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USAPhO 2014 F=ma exam #14, (rotational dynamics)

  1. Apr 14, 2014 #1
    PDF: http://www.aapt.org/physicsteam/2014/upload/exam1-2014-2-2-answers.pdf

    1. The problem statement, all variables and given/known data

    A disk of moment of inertia I, mass M, and radius R has a cord wrapped around it tightly as
    shown in the diagram. The disk is free to slide on its side as shown in the top down view. A
    constant force of T is applied to the end of the cord and accelerates the disk along a frictionless
    surface.

    After the disk has accelerated some distance, determine the ratio of the translational KE to total KE of the disk,

    KEtranslational / KEtotal =

    Answer) I / (MR2 +I)

    2. Relevant equations

    Torque = Iα = F x R

    KE = 1/2 mv2 or 1/2 Iω2

    3. The attempt at a solution

    My conceptual understanding is the problem here, I think. If there is absolutely no friction, then would there be any rotation? I think there would be, but I'm not sure.

    Here's an attempt:

    KEtranslational / KEtotal = Mv2 / (Mv2 + Iω2)

    Substituting v = Rω, cancelling the ω2 terms

    = MR2 / (MR2 + I)

    So I get an answer close but not exactly the correct answer. The answer I got is choice E on the actual exam, meaning my attempt probably has a common mistake.
     
    Last edited: Apr 14, 2014
  2. jcsd
  3. Apr 14, 2014 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    v = Rω isn't true. That's only true if you are rolling without slipping. Since you have no friction, that's not true. Use F=ma=mdv/dt and a similar form for the rotational motion to compute both forms of energy as a function of time, t.
     
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