Height of small ball rolling inside spinning hollow ball

[devanlevin]

as seen in the following diagram,
a small ball is placed inside a larger hollow ball, which is spun at a frequency of "f" around an axis passing through the centre(axis from 12 oclock position to 6) there is no notable friction. what is the height "h" that the small ball will rise to inside the large one.
use r,f,g to define the height

http://picasaweb.google.com/devanlevin/DropBox?authkey=sbH95pBl_D8#5273423507587078018

excuse my artwork

what i did was:

defined the height==> h=r-rcos$$\alpha$$=r(1-cos$$\alpha$$)
define the velocity==>v=2$$\Pi$$r*f

broke the vectors into radial and tangential components, then said
$$\sum$$F(radial)=N-mgcos$$\alpha$$=m$$\frac{v^{2}}{r}$$
N-mgcos$$\alpha$$=mr(2$$\Pi$$f)$$^{2}$$

cos(alpha)=N/mg - (r/g)(2pi*f)^2

h=r(1-=(N-mr(2$$\Pi$$f)^2)/mg)

is this correct up to here?
now the part I am not sure of at all,
how to get rid of N?

the correct answer is
h=r(1-g/(r(2pi*f)^2)
when f>(1/2pi)*sqrt(g/r)

and h=0
when f<(1/2pi)*sqrt(g/r)
how do i get these conditions, and where have i gone wrong finding the expression for h

 

[naresh]

Just consider the small ball alone - about what axis does it rotate? What is the radius of its motion, and what direction does the centripetal force point in?

 

[devanlevin]

thats what i did, isn't it??

 

[naresh]

Not quite. In your solution, the centripetal force is along N.

The small ball moves in a circle, whose center is not the center of the bigger ball. The radius of this circle is not r.

Also, in solving force balance equations, you usually write down equations along two perpendicular axes. What is your second equation?

 

[devanlevin]

i don't really see what you mean? how would you solve this problem?

 

[naresh]

Hmm. The small ball moves in a horizontal circle. Agree?

The radius of this circle is r sin \alpha. The centripetal force is in the horizontal direction. Can you work it out from there?

 

[devanlevin]

whats the centripetal force? mv^2/r*sin(alpha)??
then from there i find what sin(alpha) is and then, use that to find the height h=r(1-cos(alpha))
is that right??

 

[devanlevin]

$$\sum$$F=m$$\frac{v^{2}}{rsin\alpha}$$=m((2pi*f)^2)r*sin(alpha)

now what?? what do i compare the force to??

 

[devanlevin]

dont worry about it, i managed,, thanks for the help