# How to find tangential/radial/angular velocity for motion in any curve? [closed]

Is the radial velocity responsible only for changing distance between objects and the component perpendicular to it only for change in direction? If so why?

Please try to give a different explanation than saying that the radial velocity points in the line of sight can only increase the distance, and radial velocity is not affected by the component perpendicular to it, because I find this difficult to understand as velocity can be decomposed into two vectors that are not perpendicular, by using non-perpendicular coordinate axes.

What is the proof of the relation between tangential and angular velocity along any curve?

The formula for angular velocity (I am referring to proof of this relation) is given by $v \cos(\beta)/R$ (where $v$ is the speed and $R$ distance from the origin or observer). $v \sin(\beta)$ is the radial velocity. Is there a specific name for the $v \cos(\beta)$ component?

• How do you defina radial and tangential velocity "for any curve"? That may sound like a flippant question but it is not intended as such. Is this relative to a fixed observer, or the instantaneous center of rotation? – Floris Jun 15 '15 at 21:40
• The velocity tangential to a curve is simply the speed. It is the velocity in direction of the tangent to the curve, which is the same as the direction of the velocity, i.e. the velocity component in the local direction of the velocity. – Walter Jun 15 '15 at 22:01
• @walter Correct me if i am wrong but i think that it is that way for circular motion , if something moves in a curve other than circle than component of velocity perpendicular to the line of sight is what i am referring to as tangential, and parallel component as radial. – Robin Hood Jun 16 '15 at 5:17
• @Floris I have seen some formulas for it other than circular motion. Relative to a fixed observer.Maybe the term isn't exactly tangential , is there a specific name for the component of velocity perpendicular to the line of sight ? – Robin Hood Jun 16 '15 at 5:18
• Do you not know how to decompose a vector into two perpendicular components or is this question somehow more subtle? Perhaps a diagram would be helpful - because right now I am struggling to understand what you are struggling with. Is your curve described as a parametric equation or a function y=f(x)? – Floris Jun 16 '15 at 12:39

Forget about vectors then. Just see it the intuitive physical way.

• If you push the speeder to speed up, then you accelerate forward. And speed increases.
• If you brake, then you slow down and reduce the speed. This is acceleration again, but negative. Or we could say backwards.

What if you have a rocket engine mounted on the car pointing to the side? Starting that one is not helping the motion of the car on the road. It is not speeding up the car or braking it on the road.

Rather it pushes the car sideways. Accelerating it sideways. Which means that you get a sideways speed now as well. So the total velocity is now suddenly the forward one as well as the sideways one; combining them is like adding them as vectors. The result is a velocity at an angle!

And now, which of them was the one that turned the car. Since velocity is now slightly in another direction, something must have turned it. And the forward or backwards acceleration did not turn it - it only speeded it up or slowed it down, as we discussed to begin with. So the turning only happens because of the sideways component that appeared. Which appeared from the sideways acceleration that were present.

The overall conclusion is now that forward or backwards acceleration (let's call it tangential, since it is in the same direction as the speed) changes the speed, while sideways acceleration (let's call it radial, because why not) causes turning.

(It sounds like your naming is different from mine here - radial is usually the name for the perpendicular component since it points along the radius of this imagined circle you would move around if you continued turning. And tangential is because of being tangential to this same imagined circle.)

If there was an angled acceleration, then you could split this into components to show that it actually just consists of a bit being tangential and a bit being radial. So it both speeds up (or brakes) and turns at the same time.

Consider an arbitrary trajectory $\vec{r}(t) = r \hat{r}$ measured from the origin, in polar coordinates (https://en.wikipedia.org/wiki/Polar_coordinate_system).

The velocity is then: $\vec{v} = \frac{d}{dt}\vec{r} = \frac{dr}{dt}\hat{r} + r \frac{d\theta}{dt}\hat{\theta}$. Relative to the origin, the radial part of the velocity is thus just $\frac{dr}{dt}$ which is the change in the distance of the object from the origin; the tangential part is $r \frac{d\theta}{dt}$ which is just the change in the direction of the object, with the $r$ determining the arc-length swept out by the object as it changes direction.

