My Math Forum  

Go Back   My Math Forum > Science Forums > Physics

Physics Physics Forum


Thanks Tree7Thanks
Reply
 
LinkBack Thread Tools Display Modes
March 18th, 2017, 04:00 PM   #11
Math Team
 
Joined: Jul 2011
From: Texas

Posts: 2,625
Thanks: 1306

$L = m |r|\,|v| \sin{\theta}$

$\theta = \arccos\left(-\dfrac{4}{5}\right) \implies \sin{\theta} = \dfrac{3}{5}$

$|L| = 9 \cdot 4\sqrt{5} \cdot 3\sqrt{5} \cdot \dfrac{3}{5} = 324$

The above calculation gives the magnitude of the angular momentum. To understand the direction for $\vec{L}$, reference the attached graph of the position and velocity vectors on the x-z plane ...

Looking at the graph, the point of view is from the negative side of the y-axis. As $\vec{r}$ rotates CCW toward $\vec{v}$, the direction of $\vec{L}$ is out of the page in the negative direction of the y-axis, hence $\vec{L} = -324 \, kg\, m^2/s$

This is nice thing about using the 3x3 determinant to get the cross-product ... it provides both magnitude & direction in one calculation.

Thanks from topsquark and The_Ys_Guy
skeeter is offline  
 
Reply

  My Math Forum > Science Forums > Physics

Tags
angular, determine, momentum



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Momentum Problem taylor_1989_2012 Physics 2 December 9th, 2015 11:14 AM
Rotation matrices and exponential map for angular momentum lockdown Physics 1 March 1st, 2013 08:41 PM
orbital angular momentum (OAM) of EM wave, is it possible. trapezodial Physics 0 March 2nd, 2012 02:14 PM
Momentum? MathBane Algebra 1 October 9th, 2009 07:28 PM
Momentum johnny Physics 5 February 9th, 2008 05:28 PM





Copyright © 2017 My Math Forum. All rights reserved.