Before I got what I thought was the correct answer here I posted in alt.sci.physics newsgroup, and I just looked there for replies and to my surprise, I found out that I was still quite wrong! I neglected to think of the fact that there is a rotational component of additional (or reduced, depending on whether the ball is rolling down or up) component of kinetic energy that contributes to a change in the speed or rotation of the ball. Here's the relevant post:
> IOW, not counting for rolling resistance and wind/air resistance, how
> fast do I need to get a ball rolling so that it will come to rest after
> rising one foot up an incline?
A ball mass m and radius r rolling at speed v has translational kinetic
energy KE = (1/2) m*v^2 and rotational kinetic energy = (1/2) I*w^2 where I
is the moment of inertia (a function of the mass and radius (for a uniform
sphere I = (2/5)*m*r^2 and w is the angular velocity = v/r).
The total kinetic energy for a rolling uniform sphere is then K(tot) =
(9/10)*m*v^2.
This energy will be exerted climbing uphill until its increase in
gravitational kinetic energy m*g*h consumes it all:
(9/10)*m*v^2 = m*g*h
The m's conveniently drop out (meaning that the result is the same for all
balls regardless of mass). Solving for v as a function of h we get:
v = sqrt[(10/9)*g*h]
In the archaic units familiar to golfers, g = 32 ft/(sec)^2 and h = 1 foot,
so v = sqrt(32*1/0.9) = 5.96 ft/sec.
Tom Davidson
Brighton, CO
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