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micahwood50

  • one year ago

Physics Rolling Help Needed

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

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

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

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

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

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

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

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

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

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

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

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

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

  15. Mashy
    • one year ago
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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 :(

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

  17. Mashy
    • one year ago
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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

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

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

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

  21. Mashy
    • one year ago
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    ok ok wait

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

  23. Mashy
    • one year ago
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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!

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

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

  26. Mashy
    • one year ago
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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!

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

  28. Mashy
    • one year ago
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    your welcome :)

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

  30. micahwood50
    • one year ago
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    I will.

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

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