Why is tangential acceleration of an object moving in circular motion given by $\dfrac{d \vec{v}}{dt}$

I don't know why everywhere it is writen $$\vec{a_{net}}=\dfrac{\vec{v²}}{r} + \dfrac{\vec{dv}}{dt} \ ,$$ where $a_{net}$ is the net acceleration. But $\dfrac{d \vec{v}}{dt}$ is supposed to be the net acceleration.


You seem to be confusing vector notation and scalar notation.

In general $\vec{a}=\frac{d \vec{v}}{dt}$.

In circular motion with tangential acceleration, total acceleration is composed of tangential acceleration and radial acceleration.



The last two equation are concerned with specific components and are therefore scalar equations. These equations are derived from the first, vector-based, equation.

  • $\begingroup$ Not quite clear what it is that you are asking. The derivative of a vector is well defined. The results of this derivative for the circular motion were provided on the answer. If you want the result in a vector form, you can write it down as: $-a_r\hat{r} + a_t\hat{n}$. $\endgroup$ – npojo Sep 7 '18 at 12:47
  • $\begingroup$ @npojo Shouldn't the vector form be $-a_r\mathbf{n}+a_t\frac{\mathbf{v}}{|v|}$where n is the normal to the trajectory (the second vector is tangent to the trajectory). $\endgroup$ – Chet Miller Sep 8 '18 at 0:11
  • $\begingroup$ @ChesterMiller I believe we are saying the same thing, just different nomenclature. My $\hat{r}$ is your n. My $\hat{n}$ is normal to $\hat{r}$ which is along the trajectory. $\endgroup$ – npojo Sep 8 '18 at 7:14
  • $\begingroup$ @npojo. Oh OK. I thought your n stood for "unit normal." $\endgroup$ – Chet Miller Sep 8 '18 at 12:08

$A_\text{net}$ in translational motion is always defined as the time derivative of the velocity of the centre of mass.

But now in the case of circular motion, a whole new force called the centripetal force comes into existence.

This creates an acceleration , known as centripetal acceleration, which is always directed towards the centre of the circular path.

This $A_\text{centripetal}$ has a magnitude of $\vec{v^2}/r$.

Finally the $A_\text{net}$ of a body is the vector sum of all the accelerations. Therefore,

$$A_\text{net}=\frac{\vec{v^2}}{r}+\frac{\mathrm d\vec v}{\mathrm dt} $$ where the first term is due to circular motion and the last term due to translation which is tangential.

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    $\begingroup$ I know these old school stuff, tell me something new. So according to your explanation for a uniform circular motion $\dfrac{d \vec{v}}{dt}=0$. $\endgroup$ – user203191 Sep 7 '18 at 11:49
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    $\begingroup$ The if you derive an expression for centripetal acceleration in a uniform circular motion, you will find that centripetal acceleration is actually $\dfrac{d \vec{v}}{dt}$. This means that centripetal acceleration is a part of $\dfrac{d \vec{v}}{dt}$ in circular motion. $\endgroup$ – user203191 Sep 7 '18 at 11:51

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