Double gear on rack, accelerations

The lower rack is stationary and the gear rolls on it without slipping, so the contact point $C$ is the instantaneous centre. Take $x$ to the right, $y$ up. $B$ is on the inner radius ($r_2=0.100$ m) above $A$; $C$ is at $r_1 = 0.150$ m below $A$; $D$ is taken on the outer radius, $0.150$ m to the left of $A$.

(a) Angular acceleration

$$\omega = \frac{v_A}{r_1} = \frac{1.2}{0.150} = 8.00\ \text{rad/s (clockwise)}, \qquad \alpha = \frac{a_A}{r_1} = \frac{3}{0.150} = 20.00\ \text{rad/s}^2\ \text{(clockwise)}$$

(b) Accelerations of B, C and D

Use $\vec a_P = \vec a_A + \vec\alpha\times\vec r_{P/A} - \omega^2\vec r_{P/A}$ with $\vec\alpha = -20.00\,\hat k$, $\omega^2 = 64.00$ rad²/s², $\vec a_A = 3\,\hat i$.

Point B, $\vec r_{B/A} = 0.100\,\hat j$:

$$\vec a_B = 3\hat i + (-20.00\hat k)\times(0.1\hat j) - 64.00(0.1\hat j) = (5.00\,\hat i -6.40\,\hat j)\ \text{m/s}^2$$

$a_B = 8.12$ m/s², directed 52.0° below the horizontal, to the right.

Point C, $\vec r_{C/A} = -0.150\,\hat j$:

$$\vec a_C = 3\hat i + (-20.00\hat k)\times(-0.15\hat j) - 64.00(-0.15\hat j) = (0.00\,\hat i +9.60\,\hat j)\ \text{m/s}^2$$

$a_C = 9.60$ m/s², directed upward towards $A$. This is the centripetal acceleration of the contact point; its horizontal part is zero.

Point D, $\vec r_{D/A} = -0.150\,\hat i$:

$$\vec a_D = 3\hat i + (-20.00\hat k)\times(-0.15\hat i) - 64.00(-0.15\hat i) = (12.60\,\hat i +3.00\,\hat j)\ \text{m/s}^2$$

$a_D = 12.95$ m/s², directed 13.4° above the horizontal, to the right.

Answer: $\alpha = 20.00$ rad/s² clockwise; $a_B = 8.12$ m/s² (52.0° below horizontal), $a_C = 9.60$ m/s² (upward), $a_D = 12.95$ m/s² (13.4° above horizontal).