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ParthKohli

  • one year ago

Physics...

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  1. ParthKohli
    • one year ago
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    @imqwerty

  2. ParthKohli
    • one year ago
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    OK, so I have a really specific question. Whenever we study circular motion and aspects of circular motion, I see some solutions including the extra rotational kinetic energy when applying the work-energy theorem and some don't do that. When is it OK to not include the rotational kinetic energy?

  3. ParthKohli
    • one year ago
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    I'll do a followup question as an example to explain why I'm confused here.

  4. ganeshie8
    • one year ago
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    If I remember correctly, work-energy theorem is like universal... it works for all kinds of forces; It doesn't matther if the forces are conservative or not. Conservation of energy requires the forces to be conservative, though.

  5. ganeshie8
    • one year ago
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    Could you provide examples to contrast where rotational kinetic energy is included/excluded in work energy equation

  6. ParthKohli
    • one year ago
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    This isn't the real question yet, but here's where the confusion arises.|dw:1443076050348:dw|\[mgL = \frac{1}{2}mv^2\tag{1} \]\[mgL = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mL^2 \frac{v^2}{L^2} = mv^2\tag{2}\]I think \(\tag 1\) is wrong and \(\tag 2\) is right.

  7. ParthKohli
    • one year ago
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    |dw:1443076205726:dw|

  8. ParthKohli
    • one year ago
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    Wow, sorry about the formatting in that post. I think that 1 is wrong and 2 is right.

  9. ganeshie8
    • one year ago
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    Work done "by a force" = change in kinetic energy "due to that force"

  10. ParthKohli
    • one year ago
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    Oh, so what exactly is happening here? Is it not the gravitational force that is also assisting it to gain the rotational kinetic energy?

  11. ganeshie8
    • one year ago
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    |dw:1443076331186:dw|

  12. ganeshie8
    • one year ago
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    Is that thing spinning and falling freely under gravitation ?

  13. ParthKohli
    • one year ago
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    Arrey... I meant conservation of energy, not the work-energy theorem. It's nothing, really. A particle is just fixed at the end of a massless rod/string and it starts to go down. Its path is circular.

  14. ParthKohli
    • one year ago
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    Yes.

  15. ganeshie8
    • one year ago
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    |dw:1443076456909:dw|

  16. ParthKohli
    • one year ago
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    \[E_i = E_f\]\[mgL = \frac{1}{2}mv_f^2\]or\[mgL = mv_f^2\]Which one do you think is correct? The second one also includes rotational kinetic energy.

  17. ganeshie8
    • one year ago
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    This is like pendulum/SHO right ?

  18. ganeshie8
    • one year ago
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    SHO = simple harmonic oscillation

  19. ParthKohli
    • one year ago
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    Yes, I drew that arrow in the diagram too...

  20. ParthKohli
    • one year ago
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    Yup.

  21. Jhannybean
    • one year ago
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    Kind of like how a gransfather clockworks.

  22. Empty
    • one year ago
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    I think the problem is these are both two optional ways of representing the same thing:\[\frac{1}{2}mv^2 = \frac{1}{2}I\omega^2 \]

  23. ganeshie8
    • one year ago
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    good question, the situation as same as that of earth-sun system : earth spins and orbits around the sun

  24. ParthKohli
    • one year ago
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    Exactly, Empty. I see that some solutions add both rotational and translational kinetic energies, and others say that they're the same thing.

  25. ParthKohli
    • one year ago
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    I personally feel that you should be adding those two. They're not the same, and are two components of rotation.

  26. imqwerty
    • one year ago
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    do u want to apply this kinda concept - \[\omega _{f}^2 =\omega_{i}^2 +2\alpha \theta\]?

  27. Empty
    • one year ago
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    Ok, show me an example where they have used both like this, so we can determine what's going on. For this problem you've given, it's clear that only one is necessary because there's only one degree of freedom here.

