Maximum height

1. Nov 27, 2008

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

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 im not sure of at all,
how to get rid of N?

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

Last edited by a moderator: May 3, 2017
2. Nov 27, 2008

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?

3. Nov 28, 2008

devanlevin

thats what i did, isnt it??

4. Nov 28, 2008

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?

5. Nov 28, 2008

devanlevin

i dont really see what you mean? how would you solve this problem?

6. Nov 28, 2008

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?

7. Nov 28, 2008

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??

8. Nov 28, 2008

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??

9. Nov 28, 2008

devanlevin

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