslover I would be amazed and thankful for COMPLETE explanation why do you claim the first equation - that the angle is a simple quadratic function of time. As you know the angular acceleration IS NOT CONSTANT HERE

  80. Coolsector
    • 3 years ago
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    i dont understand why alpha is what you said.. but maybe its just me

  81. Coolsector
    • 3 years ago
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    yes .. same problem here

  82. Mikael
    • 3 years ago
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    I think @mathslover is NOT correct in HIS FIRST STATEMENT on that page on which all is based upon

  83. Mikael
    • 3 years ago
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    So we proceed

  84. Algebraic!
    • 3 years ago
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    there's no g in your soln. @mathslover

  85. mathslover
    • 3 years ago
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    Oh I made it for horizontal plane... vertical plane hmn... well guys wait

  86. Mikael
    • 3 years ago
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    \[\int\limits_{\xi_0}^{0} d \xi \frac{1}{\sqrt{(1-\xi^2)(A- \xi)}}\]

  87. mathslover
    • 3 years ago
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    Mikael whts ur age? grade? so that I can understand the level of this

  88. Mikael
    • 3 years ago
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    HERE Xi is rescaled variable h, and A is some constant containing fractions of E_0 and g. Outside of the integral is of course a Multiplier (not written) which is physically crucial and mathematically trivial

  89. mathslover
    • 3 years ago
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    I think this is \(\large{\textbf{above}}\) my level sorry.... I am in 9th grade and did my best.... sorry

  90. Mikael
    • 3 years ago
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    This is the level of a good Physical-Course on analytival mechanics in Cambridge/Berkeley/MIT 1-st or second year. Also Ms. @Algebraic! is probably generally pointed to a more Engineering diection of solution - using torques and angular accelerations. Aand Mr. @radar can I send you a message unrelated to math but related to this site ?

  91. mathslover
    • 3 years ago
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    Sorry.... above my level...

  92. Algebraic!
    • 3 years ago
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    Ms. I like that snark.

  93. Mikael
    • 3 years ago
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    I disagree with you @mathslover - you saw it is almost school level

  94. mathslover
    • 3 years ago
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    What do you mean mikael? well leave ti whatever do you mean........ continue your work..

  95. mathslover
    • 3 years ago
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    *it

  96. Mikael
    • 3 years ago
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    Well 1) This is all - please help me by some Tables / Wolframalpha to find what type of Integral this belongs to 2) I suspect it is elliptic Integral

  97. Mikael
    • 3 years ago
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    Aand please appreciate the effort by medals - you are welcome. The question itself was supplied by Mr. @henpen here http://openstudy.com/study#/updates/50549e72e4b0a91cdf445d5b

  98. Mikael
    • 3 years ago
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    Here is the answer as supplied by Wolframalph

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  99. Mikael
    • 3 years ago
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    Sorry here in complete - as I thought elliptic integral

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  100. Mikael
    • 3 years ago
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    Can see the soln @mukushla

  101. mukushla
    • 3 years ago
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    i just came here :)

  102. Mikael
    • 3 years ago
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    @mukushla The original challenge was for 2 sets of people 1) You, @sauravshakya, and other olympics-math types 2) Physicists So try solving without any reading the soln

  103. Mikael
    • 3 years ago
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    So I guesses right - about IMO participation ?

  104. mukushla
    • 3 years ago
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    sorry lost connection Mr.Mikael :) man im a chemical engineer :) what is the math part of question ?

  105. Mikael
    • 3 years ago
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    You need to find how long it takes for a spherical bead of polymer to slied without friction though a perfectly circular half circular tube to the sovent chamber if initially the small bead has v_0 velocity and starts from the top of the tube

  106. Mikael
    • 3 years ago
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    solvent chamber (i meant)

  107. mukushla
    • 3 years ago
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    out of my knowledge :)

  108. Mikael
    • 3 years ago
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    You should use the simple mechanical conservation of energy and definition of velocity. Some geometry too

  109. Mikael
    • 3 years ago
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    Well - see the solution and dissolve your sorrows...

  110. sauravshakya
    • 3 years ago
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    Actually this question is also out of my reach.

  111. sauravshakya
    • 3 years ago
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    Is this the question from a high school level?

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