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I'm afraid that I really don't understand the answer given below!....
It seems to me that the problem is simply one of conservation of energy.
A ball that "Rolls without Slipping" has two parts to its total
kinetic energy when moving along a horizontal surface:
(1/2) m v^2 <=> translational kinetic energy
(1/2) I w^2 <=> rotational kinetic energy ("w" is omega)
Since the ball "Rolls without Slipping" there is NO ENERGY LOST TO
FRICTION because there is ZERO MOTION of the instantaneous point of
contact in the direction of the frictional force (leading to a zero work
integral).
When the ball encounters an incline and assuming it still "Rolls without
Slipping", it goes up the incline with NO LOSS TO FRICTION, transferring
ALL of both its translational and rotational kinetic energies directly to
gravitational potential energy. Since TOP is by definition the "highest
point" to which it rises, it must get to a TOP. And, I maintain that this
is the highest TOP to which it may rise.
If on the other hand, when the ball gets to the incline, it "Slips without
Rolling" up the hill, it can only convert its translational kinetic energy
to gravitational potential energy. The rotational kinetic energy is, if
you will, "trapped" due to the lack of friction to couple it to the motion
of the ball. It should, as previously described, rise to a maximum height
(which MUST BE LESS than that of the "Rolls w/o Slipping" case) and sit
spinning at this lower TOP.
ERTEL SENDS.
/^\-/^\
/ \
| * |
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/ Prof. John P. Ertel \
/ veteran Eagle Scout \
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