Rolling as Translation and Rotation Combined — Question 2

Question 2

A stationary contact point can have nonzero acceleration

A wheel rolls right without slip at constant center speed $4\,\mathrm{m/s}$ and radius $0.50\,\mathrm m$. Track the particular material point currently touching the ground.

1. Find its instantaneous velocity and acceleration.

2. Explain why zero velocity does not imply zero acceleration or a permanently fixed pivot.

Question 2 - Solution

A stationary contact point can have nonzero acceleration

(a) Derivation and calculation

No slip gives $\omega=-V/R=-8$ rad/s. At the bottom, $\vec r=-R\hat y$: $$\vec v_b=V\hat x+\omega\hat z\times(-R\hat y)=\boxed{\vec0}.$$ Because $\vec a_C=0$ and $\alpha=0$, $$\vec a_b=-\omega^2\vec r=\frac{V^2}{R}\hat y =\boxed{32\hat y\ \mathrm{m/s^2}}.$$

(b) Result and interpretation

Zero instantaneous velocity does not imply zero velocity derivative: $$\vec v_b(t_0)=0,\qquad\left.\frac{d\vec v_b}{dt}\right|_{t_0}=32\hat y\ne0.$$ For a short interval $\Delta t$, $$\Delta y\simeq\tfrac12(32)(\Delta t)^2>0.$$ The material point rises away from the surface as another point becomes the contact; it is not a permanently fixed pivot.

Physical check

The acceleration is upward because the centripetal component points from the bottom toward the center.