Question 15.SP.6: The double gear shown rolls on the stationary lower rack; th...
The double gear shown rolls on the stationary lower rack; the velocity of its center A is 1.2 m/s directed to the right. Determine (a) the angular velocity of the gear, (b) the velocities of the upper rack R and of point D of the gear.
STRATEGY: The double gear is undergoing general motion, so use rigid body kinematics. Resolve the rolling motion into two component motions: a translation of point A and a rotation about the center A (Fig. 1). In the translation, all points of the gear move with the same velocity \mathbf{V}_A. In the rotation, each point P of the gear moves about A with a relative velocity \mathbf{v}_{P / A}=\omega \mathbf{k} \times \mathbf{r}_{P / A}, where \mathbf{r}_{P / A} is the position vector of P relative to A.
MODELING and ANALYSIS:
a. Angular Velocity of the Gear. Since the gear rolls on the lower rack, its center A moves through a distance equal to the outer circumference 2 \pi r_1 for each full revolution of the gear. Noting that 1 \mathrm{rev}=2 \pi rad, and that when A moves to the right \left(x_A>0\right), the gear rotates clockwise (θ < 0), you have
\frac{x_A}{2 \pi r_1}=-\frac{\theta}{2 \pi} \quad x_A=-r_1 \theta
Differentiating with respect to the time t and substituting the known values v_A=1.2 m/s and r_1=150 \mathrm{~mm}=0.150 \mathrm{~m} \text {, } you obtain
v_A=-r_1 \omega \quad 1.2 \mathrm{~m} / \mathrm{s}=-(0.150 \mathrm{~m}) \omega \quad \omega=-8 \mathrm{rad} / \mathrm{s}
\omega=\omega \mathbf{k}=-(8 \mathrm{rad} / \mathrm{s}) \mathrm{k}
where k is a unit vector pointing out of the page.
b. Velocity of Upper Rack. The velocity of the upper rack is equal to the velocity of point B; you have
\begin{aligned}\mathbf{v}_R &=\mathbf{v}_B=\mathbf{v}_A + \mathbf{v}_{B / A}=\mathbf{v}_A + \omega \mathbf{k} \times \mathbf{r}_{B / A} \\&=(1.2 \mathrm{~m} / \mathrm{s}) \mathbf{i} – (8 \mathrm{rad} / \mathrm{s}) \mathbf{k} \times(0.100 \mathrm{~m}) \mathbf{j} \\&=(1.2 \mathrm{~m} / \mathrm{s}) \mathbf{i} + (0.8 \mathrm{~m} / \mathrm{s}) \mathbf{i}=(2 \mathrm{~m} / \mathrm{s}) \mathbf{i}\end{aligned}
\mathrm{v}_R=2 \mathrm{~m} / \mathrm{s} \rightarrow
Velocity of Point D. The velocity of point D has two components (Fig. 2):
\begin{aligned}\mathbf{v}_D &=\mathbf{v}_A + \mathbf{v}_{D / A}=\mathbf{v}_A + \omega \mathbf{k} \times \mathbf{r}_{D / A} \\&=(1.2 \mathrm{~m} / \mathrm{s}) \mathbf{i} – (8 \mathrm{rad} / \mathrm{s}) \mathbf{k} \times(-0.150 \mathrm{~m}) \mathbf{i} \\&=(1.2 \mathrm{~m} / \mathrm{s}) \mathbf{i} + (1.2 \mathrm{~m} / \mathrm{s}) \mathbf{j}\end{aligned}
\mathbf{v}_D=1.697 \mathrm{~m} / \mathrm{s} \text{⦨} 45^{\circ}
REFLECT and THINK: The principles involved in this problem are similar to those that you used in Sample Prob. 15.3, but in this problem, point A was free to translate. Point C, since it is in contact with the fixed lower rack, has a velocity of zero. Every point along diameter CAB has a velocity vector directed to the right (Fig. 1) and the magnitude of the velocity increases linearly as the distance from point C increases.
