How does centripetal acceleration have direction/vector and magnitude while in the formula $v^2=v\cdot v$ is scalar?

$$a_c=v^2/r$$ 1. How does centripetal acceleration have direction or vector while in the formula dot product between velocity vector is scalar (as in kinetic energy)? Radius is scalar quantity. What is the direction in the formula which shows direction toward the center of circle $$v^2=v\cdot v=\mathrm{scalar}$$ $$\mathrm{radius}=r=\mathrm{scalar}$$ 2. If the velocity magnitude is constant and only direction is changing, why does centripetal acceleration have magnitude by changing direction? How do we understand and read the formula in terms of physics?

Let's draw the vectors involved: If we write out the vectors as their $$(x,y)$$ components we get:

$$\mathbf r = (r\cos\omega t, r\sin\omega t) \tag{1}$$

where $$r$$ is the modulus or $$\mathbf r$$ and $$\omega$$ is the modulus of the angular velocity. Differentiating this gives the velocity:

$$\mathbf v = \frac{d\mathbf r}{dt} = (-r\omega\sin\omega t, r\omega\cos\omega t ) \tag{2}$$

and differentiating again gives the acceleration:

$$\mathbf a = \frac{d\mathbf v}{dt} = (-r\omega^2\cos\omega t, -r\omega^2\sin\omega t ) \tag{3}$$

And comparing equations (1) and (3) we see that equation (3) can be simplified to:

$$\mathbf a = -\omega^2 \mathbf r \tag{4}$$

And there's your vector equation for $$\mathbf a$$.

• Thanks @jhon Rennie 1. I know the expression you mentioned does it mean that in $a_c=v^2/r$ we are saying the acceleration is scalar quantity 2. how do we understand the value of $a_c=2m/s^2$ what is mean by that value, direction of acceleration is changes by this amount. what is meant by that direction of acceleration changes .
– 123
Mar 9 '19 at 8:20
• @123 the quantity $v^2/r$ is the modulus of $\mathbf a$. The direction is $-\hat{\mathbf r}$. The position vector $\mathbf r$ is a function of time so $\mathbf a$ is also a function of time. Mar 9 '19 at 8:55

In your expression you have used the magnitudes of the vectors and hence any directional properties are lost.

In fact for uniform circular motion $$\vec a = - \omega ^2 \, \vec r$$ where $$\omega$$ is the angular speed which has a magnitude $$\omega \,r$$ and $$\vec r$$ is the the radial vector of magnitude $$r$$.

Update as the result of a comment.

Perhaps a more visual approach will help?

Consider an object moving in a circle of radius $$r$$ at a constant speed $$v$$.

The object moves between two positions in a time $$\Delta t$$ as shown in the diagram below.

In that time it has has moved a distance $$\Delta s$$ along the arc of the circle.

$$\Delta s = r\, \Delta \theta \Rightarrow \dfrac{\Delta s}{\Delta t} = r\, \dfrac{\Delta \theta} {\Delta t} \Rightarrow v = r\, \omega$$ where $$\omega = \dfrac {\Delta \theta}{\Delta t}$$ is the angular speed.

Now looking at the vector triangle on the right where $$\vec v_{\rm old} +\vec v_{\rm change}= \vec v_{\rm new}$$ with the magnitudes of $$\vec v_{\rm old}$$ and $$\vec v_{\rm new}$$ being the same and equal to the speed of the object $$v$$.

The magnitude of the change in velocity $$|\vec v_{\rm change}| \approx v\, \Delta \theta \Rightarrow \dfrac{|\vec v_{\rm change}|}{\Delta t} \approx v\, \dfrac{\Delta \theta}{\Delta t} \Rightarrow a = v \, \omega$$ as $$\Delta t$$ and so $$\Delta \theta$$ tend to zero.

$$a$$ is the magnitude of the centripetal acceleration and although the speed does not change, the acceleration (change in velocity) does have a magnitude.

And finally as $$\Delta t$$ tends to zero, $$\Delta \theta$$ tends to zero, and $$\alpha$$ tends to $$90^\circ$$ so the direction of the change in velocity (acceleration) is at right angles to the initial velocity which is along a tangent to the centre, is towards the centre of the circle.

$$\Rightarrow \vec a = - \omega^2 \,\vec r$$

• Thanks for the reply @farcher 1. i know the expression you mentioned does it mean that in $a_c= v^2/r$ we are saying the acceleration is scalar quantity 2. how do we understand the value of $a_c=2m/s^2$ what is mean by that value, direction of acceleration is changes by this amount. what is meant by that direction of acceleration changes .
– 123
Mar 9 '19 at 8:17
• Pls correct if i am wrong. I can say if object want to move with $v=2m/s$ and radius $r=1m$ then we can say $a_c=v^2/r=4m/s^2$ required to move a object in circle with direction toward the center. Or in term of force with $m=1kg$ , $\vec F_c=4N$ needed from the center to keep object in circle. What if object also has tangent acceleration $a_t$ then $a_c$ continuously needed to change to keep object in circle or $a_t$ and $a_c$ are independent.
– 123
Mar 9 '19 at 17:20