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 APart completeA uniform solid cylinder.0.333SubmitPrevious Answers CorrectCorrect answer is shown. Your answer 1313 = 0.33333 was either rounded differently or used a different number of significant figures than required for this part.IDENTIFY: Apply K=Kcm+Krot𝐾=𝐾𝑐𝑚+𝐾𝑟𝑜𝑡.SET UP: For an object that is rolling without slipping, vcm=Rω𝑣𝑐𝑚=𝑅𝜔.EXECUTE: The fraction of the total kinetic energy that is rotational is(1/2)Icmω2(1/2)Mv2cm+(1/2)Icmω2=11+(M/Icm)v2cm/ω2=11+(MR2/Icm)(1/2)𝐼𝑐𝑚𝜔2(1/2)𝑀𝑣𝑐𝑚2+(1/2)𝐼𝑐𝑚𝜔2=11+(𝑀/𝐼𝑐𝑚)𝑣𝑐𝑚2/𝜔2=11+(𝑀𝑅2/𝐼𝑐𝑚)Icm=(1/2)MR2𝐼𝑐𝑚=(1/2)𝑀𝑅2, so the above ratio is 1/31/3.Part BPart completeA uniform sphere.0.286SubmitPrevious Answers CorrectCorrect answer is shown. Your answer 2727 = 0.28571 was either rounded differently or used a different number of significant figures than required for this part.Icm=(2/5)MR2𝐼𝑐𝑚=(2/5)𝑀𝑅2 so the above ratio is 2/72/7.Part CA thin-walled hollow sphere.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 DA hollow, cylinder with outer radius R and inner radius R/2.
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 APart completeA uniform solid cylinder.0.333SubmitPrevious Answers CorrectCorrect answer is shown. Your answer 1313 = 0.33333 was either rounded differently or used a different number of significant figures than required for this part.IDENTIFY: Apply K=Kcm+Krot𝐾=𝐾𝑐𝑚+𝐾𝑟𝑜𝑡.SET UP: For an object that is rolling without slipping, vcm=Rω𝑣𝑐𝑚=𝑅𝜔.EXECUTE: The fraction of the total kinetic energy that is rotational is(1/2)Icmω2(1/2)Mv2cm+(1/2)Icmω2=11+(M/Icm)v2cm/ω2=11+(MR2/Icm)(1/2)𝐼𝑐𝑚𝜔2(1/2)𝑀𝑣𝑐𝑚2+(1/2)𝐼𝑐𝑚𝜔2=11+(𝑀/𝐼𝑐𝑚)𝑣𝑐𝑚2/𝜔2=11+(𝑀𝑅2/𝐼𝑐𝑚)Icm=(1/2)MR2𝐼𝑐𝑚=(1/2)𝑀𝑅2, so the above ratio is 1/31/3.Part BPart completeA uniform sphere.0.286SubmitPrevious Answers CorrectCorrect answer is shown. Your answer 2727 = 0.28571 was either rounded differently or used a different number of significant figures than required for this part.Icm=(2/5)MR2𝐼𝑐𝑚=(2/5)𝑀𝑅2 so the above ratio is 2/72/7.Part CA thin-walled hollow sphere.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 DA hollow, cylinder with outer radius R and inner radius R/2.
Solution
Part A For a uniform solid cylinder, the moment of inertia (I) is given by I = (1/2)MR^2, where M is the mass and R is the radius of the cylinder. The total kinetic energy (K) of a rolling object is the sum of its translational kinetic energy and its rotational kinetic energy, given by K = (1/2)Mv^2 + (1/2)Iω^2. Here, v is the velocity of the center of mass, and ω is the angular velocity. For an object rolling without slipping, v = Rω. Substituting these values into the equation for K, we find that the fraction of the total kinetic energy that is rotational is (1/2)Iω^2 / [(1/2)Mv^2 + (1/2)Iω^2] = 1 / [1 + (M/I)v^2/ω^2] = 1 / [1 + (MR^2/I)]. Substituting I = (1/2)MR^2, we find that this fraction is 1/3.
Part B For a uniform sphere, the moment of inertia is I = (2/5)MR^2. Substituting this into the equation for the fraction of the total kinetic energy that is rotational, we find that this fraction is 2/7.
Part C For a thin-walled hollow sphere, the moment of inertia is I = (2/3)MR^2. Substituting this into the equation for the fraction of the total kinetic energy that is rotational, we find that this fraction is 2/5.
Part D For a hollow cylinder with outer radius R and inner radius R/2, the moment of inertia is I = (1/2)M(R^2 + (R/2)^2) = (3/8)MR^2. Substituting this into the equation for the fraction of the total kinetic energy that is rotational, we find that this fraction is 3/7.
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