Velocity Vector in Polar Coordinates (Kleppner p.30)

In summary, K&K say that the right term for the component of velocity directed radially outward is the component in the tangential direction. They prove this is so by proving that the quantity \frac{d\hat{r}}{dt} is in the tangential direction. This argument leads to the conclusion that \hat{r} vanishes in the limit.
  • #1
Von Neumann
101
4
In polar coordinates we have [itex] \vec{r} = r \hat{r} [/itex] [itex]\Rightarrow \vec{v} = \frac{d}{dt}({r \hat{r}}) = \dot{r}\hat{r} + r \frac{d \hat{r}}{dt} [/itex].

In the book Introduction to Mechanics, K & K says the right term is the component of velocity directed radially outward. (Surely a typo, as the left term is the velocity component associated with the direction [itex] \hat{r} [/itex].) Then he goes on to say it's a good guess that the other term is the component in the tangential [itex] \left( \hat{\theta} \right) [/itex] direction. He proves this is so in 3 ways; namely by proving [itex] \frac{d\hat{r}}{dt} [/itex] is in the [itex] \hat{\theta} [/itex] direction). The first two ways I understand - It's the third one I'm stuck on.

He starts by drawing two position vectors [itex] \vec{r} [/itex] and [itex] \vec{r} + \Delta \vec{r} [/itex] at the respective times [itex] t [/itex] and [itex] t + \Delta t [/itex], along with their respective unit vectors [itex] \hat{r}_{1} [/itex] , [itex] \hat{r}_{2} [/itex] , [itex] \hat{\theta}_{1} [/itex] , and [itex] \hat{\theta}_{2} [/itex]. From the geometry, we see that [itex] \Delta \hat{r} = \hat{\theta}_{1}sin\Delta\theta - \hat{r}_{1}(1-cos\Delta\theta) [/itex], where [itex] \Delta\theta [/itex] is the angle between the two position vectors).

From this we see that [itex] \frac{\Delta\hat{r}}{\Delta t} = \hat{\theta}_{1}\frac{sin\Delta\theta}{\Delta t} - \hat{r}_{1}\frac{(1-cos\Delta\theta)}{\Delta t} = \hat{\theta}_{1} \left( \frac{\Delta\theta - \frac{1}{6}(\Delta\theta)^3+\cdots}{\Delta t} \right) - \hat{r}_{1} \left( \frac{\frac{1}{2}(\Delta\theta)^2 - \frac{1}{24}(\Delta\theta)^4+\cdots}{\Delta t} \right)[/itex]. Almost there, just need to take the limit of this quantity as [itex] \Delta t [/itex] tends to 0. So we need to evaluate [itex] \frac{d \hat{r}}{dt} = \displaystyle \lim_{\Delta t \to 0} \frac{\Delta\hat{r}}{\Delta t}[/itex].

He make the following argument which concludes the proof:

"In the limit [itex] \Delta t \to 0 [/itex], [itex]\Delta\theta [/itex] approaches zero, but [itex]\Delta\theta/\Delta t[/itex] approaches the limit [itex]d\theta/dt[/itex]. Therefore, [itex] \displaystyle \lim_{\Delta t \to 0} \frac{\Delta\theta}{\Delta t}(\Delta\theta)^n [/itex] for [itex] n>0 [/itex]. The term in [itex] \hat{r} [/itex] entirely vanishes in the limit and we are left with [itex] \frac{d \hat{r}}{dt}= \dot{\theta} \hat{\theta} [/itex]."

I understand that [itex] \Delta\theta/\Delta t[/itex] approaches [itex]d\theta/dt[/itex] as [itex] \Delta t \to 0 [/itex], but I'm lost after that. How does one come to the conclusion that [itex] \displaystyle \lim_{\Delta t \to 0} \frac{\Delta\theta}{\Delta t}(\Delta\theta)^n [/itex] for [itex] n>0 [/itex]? Then, how does this lead us to the conclusion that [itex] \hat{r} [/itex] entirely vanishes in the limit?

I've been trying my hardest to work through this text, but I tend to get snagged for quite some time on explanations like the above. Usually I can fill in the missing steps myself. I feel as though I cannot thoroughly penetrate this textbook as I have others.

Thanks in advance for any guidance.
 
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  • #2
Von Neumann said:
How does one come to the conclusion that [itex] \displaystyle \lim_{\Delta t \to 0} \frac{\Delta\theta}{\Delta t}(\Delta\theta)^n [/itex] = 0 for [itex] n>0 [/itex]?

Maybe you can make use of the product law for limits if you think of ##\frac{\Delta \theta}{\Delta t}## and ##(\Delta \theta)^n## as functions of ##\Delta t##.
 
