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Homework Help: Angular momentum stone problem

  1. May 4, 2008 #1
    1. The problem statement, all variables and given/known data
    A stone falls from rest from the top of a building. Which of the following graphs' shapes best represents the stone's angular momentum L about the point P as a function of time?

    http://img413.imageshack.us/img413/1033/54161793ev0.png [Broken]

    a) L = 0
    b) L = c (constant)
    c) L = c * t (linear in t)
    d) L = t^2
    e) L = upside-down parabola with vertex at some positive x and positive y, passing through the origin

    3. The attempt at a solution
    I think I solved this problem using the definition of angular momentum as the cross product of r and p (calling point P (x_0, y_0)):

    [tex]\vec{L} = \vec{r} \times \vec{p} = \left| \begin{array}{ccc}
    \hat{i} & \hat{j} & \hat{k} \\
    r_x & r_y & 0 \\
    0 & p_y & 0 \end{array} \right| = r_x p_y \hat{k} = x_0 m g t \hat{k}[/tex]

    where [itex]r_x = x_0[/itex] and [itex]p_y = - m g t[/itex]

    So apparently linear momentum is linear in time. This question is for an AP Physics C sample multiple choice, so I have a hard time believing they want us to evaluate a cross-product to figure out this. Is there some intuitive way to understand this? Or a quick way to do it? I tried using the definition of cross product as [itex]r p \sin(\theta)[/itex] but that doesn't get me very far either.
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. May 4, 2008 #2


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    Homework Helper

    Hi awvvu,

    About your definition [itex]rp \sin(\theta)[/itex], I think that is a good way to see it. If you think about the path of the stone as it falls (vertically), notice that r and theta both change with time, but [itex]r\sin\theta[/itex] is a constant--it's just the horizontal distance from the point to the ball's path. So the only thing changing with time in [itex]rp \sin(\theta)[/itex], is p.

    There are also the slightly differently-written forms [itex]m v_{\perp} r[/itex] and [itex]m v r_{\perp}[/itex] for the angular momentum of a particle. The second form works best here: you know the velocity is vertical, so [itex]r_{\perp}[/itex] must be the horizontal distance.
  4. May 4, 2008 #3
    Ah, I was trying to look at each multiplied term separately and I couldn't figure out the relationship between r and sin(theta).
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