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acceleration means the rate of change in velocity (vector quantity) and the differentiation means to divide a certain quantity into small elements (i.e $dx$) as we do to find the acceleration at any instant we divide ($dv$ :a very small change in velocity ) by ($dt$: the small time of that change) but in normal acceleration of circular motion , I were told that normal acceleration means the change in the direction of $v$ even if magnitude $v$ normal is zero my question is how do we divide direction into small differential (d..) as we do with the displacement and time and how can we represent the rate of change in direction with a number? and what does that number mean ? the measuring unit of normal acceleration can be rad /sec^2 as the angle means direction or rad/sec?

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    $\begingroup$ What about when velocity is perpendicular to position? Do you have an issue with that too? $\endgroup$ – Aaron Stevens Oct 5 '19 at 17:11
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Okay so think about the position vector of an object moving counter-clockwise in uniform circular motion. Let's choose the origin of our coordinate system to be the centre of the circle. At time $t_1,$ let's say that the position vector $\mathbf{r}_1$ of the object is pointing "East". A short time later $t_2,$ the object will have a new position vector $\mathbf{r}_2$ which is now pointing slightly "North-East".

Since this is uniform circular motion, the magnitudes of $\mathbf{r}_1$ and $\mathbf{r}_2$ are equal to each other (and to the radius of the circle). However, their directions are different. Consider the infinitesimal quantity $d\mathbf{r}$. If we take $t_1$ and $t_2$ to be infinitesimally close together, we may write $d\mathbf{r}=\mathbf{r}_2-\mathbf{r}_1.$ This vector is not zero even though the magnitudes of $\mathbf{r}_2$ and $\mathbf{r}_1$ are equal. If you draw it out you will find that $\mathbf{r}_2-\mathbf{r}_1$ is in fact tangent to the circle.

The velocity is the vector quantity $\mathbf{v}=\frac{d\mathbf{r}}{dt}.$ In our case, we have that

$$\mathbf{v}=\frac{d\mathbf{r}}{dt}=\frac{\mathbf{r}_2-\mathbf{r}_1}{t_2-t_1}.$$

Note that $\mathbf{v}$ points in the same direction as $\mathbf{r}_2-\mathbf{r}_1$ since $t_2-t_1$ is just a number. So we can still define a velocity even though the speed is constant and only the direction is changing. Now we can consider the acceleration. If we take two velocity vectors $\mathbf{v}_1$ and $\mathbf{v}_2$ at times $t_1$ and $t_2$, we can define the acceleration in a similar way to how we defined velocity:

$$\mathbf{a}=\frac{d\mathbf{v}}{dt}=\frac{\mathbf{v}_2-\mathbf{v}_1}{t_2-t_1},$$

which is again not zero and in fact points towards the centre of the circle. Note that $\mathbf{a}$ points in the same direction as $\mathbf{v}_2-\mathbf{v}_1$ since $t_2-t_1$ is just a number.

It is a useful exercise to draw out this scenario for yourself and do the vector addition. Simply draw the vectors $\mathbf{r}_1$ and $\mathbf{r}_2$, add them 'tip-to-tail' to find $\mathbf{r}_2-\mathbf{r}_1$ which tells you the direction of $\mathbf{v}$. You should find $\mathbf{v}$ to always be tangent to the circle, so you can draw any two velocity vectors $\mathbf{v}_1$ and $\mathbf{v}_2,$ and add them 'tip-to-tail' to find $\mathbf{v}_2-\mathbf{v}_1$ which tells you the direction of $\mathbf{a}$.

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  • $\begingroup$ if we increase the tangential velocity with a constant acceleration will this affect the normal acceleration or they act independently ? $\endgroup$ – Kareem Ahmed Oct 5 '19 at 19:51
  • $\begingroup$ The equation for the magnitude of the 'normal acceleration' (centripetal acceleration) is $a=\frac{v^2}{r},$ where $v$ is the magnitude of the tangential velocity vector $\mathbf{v}$ and $r$ is the radius of the circle. So if we increase the tangential velocity, then the magnitude of the normal acceleration also increases. $\endgroup$ – aRockStr Oct 5 '19 at 20:24

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