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I'm a bit confused regarding the directions of velocities and acceleration in curvilinear motion. Assume a curvilinear motion, which is not circular. I know that tangential component of velocity and acceleration are tangential to the curve at any point on the curve. But what about normal acceleration and normal velocity?

  1. Are they same as radial acceleration and radial velocity?
  2. Are they in the same direction of the position vector r?
  3. Or are they parallel to radius of curvature of the curve?
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  • $\begingroup$ Does this link physics.stackexchange.com/questions/276823/… answer your question? $\endgroup$
    – GiorgioP
    Dec 2 '20 at 18:07
  • $\begingroup$ Thanks, I read that one. I think I'm clear on the acceleration front now. If I'm right, acceleration has a tangential component, and a normal/centripetal component which is parallel to the radius of curvature at any point in the curve. But what about radial and normal velocity? Are they parallel to the radius of curvature or the position vector, given that radial velocity is dr/dt? $\endgroup$ Dec 2 '20 at 18:19
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    $\begingroup$ By definition, there is no component of the velocity but the tangential one. $\endgroup$
    – GiorgioP
    Dec 2 '20 at 18:23
  • $\begingroup$ No, that's only when radial velocity (dr/dt) is zero. Like in uniform circular motion. Otherwise there can be a radial velocity component. $\endgroup$ Dec 2 '20 at 18:26
  • $\begingroup$ What is the definition of a tangent to a curve described by the position vector $\vec r(\alpha)$ at one point (characterized by the parameter $\alpha =\alpha_0$? the limit of $\frac{{\vec r}(\alpha_0 + \Delta \alpha )-{\vec r}(\alpha_0)}{\Delta \alpha}$ for $\Delta \alpha \rightarrow 0$. A part of the choice of the parameterization, where does it differ from the definition of velocity? $\endgroup$
    – GiorgioP
    Dec 2 '20 at 18:37
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I don't think there is anything like normal velocity. Velocity acts only along the direction tangential to the path of the object. To prove my point lets say that there is a component of velocity in direction not tangential to the path. Then after moving over a differential element we would end up at a point in the direction of the net velocity which would deviate from its original path.

enter image description here

As for the normal acceleration, it is along the radius of curvature and is responsible for rotating the velocity vector and turning the direction of motion along the path

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  • $\begingroup$ True, there is no component of velocity in the normal direction. In the radial-transverse co-ordinate system, there is a component of velocity in the radial direction (dr/dt), along with the transverse component rdθ/dt. In the normal-tangential co-ordinate system, normal component of velocity is zero. I was a bit confused because I mixed up these two. It later became clear. Thanks, anyways! $\endgroup$ Dec 23 '20 at 7:29

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