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micahwood50

Physics Rolling Help Needed

  • one year ago
  • one year ago

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  1. micahwood50
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    • one year ago
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  2. micahwood50
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    I tired to use a conservation of energy and I kept get wrong answer. Why doesn't it work?

    • one year ago
  3. micahwood50
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    \[\large E_{tot_o} = E_{tot_f}\]\[\large U_o = K_t+K_r\]\[\large mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2\]

    • one year ago
  4. Mashy
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    did you get the h right? what did you calculate h as?

    • one year ago
  5. micahwood50
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    Well, I calculated h as R - Rcosθ (R represents radius of bowl)

    • one year ago
  6. micahwood50
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    Isn't that right?

    • one year ago
  7. micahwood50
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    |dw:1360915947670:dw|

    • one year ago
  8. Mashy
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    yea yea thats right.. :)!!.. still getting it wrong? :-/.. show your work here!

    • one year ago
  9. micahwood50
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    \[mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}\left(\dfrac{2}{5}mr^2\right)\dfrac{v^2}{r^2}\] \[= \dfrac{7}{10}mv^2\] \[\dfrac{10}{7}gh = v^2 = \omega^2 r^2\]\[\omega = \sqrt{\dfrac{10gh}{7r^2}}\]\[\boxed{\omega = \sqrt{\dfrac{10gR(1 - \cos \theta)}{7r^2}}}\]

    • one year ago
  10. micahwood50
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    @Mashy @Vincent-Lyon.Fr

    • one year ago
  11. micahwood50
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    isn't that right?

    • one year ago
  12. Mashy
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    yea.. i don't see any flaw in it.. sorry computer got stuck :P

    • one year ago
  13. micahwood50
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    That's weird... Apparently I cannot use conservation of energy for some reasons....

    • one year ago
  14. micahwood50
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    Do you know another way I can solve?

    • one year ago
  15. Mashy
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    wait.. the general equation is \[( 2gh /(1 + Icm/Mr^{2}))^{1/2}\] how do you put that division line?? mine comes slanted line :(

    • one year ago
  16. Mashy
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    thats linear velocity :D

    • one year ago
  17. Mashy
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    yea there is nothing wrong in it!!.. its correct.. but do tell me how do you put that division in equation editor

    • one year ago
  18. Mashy
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    \[1 \div 2\]

    • one year ago
  19. micahwood50
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    What's that? Just type \dfrac{}{}

    • one year ago
  20. Mashy
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    HOW TO DO THAT?!?!? :XXXXX

    • one year ago
  21. Mashy
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    ok ok wait

    • one year ago
  22. micahwood50
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    general equation, what's that?

    • one year ago
  23. Mashy
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    ok got.. anwyays.. no no thats the equation for linear velocity of cm.. in a general case.. it turns out to be the same.. its correct.. i don't see any mistake.. probably the given answer is wrong!

    • one year ago
  24. micahwood50
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    Maybe using conservation of energy won't work for some reasons...

    • one year ago
  25. micahwood50
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    Thank you, though.

    • one year ago
  26. Mashy
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    no it has to work.. conservation of energy has no exception.. except when you are dealing with mass energy equivalence.. but here certainly it should hold good!! gravity is conservative.. so it really doesn't depend upon the path taken!.. so regardless of whether its a bowl or whether its just a slope.. it should hold good and give same answer!

    • one year ago
  27. micahwood50
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    That's strange. I will ask physics teacher tomorrow. Thanks.

    • one year ago
  28. Mashy
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    your welcome :)

    • one year ago
  29. Mashy
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    and please do tell here what happens ok?

    • one year ago
  30. micahwood50
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    I will.

    • one year ago
  31. yrelhan4
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    you gotta take the centre of mass of the ball here. so h=(R-r) - (R-r)costheta

    • one year ago
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