I am a maths undergraduate. I decided to start studying some basic physics trying to be rigorous with the math and uderstand the interpretations of some of the mathematical constructs and see how they are modeled to explain physics. Vectors and vector spaces are all well defined in math.

So the first interpretation that I stumble upon is seeing vectors as "arrows" with length and direction and that these arrows follow the rules of vectors (that was done also in math calculus textbooks which I think it might be confusing if you are trying to be rigorous).

At first it was ok. But now on chapter 4 (Halliday Resnick physics) on circular motion, I found that it is an accelerating motion with constant velocity which i find it very counter intuitive since if you are accelerating means you have to be speeding up, now this is a result of the the vector interpretation giving vectors directions and the direction of the velocity is changing so i have acceleration. Still the acceleration is the rate of change of the velocity with time. And if you told me acceleration is constant I would integrate to find the velocity but that doesnt seem to work on circular motion. Why is that? Trying to understand it so that it doesnt seem counter intuitive to me. It is like if you are turning something makes you loose speed so in order to keep it constant you have to accelerate every time.Is that the case? but if it is, I want a more mathematical way to see it. Sometimes it feels like these "arrows" are not so good to explain things. Sorry if the question is not appropriate for the site.

(Halliday has not mentioned any forces so far and has explained things only using basic definitions as rate of change for velocity and acceleration and has done a mathematical trick to prove the acceleration of circular motion with similar triangles and some limits.)


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  • $\begingroup$ The acceleration in circular motion with constant angular frequency is not constant, only the magnitude of the acceleration is. The same is already true for the velocity. This is all high school level math and physics, though... $\endgroup$ – CuriousOne Jul 11 '16 at 7:53
  • $\begingroup$ Still cant understand having acceleration and velocity with constant magnitude. $\endgroup$ – Manolis Lyviakis Jul 11 '16 at 7:53
  • $\begingroup$ As a mathematician you should be able to understand the difference between a vector and its magnitude, no? You should be able to differentiate a sine and a cosine and take the magnitude of the resulting vector? $\endgroup$ – CuriousOne Jul 11 '16 at 7:53
  • $\begingroup$ The magnitude of the vector of the acceleration is not zero.Say it is a constant C .If i integrate that ill get a function of time.Non constant.Why that is not the magnitude of the velocity? $\endgroup$ – Manolis Lyviakis Jul 11 '16 at 7:57
  • $\begingroup$ You aren't integrating the magnitude of the vector, but the vector itself. You can't get constant circular motion from an acceleration vector with constant magnitude to begin with. You have to start with the constraint of circular motion $x^2 + y^2 = r^2$ and then introduce that the magnitude of the velocity vector is constant, i.e. $\dot x^2 + \dot y^2 =v^2$. $\endgroup$ – CuriousOne Jul 11 '16 at 8:05

I find no trouble in thinking of velocity and acceleration vectors as arrows. First some definitions:

  • Velocity is a vector. Speed is it's magnitude.

  • Acceleration is a vector. It's magnitude has no new name.

We agree that acceleration is present if there is a change in velocity: $$\vec a= d \vec v /dt$$ That is, any change. So if magnitude (speed) and/or direction is changed, then acceleration happened. This is better seen from a components point of view, if the vectors in the expression are split into their x and y components:

$$a_x= d v_x/dt$$ $$a_y= d v_y/dt$$

Which envelopes the expression but still only contains magnitudes. No acceleration in the x direction gives no change in x velocity etc.

  • Now, if the velocity vector points North, and the object accelerates North as well, then it is natural that the velocity magnitude (the Speed) is increased.

  • Is acceleration south, then the speed will decrease (the vector is shortened).

  • Is acceleration pointing West or East, exactly Perpendicular, then nothing is pulling the velocity vector forwards or backwards, so it isn't increasing or decreasing. It is only turning or rotating. So the object's direction is rotated and in the next instant it has moved to a location slightly to the side.

If in this new instant, we turn the acceleration vector slightly as well, so that it again is perpendicular, the same happens: no change in the speed (magnitude of velocity), it only turns a bit more.