  • #3
Well, since [itex] \displaystyle \lim_{x \to a} f(x)g(x)= \lim_{x \to a}f(x) \cdot \lim_{x \to a}g(x) [/itex] then [itex]\Rightarrow \displaystyle \lim_{\Delta t \to 0} \frac{\Delta\theta}{\Delta t}(\Delta\theta)^n = \lim_{\Delta t \to 0}\frac{\Delta\theta}{\Delta t} \cdot \lim_{\Delta t \to 0} (\Delta\theta)^n = \lim_{\Delta t \to 0}\frac{\Delta\theta}{\Delta t} \cdot 0 = 0[/itex].

How does this help though?
 
  • #4
Von Neumann said:
Well, since [itex] \displaystyle \lim_{x \to a} f(x)g(x)= \lim_{x \to a}f(x) \cdot \lim_{x \to a}g(x) [/itex] then [itex]\Rightarrow \displaystyle \lim_{\Delta t \to 0} \frac{\Delta\theta}{\Delta t}(\Delta\theta)^n = \lim_{\Delta t \to 0}\frac{\Delta\theta}{\Delta t} \cdot \lim_{\Delta t \to 0} (\Delta\theta)^n = \lim_{\Delta t \to 0}\frac{\Delta\theta}{\Delta t} \cdot 0 = 0[/itex].

How does this help though?

I hope I'm not misunderstanding your original question. I am assuming you want to show that ##\displaystyle \lim_{\Delta t \to 0} \frac{\Delta\theta}{\Delta t}(\Delta\theta)^n## equals zero. If so, isn't that what you have essentially shown?
 
  • #5
The goal is to prove that [itex] \frac{d \hat{r}}{dt} = \dot{\theta}\hat{\theta} [/itex].

I do now understand why the previous limit holds. However, why is it relevant to proving that [itex] \frac{d \hat{r}}{dt} = \dot{\theta}\hat{\theta} [/itex]?

Since a limit of products is identical to the product of the separated limits, any limit containing a factor of [itex] (\Delta\theta)^n [/itex] will result in 0. Is this why [itex] \hat{r} [/itex] vanishes? We can factor out a [itex] (\theta)^2 [/itex] from the second term of [itex] \frac{\Delta\hat{r}}{\Delta t} [/itex]. And, since [itex] n=2,>0 [/itex], then the limit is 0. Right?

But this reasoning suggests that we can factor out a [itex] (\theta) [/itex] from the first term of [itex] \frac{\Delta\hat{r}}{\Delta t} [/itex]. And since [itex] n=1,>0 [/itex] then the limit is 0 for this one as well. Which is certainly not correct, since [itex] \hat{\theta} [/itex] is not supposed to vanish.

EDIT: Hold on! I figured it out, I just need to type out all the details.
 
Last edited:
  • #6
OK. Good.
 
  • #7
I think I got it:

[tex]\begin{align*}

\frac{d\hat{r}}{dt} &= \displaystyle \lim_{\Delta t \to 0} \left( \frac{\Delta\hat{r}}{\Delta t} \right) \\

&= \lim_{\Delta t \to 0} \left [ \hat{\theta}_{1} \left( \frac{\Delta\theta - \frac{1}{6}(\Delta\theta)^3+\cdots}{\Delta t} \right) - \hat{r}_{1} \left( \frac{\frac{1}{2}(\Delta\theta)^2 - \frac{1}{24}(\Delta\theta)^4+\cdots}{\Delta t} \right) \right] \\

& = \lim_{\Delta t \to 0} \hat{\theta}_{1} \cdot \left( \lim_{\Delta t \to 0} \frac{\Delta\theta}{\Delta t} - \frac{1}{6} \lim_{\Delta t \to 0} \frac{(\Delta\theta)^3}{\Delta t} + \cdots \right) - \lim_{\Delta t \to 0} \hat{r}_{1} \cdot \left( \frac{1}{2} \lim_{\Delta t \to 0} \frac{(\Delta\theta)^2}{\Delta t} - \frac{1}{24} \lim_{\Delta t \to 0} \frac{(\Delta\theta)^4}{\Delta t} + \cdots \right) \\

& = \lim_{\Delta t \to 0} \hat{\theta}_{1} \cdot \left[ \dot{\theta} - \frac{1}{6} \lim_{\Delta t \to 0} \left( \frac{\Delta\theta}{\Delta t} \cdot (\Delta\theta)^2 \right) + \cdots \right] - \lim_{\Delta t \to 0} \hat{r}_{1} \cdot \left[ \frac{1}{2} \lim_{\Delta t \to 0} \left ( \frac{\Delta\theta}{\Delta t} \cdot \Delta\theta \right) - \frac{1}{24} \lim_{\Delta t \to 0} \left( \frac{\Delta\theta}{\Delta t} \cdot (\Delta\theta)^3 \right) + \cdots \right] \\

& = \dot{\theta} \cdot \lim_{\Delta t \to 0} \hat{\theta}_{1} - \lim_{\Delta t \to 0} \hat{r}_{1} \cdot 0 \ \left( \text{since} \displaystyle \lim_{\Delta t \to 0} \frac{\Delta\theta}{\Delta t}(\Delta\theta)^n \ \text{for} \ n>0 \right) \\

&= \dot{\theta} \hat{\theta}

\end{align*}[/tex]

Correct?
 
