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This is probably a very stupid question but I can't help me.

Tangential acceleration is $\vec{a_t}=\frac{dv}{dt}\frac{\vec{v}}{v}=\frac{\vec{v} \cdot \vec{a}}{v} \frac{\vec{v}}{v}$. Since $\vec{a}$ is the time derivative of $\vec{v}$ they should always be orthogonal to each other, so $\vec{v} \cdot \vec{a}=0$ and therefore always $\vec{a_t}=0$, which is of course wrong. What is wrong with my reasoning?

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – SuperCiocia
    Commented Feb 18, 2022 at 18:57

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In general, acceleration can of course easily be parallel to the speed. Drive straight ahead and then speed up. That was an acceleration vector parallel to the velocity vector.

When using the word tangential, I can hear that you are specifically referring to circle motion.

But it seems that you are isolating your thinking to the special-case of uniform circular motion. That is when the speed remains unchanged, such as driving around a roundabout at constant speed. In that very specific scenario you are correct that there is no tangential acceleration component. There is only the perpendicular, centre-directed radial acceleration component (called centripetal acceleration).

But imagine driving around the roundabout and then speeding up. You still move in a circular motion, so there is still a radial acceleration component. But you now speed up, meaning you are changing the magnitude of your velocity vector. Only a parallel acceleration vector can change the magnitude of the velocity vector. And such parallel acceleration is tangential. So there must be a tangential acceleration present if you ever speed up while driving in circles. You are then doind non-uniform circular motion.

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$\vec{v} \cdot \vec{a}$ generally is not zero.

However, for an object travelling in a uniform circular path: $$\vec{R} = \begin{bmatrix} \cos(t) \cr \sin(t) \end{bmatrix}$$ $$\frac{\vec{dR}}{dt} = \begin{bmatrix} -\sin(t) \cr \cos(t) \end{bmatrix}$$ $$\frac{\vec{d^2R}}{dt^2} = \begin{bmatrix} -\cos(t) \cr -\sin(t) \end{bmatrix}$$ $$ \begin{bmatrix} -\sin(t) \cr \cos(t) \end{bmatrix} \cdot \begin{bmatrix} -\cos(t) \cr -\sin(t) \end{bmatrix} $$ $$\sin(t)\cos(t)-\sin(t)\cos(t) = 0$$

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  • $\begingroup$ this is not correct you should start with $$ \vec R=\left[ \begin {array}{c} \cos \left( s \left( t \right) \right) \\\sin \left( s \left( t \right) \right) \end {array} \right] $$ with r=1 $\endgroup$
    – Eli
    Commented Feb 17, 2022 at 17:11
  • $\begingroup$ Specifically for uniform motion, the dot product is zero $\endgroup$
    – jensen paull
    Commented Feb 17, 2022 at 17:13
  • $\begingroup$ uniform you mean $\frac{dv}{dt}=0~$ v is the speed $\frac {ds}{dt}$ $\endgroup$
    – Eli
    Commented Feb 17, 2022 at 17:16
  • $\begingroup$ Uniform circular motion doesn't require $\frac{dv}{dt} $ = 0. If that were the case, I would have a straight line path. It requires that |v| to be constant. whilst travelling in a circle. V is velocity. $\endgroup$
    – jensen paull
    Commented Feb 17, 2022 at 17:19
  • $\begingroup$ The acceleration in uniform circular motion is perpendicular to the velocity. Meaning no work is being done and it travels at a constant speed. $\endgroup$
    – jensen paull
    Commented Feb 17, 2022 at 17:20

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