0
$\begingroup$

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 ?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

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.

Improve this answer
$\endgroup$
3
  • $\begingroup$ In General is $\triangle \ \vec{L} = \vec{r} \ \times \triangle \ \vec{p} $? $\endgroup$
    – Dude
    Commented 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 $\Delta \vec{L} = \vec{r} \times \Delta \vec{p} + \Delta \vec{r} \times \vec{p}$. $\endgroup$
    – gleedadswell
    Commented Aug 14, 2015 at 2:49
  • $\begingroup$ ok ok I see now yeah it makes sense now ! $\endgroup$
    – Dude
    Commented Aug 14, 2015 at 4:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.