Angular momentum stone problem

[awvvu]

Homework Statement

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

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

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)):

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

where $r_x = x_0$ and $p_y = - m g t$

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 $r p \sin(\theta)$ but that doesn't get me very far either.

 

[alphysicist]

Hi awvvu,

About your definition $rp \sin(\theta)$, 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 $r\sin\theta$ 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 $rp \sin(\theta)$, is p.

There are also the slightly differently-written forms $m v_{\perp} r$ and $m v r_{\perp}$ for the angular momentum of a particle. The second form works best here: you know the velocity is vertical, so $r_{\perp}$ must be the horizontal distance.

 

[awvvu]

Ah, I was trying to look at each multiplied term separately and I couldn't figure out the relationship between r and sin(theta).