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Shadowys

  • 2 years ago

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? :)

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  1. Shadowys
    • 2 years ago
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    |dw:1350647588577:dw| hint: theta will vanish.

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

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

  4. Algebraic!
    • 2 years ago
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    wut.

  5. Shadowys
    • 2 years ago
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    you sure about the sqrt for the g?

  6. Algebraic!
    • 2 years ago
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    these are times.

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

  8. Algebraic!
    • 2 years ago
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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?

  9. Shadowys
    • 2 years ago
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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.

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

  11. Shadowys
    • 2 years ago
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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.

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