Problem — Bowling ball and the contact point

Problem

A bowling ball rolls without slipping on a horizontal surface. The point of contact with the floor has instantaneous velocity zero. Explain why, and what it means for friction.

Solution

The velocity of a point on the ball is the sum of the translation of the centre ($v_\text{CM}$, forward) and the rotation about the centre ($\omega R$, which at the contact point points backwards). In pure rolling the constraint $v_\text{CM} = \omega R$ holds, so at the contact point the two cancel: $v_\text{contatto} = v_\text{CM} - \omega R = 0$ Since the contact point does not slide on the floor, the friction involved is static, not kinetic. Static friction does no dissipative work (there is no relative sliding), so it dissipates no energy: this is what lets a rolling ball travel long distances while losing almost no speed. $\ev{v_\text{contatto}=v_\text{CM}-\omega R=0 \;\Rightarrow\; \text{static friction, no dissipation}}$

Links

Topics: Rotational dynamics Concepts: Rolling · Static friction Objects: Rolling sphere