Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 8.9
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A "Six Stripe" poker chip[*] (which can be approximated as a solid disk of 39 mm diameter and mass 10 g) rolls on its edge across a horizontal tabletop at a constant speed, without slipping. Neglect kinetic friction and drag. What fraction of the total kinetic energy is translational?
(Given: Idisk = (1/2)·M·R2.)
Correct answer (highlight to unhide): (C)
The total kinetic energy of the poker chip is the sum of its translational and rotational components:
KE = KEtr + KErot,
where KEtr = (1/2)·m·v2, and KErot = (1/2)·I·ω2. Since the poker chip can be approximated as a solid disk, I = (1/2)·m·r2, and for rolling without slipping, ω = v/r. Substituting these into the KErot term:
KE = (1/2)·m·v2 + (1/2)·(1/2)·m·r2·(v/r)2,
KE = (1/2)·m·v2 + (1/4)m·v2 = (3/4)·m·v2.
The fraction of the total kinetic energy that is translational kinetic energy is then:
KEtr/KE = ((1/2)m·v2)/((3/4)·m·v2) = (1/2)·(4/3) = (2/3) = 0.67.
Response (A) is (1/4); response (B) is (1/2); response (D) is (3/4).
Sections 70854, 70855
Exam code: quiz05L4mN
(A) : 7 students
(B) : 18 students
(C) : 18 students
(D) : 9 students
Success level: 35%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.69