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Change in radius?

  1. Dec 9, 2013 #1
    26. A 2.0-kg block starts from rest on the positive x axis 3.0 m from the origin and thereafter has an acceleration given by a = (4.0 m/s2)ˆi (3.0 m/s2)ˆj. At the end of 2.0 s its angular momentum about the origin is:
    A. 0
    B. (36kg·m2/s)kˆ
    C. (+48kg·m2/s)kˆ
    D. (96kg·m2/s)kˆ
    E. (+96kg·m2/s)kˆ
    ans: B

    For this question, they essentially did:

    L = | r x p |
    L = (r)(m)(v)
    L = (3m)(2kg)(6m/s) * I calculated v = 6 m/s to be the tangential velocity

    Now, I was wondering why they didn't change the radius since it accelerates for 2 seconds. During this time period, it moves across the axis, parallel to the original rotation axis.

    Δx = (1/2)(4m/s2)(2s)2 and I added this length to my attempt, but it doesn't seem like the solution did. I was wondering if this is correct or if it's incorrect for future reference...
     
  2. jcsd
  3. Dec 9, 2013 #2

    haruspex

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    If you want comments on the book(?) solution, please post the complete text of that.
    But I would have thought the appropriate method here would be a purely vectorial one, which turns out to be very easy.
    ##\vec {\ddot x} = \vec c##
    ##\vec { \dot x} = \vec c t + \vec {\dot {x_0}} = \vec c t##
    ##\vec x = \frac 12 \vec c t^2 + \vec x_0##
    ##\vec L = m \vec x × \dot {\vec x} = m (\frac 12 \vec c t^2 + \vec x_0)× \vec c t = m \vec x_0 × \vec c t##
     
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