If this is repeated and done continuously, the path will be perfectly circular and at no point did we have any parallel acceleration component, so speed is not changed anywhere.

  • 2
    $\begingroup$ thank you.So say the velocity vector points North.If the acceleration vector points anywhere with an angle to the velocity different than 90 i will have a change in the speed. $\endgroup$ – Manolis Lyviakis Jul 11 '16 at 9:01
  • $\begingroup$ @ManolisLyviakis Yes, exactly. If you always split vectors into their Perpendicular and parallel components, then this becomes clear. See my addition to the answer. Because then you can always look for Perpendicular and parallel acceleration components. And if there are any components parallel to the velocity, then speed is changed. If there any components Perpendicular, then the direction of the velocity vector is changed. And then things are usually easier to overview. $\endgroup$ – Steeven Jul 11 '16 at 9:06
  • $\begingroup$ and in order for the acceleration to be perpeticular to velocity when i am moving on circle meaning acceleration must point to the center.In order to do that must be a force or something.also when a car is moving on a circle with no force acting to make the acceleration perpenticular to the velocity is it doing a circular motion as mentioned above?same thing when i run around in a circle .i run straight but just for a short distance so it is not a circular movement but a movement with repeated turns right? $\endgroup$ – Manolis Lyviakis Jul 11 '16 at 9:09
  • $\begingroup$ @ManolisLyviakis "acceleration must point to the center" Exactly! "must be a force" Exactly! In general, any turn (any change of direction) requires a Perpendicular acceleration component and thus a Perpendicular force as well. It might not seem intuitive and clear, but in both examples you mentioned, there is indeed a Perpendicular force. And that force is friction. $\endgroup$ – Steeven Jul 11 '16 at 9:42
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    $\begingroup$ No more questions, thanks, too simple to ask new questions, but you confirmed we join a v and a a vector , and is should not be right. $\endgroup$ – user104372 Jul 11 '16 at 10:48

I want a more mathematical way to see it.

Acceleration is defined as the rate of change of velocity with time. Velocity, being vector, can change just by changing direction keeping the magnitude constant. In a circular motion if the angular frequency $\omega$ is constant, then the magnitude of the velocity i.e., speed is constant but the velocity changes due to change in direction of the velocity vector.

Similarly the acceleration being vector can change keeping its magnitude constant. Expressing in circular polar coordinates the equation of motion are $$\vec{v} = \dot{r}\hat{r} + r\omega \hat{\theta}$$ $$\vec{a} = \left( \ddot{r} - r^2\omega\right)\hat{r} + \left( 2\dot{r}\omega + r\dot{\omega} \right)\hat{\theta}$$

As you can clearly see if the motion is uniform$(\dot{\omega} = 0)$ circular $(\dot{r} = 0)$ Then the expression for velocity and acceleration respectively reduce to $$\vec{v} = r\omega \hat{\theta}$$ $$\vec{a} = -r^2\omega \hat{r}$$ both of whose magnitudes are constant (both $r$ and $\omega$ constant) but the vector is changing because $\hat{\theta}$ and $\hat{r}$ are changing.

It is clear from the expression of velocity and accleration that they are perpendicular because $\hat{\theta}$ and $\hat{r}$ are always perpendicular. So the component of acceleration in the direction of velocity is zero the speed doesn't change but the velocity changes due to radial acceleration. So accleration is not necessarily speeding up or slowing down, it may be changing direction too, which is what happens in circular motion.

  • $\begingroup$ suppose $V=(v_x,v_y)$ since it is a vector (velocity) now those 2 components of the velocity can change.Meaning the points V gives as a function of$ R--R^2$ with variable t change .so $v(t_1)_x$ is different with $v(t_2)_x$ when $t_1$,$t_2$ different.But these points on $R^2$ they may be different and change with time but in a way that the quantity $\sqrt{(v_x)^2+(v_y)^2}$ remains the same .That is how i understood the change of the velocity. I thought constant velocity means the same points in $R^2$ if i see velocity as a function but the points (2,-1) ,(-2,1) have the same modular(speed). $\endgroup$ – Manolis Lyviakis Jul 11 '16 at 12:58

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