My Math Forum How magnitude of L is calculated here?

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 July 10th, 2019, 09:34 PM #1 Senior Member   Joined: Aug 2014 From: India Posts: 470 Thanks: 1 How magnitude of L is calculated here? When the Sun disappears the Earth heads off in a straight line at constant velocity as shown by the horizontal dashed line, so after some time t it has moved a distance $\displaystyle x = vt$ as I've marked on the diagram. The question is now how the angular momentum can be conserved. The answer is that angular momentum is given by the vector equation: $\displaystyle \mathbf L = \mathbf r \times m\mathbf v$ where $\displaystyle \mathbf r$ is the position vector, $\displaystyle \mathbf v$ is the velocity vector and $\times$ is the cross product. We are going to end up with the vector $\displaystyle \mathbf L$ pointing out of the page and the magnitude of L is given by: $\displaystyle |\mathbf L| = m\,|\mathbf r|\,|\mathbf v|\,\sin\theta \tag{1}$ but looking at our diagram, we see that: $\displaystyle \sin\theta = \large\frac{d}{|\mathbf r|}$ and if we substitute this into our equation (1) for the angular momentum, we get: $\displaystyle |\mathbf L| = m\,|\mathbf r|\,|\mathbf v|\,\large\frac{d}{|\mathbf r|}\normalsize = m\,|\mathbf v|\,d \tag{2}$ And this equation tells us that the angular momentum is constant i.e. it depends only on the constant velocity \mathbf v and the original orbital distance d. I didn't understand how$\displaystyle |\mathbf L| = m\,|\mathbf r|\,|\mathbf v|\,\sin\theta$ is written? Why $\displaystyle \sin$ instead of $\displaystyle \cos$ or [MATH]\tan[/MATH? I only know one formula for angular momentum; $\displaystyle L = mvr$. Last edited by skipjack; July 11th, 2019 at 01:04 AM.
 July 10th, 2019, 11:58 PM #2 Senior Member   Joined: Apr 2014 From: UK Posts: 963 Thanks: 342 Does this help? Angular Momentum Please say "If the sun disappears", saying "when" implies some prophecy in which we all die of extreme cold! Thanks from topsquark
 July 11th, 2019, 07:19 PM #3 Senior Member   Joined: Aug 2014 From: India Posts: 470 Thanks: 1 Why we have to use vectors here? Can't we derive Earth's angular momentum without vectors?
 July 11th, 2019, 07:54 PM #4 Senior Member   Joined: Jun 2019 From: USA Posts: 196 Thanks: 78 Angular momentum is a vector quantity, and is defined using the cross product. In this case, since the motion is in a plane and we therefore know the direction of the momentum, we can solve it without vector maths. Note that, once you have the sine term in your solution, everything else is scalar maths. Without using the cross product, you could derive the angular momentum by using the tangential component of the velocity (times mass times radius). This would give you the same answer (because you're doing the exact same thing, really, just from a geometry and trigonometry point of view instead of a vector point of view). Thanks from topsquark and Ganesh Ujwal
 July 11th, 2019, 10:00 PM #5 Senior Member   Joined: Aug 2014 From: India Posts: 470 Thanks: 1 Using geometry and trigonometry point of view, How does this equation $\displaystyle |\mathbf L| = m\,|\mathbf r|\,|\mathbf v|\,\sin\theta$ look like from given diagram?
July 12th, 2019, 06:17 AM   #6
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See the attached image.

The dot product $\displaystyle d = \vec{a} \cdot \vec{b}$ basically says c is the magnitude of a times the magnitude of the projection of b onto the direction of a (or vice versa).

For the cross product $\displaystyle c = \vec{a} \times \vec{b}$, c will be a vector in the direction normal to the plane containing a and b (if a and b are parallel, no unique plane is defined, but c will be zero). If we pick a direction j such that $\displaystyle (\vec{a}/|\vec{a}|, j, \vec{c}/|\vec{c}|)$ is a right-handed system, then the magnitude of c will be the magnitude of a times the magnitude of the projection of b onto direction j.

Somewhat more classically, you can also visualise the cross product as having the magnitude of the area of a parallelogram defined by a and b (Google or wiki for pictures and more discussion).
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July 12th, 2019, 06:52 PM   #7
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Quote:
 Originally Posted by DarnItJimImAnEngineer See the attached image. The.......... discussion).
I am still seeing vectors in your reply. I want without vectors equations.

 July 13th, 2019, 01:00 PM #8 Senior Member   Joined: Jun 2019 From: USA Posts: 196 Thanks: 78 For motion in the $\displaystyle (r,\theta)$ plane, angular momentum will be in the z-plane, and will have magnitude of radial distance ($\displaystyle r$) times tangential momentum ($\displaystyle v sin(\theta)$ in the given situation, as I showed in the diagram). To answer your question, by looking at the problem, and by knowing how angular momentum is calculated, we can figure it out for simple cases without necessarily using vector symbols and the cross product symbol. We can't completely get rid of the concepts of vectors (direction and orthogonality), since, as I said, angular momentum is a vector quantity, and both its direction and magnitude depend on the directions of the vector quantities involved.

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