Posted by Erin on Sunday, April 6, 2008 at 10:56pm.
For ice, you can use conservation of energy. If h is the height that the block descends,
(1/2)MV^2 = M g h
V = sqrt (2gH)
For the marble block, since it rolls, I assume it is a cylinder. Static friction is what helps it roll instead of slip, but it does not result in a of of energy, since there is no relative motion at the line of contact. Equate the potential energy change to the increase in both translational and rotational kinetic energy. V = r w, where w is the angular velocity of the rolling cyliner. The moment of inertia is I = (1/2)M r^2
M g H = (1/2) M V^2 + (1/2)I w^2
= (1/2) M V^2 + (1/2)(1/2)Mr^2(V/r)^2
= (3/4) M V^2
V = sqrt[(4/3)gH]
sd
actually for the second part, it wouldn't be a cylinder but a sphere. thus beta = 2/5. the velocity equation in this case would be V = sqrt[(2gH)/(1+beta)]
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