|
|
SECTION 10.9 • Rolling Motion of a Rigid Object 317 R s θ s = R θ Figure 10.27 For pure rolling motion, as the cylinder rotates through an angle ! , its center moves a linear distance s " R ! . P CM Q P ′ 2v CM v CM Figure 10.28 All points on a rolling object move in a direction perpendicular to an axis through the instantaneous point of contact P. In other words, all points rotate about P. The center of mass of the object moves with a velocity v CM , and the point P 1 moves with a velocity 2 v CM . The magnitude of the linear acceleration of the center of mass for pure rolling (10.26) where ( is the angular acceleration of the cylinder. The linear velocities of the center of mass and of various points on and within the cylinder are illustrated in Figure 10.28. A short time after the moment shown in All points on the cylinder have the same angular speed. Therefore, because the dis- tance from P1 to P is twice the distance from P to the center of mass, P1 has a speed CM " 2R&. To see why this is so, let us model the rolling motion of the cylinder in Fig- ure 10.29 as a combination of translational (linear) motion and rotational motion. For CM . For the pure rota- tional motion shown in Figure 10.29b, imagine that a rotation axis through the center CM ) R& " v CM ) v CM " 2v CM , which is greater than the linear speed of any other point on the cylinder. As mentioned earlier, the center of mass moves with linear speed CM while the contact point between the surface and cylinder has a linear speed of zero. We can express the total kinetic energy of the rolling cylinder as (10.27) where I P is the moment of inertia about a rotation axis through P. Applying the parallel-axis theorem, we can substitute I P " I CM ) MR 2 into Equation 10.27 to obtain or, because v CM " R&, (10.28) K " 1 2
I CM &
2 ) 1 2 Mv CM
2 K " 1 2
I CM &
2 ) 1 2 MR
2 & 2 K " 1 2
I P
&
2 a
CM " dv CM dt " R
d& dt " R( Total kinetic energy of a rolling object |