Radial Acceleration in Polar/Cylindrical Coordinates

[BrandonUSC]

My question is why isn't the radial component e^→_r of acceleration in cylindrical coords simply r'' ?
If r'' is the rate at which the rate of change of position is changing in the radial direction, wouldn't that make it the radial acceleration? I.e, the acceleration of the radius is the acceleration in the radial direction, no?

 

[BvU]

Hi Brandon,

If you can say it isn't, you can provide the other contribution(s). Which of them don't you understand ?

 

[BrandonUSC]

Hi Brandon,

If you can say it isn't, you can provide the other contribution(s). Which of them don't you understand ?

I'm basing this off the equation for acceleration in cylindrical coords: a_r = r''-r*θ'²

 

[Chestermiller]

I'm basing this off the equation for acceleration in cylindrical coords: a_r = r''-r*θ'²

You are aware that acceleration is the rate of change of the velocity vector with respect to time, right? And vectors have both magnitude and direction, right? And, even if the magnitude of a vector is constant, its direction can be changing, and this contributes to the acceleration, correct?

 

[BrandonUSC]

You are aware that acceleration is the rate of change of the velocity vector with respect to time, right? And vectors have both magnitude and direction, right? And, even if the magnitude of a vector is constant, its direction can be changing, and this contributes to the acceleration, correct?

Ignoring your pedantic and condescending tone, yes I am aware of these facts. As you can see in my post if you actually read it, I refer to r'' as the rate at which the rate is changing. In cylindrical coords the velocity vector along the radius is always in the r direction. If you have nothing to add please go away.

 

[Chestermiller]

Ignoring your pedantic and condescending tone, yes I am aware of these facts. As you can see in my post if you actually read it, I refer to r'' as the rate at which the rate is changing. In cylindrical coords the velocity vector along the radius is always in the r direction. If you have nothing to add please go away.

If you understand all this, then it isn't clear (at least to me) why you think that the acceleration in the radial direction should only be r''. Can you please explain in more detail? And please forgive me for my earlier insulting tone. I promise to be more respectful.

 

[BrandonUSC]

If you understand all this, then it isn't clear (at least to me) why you think that the acceleration in the radial direction should only be r''. Can you please explain in more detail? And please forgive me for my earlier insulting tone. I promise to be more respectful.

Okay here's my thought process:
The radial position is r.
The second time derivative of position is acceleration.
Then if the second time derivative of the radial position is r''
Then shouldn't the acceleration of the radial component of position be r''?

 

[Chestermiller]

In cylindrical coordinates, the position vector from the origin to an arbitrary location is given by:
$$\mathbf{r}=r\mathbf{i_r}$$where $\mathbf{i_r}(\theta)$ is the unit vector in the radial direction (which depends on $\theta$). So the velocity of the particle is $$\mathbf{v}=\frac{d\mathbf{r}}{dt}=\frac{dr}{dt}\mathbf{i_r}+r\frac{d\mathbf{i_r}}{d\theta}\frac{d\theta}{dt}$$But, kinematically, $$\frac{di_r}{d\theta}=\mathbf{i_{\theta}}$$Therefore, $$\mathbf{v}=\frac{dr}{dt}\mathbf{i_r}+r\frac{d\theta}{dt}\mathbf{i_{\theta}}$$
Now, assuming that you know that $$\frac{d\mathbf{i_{\theta}}}{d\theta}=-\mathbf{i_r}$$what do you get if you differentiate the velocity vector with respect to time.

 

[BrandonUSC]

In cylindrical coordinates, the position vector from the origin to an arbitrary location is given by:
$$\mathbf{r}=r\mathbf{i_r}$$where $\mathbf{i_r}(\theta)$ is the unit vector in the radial direction (which depends on $\theta$). So the velocity of the particle is $$\mathbf{v}=\frac{d\mathbf{r}}{dt}=\frac{dr}{dt}\mathbf{i_r}+r\frac{d\mathbf{i_r}}{d\theta}\frac{d\theta}{dt}$$But, kinematically, $$\frac{di_r}{d\theta}=\mathbf{i_{\theta}}$$Therefore, $$\mathbf{v}=\frac{dr}{dt}\mathbf{i_r}+r\frac{d\theta}{dt}\mathbf{i_{\theta}}$$
Now, assuming that you know that $$\frac{d\mathbf{i_{\theta}}}{d\theta}=-\mathbf{i_r}$$what do you get if you differentiate the velocity vector with respect to time.

I appreciate your effort in this answer, but the problem is I can follow the derivations, however I don't understand it intuitively.
If I'm sitting on a large rotating rod, and the end of the rod is changing in length, my perception of this change in length is independent of the rate at which we are rotating, right? So the acceleration at which it moves in my frame of reference is r''. This direction is also the radial direction I believe. So then it makes sense to me that the radial acceleration

a_r = r''​

I KNOW this isn't true, I can follow the math, but it doesn't 'click'

 

[Chestermiller]

I appreciate your effort in this answer, but the problem is I can follow the derivations, however I don't understand it intuitively.
If I'm sitting on a large rotating rod, and the end of the rod is changing in length, my perception of this change in length is independent of the rate at which we are rotating, right? So the acceleration at which it moves in my frame of reference is r''. This direction is also the radial direction I believe. So then it makes sense to me that the radial acceleration

a_r = r''​

I KNOW this isn't true, I can follow the math, but it doesn't 'click'

When you travel in a circle in your car, does the car seat have to exert a force on you in the radial direction (perpendicular to the direction that the car is traveling) to keep you moving in a circle? If so, then you are accelerating in the radial direction. Is your radial distance from the center of the circle changing with time? If not, then the 2nd derivative of the radial distance with time is also zero.

 

[BrandonUSC]

okay when you say it that way I understand, I was confused by the fact that r is a 1dimensional measurement while radial direction is changing.
Thanks