Rolling as Translation and Rotation Combined — Question 8

Question 8

The instantaneous center can lie outside a slipping wheel

A wheel with $R=0.40\,\mathrm m$ has center velocity $3\hat x\,\mathrm{m/s}$ and angular velocity $-5\hat z\,\mathrm{rad/s}$. Consider the instantaneous rigid velocity field extended beyond the wheel.

1. Locate the point of zero velocity relative to the center.

2. Find contact slip speed and explain why the zero-velocity point need not be material.

Question 8 - Solution

The instantaneous center can lie outside a slipping wheel

(a) Derivation and calculation

For $\vec r=x\hat x+y\hat y$ measured from the center, $$\vec v=3\hat x+(-5\hat z)\times(x\hat x+y\hat y) =(3+5y)\hat x-5x\hat y.$$ Setting each component to zero gives $$\boxed{x=0},\qquad\boxed{y=-3/5=-.600\ \mathrm m}.$$

(b) Result and interpretation

The bottom is at $(0,-.40)$ m, so $$\boxed{\vec v_b=[3+5(-.40)]\hat x=1.00\hat x\ \mathrm{m/s}}.$$ The zero-velocity point lies $.60-.40=.20$ m below the rim. It belongs to the extended rigid velocity field and need not be a material point of the wheel.

Physical check

The velocity is proportional to distance from the instantaneous zero-velocity point, giving $5(.20)=1$ m/s at contact.