Is centripetal acceleration almost perpendicular to velocity or it is exactly perpendicular to velocity?

In all the derivations of centripetal acceleration that I have seen so far, the direction of acceleration is said to be perpendicular to velocity but I think it's not exactly perpendicular to velocity rather it is a bit slanted away (almost perpendicular) to it.

Here's why I think so, as we all know that motion in mutually perpendicular direction are always independent of each other so we can say:

• If the Centripetal acceleration ( say $$a_c$$ ) is perpendicular to $$u$$ then after a small interval of time $$\delta t$$ the resultant velocity $$\vec{v}=u \hat{i}+a_{c} \delta t \hat{j}$$ where $$\hat{i}$$ and $$\hat{j}$$ represents the perpendicular directions one along $$u$$ and other along $$a_c$$.

But the point to note here is that the magnitude of the velocity will increase in the above scenario and the only way to keep the speed same is when $$a_c$$ is slightly slanted to $$u$$.

Is there something missing in my reasoning?

• Have you calculated the magnitude of such angle?
– user65081
Sep 8 '21 at 3:59
• I think you mean "where $\hat{i}$ and $\hat{j}$" (you wrote "i" both times) Sep 8 '21 at 15:59
• According to Pithagorean theorem any addition of orthogonal vectors would lead to increase of magnitude of the resulting vector represented by the hypothenuse... Sep 8 '21 at 16:15
• This answer will likely be helpful to look at. The centripetal acceleration is exactly the component of the acceleration that is perpendicular to the velocity and it is this component that is responsible for changes in the object's direction and hence the acceleration that corresponds to "moving in a circle" (instantaneously). Sep 8 '21 at 16:58
• Yes you are right that the equation says that centripetal acceleration changes magnitude of velocity but in reality ( or in limiting case), it doesn't get enough time to change magnitude and changes only direction Sep 9 '21 at 1:13

Is there something missing in my reasoning ?

Yes. You are treating the velocity as though it were constant over a finite time and it is not. The velocity varies over time and can only be treated as constant over an infinitesimal time. But infinitesimals do not work the way you wrote.

Since $$|\vec v|=\sqrt{\vec v^2}=\sqrt{\vec v \cdot \vec v}$$ we can write: $$d|\vec v|=|\vec v + d\vec v|-|\vec v|$$ $$= \sqrt{\left(\vec v + d \vec v \right)^2} - \sqrt{ \vec v^2}$$ the products of infinitesimals drop out and we have $$= \sqrt{\vec v^2 + 2 \vec v \cdot d\vec v} - \sqrt{\vec v^2}$$ we can series expand the first radical to get $$= \sqrt{\vec v^2} + \frac{\vec v \cdot d\vec v}{\sqrt{\vec v^2}} - \sqrt{\vec v^2}$$ $$d|\vec v|=\hat v \cdot d\vec v$$ So, $$d|\vec v|$$ can be zero if $$d\vec v$$ is perpendicular to $$\vec v$$.

Note, in all of the above derivations the vector $$\vec v+ d\vec v$$ is obtained from the standard Euclidean vector addition. $$\vec v$$ is infinitesimally different from $$\vec v + d\vec v$$ in the usual way.

Edit: for more information on how infinitesimals, differentials and so forth work see: https://people.math.wisc.edu/~keisler/foundations.pdf especially p. 34.

In the above I have made a slight abuse of notation by writing $$dy=y(x+dx)-y(x)$$ instead of the more complete and correct $$dy=\text{st}\left( \frac{y(x+dx)-y(x)}{dx} \right) dx$$ Dividing by $$dx$$, taking the standard part, $$\text{st}$$, and multiplying by $$dx$$ is what removes the products of infinitesimals.

I thought that it was not a major abuse of notation since in other contexts $$dx\ne 0$$ and $$dx^2=0$$ is a defining property of infinitesimals, but these basic properties of infinitesimals and differentials may not be understood by all readers.

• Comments are not for extended discussion; this conversation has been moved to chat. Sep 9 '21 at 3:19

You are right in one thing: if a constant force $$\vec F$$ is exerted on an object with velocity $$\vec v$$ over a period of time $$\Delta t$$, then if this force is perpendicular to $$\vec v$$, the speed $$\vert \vec v \vert$$ after the time $$\Delta t$$ will not be the same as before.

Knowing that during a uniform circular motion the speed does not change, you propose that to salvage the situation, the force must point slightly slanted backwards. This is close to what's actually happening, but also wrong. You should observe that the centripetal force is not a constant force, but is instead dependent on the object's position on the circle. So this is what happens:

At any given moment $$t$$ in time, the centripetal force $$\vec F(t)$$ is perpendicular to the velocity $$\vec v(t)$$. At any later moment in time $$t+\Delta t$$, the centripetal force $$\vec F(t+\Delta t)$$ is still perpendicular to the velocity $$\vec v(t+\Delta t)$$. But $$\vec F(t+\Delta t)$$ is not perpendicular to $$\vec v(t)$$ anymore. Instead, if $$\Delta t$$ is sufficiently small, then $$\vec F(t+\Delta t)$$ will point almost perpendicularly but slightly slanted backwards with respect to $$\vec v(t)$$, similar to what you suggest.

So the solution to your dilemma isn't that the centripetal force at a given moment points slanted backwards with respect to the velocity at a given moment. It's that the centripetal force at later moments points slanted backwards with respect to velocities at earlier moments. But if we compare force and velocity at the same moment, they will be perpendicular.

Or another way to look at it: the total impulse or the average force over the period of time $$\Delta t$$ do point slightly slanted backwards. The force at the exact beginning of the time interval $$\Delta t$$ does not.

When we write $$\mathbf v(t+\delta t)=\mathbf v(t)+\mathbf a_c(t)\cdot\delta t$$ the $$\mathbf a_c(t)$$ is perpendicular to $$\mathbf v(t)$$, and $$\mathbf a_c(t+\delta t)$$ is perpendicular to $$\mathbf v(t+\delta t)$$ (and so on). For finite $$\delta t$$, $$\mathbf a_c(t)$$ is not perpendicular to $$\mathbf v(t+\delta t)$$, but that's not an issue as these exist at different times.

To further the issue, we then take the limit as $$\delta t\to0$$, so really there isn't going to be a distinction between these different time points for infinitesimal time differences. The velocity and acceleration change together.