Question
A solid uniform marble and a block of ice, each with the same mass, start from rest at the same height above the bottom of a hill and move down it. The marble rolls without slipping, but the ice slides without friction.
Find the speed of each of these objects when it reaches the bottom of the hill.
V_ice=
V_marble=
Find the speed of each of these objects when it reaches the bottom of the hill.
V_ice=
V_marble=
Answers
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]
(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]
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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