A space platform in the form of a thin circular disc of radius a and mass m (flying saucer) is initially rotating with angular velocity w about its symmetry axis. A meteorite strikes the edge of the platform imparting an implies p to the platform. The direction P is parallel to the axis of the platform, and the magnitude of P is $\frac{maw}{4}$. Find the resulting values of the precessional rate $\Omega$ and the wobble rate $\dot{\phi}$, and the angle $\alpha$ between the symmetry axis and new axis of rotating.

So professor answered this question, however I have one thing I am not quite sure how he did. First of all of have $\vec{P} = \frac{maw}{4}\ \hat{e_3}$ and we have $w_1 = w\ \hat{e_3}$, where $w_1$ is the initial angular velocity. now This thing I don't understand

"Suppose point of impact is through the 2-axis so we have during collision." $\int{ N dt} = \vec{r}\ \times \vec{p} = a\hat{e_2} \ \times \frac{maw}{4}\hat{e_3}$

How did he get that $\int{ N dt} = \vec{r}\ \times \vec{p}$ ? Isn't that angular impluse which is change of angular momentum not angular momentum itself ?


1 Answer 1


Yes, the angular impulse should be the change in angular momentum as you say. Your professor just forgot to include the $\Delta$. In this case it won't likely cause a mistake since $\omega_1$ is in the $\hat{e}_3$ direction but you can probably see that $\Delta \omega$ is in the $\hat{e}_1$ direction.

  • $\begingroup$ In General is $\triangle \ \vec{L} = \vec{r} \ \times \triangle \ \vec{p} $? $\endgroup$
    – Dude
    Aug 14, 2015 at 2:24
  • $\begingroup$ Well, your professor defined the imparted momentum as $p$, though it is really a change so I guess that's the source of the confusion. It doesn't always have to be $\Delta \vec{L} = \vec{r} \times \Delta \vec{p}$. For example, when a figure skater pulls in their arms it is $\vec{r}$ that is changing. So more generally there is a product rule $\Delta \vec{L} = \vec{r} \times \Delta \vec{p} + \Delta \vec{r} \times \vec{p}$. $\endgroup$ Aug 14, 2015 at 2:49
  • $\begingroup$ ok ok I see now yeah it makes sense now ! $\endgroup$
    – Dude
    Aug 14, 2015 at 4:32

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