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anonymous
 3 years ago
Physics Rolling Help Needed
anonymous
 3 years ago
Physics Rolling Help Needed

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I tired to use a conservation of energy and I kept get wrong answer. Why doesn't it work?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\large E_{tot_o} = E_{tot_f}\]\[\large U_o = K_t+K_r\]\[\large mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0did you get the h right? what did you calculate h as?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well, I calculated h as R  Rcosθ (R represents radius of bowl)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1360915947670:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yea yea thats right.. :)!!.. still getting it wrong? :/.. show your work here!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}\left(\dfrac{2}{5}mr^2\right)\dfrac{v^2}{r^2}\] \[= \dfrac{7}{10}mv^2\] \[\dfrac{10}{7}gh = v^2 = \omega^2 r^2\]\[\omega = \sqrt{\dfrac{10gh}{7r^2}}\]\[\boxed{\omega = \sqrt{\dfrac{10gR(1  \cos \theta)}{7r^2}}}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Mashy @VincentLyon.Fr

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yea.. i don't see any flaw in it.. sorry computer got stuck :P

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That's weird... Apparently I cannot use conservation of energy for some reasons....

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Do you know another way I can solve?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wait.. the general equation is \[( 2gh /(1 + Icm/Mr^{2}))^{1/2}\] how do you put that division line?? mine comes slanted line :(

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thats linear velocity :D

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yea there is nothing wrong in it!!.. its correct.. but do tell me how do you put that division in equation editor

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What's that? Just type \dfrac{}{}

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0HOW TO DO THAT?!?!? :XXXXX

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0general equation, what's that?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok 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
 3 years ago
Best ResponseYou've already chosen the best response.0Maybe using conservation of energy won't work for some reasons...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no 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
 3 years ago
Best ResponseYou've already chosen the best response.0That's strange. I will ask physics teacher tomorrow. Thanks.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and please do tell here what happens ok?

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.2you gotta take the centre of mass of the ball here. so h=(Rr)  (Rr)costheta
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