Question 15.2: 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.
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 2pr_1 for each full revolution of the gear. Noting that 1 rev = 2p rad, and that when A moves to the right (x_A > 0) the gear rotates clockwise (u < 0), we write
\frac{x_A}{2 p r_1}=-\frac{ u }{2 p } \quad\quad\quad x_A=-r_1 uDifferentiating with respect to the time t and substituting the known values
v_A=1.2~ m / s ~\text { and }~ r_1=150~ mm =0.150~ m, we obtain
v_A=-r_1 \text v \quad 1.2~ m / s =-(0.150~ m ) \text v \quad\quad\quad \text v =-8 ~rad / s \\ \\ \text V = \text v k =-(8 ~rad / s ) kwhere k is a unit vector pointing out of the paper.
b. Velocities. The rolling motion is resolved into two component motions: a translation with the center A and a rotation about the center A. In the translation, all points of the gear move with the same velocity \text v_A. In the rotation, each point P of the gear moves about A with a relative velocity \text v _{P / A}=\text v k \times r _{P / A}~ \text {, where }~ r _{P / A} is the position vector of P relative to A.
Velocity of Upper Rack. The velocity of the upper rack is equal to the velocity of point B; we write
\text v _R= \text v _B=\text v _A+ \text v _{B / A}=\text v _A+\text v k \times r _{B / A}= (1.2 m/s)i – (8 rad/s)k × (0.100 m)j
= (1.2 m/s)i + (0.8 m/s)i = (2 m/s)i
\text v _R=2~ m / s~\text yVelocity of Point D
\text v _D=\text v _A+\text v _{D / A}=\text v _A+\text v k \times r _{D / A}= (1.2 m/s)i – (8 rad/s)k × (-0.150 m)i
= (1.2 m/s)i + (1.2 m/s)j
\text v _D=1.697~ m / s ~\text { a }~ 45^{\circ}
