The speed of circular motion

Question

This question comes from my previous question asked here

$$\vec v_2 = \vec v_1 + \vec v_\perp.$$

I think the logic I'm making a mistake is I presume that $\vec v_\perp = adt$ in x-direction. This presumes that during $dt$ time interval, $a$ is constant, hence this gives us > 0 value which means $\vec v_2 > \vec v_1$.

After seeing another question, I realize $a$ is not constant during $dt$ but even if it's not and changing, then we could split $dt$ again in even smaller intervals($dt_1, dt_2, dt_3,...$) and say that $v_\perp = a_1dt_1 + a_2dt_2 + a_3dt_3+...$. This way, $a$ is not constant over $dt$. but now this should give us 0 and the only way this is possible is if over small $dt$ interval, $a$ increases, then decreases and overally results in 0. What's the non-intuitive, solid proof that this is what happens?

Another way proving this is $$ v_2 = v_1\sqrt{1 + (\frac{adt}{v_1})^2} = v_1(1+\frac{1}{2}(\frac{adt}{v_1})^2 - \frac{1}{8}(\frac{adt}{v_1})^4 + \frac{1}{16}(\frac{adt}{v_1})^6)$$ and if $dt -> 0$, this becomes $v_2 = v_1$, but I don't understand this - what we gain with $dt -> 0$ intuitively? even if we take the interval that approaches zero, it doesn't say anything about acceleration and making the interval 0 doesn't seem proof to me.

Answer 1

Questions of this type that you are asking is really something you just have to prove for yourself and work it out.

In particular, you need to draw pictures of the argument until you can convince yourself that the zero time limit is necessary.

You should be starting from $$\vec r=R\begin{pmatrix}\cos\omega t\\\sin\omega t\end {pmatrix}\tag1$$ and working your way to getting centripetal acceleration formula $\vec a=-\omega^2\vec r$

But your problem at hand is with the velocity vector. Draw the pizza slice for the $2\mathrm dt$ time slice centred around $t=0$, i.e. $$\tag2\vec v(0)=u\begin{pmatrix}0\\1\end {pmatrix}\qquad\bigwedge\qquad\vec v(\pm\mathrm dt)=u\begin{pmatrix}\mp\sin(\omega\mathrm dt)\\\cos(\omega\mathrm dt)\end {pmatrix}$$

1. Show that $2\vec{\mathrm dv}\equiv\vec v(+\mathrm dt)-\vec v(-\mathrm dt)=-2u\begin{pmatrix}\sin(\omega\mathrm dt)\\0\end {pmatrix}$ and draw this on the pizza slice. You may choose the equivalent form using angles rather than times.

2. Using the definiton of acceleration as $\vec a=\frac{2\vec{\mathrm dv}}{2\mathrm dt}$ and the mathematical result that $\lim_{\varphi\to0}\frac{\sin\varphi}{\varphi}=1$, show that $\vec a=-\omega u\begin{pmatrix}1\\0\end {pmatrix}$

3. Now, explore what happens if you wish not to take the limit $\mathrm dt\to0$; how much should $|\vec a|$ shrink if you want to preserve the speed if you take $\mathrm dt$ to be finite? What direction should the vectors be in relation to each other?

4. Take a step back and start from position $\vec r$ and obtain the same results relating velocity of uniform circular motion and position.

5. Using the earlier relationship between velocity and acceleration, and now velocity and position, show that $\vec a=-\omega^2\vec r$

6. Repeat all the arguments with arbitrary time $t$ rather than $t=0$, all with taking $\mathrm dt\to0$

Answer 2

Velocity is a vector quantity which has both magnitude and direction $$\vec{V}=V\vec{i}_v$$where $\vec{i}_v$ is a unit vector in the direction of motion (which can vary along the path of motion). The acceleration is equal to the vector time derivative of the velocity:$$\vec{a}=\frac{dV}{dt}\vec{i}_v+V\frac{d\vec{i}_v}{dt}$$In circular motion with constant speed, the magnitude of the velocity vector V is constant, and the acceleration vector reduces to $$\vec{a}=V\frac{d\vec{i}_v}{dt}$$The component of the acceleration vector in the direction tangent to the circular particle path is obtained by dotting the acceleration vector with the unit vector in the direction of motion: $$a_v=\vec{i}_v\centerdot\vec{a}=V\vec{i}_v\centerdot\frac{d\vec{i}_v}{dt}=\frac{1}{2}V\frac{(\vec{i}_v\centerdot \vec{i}_v)}{dt}=\frac{1}{2}V\frac{d1}{dt}=0$$Thus, the tangential component of the acceleration is zero, which is, of course, consistent with the tangential velocity (speed) being constant.