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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.

Qmechanic
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dts
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2 Answers2

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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.

spaceisdarkgreen
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  • The perpendicular acceleration adds a new component of motion to the velocity. So, while the tangential velocity won't change, the net velocity would. – dts Sep 01 '17 at 16:15
  • Of course the velocity changes, there is acceleration. There is no such thing as net velocity or tangential velocity. – spaceisdarkgreen Sep 01 '17 at 16:23
  • 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? – dts Sep 01 '17 at 16:26
  • 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? – spaceisdarkgreen Sep 01 '17 at 16:27
  • 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.) – spaceisdarkgreen Sep 01 '17 at 16:30
  • @dts here's another way to think about it: you are assuming a finite time in your argument. But you're also fixing the acceleration vector. There will be a change in speed after a small but finite time, but only because the velocity vector rotated and did not remain perpendicular to the fixed acceleration vector. If the acceleration vector also rotates to always stay perpendicular (as in uniform circular motion) then the speed will never change. – spaceisdarkgreen Sep 01 '17 at 19:28
  • Thank you! I think I understand the point you are making. My calculus knowledge is limited (I'm still learning), but it there a way to "prove" this? Is there a way to integrate vectors that I will eventually learn about (integrating the sum of the velocity and the acceleration*dt)? – dts Sep 01 '17 at 19:32
  • @dts the last part of my answer amounts to a proof (I updated it to be a little more explicit so hopefully it's followable even though the derivative is a little bit complicated for a beginner). To do it your way, let's have the velocity is in the $x$ direction and the acceleration in the $y$ direction. Then $$\vec v' =\vec v + \vec a dt = (v,adt).$$ Then $$|\vec v'|^2 = v^2 + a^2 dt^2= |v|^2 + |a|^2dt^2.$$ So the change in speed squared is 2nd order in $dt$ so $$\lim_{dt\to 0}\frac{|v+dv|^2-|v|^2}{dt} = 0.$$ – spaceisdarkgreen Sep 01 '17 at 21:05
  • @dts I did speed squared for simplicity to avoid needing to deal with the square root... it would hold with speed too. And hopefully you can show that if instead you take $v$ and $a$ pointing along the $x$ axis, the change in speed or speed squared is first order in $dt$ and thus its derivative is nonzero. (And unrelated, but to answer one of your questions, yes you can integrate a vector wrt time. Just integrate each of its components. This is the sense in which $\Delta \vec x = \int \vec v dt$, etc.) – spaceisdarkgreen Sep 01 '17 at 21:08
  • Thank you so much! I tried to do a similar proof to show that, although the derivative of the speed is 0, the derivative of the direction (theta) of the vector is not 0. I took the derivative of the inverse tangent of vy over vx and it worked (i.e. it was not zero)! Perhaps I don't have enough calculus intuition to see why this is true, but I do see that it is true. Thank you again! – dts Sep 02 '17 at 04:56
  • @spaceisdarkgreen In the last sentence, I suppose that the velocity and acceleration should be perpendicular? – dmn Jul 20 '19 at 10:39
  • @macnguyen yes, that was a typo, thank you. – spaceisdarkgreen Jul 20 '19 at 15:20
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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.

Farcher
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