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I have seen a question similar to this asked online, but the answers do not make complete sense to me. In studying centripetal acceleration, the textbook I am using argues that the centripetal acceleration must be perpendicular to the tangential velocity, otherwise the speed of the object would change. However, wouldn't the speed still change if the acceleration were perpendicular (see this post).

Now, I have seen the arguments online that this is just a misconception because the perpendicular acceleration is lasting for an infinitesimal (not finite) amount of time so it will not change the speed (significantly). However, if the centripetal acceleration is lasting for such a short period of time that it doesn't change the speed of the object, won't it also not change the direction of the object at all (since it is "so small")? Of course, you can integrate the infinitesimal accelerations to "eventually" change the direction, but in doing this aren't you also integrating the "negligible" errors? Am I misunderstanding something? Thank you!

Edit: I know that the question I linked to is very similar, however I do not think the answers to that question fully answer my question. The question linked to started out without acknowledging the infinitesimal duration of the centripetal acceleration. This is the main focus of my question.

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The speed is the length of the velocity vector. An acceleration vector that has a positive component along the direction of the velocity will increase the length and thus the speed. Similarly, an acceleration vector with a negative component along the direction of the velocity will decrease the speed. If the acceleration vector is perpendicular to the velocity, it's component along the direction of the velocity vector is zero. Based on the two statements above, what effect would you expect that acceleration vector to have on the speed?

We can quantify the above by calculating the derivative of speed using the product rule (I'll do it in two-dimensions rather than three for simplicity). The speed is $$ s = |\vec v| = \sqrt{v_x^2+v_y^2}$$ where $v_x$ and $v_y$ are the components of the velocity vector. Using the chain rule and then the product rule and then the chain rule again gives $$\frac{ds}{dt} = \frac{1}{2\sqrt{v_x^2+v_y^2}} \left(2 v_x\frac{d v_x}{dt}+2 v_y\frac{d v_y}{dt}\right)\\=\frac{1}{s}\vec v\cdot \vec a $$ where we used the fact that the acceleration vector has components $$ \vec a = \left(\frac{dv_x}{dt},\frac{dv_y}{dt}\right).$$

From this equation we see that if the velocity and acceleration are perpendicular , $\vec v\cdot \vec a=0$ so the speed is not changing.

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  • $\begingroup$ The perpendicular acceleration adds a new component of motion to the velocity. So, while the tangential velocity won't change, the net velocity would. $\endgroup$
    – dts
    Commented Sep 1, 2017 at 16:15
  • $\begingroup$ Of course the velocity changes, there is acceleration. There is no such thing as net velocity or tangential velocity. $\endgroup$ Commented Sep 1, 2017 at 16:23
  • $\begingroup$ Perhaps I am not using the right vocabulary (or I am confused by something). Because the perpendicular acceleration increases the velocity vector's perpendicular component, wouldn't the sum of the perpendicular and tangential velocities (which is the same as the velocity) increase? $\endgroup$
    – dts
    Commented Sep 1, 2017 at 16:26
  • $\begingroup$ Also your comment has nothing to do with my argument which I'll reiterate: since an obtuse angle between velocity and acceleration clearly means it's slowing down and an acute angle clearly means it's speeding up, what about a right angle? $\endgroup$ Commented Sep 1, 2017 at 16:27
  • $\begingroup$ Yes but only to second order in $dt$, so no instantaneous change. (Also you mean the sum of squared velocities, and that adds up to the speed squared,not the velocity which is a vector.) $\endgroup$ Commented Sep 1, 2017 at 16:30
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If the speed of a body changes then so does the kinetic energy of the body which would mean that work had been done on the body.

For a mass moving at constant speed in a circle the force which is acting on the body (causing an acceleration towards the centre of the circle) is at right angles to the displacement of the body so no work is done on the body.
Thus the kinetic energy of the body does not change and hence the speed of the body does not change.

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