micahwood50
Physics Rolling Help Needed
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micahwood50
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micahwood50
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I tired to use a conservation of energy and I kept get wrong answer. Why doesn't it work?
micahwood50
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\[\large E_{tot_o} = E_{tot_f}\]\[\large U_o = K_t+K_r\]\[\large mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2\]
Mashy
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did you get the h right? what did you calculate h as?
micahwood50
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Well, I calculated h as R - Rcosθ (R represents radius of bowl)
micahwood50
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Isn't that right?
micahwood50
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|dw:1360915947670:dw|
Mashy
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yea yea thats right.. :)!!.. still getting it wrong? :-/.. show your work here!
micahwood50
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\[mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}\left(\dfrac{2}{5}mr^2\right)\dfrac{v^2}{r^2}\]
\[= \dfrac{7}{10}mv^2\]
\[\dfrac{10}{7}gh = v^2 = \omega^2 r^2\]\[\omega = \sqrt{\dfrac{10gh}{7r^2}}\]\[\boxed{\omega = \sqrt{\dfrac{10gR(1 - \cos \theta)}{7r^2}}}\]
micahwood50
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@Mashy @Vincent-Lyon.Fr
micahwood50
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isn't that right?
Mashy
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yea.. i don't see any flaw in it.. sorry computer got stuck :P
micahwood50
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That's weird...
Apparently I cannot use conservation of energy for some reasons....
micahwood50
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Do you know another way I can solve?
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 :(
Mashy
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thats linear velocity :D
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
Mashy
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\[1 \div 2\]
micahwood50
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What's that?
Just type \dfrac{}{}
Mashy
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HOW TO DO THAT?!?!? :XXXXX
Mashy
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ok ok wait
micahwood50
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general equation, what's that?
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!
micahwood50
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Maybe using conservation of energy won't work for some reasons...
micahwood50
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Thank you, though.
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!
micahwood50
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That's strange. I will ask physics teacher tomorrow. Thanks.
Mashy
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your welcome :)
Mashy
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and please do tell here what happens ok?
micahwood50
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I will.
yrelhan4
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you gotta take the centre of mass of the ball here. so h=(R-r) - (R-r)costheta