How can the linear velocity under rotational motion be along the tangent?

Question

Under rotational motion, an object moves along the circumference of the circle, so, it's linear velocity should be the linear displacement between two points over a period of time.So,the linear velocity should be along a secant of the circle, with both the ends of the vector lying at points which the moving object actually moves through. But, instead, it is taken to be along the tangent to circle and the head of the vector ends at a point which the object never passes through. So, why is that the case ?

Answer 1

It is true that we can approximate the motion of a particle by considering it's position at two successive times and taking the displacement between them as our approximate value of velocity.

Here is a diagram showing a circle traversed by our assumed particle in rotational motion, and a series of successively more accurate approximations to its velocity:

Note that each of these "velocity" arrows is in an incorrect direction. The largest one is obviously wrong, since the particle will move to the left along the circle, not straight down. But you should be able to see that in every case the trajectory the particle moves is always to the above/left side of the arrow; every one of these estimates is "too far down".

I haven't drawn any more arrows because they're already getting unreadable at this scale, but you should be able to to see that for any given arrow I can always improve my accuracy by taking a still-closer point along the circle as an alternative endpoint for the arrow. So ideally we would get the most accurate direction if we could apply that process (of improving the accuracy by taking a closer endpoint) an infinite number of times, getting an endpoint that is infinitely close to the top of the circle. The mathematical notion of limits (which I won't go into in detail here) gives us a way of formalising this, but hopefully you should be able to see that the series of ever-more-accurate-direction arrows is getting closer and closer to an arrow pointing directly left - a tangent to the top of the circle - and it turns out this is what the mathematical limit would actually give us.

That's the direction of the velocity. What about its magnitude? Here, there is a subtle conceptual error you're making that I need to correct. We took another point on the circle as the endpoint of our approximation arrows, so the head of our arrow was on a point the particle would reach in the future. But what if we consider another particle travelling twice as fast around the same circle as the original one? If its speed is twice as fast, its velocity arrow should be twice as large. But if we apply the same approximation process we're still drawing arrows going to points on the same circle! So how can this be?

The truth is drawing your arrows this way is implicitly setting the time unit you're using. Consider just the biggest/worst approximation arrow again, straight down to the other side of the circle. If that's 1 metre and it took the particle 1 second to get there (around the circle), then our "approximate" velocity is 1 metre per second. But our faster particle gets there in half a second, so that same length of velocity arrow is 1 metre per half-second (or 2 metres per second). Rescale the arrow to use the same units as the first arrow (so that we can directly compare the magnitudes of the velocities by the length of the arrow) and now the arrow extends 1 m below the bottom of the circle. But that's not saying anything about a position the particle will actually reach!

The head of the arrow actually has nothing to do with any future position of the particle. The instantaneous velocity arrow is not a displacement arrow between two points on the path. If it does happen to look like it lies on the path in one particular diagram it's a total coincidence of the units chosen in that diagram. The approximation method we're using is implicitly choosing units of time to force that coincidence, but if we draw the arrows with a consistent length corresponding to a more convenient choice of time unit (like 1s), then they will usually not match up with a diagram of the particle's path at all.

Also note that in my series of approximation arrows, it's not that there's one special unit of time that makes them all line up on the circle like this (if there were we would be justified in saying that is the "correct" unit to use). Each arrow implicitly sets a totally different unit of time. This should be clear if you think about it, as we are drawing a series of arrows giving us an ever-more-accurate approximation for the particle's velocity but these arrows are getting shorter and shorter (and in the "infinitely accurate" case would go to length zero)! If they were all approximations to the same velocity drawn at the same scale they would be vaguely similar in length, converging to a length that actually represents the magnitude of the particle's velocity at that scale.

We can interpret the "length is converging to zero" feature of this series of arrows, though. Each arrow is drawn to the position the particle will be at after an ever-shorter time interval; that's why the arrows are getting more accurate. So the arrows are covering a distance that approaches zero, over a time that approaches zero. The magnitude of the velocity is distance divided by time, so naively we're approaching $\frac{0}{0}$. The mathematics of limits gives us a well defined way to calculate a value for this ratio, but I won't get into that here. Note that this also means that in the limit where the length of the arrow approaches zero, the head of the arrow approaches zero distance away so the "tangent arrow" and the "secant arrow" are converging on the same thing, if that helps.

Answer 2

the velocity $~\mathbf v~$ can obtain from this equation

\begin{align*} &\mathbf{v}=\lim_{\Delta t=0}\left[\frac{\mathbf{R(t+\Delta t)}-\mathbf{R(t)}}{\Delta t}\right]\tag 1 \end{align*} with the position vector $~\mathbf{R}~$ for a point on a circle ($~r~$ the circle radius). \begin{align*} &\mathbf{R}=r\begin{bmatrix} \cos\left(\frac{s(t)}{r}\right) \\\\ \sin\left(\frac{s(t)}{r}\right) \\\ \end{bmatrix} \end{align*} where $~s~$ is the circle line element ($~s=r\,\phi~$) you obtain from equation (1) that

\begin{align*} &\mathbf{v}=\underbrace{\frac{d}{dt}\,s(t)}_{v}\, \underbrace{\begin{bmatrix} -\sin\left(\frac{s(t)}{r}\right) \\\\ \cos\left(\frac{s(t)}{r}\right) \\\ \end{bmatrix}}_{\mathbf{t}} =v\,\mathbf{t} \end{align*}

where $~\mathbf t~$ is the tangent vector at $~s(t)~$ with $~\mathbf{t}\cdot\mathbf{t}=1$

Answer 3

Suppose you have a ball on a thin, light string and you swing the ball in a circular orbit, then you suddenly let loose of the string. In which direction does it fly away?

Answer 4

The object does not have linear velocity. It does not follow a straight line path. It does have a tangential velocity. It has a velocity at one instant in time that points in the direction of the tangent line. The two are not the same. The object has an angular velocity.