  28. Jhannybean
    • one year ago
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    |dw:1443076723125:dw|

  29. Jhannybean
    • one year ago
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    helps if you needed to understand the relationships I guess? Haha.

  30. ganeshie8
    • one year ago
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    I think so, if there is a rotation, total kinetic energy equals the sum of translational KE and rotational KE

  31. ParthKohli
    • one year ago
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    OK, so (2) is right, right?

  32. ganeshie8
    • one year ago
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    but with pendulums, we assume the balls are not spinning

  33. ganeshie8
    • one year ago
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    depends, if you assume the ball is spinning, add \(\frac{1}{2}I\omega^2\) otherwise forget it

  34. ParthKohli
    • one year ago
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    The rotational component isn't about the balls spinning, it's about the ball rotating about the hinge point.

  35. ParthKohli
    • one year ago
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    Does circular motion qualify as rotational motion? Great, I've reduced this conversation to a single question.

  36. ganeshie8
    • one year ago
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    that is accounted as translational kinetic energy along circumferencce of circle, right ?

  37. ParthKohli
    • one year ago
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    That's the point - it's not...

  38. ParthKohli
    • one year ago
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    Or is it?

  39. imqwerty
    • one year ago
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    it depends upon the motion of system... but if the particle is jst doing circular motion then we cant say its undergoing rotational motion too..

  40. Jhannybean
    • one year ago
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    |dw:1443076982465:dw| the circular movement would just be applied to the pulley/lever thing holding it, but not to the ball itself, that's what I think..

  41. ParthKohli
    • one year ago
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    Try doing this question, for example. Find the value of \(\theta\). Here, we have a rod instead of a point mass. If we do not account for the rotational kinetic energy, we'll be in trouble (you'll get \(\cos \theta > 1\) if I recall corrrectly). If we account for the rod's rotation, why not a point mass'? |dw:1443077039882:dw|

  42. ganeshie8
    • one year ago
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    A rotating ball has a speed, so it clearly has a momentum and kinetic energy. Since it is in circular motion, it is under the presense of centripetal acceleration which always points "perpendicular" to the direction of motion. So you can forget about the rotational part of the kinetic energy.

  43. Empty
    • one year ago
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    For your question, you're just writing the same energy twice, so (1) is correct.\[\frac{1}{2}mv^2 = \frac{1}{2}I\omega^2 \]

  44. ganeshie8
    • one year ago
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    If the force is perpendicular to the motion, it does no work. so you can forget about it

  45. Empty
    • one year ago
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    This is pointless to talk about unless you show us a real problem where you see both arising, then we can explain why both arise.

  46. ParthKohli
    • one year ago
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    This is what you guys suggest. Oh, and also, \(L = 25 ~cm\).\[mg\left(\frac{L}{2} + \frac{L}{2}\cos \theta \right)=\frac{1}{2}mv^2 \]

  47. ParthKohli
    • one year ago
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    \[gL (1 + \cos \theta) = v^2\]\[\Rightarrow 9 = 10\cdot 0.25 \cdot (1 + \cos \theta) \]Lo and behold...

  48. ParthKohli
    • one year ago
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    But there has to be a theta, right? According to this, there is none.

  49. Jhannybean
    • one year ago
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    |dw:1443077457196:dw| lmao... it's an easy one I guess.

  50. ParthKohli
    • one year ago
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    \[mg\left(L + \frac{L}{2}\cos \theta \right) = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2\]\[3mgL(1 + \cos \theta) = mv^2 + \frac{mL^2}{3}\cdot \frac{v^2}{L^2} = \frac{4mv^2}{3}\]\[\Rightarrow 3\cdot 10 \cdot 0.25 (1 + \cos \theta) = 4\cdot 9/3\]

  51. Jhannybean
    • one year ago
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    Forgot something in it

  52. Jhannybean
    • one year ago
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    |dw:1443077907086:dw|

  53. ganeshie8
    • one year ago
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    are you taking 0 for potential energy at L/2 ?

