20121109

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: Idisk = (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 = 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

No comments: