- anonymous

Physics Rolling Help Needed

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

##### 1 Attachment

- anonymous

I tired to use a conservation of energy and I kept get wrong answer. Why doesn't it work?

- anonymous

\[\large E_{tot_o} = E_{tot_f}\]\[\large U_o = K_t+K_r\]\[\large mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

did you get the h right? what did you calculate h as?

- anonymous

Well, I calculated h as R - Rcosθ (R represents radius of bowl)

- anonymous

Isn't that right?

- anonymous

|dw:1360915947670:dw|

- anonymous

yea yea thats right.. :)!!.. still getting it wrong? :-/.. show your work here!

- 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

isn't that right?

- anonymous

yea.. i don't see any flaw in it.. sorry computer got stuck :P

- anonymous

That's weird...
Apparently I cannot use conservation of energy for some reasons....

- anonymous

Do you know another way I can solve?

- 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

thats linear velocity :D

- anonymous

yea there is nothing wrong in it!!.. its correct.. but do tell me how do you put that division in equation editor

- anonymous

\[1 \div 2\]

- anonymous

What's that?
Just type \dfrac{}{}

- anonymous

HOW TO DO THAT?!?!? :XXXXX

- anonymous

ok ok wait

- anonymous

general equation, what's that?

- 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

Maybe using conservation of energy won't work for some reasons...

- anonymous

Thank you, though.

- 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

That's strange. I will ask physics teacher tomorrow. Thanks.

- anonymous

your welcome :)

- anonymous

and please do tell here what happens ok?

- anonymous

I will.

- yrelhan4

you gotta take the centre of mass of the ball here. so h=(R-r) - (R-r)costheta

Looking for something else?

Not the answer you are looking for? Search for more explanations.