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ParthKohli

  • one year ago

Good question, I guess.

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  1. ParthKohli
    • one year ago
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    |dw:1443975149240:dw|

  2. ParthKohli
    • one year ago
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    Find out the time \(t_0\) in which the string rotates by \(\theta = \pi \). Find \(\vec{d_1} \) at \(t = \pi \ell / v_0 \). Find the distance travelled by \(1\) from \(t = 0 \) to \(\dfrac{\pi \ell }{v_o}\) Find the speed of \(2\) as a function of time.

  3. ParthKohli
    • one year ago
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    |dw:1443975353793:dw|

  4. ParthKohli
    • one year ago
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    @ganeshie8

  5. ParthKohli
    • one year ago
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    @imqwerty

  6. ganeshie8
    • one year ago
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    the center of mass is stationary and that thing is rotating about the center of mass with uniform velocity of \(v_0\) is it ?

  7. ganeshie8
    • one year ago
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    am i interpreting it correctly..

  8. ParthKohli
    • one year ago
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    oh, we're given that two masses are connected with a string and at \(t=0\) the mass 1 is given the velocity \(v_0\) (the other mass is not). if you think about it, the center of mass will not be stationary. \(v_{CM} = \frac{m_1v_1+m_2v_2}{m_1+m_2}\)

  9. ParthKohli
    • one year ago
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    so \(v_{CM} = \frac{v_0}2 \hat i \)

  10. ParthKohli
    • one year ago
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    and I think we're getting very close to the crux of the problem

  11. ganeshie8
    • one year ago
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    Ohk.. so it is also translating to the right as it rotates..

  12. ParthKohli
    • one year ago
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    exactly...

  13. ParthKohli
    • one year ago
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    if we observe the whole setting from the ground, we'll see that these things are accelerating and it'd be very hard to predict anything

  14. ParthKohli
    • one year ago
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    in fact even the acceleration is variable as the direction of force varies

  15. ganeshie8
    • one year ago
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    so we got velocity of center of mass pretty cheaply next i guess we want to find the angular velocity

  16. ganeshie8
    • one year ago
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    \(\omega = \dfrac{2v_0}{l}\) \(d_1(t) = \text{translation + rotation}=\dfrac{v_0}{2}i + \frac{l}{2}\cos(\omega t)i+\frac{l}{2}\sin(\omega t)j\)

  17. ganeshie8
    • one year ago
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    so \(d_1(\pi l/v_0) = (\dfrac{v_0}{2} + \dfrac{l}{2})i\)

  18. ParthKohli
    • one year ago
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    i mean there's one little thing: one part of the velocity helps it to translate, and the other to rotate.

  19. ParthKohli
    • one year ago
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    so \(v_0/2\) is being used in translation and the rest is used for rotation

  20. ParthKohli
    • one year ago
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    \[\omega = v_o/l \]

  21. ParthKohli
    • one year ago
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    so I think we're finally in the solution of the problem: try doing calculations in the frame of the center of mass.

  22. ganeshie8
    • one year ago
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    |dw:1443977480377:dw|

  23. ParthKohli
    • one year ago
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    in the frame of the center of mass|dw:1443977558039:dw|

  24. ParthKohli
    • one year ago
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    since tension is perpendicular to the direction of motion, this is uniform circular motion

  25. ganeshie8
    • one year ago
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    \(\omega = \dfrac{v_0}{l}\) \(d_1(t) = \text{translation + rotation}=\dfrac{v_0}{2}i + \frac{l}{2}\cos(\omega t-\pi/2)i+\frac{l}{2}\sin(\omega t-\pi/2)j\) so \(d_1(\pi l/v_0) = \dfrac{v_0}{2}i + \dfrac{l}{2}j\)

  26. ganeshie8
    • one year ago
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    In frame of reference of center of mass, it should be \(d_1(\pi l/v_0) = 0i + \dfrac{l}{2}j\)

  27. ParthKohli
    • one year ago
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    yes, correct!

  28. ganeshie8
    • one year ago
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    distance travelled = \(\pi l/2 + \frac{v_0}{2}*\pi l/v_0 = \pi l\)

  29. ParthKohli
    • one year ago
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    \[s = \int |\vec v| dt\]

  30. ganeshie8
    • one year ago
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    should get the same answer

  31. ganeshie8
    • one year ago
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    right ?

  32. ParthKohli
    • one year ago
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    I'm not sure.

  33. ganeshie8
    • one year ago
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    |dw:1443981626417:dw| not necessarily, better to evaluate the integral yeah

  34. ParthKohli
    • one year ago
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    indeed

  35. ParthKohli
    • one year ago
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    hey, could you go back to the question i asked earlier yesterday about the multinomial interpretation of the stars-and-bars problem i understood why it holds but how do i solve a sample problem there

  36. ganeshie8
    • one year ago
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    Okay.. so it sems we need to split that into two integrals : \(0 \to 3\pi l/(4v_0)\) \(3\pi l/(4v_0)\to \pi l/v_0\)

  37. ganeshie8
    • one year ago
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    \(\color{red}{*}\) \(\omega = \dfrac{v_0}{l}\) \(d_1(t) = \text{translation + rotation}=\dfrac{v_0}{2}\color{red}{t}i + \frac{l}{2}\cos(\omega t-\pi/2)i+\frac{l}{2}\sin(\omega t-\pi/2)j\)

  38. ganeshie8
    • one year ago
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    distance travelled = \(|d_1(3\pi l/(4v_0)) - d_1(0)|+|d_1(\pi l/v_0) - d_1(3\pi l/(4v_0))|\)

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