Rolling as Translation and Rotation Combined — Question 7

Question 7

A camera frame can hide translation but cannot remove spin

A radius-$0.50\,\mathrm m$ wheel rolls right at $2\,\mathrm{m/s}$. A camera translates right at $2\,\mathrm{m/s}$ without rotating.

1. Find top and bottom velocities in the camera frame.

2. Is the bottom point the instantaneous center of zero velocity in that frame?

Question 7 - Solution

A camera frame can hide translation but cannot remove spin

(a) Derivation and calculation

Ground-frame rolling gives $$\omega=-V/R=-2/.50=-4\ \mathrm{rad/s},\quad v_t=2V=4,\quad v_b=0.$$ Subtract the camera velocity $U=2$ m/s: $$\boxed{v_t'=4-2=+2\ \mathrm{m/s}},\qquad \boxed{v_b'=0-2=-2\ \mathrm{m/s}}.$$

(b) Result and interpretation

The center is stationary in the camera frame: $$V'=V-U=0,\qquad\vec v_P'=\vec\omega\times\vec r_{P/C}.$$ Thus $$\boxed{\text{zero-velocity point in the wheel plane: }C}.$$ The bottom is not stationary in this frame. Translating the reference frame changes the instantaneous zero-velocity point without changing $\omega$.

Physical check

Top and bottom speeds are equal and opposite in a frame translating with the center.