  • #8
Von Neumann said:
I think I got it:

[tex]\begin{align*}

\frac{d\hat{r}}{dt} &= \displaystyle \lim_{\Delta t \to 0} \left( \frac{\Delta\hat{r}}{\Delta t} \right) \\

&= \lim_{\Delta t \to 0} \left [ \hat{\theta}_{1} \left( \frac{\Delta\theta - \frac{1}{6}(\Delta\theta)^3+\cdots}{\Delta t} \right) - \hat{r}_{1} \left( \frac{\frac{1}{2}(\Delta\theta)^2 - \frac{1}{24}(\Delta\theta)^4+\cdots}{\Delta t} \right) \right] \\

& = \lim_{\Delta t \to 0} \hat{\theta}_{1} \cdot \left( \lim_{\Delta t \to 0} \frac{\Delta\theta}{\Delta t} - \frac{1}{6} \lim_{\Delta t \to 0} \frac{(\Delta\theta)^3}{\Delta t} + \cdots \right) - \lim_{\Delta t \to 0} \hat{r}_{1} \cdot \left( \frac{1}{2} \lim_{\Delta t \to 0} \frac{(\Delta\theta)^2}{\Delta t} - \frac{1}{24} \lim_{\Delta t \to 0} \frac{(\Delta\theta)^4}{\Delta t} + \cdots \right) \\

& = \lim_{\Delta t \to 0} \hat{\theta}_{1} \cdot \left[ \dot{\theta} - \frac{1}{6} \lim_{\Delta t \to 0} \left( \frac{\Delta\theta}{\Delta t} \cdot (\Delta\theta)^2 \right) + \cdots \right] - \lim_{\Delta t \to 0} \hat{r}_{1} \cdot \left[ \frac{1}{2} \lim_{\Delta t \to 0} \left ( \frac{\Delta\theta}{\Delta t} \cdot \Delta\theta \right) - \frac{1}{24} \lim_{\Delta t \to 0} \left( \frac{\Delta\theta}{\Delta t} \cdot (\Delta\theta)^3 \right) + \cdots \right] \\

& = \dot{\theta} \cdot \lim_{\Delta t \to 0} \hat{\theta}_{1} - \lim_{\Delta t \to 0} \hat{r}_{1} \cdot 0 \ \left( \text{since} \displaystyle \lim_{\Delta t \to 0} \frac{\Delta\theta}{\Delta t}(\Delta\theta)^n \ \text{for} \ n>0 \right) \\

&= \dot{\theta} \hat{\theta}

\end{align*}[/tex]

Correct?

Yes, that looks OK. In going from the second to third line you don't really need to take the limits of ##\hat{\theta}_1## and ##\hat{r}_1## since they are fixed vectors that don't depend on ##\Delta t##. But, what you did is fine.
 
  • #9
TSny said:
Yes, that looks OK. In going from the second to third line you don't really need to take the limits of ##\hat{\theta}_1## and ##\hat{r}_1## since they are fixed vectors that don't depend on ##\Delta t##. But, what you did is fine.

I meant it implicitly in going from the 5th to the 6th line. Thanks TSny!
 

1. What is the velocity vector in polar coordinates?

The velocity vector in polar coordinates represents the speed and direction of an object in a two-dimensional space. It is defined by two components: magnitude (speed) and direction (angle).

2. How is the velocity vector calculated in polar coordinates?

The magnitude of the velocity vector in polar coordinates is calculated using the Pythagorean theorem, while the direction is determined by the inverse tangent function. The resulting vector is represented as (r, θ), where r is the magnitude and θ is the direction.

3. How is the velocity vector different from the displacement vector in polar coordinates?

The velocity vector in polar coordinates represents the rate of change of an object's position, while the displacement vector represents the change in position itself. The velocity vector takes into account the direction of motion, while the displacement vector only considers the distance between the initial and final positions.

4. Can the velocity vector in polar coordinates be negative?

Yes, the velocity vector in polar coordinates can have negative components. This indicates that the object is moving in the opposite direction of the positive angle (counterclockwise) or radius (outward) in the coordinate system.

5. How is the velocity vector in polar coordinates converted to Cartesian coordinates?

The velocity vector in polar coordinates can be converted to Cartesian coordinates using the following equations:
vx = rcos(θ)
vy = rsin(θ)
Where vx and vy are the Cartesian components of the velocity vector, r is the magnitude, and θ is the direction in polar coordinates.

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