Physics quiz question: rolling poker chip

Physics 205A Quiz 5, fall semester 2012
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 8.9

"Stock Video of rolling Poker Chips"
Free Stock Video resources (KozziImages)
youtu.be/o6DVWSXSx9w

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: I_disk = (1/2)·M·R².)

(A) 0.25.
(B) 0.50.
(C) 0.67.
(D) 0.75.

[*] vegassuppliesandgifts.com/page/VSAG/CTGY/10-1051/.

Correct answer (highlight to unhide): (C)

The total kinetic energy of the poker chip is the sum of its translational and rotational components:

KE = KE_tr + KE_rot,

where KE_tr = (1/2)·m·v², and KE_rot = (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 KE_rot 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:

KE_tr/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