What is the derivation of the second equation for area velocity?

  • Thread starter Thread starter Shreya
  • Start date Start date
  • Tags Tags
    Formula Velocity
AI Thread Summary
The discussion focuses on the derivation of the second equation for area velocity, building on an understanding of the first equation. The area swept out by a particle in time δt is approximated using the cross-product of position and velocity vectors, leading to the first formula. For the second equation, the discussion explores calculating the area in both Cartesian and polar coordinates, emphasizing the relationship between the radius and angular velocity. The participants confirm that the derived expression for the area velocity is correct. The conversation highlights the importance of understanding vector relationships in deriving these equations.
Shreya
Messages
187
Reaction score
64
Homework Statement
I want to understand the second equation for Areal velocity as given below.
Relevant Equations
Refer below.
Screenshot from 2023-01-19 20-29-43.png
Screenshot from 2023-01-19 20-30-05.png

I think I understand how the first equation comes about.

Screenshot from 2023-01-19 20-43-16.png

In ##dt## the particle travels by ##dr##, I considered it as a triangle with altitude ##r## and base ##dr##. On dividing the area travelled in ##dt## by ##dt## we get the above equation.
A similar argument can be applied to ##\frac 1 2 \rho^2 \frac {d\phi} {dt}## as ##\rho## is same as r and ##\rho \frac {d\phi} {dt}## is same as ##\frac {dr} {dt}##

But, I am not able to understand the 2nd equation. I can provide a similar argument for ##xv_y## but can,t seem to reason any further.
Please be kind to help.
 

Attachments

  • Screenshot from 2023-01-19 20-39-16.png
    Screenshot from 2023-01-19 20-39-16.png
    3.8 KB · Views: 133
Physics news on Phys.org
The unsigned area swept out in time \delta t is approximately |\delta S| \approx \tfrac12 <br /> \|\mathbf{r} \times (\mathbf{r} + \mathbf{v}\delta t)\| = <br /> \tfrac12 \|\mathbf{r} \times \mathbf{v}\| |\delta t|. This is because the magnitude of the cross-product gives the area of the parallelogram bounded by \mathbf{r} and \mathbf{r} + \mathbf{v} \delta t, and the area swept out is approximately the area of a triangle which is one half of the parallelogram. Dividing by \delta t and taking the limit \delta t \to 0 gives <br /> \left|\frac{dS}{dt}\right| = \tfrac12 \|\mathbf{r} \times \mathbf{v}\|. This is essentially the derivation of the first formula.

To derive the second, in the case of motion in the (x,y,0) plane we can calculate \frac12\|\mathbf{r} \times \mathbf{v}\| either in cartesians or in plane polars using the standard formulae \begin{split}<br /> \mathbf{r} &amp;= r\mathbf{e}_r(\theta) \\<br /> \mathbf{v} &amp;= \dot r \mathbf{e}_r(\theta) + r\dot \theta \mathbf{e}_{\theta}(\theta) \end{split} where <br /> \mathbf{e}_r(\theta) \times \mathbf{e}_\theta(\theta) \equiv \mathbf{e}_z. We can then adopt the convention that S increases in the direction of increasing \theta to remove the absolute value signs.
 
pasmith said:
To derive the second, in the case of motion in the (x,y,0) plane we can calculate 12‖r×v‖ either in cartesians or in plane polars using the standard formulae r=rer(θ)v=r˙er(θ)+rθ˙eθ(θ) where er(θ)×eθ(θ)≡ez. We can then adopt the convention that S increases in the direction of increasing θ to remove the absolute value signs.
Thank You so much @pasmith !
##\vec r = x \hat i + y \hat j##
##\vec v = v_x \hat i + v_y \hat j##

##\vec r \times \vec v = (x v_y - y v_x) \hat k##
##\frac {dS} {dt} = \frac 1 2 (xv_y-yv_x)##
Is the above right?
 
Yes, the above is right
 
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Correct statement about a reservoir with an outlet pipe'
The answer to this question is statements (ii) and (iv) are correct. (i) This is FALSE because the speed of water in the tap is greater than speed at the water surface (ii) I don't even understand this statement. What does the "seal" part have to do with water flowing out? Won't the water still flow out through the tap until the tank is empty whether the reservoir is sealed or not? (iii) In my opinion, this statement would be correct. Increasing the gravitational potential energy of the...
Back
Top