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ParthKohli

  • one year ago

hmm

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  1. ParthKohli
    • one year ago
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    |dw:1443892871026:dw|

  2. ParthKohli
    • one year ago
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    \[dW = \vec F \cdot \vec {dr} = (F \hat i ) \cdot ??? \]

  3. ganeshie8
    • one year ago
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    let \(F=\langle f(x),0 \rangle \) \[W = \int\limits_{P1}^{P2} F.dr = \int\limits_{P1}^{P2} \langle f(x),0 \rangle.\langle dx,dy \rangle = \int\limits_{P1}^{P2} f(x)dx \]

  4. ParthKohli
    • one year ago
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    \[|dr| = l|d\theta| \]

  5. ganeshie8
    • one year ago
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    Once you setup the work integral, it is just a normal integral, you can evaluate it using any of the tricks that you're familiar with

  6. ParthKohli
    • one year ago
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    you know, it contains \(mg\) in it. I took a good look at the choices and the only one where I'm getting \(W = 0\) putting \(\theta =0\) is \(mg \ell (1 - \cos \theta)\), so some clever work needed right there.

  7. ganeshie8
    • one year ago
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    whats the value of F ?

  8. ParthKohli
    • one year ago
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    it can vary.

  9. ParthKohli
    • one year ago
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    honestly it doesn't even matter because I think they've calculated it using initial and final potential energies.

  10. ParthKohli
    • one year ago
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    and all these users visit this thread only when I create my dumbest thread...

  11. ParthKohli
    • one year ago
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    I don't see why all the credit for changing the potential energy is going to \(F\) when some of the contribution is made by \(T\)

  12. ganeshie8
    • one year ago
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    then simply use that, its just a geometry problem

  13. ganeshie8
    • one year ago
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    the force T is always there, but i think we ignore it because it is always perpendicular to the direction of bob

  14. ParthKohli
    • one year ago
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    still has a horizontal component no?

  15. ganeshie8
    • one year ago
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    |dw:1443893772125:dw|

  16. ganeshie8
    • one year ago
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    it haas no component along the direction in which the bob is moving

  17. ParthKohli
    • one year ago
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    hmm, of course, yeah. that's true. in absence of other forces, if we see a bob undertaking pendulum motion, we can say that the potential energy which is increased is actually a consequence of previously possessed kinetic energy and not a force. cool.

  18. ParthKohli
    • one year ago
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    nice. alright. good. great. understood.

  19. ganeshie8
    • one year ago
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    looks good to me too initially when you release it from an angle, it has highest potentil energy, which transforms to kinetic energy when the bob goes to its lowerst point, then transforms back to potential energy when the bob goes goes back to its highest point..

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