A particle following a prescribed path has its velocity vector parameterized as

$$\vec{v} = \vec{e} \,v$$

where $\vec{e}$ is the tangent vector and $v$ is the speed at that instant. This is kind of obvious. But you use the above to find the tangent vector if you know that radial vector $\vec{r}$. Use $\vec{v} = \frac{{\rm d}}{{\rm d}t} \vec{r} = \vec{e} v$. So for example if the position is a function of an angle $\theta$ (like with polar coordinates) you have

$$\frac{\partial \vec{r}}{\partial \theta} \dot{\theta} = \vec{e}\,v$$

or $$v = \dot{\theta} \| \frac{\partial \vec{r}}{\partial \theta} \|$$

and

$$\vec{e} = \frac{ \frac{\partial \vec{r}}{\partial \theta} }{ \| \frac{\partial \vec{r}}{\partial \theta} \| }$$

Now here is the fun part. The acceleration is parametrized as

$$\vec{a} = \vec{e}\,\dot{v} + \vec{n}\, \frac{v^2}{\rho}$$

where $\vec{n}$ is a normal direction to the path and $\rho$ the radius of curvature of the path. The the part of the acceleration along $\vec{n}$ goes towards changing the direction of motion and the part along $\vec{e}$ changes the speed.

A planar particle moving with speed $(\dot{x},\dot{y})$ would have radius of curvature equal to

$$\frac{1}{\rho} = \frac{\dot{y} \ddot{x} - \ddot{y} \dot{x}}{\left( \dot{x}^2+\dot{y}^2\right)^\frac{3}{2}}$$

This means that in a coordinate system where the particle is towards the +x axis the velocity vector is

$$\vec{v} = \begin{pmatrix} \dot{r} & r \dot{\theta} \end{pmatrix}$$

$$v = \sqrt{\dot{r}^2 + r^2 \dot{\theta}^2}$$

and the acceleration vector

$$\vec{a} = \begin{pmatrix} \frac{\dot{r}}{v} & \frac{r \dot{\theta}}{v}\end{pmatrix} \frac{r^2\dot{\theta}\ddot{\theta}+r \dot{r}\dot{\theta}^2+\dot{r} \ddot{r}}{v} + \begin{pmatrix}\frac{r \dot{\theta}}{v} & -\frac{\dot{r}}{v}\end{pmatrix} \frac{r^2\dot{\theta}^3+r(\dot{r}\ddot{\theta}-\ddot{r}\dot{\theta})+2\dot{r}^2\dot{\theta}}{v}$$

$$\dot{v} = \frac{r^2\dot{\theta}\ddot{\theta}+r \dot{r}\dot{\theta}^2+\dot{r} \ddot{r}}{v}$$ $$\frac{v^2}{\rho} = \frac{r^2\dot{\theta}^3+r(\dot{r}\ddot{\theta}-\ddot{r}\dot{\theta})+2\dot{r}^2\dot{\theta}}{v}$$

So the radial acceleration $\ddot{r}$ enter is both terms (change in speed and change in direction) as you can see above.

You would be pretty accurate in saying: "the radial velocity responsible only for changing distance between objects and the component perpendicular to it only for change in direction"

Why?

Consider each case individually:

1) An object has only "radial velocity" that is, it points directly away (or towards) the observer: $\mathbf v = v_0 \hat {\mathbf r}$.

Well, it moving directly towards or away from you, as 'directly' as possible.

2) An object has only "angular velocity" which we'll take to be in the $\hat \theta$ direction (imagine a 2D world), so $\mathbf v = v_0 \hat {\mathbf \theta}$.

This is the case for something whose radius from you never changes, as it's always moving in the 'theta' direction from the observer. This is circular motion. Note, this changes with position! (and for this reason is unlike Cartesian coordinates, that are time/orientation independent).