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Shadowys

Here's a fun one: Consider an incline, which is one part of a circle, radius 'r' m. From one end of an incline(A), to the center O, there is a frictionless plane inclined at an angle of theta. A mass (A) slided down the striaght. slope. Another mass (B) which has the same mass of A, slides down the circular incline from the other end. Determine which mass reaches the center O first. It's better if you show your calculations...You're not guessing, are you? :)

  • one year ago
  • one year ago

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  1. Shadowys
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    |dw:1350647588577:dw| hint: theta will vanish.

    • one year ago
  2. Algebraic!
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    \[2\sqrt{\frac{ r }{ g }}\]vs\[\frac{ \pi }{ 2 } \sqrt{\frac{ r }{ g }} \] \[...\huge ?\]

    • one year ago
  3. Shadowys
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    lol calc then one of them is the right one.

    • one year ago
  4. Algebraic!
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    wut.

    • one year ago
  5. Shadowys
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    you sure about the sqrt for the g?

    • one year ago
  6. Algebraic!
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    these are times.

    • one year ago
  7. Shadowys
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    yup. doesn't the g during the derivation has a square?

    • one year ago
  8. Algebraic!
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    don't think so. I see mine is wrong for a different reason though: used the wrong angle for chord length.... so what's the simple way to do it?

    • one year ago
  9. Shadowys
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    hmm..I'm used the work-energy theorem and striaght-line motion equations but let's see if others have more elegant answers.

    • one year ago
  10. vannayen
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    |dw:1350732149436:dw| compare v1 and v2 when it reach center O |dw:1350733431967:dw|

    • one year ago
  11. Shadowys
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    Would you mind labelling the graph? lol Didn't get 'l' because I can't the see the whole pic. And who is 1 and 2?\ Hint: Find the time taken. The velocity is just part of it. Because the distance taken might(might!!) just cancel them out.

    • one year ago
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