0
$\begingroup$

In The Feynman lectures, under the chapter entitled Vectors, Feynman writes:

enter image description here

enter image description here

My two intimately related questions are: 1) What does he mean by the magnitude of velocity? is he talking about the magnitude of $v_1$, $v_2$ or $\Delta v$?

2)I have struggled a lot on my own, trying to figure out where this equation came from, but I didn't succeed. Can someone explain to me where it did come from?

|cite|improve this question
$\endgroup$
  • 1
    $\begingroup$ I will say it like $\dfrac{\Delta v_\perp}{v_1} = \sin\Delta \theta \approx \Delta\theta$. So, $v_1 = v.$ Feynman here divided the component of the change in velocity in two parts so as to find the centripetal acceleration & tangential acceleration separately; it's quite simple! $\endgroup$ – user36790 Jul 11 '15 at 20:13
1
$\begingroup$

$v$ isn't referring to either $v_1$ or $v_2$, necessarily; $v_1$ is representing the vector before it moves, and $v_2$ is the vector after this movement. If we are working in polar coordinates (the reason he is using $v_\perp$ and $v_\parallel$), then let's suppose this small movement isn't changing the magnitude of the vector, it is just changing the direction (hence why we just care about $v_\perp$).

In very small rotations, we can approximate the distance moved in the rotation simply by the magnitude of the vector multiplied by the angle moved. Hence we get

$$\Delta v_\perp\approx v\Delta\theta$$

In other words, look at figure 11-8. Imagine that the vector $v_\perp$ is very small. Then you will notice that $v_1$ and $v_2$ will converge to the same length, we call this $v$. Then try to see if you can convince yourself that what I said before is true.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ I know that $\lim_{\theta\to 0}sin(\theta)=\theta$, and since the triangle bounded by $v_1$ , $v_2$ and $\Delta v_\perp$ is a right triangle, therefore $\frac{\Delta v_\perp}{v_1}=$ $\lim_{\theta\to 0}sin(\theta)=\theta$. is this what you mean? $\endgroup$ – Omar Nagib Jul 11 '15 at 20:21
  • $\begingroup$ Yes that's a great way to think about it! $\endgroup$ – Alex Mathers Jul 11 '15 at 20:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.