The kinetic energy of a rolling solid sphere is
(1/2)MV^2 + (1/2)I w^2
= (1/2)M*V^2 + (1/2)(2/5)M(R*w)^2
= (7/10)M*V^2
Set that equal to M*g*H and solve for the height that it rises, H.
H = (0.7) V^2/g
The radius doesn't matter. Neither does the material.
It will only rise about 1.1 cm
20 cm/s is quite slow
C) A steel ball bearing the radius 1 cm is rolling along a table with speed 20 cm/sec when it starts to roll up an incline. How high above the table level will it rise before stopping? Ignore friction losses
2 answers
it actually equals 0.28 cm I have no clue how you got 1.1 cm