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        riddles >> putnam exam (pure math) >> Ball-bearing on Sphere
        
(Message started by: ThudanBlunder on Nov 26^th, 2007, 12:01pm)

Title: Ball-bearing on Sphere
Post by ThudanBlunder on Nov 26^th, 2007, 12:01pm

I have seen this one before, hopefully not here:

A slippery ball-bearing is placed in equilibrium on the top of a sphere. It is given a negligibly small impulse which makes it slide down the sphere. Where does it leave the surface of the sphere?

Pre-emptive assumptions for nitpickers:
1) The radius of the ball-bearing is negligible compared with the sphere.
2) The ball-bearing slides without rolling - ie. there is no friction.

What can we say if the ball-bearing rolls without slipping?

Title: Re: Ball-bearing on Sphere
Post by SMQ on Nov 26^th, 2007, 12:28pm

[hide]Let the ball-bearing have slid through an angle http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif around a sphere or radius R. From conservation of energy we have mgh = mv²/2 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif v² = 2gR(1 - cos http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif). The ball-bearing will leave the sphere when the centrifugal force due to its velocity along the surface of the sphere is equal to the radial component of the acceleration due to gravity: v²/R = g cos http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif. Substituting for velocity we have 2gR(1 - cos http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif)/R = g cos http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif 2 - 2 cos http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif = cos http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif= cos^-1 (2/3) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/approx.gif 48.2^o (declination from vertical), regardless of the radius of the sphere or the acceleration of gravity![/hide]

The case where the ball-bearing rolls should be similar, with the addition of a term for [hide]energy stored as internal angular momentum of the ball-bearing[/hide]. In general, we can say that the rolling ball bearing [hide]will leave the surface of the sphere later than in the sliding case, as some of the gravitational potential energy (which was converted entirely to kinetic energy in the sliding case) will now be stored in the internal rotational momentum of the bearing, and so the rolling bearing will be traveling slightly slower than an equivalent sliding bearing at any given point.[/hide]

--SMQ