Radial and tangential components of acceleration

[fog37]

Hello,
In 2D kinematics, the acceleration vector $a(t)$ can be expressed either in Cartesian coordinates $a_x$ and $a_y$ or in polar coordinates $r$ an $\theta$. It depends on the problem.
But it is also possible to express the acceleration $a(t)$ in the so called "intrinsic coordinates" using the intrinsic unit vectors $T$ (the tangential vector), $N$ (the normal vector).

The unit vectors $T$ and $N$ form an orthonormal basis. Each vector changes direction as the object travels its trajectory since the vectors from a moving basis. The component of the acceleration along the $N$ vector is called the centripetal component while the component along the $T$ vector is the tangential acceleration, correct?

Can we talk about tangential and centripetal acceleration components even when we express the acceleration vector in polar coordinates? I don't think so since the radial unit vector is not always perpendicular to the trajectory curve. Just checking...

What are the benefits of intrinsic coordinates?

thanks.

 

[Chestermiller]

I think this can be done with cylindrical coordinates. In cylindrical coordinates, the velocity vector is given by $$\mathbf{v}=\frac{dr}{dt}\mathbf{i_r}+r\frac{d\theta}{dt}\mathbf{i_{\theta}}$$From this it follows that the unit vector in the tangential direction is
$$\mathbf{t}=\frac{\mathbf{v}}{|\mathbf{v}|}=\frac{\frac{dr}{dt}\mathbf{i_r}+r\frac{d\theta}{dt}\mathbf{i_{\theta}}}{\sqrt{\left(\frac{dr}{dt}\right)^2+\left(r\frac{d\theta}{dt}\right)^2}}$$and the unit vector in the normal direction is:$$\mathbf{n}=\frac{-r\frac{d\theta}{dt}\mathbf{i_r}+\frac{dr}{dt}\mathbf{i_{\theta}}}{\sqrt{\left(\frac{dr}{dt}\right)^2+\left(r\frac{d\theta}{dt}\right)^2}}$$In addition, the acceleration vector is equal to the time derivative of the velocity vector: $$\mathbf{a}=\left[\frac{d^2r}{dt^2}-r\left(\frac{d\theta }{dt}\right)^2\right]\mathbf{i_r}+\left[2\frac{dr}{dt}\frac{d\theta}{dt}+r\frac{d^2\theta}{dt^2}\right]\mathbf{i_{\theta}}$$
The tangential component of the acceleration is obtained by dotting the acceleration vector with the unit vector in the tangential direction. What do you get when you do this? The centripetal component of the acceleration is obtained by dotting the acceleration vector with the unit vector in the normal direction. What do you get when you do this?

 

[Chestermiller]

Did you mean polar coordinates? Cylindrical coordinates would include a Z term (r, θ, z). I'll delete this post later.

Yes. I usually use the terms interchangeably.

 

[fog37]

In 2D, polar and cylindrical are indeed the same unit basis vectors.

I think the decomposition of the acceleration vector $a(t)$ into a tangential component (locally tangent to the path) and centripetal component (perpendicular to the path) truly derivies from using the intrinsic basis: $a(t) = a_centripetal(t) \hat{N} + a_tangential \hat{T}$

and considering the local osculating plane. For example, at a specific point $P$ on the trajectory, the unit basis vector $\hat{N}$ is exactly in the direction of the radius of the local osculating circle and directed towards its center while, when using polar coordinates, the polar unit vector $\hat{r}$ is not directed to the center of the osculating circle.

As Chestermiller suggests, we can always with the dot product between the polar unit vector and the tangential unit vector but that is an extra step.

 

[Chestermiller]

As Chestermiller suggests, we can always with the dot product between the polar unit vector and the tangential unit vector but that is an extra step.

I didn't say polar unit vector. I said normal unit vector.

 

[fog37]

You did. But the unit radial vector $\textbf{i}_r$ is not necessarily normal to the trajectory of the moving particle, correct? The vector $\textbf{i}_r$ is normal to the angular unit vector $\textbf{i}_\theta$.

 

[Chestermiller]

You did. But the unit radial vector $\textbf{i}_r$ is not necessarily normal to the trajectory of the moving particle, correct? The vector $\textbf{i}_r$ is normal to the angular unit vector $\textbf{i}_\theta$.

That’s what my equations indicate. So?

 

[fog37]

Nothing wrong mathematically indeed, just noticing that when many intro physics book talk/introduce the concepts of centripetal and tangential accelerations as vector components of the acceleration vector they implicitly use the local intrinsic unit vectors and intrinsic basis $(T,N,B)$ to express them and not other coordinate systems like cylindrical, spherical or rectangular.

The rate of change of the three unit vectors $(T,N,B)$ are called the Frenet equations...