  54. ParthKohli
    • one year ago
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    Rotational and translational kinetic energies are not the same. They just turn out to be the same here because the moment of inertia is mL^2, which could have been anything.

  55. ParthKohli
    • one year ago
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    Yeah, and I'm concentrating the entire system at its center of mass which is legal in physics.

  56. ganeshie8
    • one year ago
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    |dw:1443078285888:dw|

  57. ganeshie8
    • one year ago
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    It's been so many years, let me try and catch up on these..

  58. ParthKohli
    • one year ago
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    Oh, great catch here... the 3m/s is for the bottom-most point, not the middle one. For the middle one, it's should be 3/2. Sorry 'bout that.

  59. ganeshie8
    • one year ago
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    potential energy = \(mg\left(\dfrac{L}{2} - \dfrac{L}{2}\cos\phi\right) \) |dw:1443078517457:dw|

  60. ParthKohli
    • one year ago
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    Fixed:\[mgL/2 (1 + \cos \theta) = 1/2 m(3/2)^2 + 1/2 \cdot mL^2/3 \cdot 3^2/L^2 \]

  61. ParthKohli
    • one year ago
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    That looks ugly.\[10\cdot 0.25 \cdot (1 + \cos \theta) = 9/8 + 3\]\[1 + \cos \theta = 33/(8\cdot2.5) = 33/20 \]

  62. ParthKohli
    • one year ago
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    Still dunno.

  63. ParthKohli
    • one year ago
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    Something doesn't feel right about this. Ugh.

  64. ganeshie8
    • one year ago
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    \(\phi =\theta+90 \) so shouldn't the potential energy be \[mg\left(\dfrac{L}{2} - \dfrac{L}{2}\cos\phi\right) = mg\left(\dfrac{L}{2} - \dfrac{L}{2}\cos(\theta+90)\right)\\~\\= mg\left(\dfrac{L}{2} +\dfrac{L}{2}\color{red}{\sin}\theta\right)\] ?

  65. ParthKohli
    • one year ago
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    Absolutely. Sorry, I was too busy worrying about the other aspect of the problem that I actually forgot to solve it correctly.

  66. ParthKohli
    • one year ago
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    That gives \(\sin \theta = 0.65\)

  67. ganeshie8
    • one year ago
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    Ok so we're good with potential energy let me try and understand whats going with rotational ke

  68. ganeshie8
    • one year ago
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    are you sure the moment of inertia of that rod is \(\dfrac{mL^2}{3}\) ?

  69. ParthKohli
    • one year ago
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    If you try to solve it like this...\[\rm mgL/2 (1 + \cos \theta) = \frac{1}{2}mv^2\]then \(v = 3/2 \), \(L = 0.25 m\) and so on.

  70. ParthKohli
    • one year ago
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    Yes.

  71. ganeshie8
    • one year ago
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    also the speed of center of mass will not be 3 m/s it should be way less than that

  72. ParthKohli
    • one year ago
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    Yeah, I wrote that correction above. It's 3/2 and not 3.

  73. ParthKohli
    • one year ago
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    And I wrote \(\cos \theta\) again <_<\[10\cdot 0.25 \cdot (1 + \sin \theta) = 9/4\]\[\Rightarrow 1 + \sin \theta = 9/10\]

  74. ParthKohli
    • one year ago
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    And so \(\sin \theta = -1/10\) Which is invalid.

  75. ParthKohli
    • one year ago
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    So we definitely have to add rotational kinetic energy to that!

  76. ganeshie8
    • one year ago
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    yes replace \(m\) by \(I\) and \(v\) by \(\omega\)

  77. ParthKohli
    • one year ago
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    so is \(mgL = mv^2\) correct in my first question?

  78. ganeshie8
    • one year ago
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    Total mechanical energy, \(mgL/2(1+\sin\theta) + \frac{1}{2} I\omega^2\), is conserved

  79. ParthKohli
    • one year ago
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    \[mgL/2(1 + \sin \theta) = \frac{1}{2}I \omega^2 + \frac{1}{2}mv^2\]

  80. ParthKohli
    • one year ago
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    Another question: if I take the whole rod to be concentrated at the center of mass, I still take \(I \) to be the moment of inertia of the whole rod, right? And not just the point-mass?

