## micahwood50 2 years ago Physics Rolling Help Needed

1. micahwood50

2. micahwood50

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

3. micahwood50

$\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$

4. Mashy

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

5. micahwood50

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

6. micahwood50

Isn't that right?

7. micahwood50

|dw:1360915947670:dw|

8. Mashy

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

9. micahwood50

$mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}\left(\dfrac{2}{5}mr^2\right)\dfrac{v^2}{r^2}$ $= \dfrac{7}{10}mv^2$ $\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}}}$

10. micahwood50

@Mashy @Vincent-Lyon.Fr

11. micahwood50

isn't that right?

12. Mashy

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

13. micahwood50

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

14. micahwood50

Do you know another way I can solve?

15. Mashy

wait.. the general equation is $( 2gh /(1 + Icm/Mr^{2}))^{1/2}$ how do you put that division line?? mine comes slanted line :(

16. Mashy

thats linear velocity :D

17. Mashy

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

18. Mashy

$1 \div 2$

19. micahwood50

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

20. Mashy

HOW TO DO THAT?!?!? :XXXXX

21. Mashy

ok ok wait

22. micahwood50

general equation, what's that?

23. Mashy

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!

24. micahwood50

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

25. micahwood50

Thank you, though.

26. Mashy

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!

27. micahwood50

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

28. Mashy

29. Mashy

and please do tell here what happens ok?

30. micahwood50

I will.

31. yrelhan4

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