anonymous
  • anonymous
Physics Rolling Help Needed
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
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anonymous
  • anonymous
I tired to use a conservation of energy and I kept get wrong answer. Why doesn't it work?
anonymous
  • anonymous
\[\large E_{tot_o} = E_{tot_f}\]\[\large U_o = K_t+K_r\]\[\large mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2\]

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anonymous
  • anonymous
did you get the h right? what did you calculate h as?
anonymous
  • anonymous
Well, I calculated h as R - Rcosθ (R represents radius of bowl)
anonymous
  • anonymous
Isn't that right?
anonymous
  • anonymous
|dw:1360915947670:dw|
anonymous
  • anonymous
yea yea thats right.. :)!!.. still getting it wrong? :-/.. show your work here!
anonymous
  • anonymous
\[mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}\left(\dfrac{2}{5}mr^2\right)\dfrac{v^2}{r^2}\] \[= \dfrac{7}{10}mv^2\] \[\dfrac{10}{7}gh = v^2 = \omega^2 r^2\]\[\omega = \sqrt{\dfrac{10gh}{7r^2}}\]\[\boxed{\omega = \sqrt{\dfrac{10gR(1 - \cos \theta)}{7r^2}}}\]
anonymous
  • anonymous
@Mashy @Vincent-Lyon.Fr
anonymous
  • anonymous
isn't that right?
anonymous
  • anonymous
yea.. i don't see any flaw in it.. sorry computer got stuck :P
anonymous
  • anonymous
That's weird... Apparently I cannot use conservation of energy for some reasons....
anonymous
  • anonymous
Do you know another way I can solve?
anonymous
  • anonymous
wait.. the general equation is \[( 2gh /(1 + Icm/Mr^{2}))^{1/2}\] how do you put that division line?? mine comes slanted line :(
anonymous
  • anonymous
thats linear velocity :D
anonymous
  • anonymous
yea there is nothing wrong in it!!.. its correct.. but do tell me how do you put that division in equation editor
anonymous
  • anonymous
\[1 \div 2\]
anonymous
  • anonymous
What's that? Just type \dfrac{}{}
anonymous
  • anonymous
HOW TO DO THAT?!?!? :XXXXX
anonymous
  • anonymous
ok ok wait
anonymous
  • anonymous
general equation, what's that?
anonymous
  • anonymous
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!
anonymous
  • anonymous
Maybe using conservation of energy won't work for some reasons...
anonymous
  • anonymous
Thank you, though.
anonymous
  • anonymous
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!
anonymous
  • anonymous
That's strange. I will ask physics teacher tomorrow. Thanks.
anonymous
  • anonymous
your welcome :)
anonymous
  • anonymous
and please do tell here what happens ok?
anonymous
  • anonymous
I will.
yrelhan4
  • yrelhan4
you gotta take the centre of mass of the ball here. so h=(R-r) - (R-r)costheta

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