Two rigid hemispheres A and B with uniform volume density p have radii a and b, respectively. Hemisphere B has its flat face glued to a plane. Hemisphere A is then balanced on top of hemisphere B such that their curved surfaces are in contact.
Naturally, A is in equilibrium when its flat face lies parallel to the flat face of B. However, if given a small nudge, A rolls without slipping on the curved surface of B and will either oscillate about the equilibrium position or fall.
The constraint on aa such that A can oscillate is given to be kb>a, where k is some positive real number.
Find the value of k.
Assume that gravity points down, perpendicular to the plane of B's flat face.
If theta is the arc through which the rolling occurs with respect
to the lower sphere "A", then the angle through which the upper sphere
"B" rotates is (A/B)* theta, because the non-slipping arc length must be
A*theta for both objects.
If we rotate B
exactly to the tipping point, there will be a triangle formed by the
radius of B (center of flat surface to point of contact), a distance of
(3/8)*B (center of flat surface to CG of object B), and an unknown
length (CG of object B to point of contact), and this 3rd side will be
pointing straight down in line with gravity. We know we have the right
amount of rotation if the unknown length is equal to:
sqrt{B^2 + (3/8)^2 B^2 - 2(B)((3/8)B*cos((A/B)*theta)}
or B* sqrt{1 + 9/64 - (3/4)*cos((A/B)*theta)}
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Posted by Larry
on 2019-11-18 14:47:38 |
Here is a try ...
