My Math Forum Determine the angular momentum

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 March 16th, 2017, 04:29 PM #1 Member   Joined: Feb 2014 Posts: 91 Thanks: 1 Determine the angular momentum Well finally got to the last question. I thought this one would be easy but I'm unsure. I've looked into this question. I have the mass, the velocity, and I think radius. If I manage to get the last one right that would be awesome.
March 16th, 2017, 05:08 PM   #2
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Quote:
 Originally Posted by The_Ys_Guy Well finally got to the last question. I thought this one would be easy but I'm unsure. I've looked into this question. I have the mass, the velocity, and I think radius. If I manage to get the last one right that would be awesome.
The formula for angular momentum is $\displaystyle \vec{L} = m \vec{r} \times \vec{v} = m |v| |r| ~ sin(\theta)$
where $\displaystyle \theta$ is the angle between $\displaystyle \vec{r}$ and $\displaystyle \vec{v}$.

How do you find that angle? (Hint: Use the definition of the cross product.)

-Dan

 March 16th, 2017, 05:18 PM #3 Math Team     Joined: Jul 2011 From: Texas Posts: 3,005 Thanks: 1588 you've assumed the angle between the linear momentum and position vectors is 90 degrees ... it's not. $\theta = \arccos\bigg[\dfrac{\vec{r} \cdot \vec{v}}{|\vec{r}||\vec{v}|}\bigg]$ $\vec{L} = \vec{r} \times m \vec{v} = |\vec{r}| \cdot m|\vec{v}| \sin{\theta}$ Thanks from topsquark and The_Ys_Guy Last edited by skeeter; March 16th, 2017 at 05:50 PM.
 March 17th, 2017, 09:20 AM #4 Member   Joined: Feb 2014 Posts: 91 Thanks: 1 Okay here is attempt two. I look all over the web to find anything I could and I think I got it right. It appears the cross product of vectors with only 2 dimension is the determinent. The cross product definition had an n variable at the end.
 March 17th, 2017, 01:11 PM #5 Math Team     Joined: Jul 2011 From: Texas Posts: 3,005 Thanks: 1588 $\vec{r} \times m\vec{v}=9\begin{vmatrix} i & j & k\\ -8& 0& 4\\ 3 & 0 & -6 \end{vmatrix}=9\bigg[0\vec{i}-(48-12)\vec{j}+0\vec{k}\bigg]=-324\vec{j}$ Thanks from topsquark and The_Ys_Guy
 March 18th, 2017, 09:40 AM #6 Member   Joined: Feb 2014 Posts: 91 Thanks: 1 I hope I followed instruction correctly this time. Also cross product is way more complex than I thought.
March 18th, 2017, 10:01 AM   #7
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matrix calculations were not meant for humans to do by hand, imho ...
Attached Images
 matrixA.png (1.4 KB, 8 views) cross_product.png (930 Bytes, 8 views)

 March 18th, 2017, 10:21 AM #8 Member   Joined: Feb 2014 Posts: 91 Thanks: 1 Did I finally calculate theta right? I'll look up the matrix functionality of my calculator but I really need to know if I'm on the right path.
 March 18th, 2017, 11:25 AM #9 Math Team     Joined: Jul 2011 From: Texas Posts: 3,005 Thanks: 1588 $|r| = \sqrt{(-8 )^2+4^2} = \sqrt{80} = 4\sqrt{5}$ $|v| = \sqrt{3^2+(-6)^2} = \sqrt{45} = 3\sqrt{5}$ $|r| \, |v| = 60$ try again ...
 March 18th, 2017, 03:10 PM #10 Member   Joined: Feb 2014 Posts: 91 Thanks: 1 this is depressing now.....I really hope there aren't a lot of these on the next test. Please tell me I got it this time.

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