# In Uniform Circular Motion, why does the normal accelaration not increase the magnitude of velocity?

This very simple question was posed by a high-school student in the class.

Consider a particle going in a uniform circular motion (uniform implies that the speed is constant). We know that there is a centripetal acceleration at every point in the motion, which changes the direction of the velocity.

Now, consider an infinitesimal period of time dt for this motion at some given instant. Let's say that the initial velocity is $v_1\vec{i}$ in X-direction. Then, the acceleration will add up a component $a.dt\vec{j}$ to this velocity. Adding these 2 quantities should give us our final velocity $\vec{v_2}$. See this diagram:

Now, it is clear from this reasoning that $|\vec{v_2}| > v_1$. So, the magnitude of the velocity should increase!!

Now, I know that this is WRONG. I just don't know where the fault lies. Clearly, there is some fault in the mathematical modeling done here.

I am thinking that there must be some fault in the causation modeling. Acceleration is not CAUSING the change in velocity. It is an EFFECT of the change in velocity. That is one probably answer. But I'd like you to shed more light on this conundrum. How should one go about explaining this to a high-school kid?

• $dt$ is negligibly small. And so is this speed difference. It simply disappears if you shrink $dt$ to become very, very small. Commented Dec 6, 2016 at 10:21
• Or, to say @Steeven's comment a different way. $v_2$ is bigger that $v_1$. But the amount that it is bigger by varies like $\delta t^2$, so that when you divide the difference by $\delta t$, the average speed change varies like $\delta t$. So it vanishes in the limit. How to explain this to a high schooler? Just as above. It is very weird at first, so encourage them to look carefully at the definition of the limit, not just go by diagramatic gut feel. The comment to make is this is how it works when we define the derivative like we do - it's just something one needs to get used to ... Commented Dec 6, 2016 at 11:06
• ... in developing an intuition for differential calculus. +1 for a question that often comes up from teenagers learning calculus - every teacher should be ready for this one and have their own understanding sorted for it. Commented Dec 6, 2016 at 11:10
• An alternative to explaining the calculus correctly would be to simply represent the velocity vector in a polar coordinate system, in which case the result becomes obvious without calculus.
– Pirx
Commented Dec 6, 2016 at 12:04

After time $dt$, the particle will have moved a distance $v_1 dt$ so the two velocity vectors should not start at the same point. If you do a proper drawing and take the limit $dt\to 0$, as you should, if you are talking about infinitesimal time, you see that $v_2 \to v_1$
$$v_2=v_1\sqrt{1+\left(\frac{adt}{v_1} \right)^2}=v_1\left(1+... dt^2+...dt^4+...\right)$$ is quadratic in $dt$. If you had the acceleration in direction of the velocity, you'd get the correction linear in $dt$. That is why in the limit $dt\to 0$, $$\frac{dv}{dt}=\frac{v_2-v_1}{dt}$$ disappears for your case, but does not disappear for acceleration parallel to the velocity.