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Suppose we have two equivalent rigid cylinders. Cylinder 1 is moving (translating) with constant velocity of v. Cylinder 2 is rotating without slipping and its center’s velocity is constant and equal to v. So, the motion equations of both centers of cylinders are same (x=vt). If we consider centers of cylinders, their kinematics are same. My questions are: 1. What is the difference between these two points (centers of cylinders)? 2. Can we define rotation for a point about an axis that crosses that point?

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    $\begingroup$ can you clarify your question? I don't know what you mean in your second question, 'can we define rotation for a point about an axis that crosses that point' $\endgroup$
    – anon01
    Commented Apr 21, 2016 at 18:11
  • $\begingroup$ I think path of motion of two points is same and their kinematics also, but 2nd is rotating while 1st is translating. $\endgroup$
    – lucas
    Commented Apr 21, 2016 at 18:15
  • $\begingroup$ ok, I see. I'm typing an answer now $\endgroup$
    – anon01
    Commented Apr 21, 2016 at 18:16

2 Answers 2

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"What is the difference between these two points (centers of cylinders)? "

1) there is no difference in the motion of the points (lines that extend down the cylinder, actually). They each translate in space with the center of the cylinder as you might expect.

2) "Can we define rotation for a point about an axis that crosses that point?"

I assume your asking if a point can rotate about its 'own axis'. Since points are mathematical objects of zero dimension, with no internal structure, it is mathematically non-sensical to talk about them rotating 'about their own axis'; there is no internal axis for them to rotate about. Actually, points and the real number line are a little pathological/non physical when you consider them carefully... they are 0 and 1 dimensional objects, but everything you've interacted with is 3D, even if small or thin. Take a look at the wikipedia page for a point for more information.

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  • $\begingroup$ What is the 1st question's answer? $\endgroup$
    – lucas
    Commented Apr 21, 2016 at 18:01
  • $\begingroup$ Thank you because of your answer. But I know what you say about 2nd question. Ok. If the points are same, then please see this: physics.stackexchange.com/questions/251061/is-this-a-paradox $\endgroup$
    – lucas
    Commented Apr 21, 2016 at 18:34
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    $\begingroup$ What about it? I don't know what you are asking. $\endgroup$
    – anon01
    Commented Apr 21, 2016 at 18:38
  • $\begingroup$ The center of cylinder 1 is rotating about an axis at infinity at each moment. But the center of cylinder 2 is rotating about the axis that crosses contact point with the ground at each moment. $\endgroup$
    – lucas
    Commented Apr 21, 2016 at 18:42
  • $\begingroup$ The first cylinder isn't rotating at all. What is the purpose of describing a rotation about infinity? $\endgroup$
    – anon01
    Commented Apr 21, 2016 at 18:55
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Rotations for infinitesimal points are not defined. A rotation is only defined when you have two or more points as a way to describe the fact that their relative distance remains constant.

Also rotational motion is shared for an entire body, meaning that all point on a body rotate the same. The idea of location for rotation only enters when linear velocity is considered as the location where such velocity is zero.

Your first case of a purely translating body, it can be said this is equivalent to a zero rotational speed located at infinity such that $v = \omega \cdot r = 0 \cdot \infty = \text{finite}$.

So the rotational motion (of the centers) is different in the two cases, and the translational motion is only equivalent at the central axis and nowhere else.

Appendix

If a cylinder is translating with velocity $v$ at the center and rotating by $\omega$ then it is said the motion is equivalent to an instantaneous rotation about a point $h = \frac{v}{\omega}$ above the center. For pure rolling to the right, the rotation is clockwise (negative) and the center of rotation is at $d=-r$, or at the contact point.

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  • $\begingroup$ Thank you because of your answer . But I couldn't understand this part of your answer: "So the rotational motion (of the centers) is different in the two cases, and the translational motion is only equivalent at the central axis and nowhere else." Because the motion equation for both points is same (v=xt). $\endgroup$
    – lucas
    Commented Apr 21, 2016 at 19:58
  • $\begingroup$ How can we calculate indeterminate form of $v = \omega \cdot r = 0 \cdot \infty $ in this case? $\endgroup$
    – lucas
    Commented Apr 21, 2016 at 20:13
  • $\begingroup$ You understand the linear velocities are a vector field. They change value as a function of position. So only at the central axis the value matches between the two cases. $\endgroup$ Commented Apr 21, 2016 at 23:37
  • $\begingroup$ $0·\infty$ can be anything. Usually we know $v$ and $\omega$ and we calculate $r$. If $\omega=0$ then $r=\infty$. $\endgroup$ Commented Apr 21, 2016 at 23:39
  • $\begingroup$ Dear ja72, I know all you have been written ever. What I am thinking about is we cannot consider center of second cylinder as a particle. I know it is a point and I know what is point mathematical concept completely. I think we should create a new definition for points rotation. The center of second cylinder is a indeterminate form. It has both of features: particle features (because it is a point) and body features (because it is rotating). I think my last sentence is the key for this question, but my mathematics isn't enough for creation of new definition. $\endgroup$
    – lucas
    Commented Apr 22, 2016 at 4:56

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