University Physics I - Classical Mechanics, 2019

208 CHAPTER 9. ROTATIONAL DYNAMICS
when the condition for rolling without slipping is satisfied:
|v cm | = R|ω| (9.34)
At this point, the object will start rolling without slipping, and losing speed at a much slower rate.
The origin of the condition (9.34) is fairly straightforward. You can imagine an object that is
rolling without slipping as “measuring the surface” as it rolls (or vice-versa, the surface measuring
the circumference of the object as its different points are pressed against it in succession). So, after it
has completed a revolution (2π radians), it should have literally “covered” a distance on the surface
equal to 2πR, that is, advanced a distance 2πR. But the same has to be true, proportionately,
for any rotation angle Δθ other that 2π: since the length of the corresponding arc is s = R|Δθ|,
inarotationoveranangle|Δθ| the center of mass of the object must have advanced a distance
|Δx cm | = s = R|Δθ|. Dividing by Δt as Δt → 0thenyieldsEq.(9.34).
v cm
θ
Figure 9.8: Left: illustrating the rolling without slipping condition. The cyan line on the surface has the
same length as the cyan-colored arc, and will be the distance traveled by the disk when it has turned through
an angle θ. Right: velocities for four points on the edge of the disk. The pink arrows are the velocities in
the center of mass frame. In the Earth reference frame, the velocity of the center of mass, ⃗v cm , in green, has
to be added to each of them. The resultant is shown in blue for two of them.
Note that, unlike Eq. (8.36), which it very much resembles, Eq. (9.34) isnot a “vector identity in
disguise”: there is nothing like Eq. (9.14) that we could substitute for it in order to make the signs
automatically come out right. You should just treat it as a relationship between the magnitudes of
⃗v cm and ⃗ω and just pick the signs appropriately for each circumstance, based on your convention
for positive directions of translation and rotation.
In fact, we could use Eq. (9.14) to find the velocity of any point on the circle, if we go to a reference
frame where the center is at rest—which is to say, the center of mass (CM) reference frame; then, to
go back to the Earth frame, we just have to add ⃗v cm (as a vector) to the vector we obtained in the
CM frame. Fig. 9.8 shows the result. Note, particularly, that the point at the very bottom of the
circle has a velocity −R|ω| in the CM frame, but when we go back to the Earth frame, its velocity
is −R|ω| + v cm = −R|ω| + R|ω| = 0 (by the condition (9.34)). Thus, as long as the condition for
rolling without slipping holds, the point (or points) on the rolling object that are momentarily in