Law of conservation of angular momentum

In summary: Well for initial momentum:sin(θ) = b/rI guess for the final momentum this doesn't change. On inspection of the figure, I conclude that it doesn't change. Thus b = rsin(θ).Thus L = m*(dΦ/dt) x r = m*(dΦ/dt)*b*sin(θ) = m*(dΦ/dt)*b^2/r?Yes, that is correct.
  • #1
Boltzman Oscillation
233
26
Given the figure, how can i arrive to this formula knowing that angular momentum is conserved?

vectorandfigure.png


I know that p = mv and L = p x r. So the initial momentum will be L1 = mV x R and the final momentum will be L2 = mv x r.

I am not sure how R will equal to b since the distance between the initial position of the electron is clearly not b distance apart from the scatterer. I am also not sure how to modify the final angular momentum to fit the formula.
 

Attachments

  • vectorandfigure.png
    vectorandfigure.png
    10.2 KB · Views: 1,653
Physics news on Phys.org
  • #2
Boltzman Oscillation said:
I know that p = mv and L = p x r. So the initial momentum will be L1 = mV x R and the final momentum will be L2 = mv x r.

I am not sure how R will equal to b since the distance between the initial position of the electron is clearly not b ...
Your b is the magnitude of the r vector component, that is perpendicular to p (or v), before and after the scattering. Do you understand the cross product?
https://en.wikipedia.org/wiki/Cross_product
 
Last edited:
  • #3
A.T. said:
Your b is the magnitude of the r vector component, that is perpendicular to p (or v), before and after the scattering. Do you understand the cross product?
https://en.wikipedia.org/wiki/Cross_product
yes I understand the cross product but I've not had enough practice with it (my math foundations class was a mess). Isnt the magnitude how large the arrow is? In that case then the magnitude b wouldn't be the same as the magnitude of vector r would it? r stretches and shortens throughout the path of movement.
 
  • #4
Boltzman Oscillation said:
In that case then the magnitude b wouldn't be the same as the magnitude of vector r would it?
Look at your picture. Do they look the same?

Boltzman Oscillation said:
r stretches and shortens throughout the path of movement.
Yes, but b is just one component of r, before and after scattering, not throughout.
 
  • #5
A.T. said:
Look at your picture. Do they look the same?Yes, but b is just one component of r, before and after scattering, not throughout.
ohh so I am only looking at before and after the scattering and not throughout. Hmm so in that case r = b in both cases. Now the velocity of the particle after the scattering can be described by position/time. Well position will be given by the angle. The derivative of position in respect to time is position right? So the velocity will equal d(angle)/dt. The final angular momentum will then equal:

L = m * d(angle)/dt x b. = m*(dΦ/dt)*bsin(90) = m*(dΦ/dt)?

Ugh why dint I learn vector calculus correctly? :(
I guess its best to take my time to learn it now so it won't impede me later.
 
  • #6
A.T. said:
b is just one component of r
Boltzman Oscillation said:
r = b
No, see above.
 
  • #7
A.T. said:
No, see above.
Okay so L = m*(dΦ/dt) x r = m*(dΦ/dt)*b*sin(angle between the two) ?
 
  • #8
Boltzman Oscillation said:
Okay so L = m*(dΦ/dt) x r = m*(dΦ/dt)*b*sin(angle between the two) ?
Which is greater, b or |r|?
 
  • #9
A.T. said:
Which is greater, b or |r|?
err I thought b was the magnitude of r? so arent they the same in errrr greatness?
 
  • #10
Boltzman Oscillation said:
err I thought b was the magnitude of r? so arent they the same in errrr greatness?
sigh, on the initial momentum then b will be the yth component of r while in the final momentum r will be greater than b.
 
  • #11
Boltzman Oscillation said:
sigh, on the initial momentum then b will be the yth component of r while in the final momentum r will be greater than b.
So what is the mathematical relation between b and r?
 
  • #12
A.T. said:
So what is the mathematical relation between b and r?
Well for initial momentum:

sin(θ) = b/r

I guess for the final momentum this doesn't change. On inspection of the figure, I conclude that it doesn't change. Thus b = rsin(θ).
Thus L = m*(dΦ/dt) x r = m*(dΦ/dt)*b*sin(θ) = m*(dΦ/dt)*b^2/r?
 

1. What is the Law of Conservation of Angular Momentum?

The Law of Conservation of Angular Momentum states that the total angular momentum of a system remains constant unless an external torque is applied. This means that the rotational motion of a system will remain constant unless acted upon by an outside force.

2. How does the Law of Conservation of Angular Momentum apply to everyday life?

This law can be observed in many everyday situations, such as when a figure skater spins faster by pulling their arms closer to their body or when a spinning top stays upright due to its angular momentum. It also applies to the motion of planets in our solar system, as their angular momentum keeps them in their orbits.

3. What is the difference between angular momentum and linear momentum?

Angular momentum is a measure of the rotational motion of an object, while linear momentum is a measure of the translational motion of an object. Angular momentum takes into account both the mass and the distribution of mass in an object, whereas linear momentum only considers the mass and velocity of an object.

4. Can angular momentum be created or destroyed?

No, according to the Law of Conservation of Angular Momentum, angular momentum cannot be created or destroyed. It can only be transferred from one object to another or changed in direction by an external torque.

5. How is the Law of Conservation of Angular Momentum related to Newton's First Law of Motion?

The Law of Conservation of Angular Momentum is essentially a rotational version of Newton's First Law of Motion, also known as the Law of Inertia. Both laws state that an object will maintain its state of motion unless acted upon by an external force. In the case of angular momentum, this applies to the rotational motion of an object.

Similar threads

Replies
3
Views
1K
Replies
4
Views
1K
Replies
9
Views
3K
  • Mechanics
Replies
2
Views
815
Replies
9
Views
1K
Replies
30
Views
1K
  • Mechanics
2
Replies
53
Views
2K
Replies
36
Views
14K
Back
Top