  81. ganeshie8
    • one year ago
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    |dw:1443079851015:dw|

  82. ganeshie8
    • one year ago
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    the object is rotating about that point A, so the moment of inertia is about that point A. it wont change right ?

  83. ganeshie8
    • one year ago
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    see, im rusty on these, im just going by intuition, i can be wrong...

  84. ParthKohli
    • one year ago
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    Yup. And \(I = mL^2/3\).

  85. ganeshie8
    • one year ago
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    so it is the property of that rotating rod about that point A it should not change

  86. ParthKohli
    • one year ago
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    Yes, indeed.

  87. ParthKohli
    • one year ago
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    \[mgL/2 ( 1 + \sin \theta ) = \frac{1}{2}m\left(3/2\right)^2 + \frac{1}{2}\frac{mL^2}{3} \left(\frac{3^2}{L^2}\right) \]

  88. ParthKohli
    • one year ago
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    \[gL ( 1 + \sin \theta ) = 9/4 + 3 = 21/4\]\[\Rightarrow 1 + \sin \theta = 21/4\cdot 10\cdot 0.25 = 21/10\]\[\sin \theta = 11/10\]JEEEEEZUZ!

  89. ParthKohli
    • one year ago
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    wait, what if this problem is wrong?

  90. ParthKohli
    • one year ago
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    let's reverse the thing and find the maximum velocity at the bottom

  91. ParthKohli
    • one year ago
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    \[mgL = \frac{1}{2}mv_{CM}^2 + \frac{1}{2}I\omega^2 \] (for max. velocity)\[2mgL =mv_{CM}^2 + mL^2/3 \cdot (v_{CM}/2)^2/ (L/2)^2 \]

  92. thadyoung
    • one year ago
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    LMAO!! We just ignored @Jhannybean

  93. ParthKohli
    • one year ago
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    \[2mgL = 4mv_{CM}^2/3\]\[3gL/2 = v_{CM}^{2}\]

  94. ParthKohli
    • one year ago
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    \[v_{bottom}^2 = 4 v^2_{CM} = 6gL = 6\cdot 10 \cdot 0.25 = \cdots\]So I guess we're fine.

  95. ganeshie8
    • one year ago
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    \(3g(1+\sin\theta) =L\omega^2\)

  96. ParthKohli
    • one year ago
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    What's that? Just the rotational kinetic energy?

  97. ParthKohli
    • one year ago
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    So \(v_{bottom} = \sqrt{15} > 3\)

  98. ParthKohli
    • one year ago
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    Yeah, that's a good idea... what if we just include the rotational kinetic energy?

  99. ParthKohli
    • one year ago
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    \[3gL/2( 1 + \sin \theta) = v_{CM}^2 = 9/4\]\[ gL(1 + \sin \theta ) = 3/2 \]\[ 1 + \sin \theta = 3/5\]Oh lol, again...

  100. ParthKohli
    • one year ago
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    So absolutely nothing works at this point. =_=_=_=_=_=

  101. ganeshie8
    • one year ago
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    Here is my best attempt : Total energy when the rod is at angle \(\theta\) is given by \(\dfrac{1}{2}I\omega^2 + mg\frac{L}{2}(1+\sin\theta) \) since the rod hasn't started rotating, \(\omega = 0\), so the total energy becomes \(mg\frac{L}{2}(1+\sin\theta)\tag{1} \) Total energy when the rod is at its final position is given by \(\dfrac{1}{2}I\omega^2 +mg*0 \tag{2}\) since the forces are conservative, the total energy is conserved : \[mg\frac{L}{2}(1+\sin\theta)=\dfrac{1}{2}I\omega^2\] we can solve \(\theta\)

  102. ParthKohli
    • one year ago
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    Could you solve the rest? I just am not able to.

