Mechanism AOB driven by gear D, tangential force

Gear $E$ and the arm $OB$ rotate together about $O$, so the tangential force $F_D$ of gear $D$ at the pitch radius $r_E = 0.120$ m must supply the torque needed to give both bodies the angular acceleration $\alpha = 40$ rad/s². The angular velocity (8 rad/s) does not enter the moment equation: both bodies rotate about the fixed axis $O$, and the centripetal terms produce no moment about $O$.

Assumptions. Arm $OB$ is a uniform slender rod of length $OB = 0.400$ m pivoted at its end $O$ (the "40 mm" in the figure is read as 400 mm). At this instant the rod is vertical, or the mechanism turns in a horizontal plane, so the weights have no moment about $O$. If the rod length differs, use $I_{OB} = \tfrac13 m L^2$ with the actual $L$.

Moments of inertia about O

$$\bar I_E = m_E\bar k_E^2 = 4(0.085)^2 = 0.0289\ \text{kg m}^2$$ $$I_{OB} = \tfrac13 m_{OB}L^2 = \tfrac13(3)(0.400)^2 = 0.160\ \text{kg m}^2$$ $$I_O = I_E + I_{OB} = 0.1889\ \text{kg m}^2$$

Equation of motion about O

$$F_D\,r_E = I_O\,\alpha$$ $$F_D = \frac{I_O\alpha}{r_E} = \frac{0.1889\times40}{0.120} = 62.97\ \text{N}$$

The moment is $M_O = I_O\alpha = 7.56$ N m.

Answer: the tangential force of gear $D$ on gear $E$ is about $62.97$ N ($\approx 63.0$ N), acting in the sense of the counter-clockwise angular acceleration.