Rolling as Translation and Rotation Combined — Question 4

Question 4

A moving conveyor changes the rolling constraint

A wheel with $R=0.20\,\mathrm m$ rolls without slip on a belt moving left at $2\,\mathrm{m/s}$. Its center moves right at $1\,\mathrm{m/s}$.

1. Find angular velocity and the velocity of the top point in the laboratory.

2. Find the center and top velocities in the belt frame.

Question 4 - Solution

A moving conveyor changes the rolling constraint

(a) Derivation and calculation

The laboratory contact condition is $$V+\omega R=U=-2\ \mathrm{m/s}.$$ Thus $$\boxed{\omega=\frac{U-V}{R}=\frac{-2-1}{.20}=-15\ \mathrm{rad/s}}.$$ The top velocity is $$\boxed{v_t=V-\omega R=1-(-15)(.20)=4\ \mathrm{m/s}}.$$

(b) Result and interpretation

Under the translating-frame change $v'=v-U$, $$\boxed{V'=1-(-2)=3\ \mathrm{m/s}},\qquad \boxed{v_t'=4-(-2)=6\ \mathrm{m/s}},$$ $$v_b'=U-U=0.$$ Angular velocity is unchanged, so $v_t'=2V'$ in the belt frame, where the surface is stationary.

Physical check

The factor-of-two top speed is relative to a stationary rolling surface, not necessarily the laboratory.