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micahwood50

  • 2 years ago

Physics Rolling Help Needed

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  1. micahwood50
    • 2 years ago
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  2. micahwood50
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    did you get the h right? what did you calculate h as?

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

  6. micahwood50
    • 2 years ago
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    Isn't that right?

  7. micahwood50
    • 2 years ago
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    |dw:1360915947670:dw|

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

  9. micahwood50
    • 2 years 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
    • 2 years ago
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    @Mashy @Vincent-Lyon.Fr

  11. micahwood50
    • 2 years ago
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    isn't that right?

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

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

  14. micahwood50
    • 2 years ago
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    Do you know another way I can solve?

  15. Mashy
    • 2 years 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
    • 2 years ago
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    thats linear velocity :D

  17. Mashy
    • 2 years 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
    • 2 years ago
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    \[1 \div 2\]

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

  20. Mashy
    • 2 years ago
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    HOW TO DO THAT?!?!? :XXXXX

  21. Mashy
    • 2 years ago
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    ok ok wait

  22. micahwood50
    • 2 years ago
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    general equation, what's that?

  23. Mashy
    • 2 years 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
    • 2 years ago
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    Maybe using conservation of energy won't work for some reasons...

  25. micahwood50
    • 2 years ago
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    Thank you, though.

  26. Mashy
    • 2 years 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
    • 2 years ago
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    That's strange. I will ask physics teacher tomorrow. Thanks.

  28. Mashy
    • 2 years ago
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    your welcome :)

  29. Mashy
    • 2 years ago
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    and please do tell here what happens ok?

  30. micahwood50
    • 2 years ago
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    I will.

  31. yrelhan4
    • 2 years 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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