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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.

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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.

$a_r=\frac{|v|^2}{r}$

$a_t=\frac{d|v|}{dt}$

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

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  • $\begingroup$ 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
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$A_\text{net}$ 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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