  103. ganeshie8
    • one year ago
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    plugin \(\omega = \dfrac{v}{L} =\dfrac{3}{25} \)

  104. ganeshie8
    • one year ago
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    I'm getting \[1+\sin\theta = \dfrac{9}{125g}\]

  105. ParthKohli
    • one year ago
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    so that's wrong, isn't it?

  106. ganeshie8
    • one year ago
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    not so sure http://www.wolframalpha.com/input/?i=solve+1%2Bsinx+%3D+9%2F%28125*10%29%2C0%3Cx%3C2pi

  107. ParthKohli
    • one year ago
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    But 0<x<pi/2

  108. BAdhi
    • one year ago
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    To answer the problem of whether the rotional energy should be applied to the linear motion energy, we should go into the point where the basic concepts of the rational parameters are defined. As Ive learned, the moment of inertia was defined as following, |dw:1443082442329:dw| So as the diagram shows the object (placed horizontally so that theres no potential energy changes when rotate) rotates around the O axis and we can take small masses \(\delta m_i\) on it at a distance of \(r\). When the object starts to rotate that small mass will have v velocity . So the kinetic energy the small mass will have is, \(\delta E _i= \frac{1}{2}\delta m_iv_i^2 \) with \(v_i = r_i \omega\) \(\delta E_i = \frac{1}{2}\delta m_ir_i^2\omega^2 \) Since we need the kinetic energy of the whole object which is rotating, \[\begin{align*} E &= \sum \limits_{i=0}^{i=N} E_i \\ &= \sum \limits_{i=0}^{i=N} \delta m_ir_i^2\omega^2\\ &=\omega^2 \underbrace{\sum \limits_{i=0}^{i=N} \delta m_ir_i^2}_{I}\\ &= I \omega^2 \end{align*}\] So I is taken as that large sum. Whats important to see here is that eventually the rotational energy is taken from the linear kinetic energy. So at the end of the day they are both the same.So There is no need to consider a rotational kintic energy if we take linear kinetic energy

  109. BAdhi
    • one year ago
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    sorry missed the 1/2 in the middle for the E, it should be \(E= \frac 1 2 I \omega^2\) :(

  110. ParthKohli
    • one year ago
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    Amazing. I wonder why I forgot that.

  111. ParthKohli
    • one year ago
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    Thank you so very much!

  112. BAdhi
    • one year ago
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    It a pleasure ;)

  113. ParthKohli
    • one year ago
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    I mean you didn't really need to type all of that, but thanks!

  114. ParthKohli
    • one year ago
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    But we do need to add the linear component in plane motion.

  115. ParthKohli
    • one year ago
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    Yup, so the verdict: - In pure rotation, rotational kinetic energy is the kinetic energy. - In pure translational, translational kinetic energy is the kinetic energy. - In plane motion (a superposition of the two) we have to add both.

  116. ParthKohli
    • one year ago
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    That's great, I wonder why I totally forgot the origin. Sometimes you just have to look at the basics. =_=

  117. BAdhi
    • one year ago
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    umm what do you mean by the plane motion ?

  118. ParthKohli
    • one year ago
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    Nothing, just rotational+translational motion.

  119. BAdhi
    • one year ago
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    Ok. Btw to which group do you take this original question?

  120. ParthKohli
    • one year ago
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    ?

  121. BAdhi
    • one year ago
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    I mean to which group among the three catagories does the scenario - "particle rotating around a point " falls

  122. ParthKohli
    • one year ago
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    Oh, it's pure rotation. I got it. :)

  123. ParthKohli
    • one year ago
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    Basically in plane motion, the axis of rotation is translating.

  124. BAdhi
    • one year ago
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    you mean the third catagory where you should sum the linear motion energy and the ration energy?

  125. ParthKohli
    • one year ago
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    Exactly. But here, the axis is stationary and the original question is a kind of pure rotation.

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