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Moment of Inertia 2

  1. Dec 2, 2009 #1
    1. The problem statement, all variables and given/known data


    a) A disk with radius R and mass M is rolling without slipping on a level surface, as shown. The moment of inertia about an axis perpendicular to the page and through the center of the disk is I. If the center of
    the disk is moving at speed v, what is the kinetic energy of the disk?
    b) For the disk described in part (a), what is the instantaneous speed vL, relative to the ground, of the lowest point on the disk? What is the instantaneous speed vH, relative to the ground, of the highest point on
    the disk?

    c) Suppose the disk described in part (b) is allowed to roll down an inclined plane, oriented at an angle θ relative to the horizontal, as shown.
    Assume that it rolls without slipping, but ignore all frictional effects other than the friction that is necessary to prevent it from slipping. Let g
    denote the acceleration of gravity. Calculate the magnitude aCM of the acceleration of the center of mass.

    2. Relevant equations



    3. The attempt at a solution

    part a)

    T = .5mv2 + .5Iw2

    is this right, its barley even a question

    part b)

    Wheight = v/R
    Wbottom = -v/R

    part c)

    mgsinQ - f = ma

    fr = .5mr2[tex]\alpha[/tex]

    mgsinQ - .5ma = ma

    gsinQ = 3/2 a

    a = 2/3 gsinQ
     

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    • 3c.pdf
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  2. jcsd
  3. Dec 2, 2009 #2

    ideasrule

    User Avatar
    Homework Helper

    Yup, that's right.

    The question asks for velocity relative to the ground, not for angular speed relative to the center of mass. If the bottom of the wheel is moving at v relative to the center of mass and the center of mass is moving at v relative to the ground, how fast is the bottom moving with respect to the ground? How about the top?

    Yeah, that's right. An easier way to do things is to consider the contact point as the "pivot" and apply the rotational equations to that. "I" would be 0.5mR^2 + mR^2 = 1.5mR^2 and torque would be mgRsinQ, so:

    mgRsinQ=1.5mR^2*alpha
    gsinQ=1.5R*a/R
    a=2/3gsinQ
     
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