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micahwood50 Group TitleBest ResponseYou've already chosen the best response.0
I tired to use a conservation of energy and I kept get wrong answer. Why doesn't it work?
 one year ago

micahwood50 Group TitleBest ResponseYou've already chosen the best response.0
\[\large E_{tot_o} = E_{tot_f}\]\[\large U_o = K_t+K_r\]\[\large mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2\]
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.0
did you get the h right? what did you calculate h as?
 one year ago

micahwood50 Group TitleBest ResponseYou've already chosen the best response.0
Well, I calculated h as R  Rcosθ (R represents radius of bowl)
 one year ago

micahwood50 Group TitleBest ResponseYou've already chosen the best response.0
Isn't that right?
 one year ago

micahwood50 Group TitleBest ResponseYou've already chosen the best response.0
dw:1360915947670:dw
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.0
yea yea thats right.. :)!!.. still getting it wrong? :/.. show your work here!
 one year ago

micahwood50 Group TitleBest ResponseYou've already chosen the best response.0
\[mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}\left(\dfrac{2}{5}mr^2\right)\dfrac{v^2}{r^2}\] \[= \dfrac{7}{10}mv^2\] \[\dfrac{10}{7}gh = v^2 = \omega^2 r^2\]\[\omega = \sqrt{\dfrac{10gh}{7r^2}}\]\[\boxed{\omega = \sqrt{\dfrac{10gR(1  \cos \theta)}{7r^2}}}\]
 one year ago

micahwood50 Group TitleBest ResponseYou've already chosen the best response.0
@Mashy @VincentLyon.Fr
 one year ago

micahwood50 Group TitleBest ResponseYou've already chosen the best response.0
isn't that right?
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.0
yea.. i don't see any flaw in it.. sorry computer got stuck :P
 one year ago

micahwood50 Group TitleBest ResponseYou've already chosen the best response.0
That's weird... Apparently I cannot use conservation of energy for some reasons....
 one year ago

micahwood50 Group TitleBest ResponseYou've already chosen the best response.0
Do you know another way I can solve?
 one year ago

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

Mashy Group TitleBest ResponseYou've already chosen the best response.0
thats linear velocity :D
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.0
yea there is nothing wrong in it!!.. its correct.. but do tell me how do you put that division in equation editor
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.0
\[1 \div 2\]
 one year ago

micahwood50 Group TitleBest ResponseYou've already chosen the best response.0
What's that? Just type \dfrac{}{}
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.0
HOW TO DO THAT?!?!? :XXXXX
 one year ago

micahwood50 Group TitleBest ResponseYou've already chosen the best response.0
general equation, what's that?
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.0
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!
 one year ago

micahwood50 Group TitleBest ResponseYou've already chosen the best response.0
Maybe using conservation of energy won't work for some reasons...
 one year ago

micahwood50 Group TitleBest ResponseYou've already chosen the best response.0
Thank you, though.
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.0
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!
 one year ago

micahwood50 Group TitleBest ResponseYou've already chosen the best response.0
That's strange. I will ask physics teacher tomorrow. Thanks.
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.0
your welcome :)
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.0
and please do tell here what happens ok?
 one year ago

micahwood50 Group TitleBest ResponseYou've already chosen the best response.0
I will.
 one year ago

yrelhan4 Group TitleBest ResponseYou've already chosen the best response.2
you gotta take the centre of mass of the ball here. so h=(Rr)  (Rr)costheta
 one year ago
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