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 March 16th, 2017, 05:29 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, 06:08 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, 06:18 PM #3 Math Team   Joined: Jul 2011 From: Texas Posts: 3,094 Thanks: 1677 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 06:50 PM. March 17th, 2017, 10:20 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, 02:11 PM #5 Math Team   Joined: Jul 2011 From: Texas Posts: 3,094 Thanks: 1677 $\vec{r} \times m\vec{v}=9\begin{vmatrix} i & j & k\\ -8& 0& 4\\ 3 & 0 & -6 \end{vmatrix}=9\bigg[0\vec{i}-(48-12)\vec{j}+0\vec{k}\bigg]=-324\vec{j}$ Thanks from topsquark and The_Ys_Guy March 18th, 2017, 10:40 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, 11:01 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, 11:21 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, 12:25 PM #9 Math Team   Joined: Jul 2011 From: Texas Posts: 3,094 Thanks: 1677 $|r| = \sqrt{(-8 )^2+4^2} = \sqrt{80} = 4\sqrt{5}$ $|v| = \sqrt{3^2+(-6)^2} = \sqrt{45} = 3\sqrt{5}$ $|r| \, |v| = 60$ try again ... March 18th, 2017, 04:10 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.   Tags angular, determine, momentum Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post taylor_1989_2012 Physics 2 December 9th, 2015 12:14 PM lockdown Physics 1 March 1st, 2013 09:41 PM trapezodial Physics 0 March 2nd, 2012 03:14 PM MathBane Algebra 1 October 9th, 2009 08:28 PM johnny Physics 5 February 9th, 2008 06:28 PM

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