It is well known that for a hollow, cylindrical shell rolling without slipping on a horizontal surface, half of the total kinetic energy is translational and half is rotational. What fraction of the total kinetic energy is rotational for the following objects rolling without slipping on a horizontal surface?For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of Race of the rolling bodies.Part AA uniform solid cylinder.Activate to select the appropriates template from the following choices. Operate up and down arrow for selection and press enter to choose the input value typeActivate to select the appropriates symbol from the following choices. Operate up and down arrow for selection and press enter to choose the input value typenothingSubmitRequest AnswerPart BA uniform sphere.
Question
It is well known that for a hollow, cylindrical shell rolling without slipping on a horizontal surface, half of the total kinetic energy is translational and half is rotational. What fraction of the total kinetic energy is rotational for the following objects rolling without slipping on a horizontal surface?For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of Race of the rolling bodies.Part AA uniform solid cylinder.Activate to select the appropriates template from the following choices. Operate up and down arrow for selection and press enter to choose the input value typeActivate to select the appropriates symbol from the following choices. Operate up and down arrow for selection and press enter to choose the input value typenothingSubmitRequest AnswerPart BA uniform sphere.
Solution
The kinetic energy of a rolling object is the sum of its translational kinetic energy and its rotational kinetic energy. The fraction of the total kinetic energy that is rotational depends on the shape of the object.
Part A: For a uniform solid cylinder, the moment of inertia I is given by (1/2)MR^2, where M is the mass and R is the radius. The rotational kinetic energy is (1/2)Iω^2 = (1/4)MR^2ω^2. The translational kinetic energy is (1/2)Mv^2, and since the cylinder is rolling without slipping, v = Rω, so the translational kinetic energy is (1/2)MR^2ω^2. Therefore, the total kinetic energy is (1/4 + 1/2)MR^2ω^2 = (3/4)MR^2ω^2, so the fraction of the total kinetic energy that is rotational is (1/4)/(3/4) = 1/3.
Part B: For a uniform sphere, the moment of inertia I is given by (2/5)MR^2. The rotational kinetic energy is (1/2)Iω^2 = (1/5)MR^2ω^2. The translational kinetic energy is (1/2)Mv^2 = (1/2)MR^2ω^2. Therefore, the total kinetic energy is (1/5 + 1/2)MR^2ω^2 = (7/10)MR^2ω^2, so the fraction of the total kinetic energy that is rotational is (1/5)/(7/10) = 2/7.
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