14
$\begingroup$

Let's say we have a particle rotating in the $xy$ plane, centered at $(0,0)$, with a radius $r$ and constant speed $v$. When it's at the point $(0,r)$, the horizontal component of the velocity is $v_x=v$ and vertical component is $v_y=0$. Moreover, the direction of total force on it, the centripetal force, is along the negative $y$ axis. After an infinitesimal time has passed, the vertical velocity increases in the negative $y$ direction by an infinitesimal amount and the horizontal component decreases by an infinitesimal amount such that the vector resultant of the new components equals $v$.

Now my question is, the vertical component changed due to the centripetal force, which was in that direction. But since at that particular moment, there was no force in the horizontal direction, what caused the change in that component?

$\endgroup$
0

4 Answers 4

18
$\begingroup$

Your question is interesting because it shows the dangers of working with infinitesimals without a careful control of their meaning.

Basically, your question is applicable to any motion, when the trajectory in phase space reaches an extremum along one of the directions. So, for simplicity let's discuss the simple one dimensional harmonic oscillator, which, in your example exactly corresponds to the $x$- component of the uniform circular motion.

The equation of motion is $$ \ddot x(t) = - \omega^2x(t), $$ valid for any time $t$. When $x=0$, acceleration is zero and the speed along $x$ direction is maximum.

So,how it happens that there is a decrease of $\dot x$?

A naïf application of infinitesimals is misleading. In the present case $$ \dot x(t+dt) \simeq \dot x(t) + \ddot x(t) dt $$ would imply $\dot x(t+dt) = \dot x(t) $. But this is just the first order result. First order approximations are the leading term of a local analysis of the behavior of a regular function provided they do not vanish. When, like in the case of an extremum of velocity at time $t_0$ (threfore $\ddot x(t_0)=0$), the first order variation is zero, it is necessary to look for the next non-zero term in an expansion in powers of $dt$. In the present case: $$ \dot x(t_0+dt) = \dot x(t_0) + \ddot x(t_0) dt + \frac12 \dddot x(t_0) dt^2 = \dot x(t_0) - \frac12 \omega^2 \dot x(t_0) dt^2, $$ where use has been done of the equation of motion (by taking a time derivative of both sides) in order to express the third derivative of $x$ with respect to time as a function of $\dot x$.

Therefore, one can see that, at the dominant non vanishing order in $dt$, $\dot x$ is correctly varying.

$\endgroup$
8
  • 1
    $\begingroup$ Side note: This kind of approximation is one of the reasons video game physics are sometimes not realistic. Because a simple simulation will calculate no movement in that time-step, and the object will only start moving on the next time-step. This sort of thing can cause what should be a circular motion to precess, and maybe increase or decrease the radius, unless the object is specifically programmed to move in a circle. $\endgroup$ Commented Feb 3, 2020 at 11:27
  • $\begingroup$ @user253751 true, but only if one is using a poor numerical algorithm. $\endgroup$ Commented Feb 3, 2020 at 13:01
  • 1
    $\begingroup$ naïf should be naive, or is that a term I should know? $\endgroup$
    – Baldrickk
    Commented Feb 3, 2020 at 14:51
  • 1
    $\begingroup$ @Baldrickk Merriam-Webster gives naïf or naif as synomyms of the more frequent (in English) naive ( merriam-webster.com/dictionary/naive#synonyms ), although in French they are just the masculine (naif) o feminine (naive) versions of the same adjective. $\endgroup$ Commented Feb 3, 2020 at 15:33
  • 1
    $\begingroup$ @ToddSewell Sure, force is for a function which mathematically encodes the force. In such a game, intuition does not help, unless one has developed an excellent mathematical intuition. However, even from the physical point of view, your example would correspond to the case of a point on the top of a very flat potential maximum such that, if at $t=0$ the point is at rest, either it will remain there forever $x(t)=0$, or it will move according to your proposed evolution. Both solutions are admissible. I think you would agree that it is a pathological case. $\endgroup$ Commented Apr 12, 2020 at 20:42
11
$\begingroup$

I'm going to take a slightly different approach compared to the other answers, although we really all are essentially saying the same thing. The issue is that you are treating the "next instant" as finite. Therefore at the next finite instant the horizontal velocity will not change at all, and it isn't until the next finite step after the first one that the horizontal velocity will change.

More specifically, you are essentially thinking in terms of solving your differential equations $$\frac{\text d\mathbf r}{\text dt}=\mathbf v$$ $$m\frac{\text d\mathbf v}{\text dt}=\mathbf F$$

by using an approximation $$\Delta \mathbf r=\mathbf v\Delta t$$ $$m\Delta \mathbf v=\mathbf F\Delta t$$

for finite $\Delta t$. Therefore, if your initial conditions are $$\mathbf r(0)=(0,R)$$ $$\mathbf v(0)=(v,0)$$ and if your force is given by $$\mathbf F(t)=(-F\sin(\omega t),-F\cos(\omega t))$$

then your first step will give you $$\mathbf r(\Delta t)=\mathbf r(0)+\Delta \mathbf r=(0,R)+(v,0)\Delta t=(v\Delta t,R)$$ $$\mathbf v(\Delta t)=\mathbf v(0)+\Delta \mathbf v=(v,0)+(0,-F/m)\Delta t=(v,-F/m\cdot\Delta t)$$

Which is what you were thinking... The horizontal velocity doesn't change, and since the object was only moving horizontally, it doesn't move down either. Once you apply this approximation again you will see the object start to move down, and the horizontal velocity will change. This is shown in the animation below for a rather large $\Delta t$ where I have paused it after the first step to show the purely horizontal motion.

enter image description here

Once you make $\Delta t$ smaller and smaller, you will see that we get a trajectory that looks more like uniform circular motion.

enter image description here

But you can never escape that the very next instant will have no change in horizontal velocity for finite $\Delta t$. This is because in the continuous limit there really is no "next moment", i.e., you cannot pick the "next real number" after $0$.

$\endgroup$
1
  • $\begingroup$ @Riz222 No problem :) $\endgroup$ Commented Feb 11, 2020 at 20:01
2
$\begingroup$

People have given detailed answers above. I will just offer a more intuitive way of thinking about it. You're right that at the instant under consideration there is no horizontal force on the object. So that tells us that its horizontal speed is not under any influence of change at that instant. Therefore, the object continues with that speed but as soon as it leaves the point under consideration there is then a horizontal component of force on the object so its speed changes.

$\endgroup$
1
$\begingroup$

It is peculiar indeed! But there is a resolution. It lies in understanding limits.

Say at $t=0$ you are at $\left(0,r\right)$ where the force is purely along $-\hat{y}$. Then at any arbitrary instant $dt$ the force is no longer purely along $-\hat{y}$. It is like saying at $t=0$ there is no force along $\hat{x}$ but at the very next instant there is. But this statement is meaningless. It’s like asking what’s the real number next to $0$.

Thus the motion is purely along $x$ only at exactly $t=0$ and has components along $y$ for $t\ne0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.