Why does the integral of radial acceleration not yield radial velocity?

Question

I was reading chapter 1 of Kleppner and Kolenkow's An Introduction to Mechanics when I came across this problem:

I am not sure if I understand this idea about radial acceleration and velocity correctly. I interpreted the last sentence as saying that the equation is not true because the direction of the unit vectors (i.e. the direction that radial refers to) changes with time. I also read this as meaning that you can only integrate the components of acceleration to get components of velocity or integrate components of velocity to get displacement components if those components always refer to the same direction (like how $\hat i$, $\hat j$ and $\hat k$ each refer to a constant direction). Is this understanding correct? If it is not, could you please explain why this equation is not true?

Answer 1

A simple example why acceleration may occur in a moving reference frame, even if the velocity in this frame is constant, is centripetal acceleration. Imagine that you drive with your car at constant linear velocity, and then you get into a bend in the road. You will feel that you get pressed laterally into your seat, which is a consequence of you being accelerated.

How can that be, since you are moving at constant tangential velocity (speedometer reading is constant) and the distance from the curve center is also constant (i.e. zero/constant radial velocity, because you are moving on a circle)? It is the consequence of acceleration in a moving reference frame being composed of the derivative of the coordinates in that system and the derivative of the system (i.e. its unit vectors) itself.

Think of it as a manifestation of the product rule of differentiation: inertial coordinates are the product of moving coordinates and moving unit vectors. Hence, inertial acceleration has a component coming from the derivative of the moving coordinates, and a component coming from the derivative of the moving unit vectors of the moving coordinate system.

Answer 2

The problem with polar coordinates is that the basis induced by the 'polar grid' is 'local'. This means that what the $\hat{r}$ and $\hat{\theta}$ depends on what point you are in space. Let us contrast this with cartesian coordinates where the velocity can be written as:

$$ \vec{v} = v_x \hat{i} + v_y \hat{j}$$

On computing the acceleration:

$$ \frac{d}{dt} \vec{v} = \hat{i} \frac{d}{dt} v_x + v_x \frac{d}{dt} \hat{i} + \hat{j} \frac{d}{dt} v_y + v_y \frac{d}{dt} \hat{j}$$

Now the important thing here is that if you are not in a rotating frame then the derivative of basis is zero. It means that the $\hat{i}$ and $\hat{j}$ at one point is the same as that in the other point.

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In the above, I summarized the issue. Now let's remedy the doubt with explicit calculations, we first find the derivative of polar basis and then move to figuring out what should be expression to find acceleration.

Now, let's compare this to the case of the radial basis vector $\hat{r}$, as I said before it's a function of $r$ and $\theta$. Upon differentiating with time, we get:

$$ \frac{d}{dt} \hat{r}(r,\theta) = \frac{\partial \hat{r} }{\partial r} \frac{dr}{d t} + \frac{ \partial \hat{r} }{\partial \theta} \frac{ d \theta}{dt}$$

Now, by some geometric calculations (I'll add a reference to this later), we find that:

$$ \frac{\partial\hat{r} }{\partial r} = 0$$ $$ \frac{ \partial \hat{\theta} }{\partial \theta} = \frac{1}{r}\hat{\theta}$$

Hence,

$$ \frac{d}{dt} \hat{r} = \frac{1}{r} \hat{\theta} \frac{d \theta}{dt}$$

Similarly, we can find:

$$ \frac{d}{dt} \hat{\theta} = - \left(\frac{d \theta}{dt} \right) \hat{r}$$

Due to this , we find that rate of change of the $\theta$ basis is controlled by the basis in the $\hat{r}$ direction and vice versa. Therefore simply taking the acceleration component in the radial direction and integrating will not give the radial velocity.

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Special cases?

If $ \frac{d \theta}{dt}=0$, then sure the idea of $\frac{dv_r}{dt} = a_r$ is indeed correct.

Answer 3

Lets write the position vector $\vec{R}$ with polar coordinates \begin{align*} &\vec{R}=r(t)\,\begin{bmatrix} \cos(\varphi(t)) \\ \sin(\varphi(t)) \\ \end{bmatrix}=r(t)\,\vec{e}_r(t)\\\\ &\text{the velocity is}\\\\ &\vec{\dot{R}}= \dot r\,\vec{e}_r+r\,\vec{\dot{e}}_r=v_r\,\vec e_r+r\,\dot{\varphi}\,\vec{e}_\varphi\\\\ &\text{and the acceleration}\\\\ &\vec{\ddot{R}}=\underbrace{\left(\dot{v}_r-r\,\dot{\varphi}^2\right)}_{a_r}\,\vec{e}_r+ \left(r\,\ddot{\varphi}+2\,\dot{r}\,\dot{\varphi}\right)\,\vec{e}_\varphi\\ &\Rightarrow\\ &a_r\ne \frac{dv_r}{dt} \end{align*} only in case that $~\varphi(t)=~$ const ,you obtain \begin{align*} &a_r=\frac{dv_r}{dt} \end{align*}