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In my book the following statement was written and I didn't understand it well. Can anyone explain it in a more simple way?

Figure 11-6 suggests another way to look at the rolling motion of a wheel - namely as pure rotation about an axis that always extends through the point,P where the wheel contacts the street as the wheel moves. We consider the rolling motion to be pure rotation about an axis passing through point P and perpendicular to the plane of the figure.

I don't understand how the rolling motion of wheel which is the combination of translational and rotational motion of wheel can be expressed as pure rotation about an axis that always extends through the point,P where the wheel contacts the street as the wheel moves?

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  • $\begingroup$ Do you happen to have the figure available to see? $\endgroup$
    – Involute
    Jan 17, 2015 at 14:41
  • $\begingroup$ @KnowledgeisFdotv I don't know how to create those figures, if I had known that I would have added it along with the question. $\endgroup$
    – pcforgeek
    Jan 17, 2015 at 15:29

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The point of contact of the wheel with the street is actually stationary, provided the wheel does not slip. This is due to assuming the friction the street exerts on the wheel at the point of contact is sufficiently high.

However, this is only true for a particular instance! The very next moment, the point of contact, P, is no longer is contact with the street, and a new point of contact is elected. That is to say, the point of contact is not a real point on either the wheel or the street over an finite period of time. Basically, the wheel is in pure rotation about P, but only in that instant in time. If it were in pure rotation about P continually, the axis would not move, and the wheel would orbit into and out of the ground about that point, which is rather physically impossible.

Because P is instantaneously stationary, P is the instantaneous centre of rotation, and hence, there is instantaneous pure rotation about P.

In my opinion, one of the best ways to visualise the point P being stationary for a certain point in time is to consider the velocities of points on the wheel, and to consider them as two components: the component of velocity that is due to the translative motion of the wheel, and the component of velocity due to the wheels rotation. Then, add the components. Here is a diagram illustrating that:

enter image description here enter image description here enter image description here

Note how the actual velocities of points on the wheel are perpendicular to the line connecting that point and the point of contact. Hence, the wheel is (instantaneously) in pure rotation about the point of contact.

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  • $\begingroup$ Your diagrams assume a nonzero initial translational velocity, though. I have always had trouble understanding how a torque on a stationary wheel (with tire) produces a lateral translation, and the "rolling as pure rotation" perspective is the closest thing I've discovered to explaining it. Do you have a better explanation? $\endgroup$
    – BlueBomber
    Apr 11, 2017 at 18:34
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This is the crux of Euler's rotation theorem. Per this theorem, the instantaneous motion of a rigid body in two or three dimensional space can always be described as a pure rotation about some point. This theorem is provably true in two or three dimensional Euclidean space (but only in two or three dimensional Euclidean space). We appear to live in three dimensional Euclidean space, so long as you ignore relativistic effects.

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    $\begingroup$ The Wikipedia article that you link to doesn't support your statement (since instead of discussing instantaneous motion, it discusses total displacement over a finite time with the restriction that "a point on the rigid body remains fixed"). And your statement is not quite correct as presented, since rectilinear motion cannot be so described. $\endgroup$
    – ruakh
    Jan 17, 2015 at 17:32

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