Question 16.SP.7: The portion AOB of a mechanism consists of a 400-mm steel ro...
The portion AOB of a mechanism consists of a 400-mm steel rod OB welded to a gear E with a radius of 120 mm that can rotate about a horizontal shaft O. It is actuated by a gear D and, at the instant shown, has a clockwise angular velocity of 8 rad/s and a counterclockwise angular acceleration of 40 rad/s². Knowing that rod OB has a mass of 3 kg and gear E has a mass of 4 kg and a radius of gyration of 85 mm, determine (a) the tangential force exerted by gear D on gear E, (b) the components of the reaction at point O on the shaft.
STRATEGY: Since you are asked to determine forces, use Newton’s second law.
MODELING: For your system, choose the single object that consists of the steel rod OB and the gear E. Since these two objects are welded together, they have the same angular velocity and angular acceleration. Rather than finding the center of mass for this object, use the center of mass for gear E and for rod OB separately in your kinetic diagram. Therefore, first determine the components of the acceleration of the mass center G_{OB} of the rod (Fig. 1) as
\begin{aligned}\left(\bar{a}_{O B}\right)_t &=\bar{r} \alpha=(0.200 \mathrm{~m})\left(40 \mathrm{rad} / \mathrm{s}^2\right)=8 \mathrm{~m} / \mathrm{s}^2 \\\left(\bar{a}_{O B}\right)_n &=\bar{r} \omega^2=(0.200 \mathrm{~m})(8 \mathrm{rad} / \mathrm{s})^2=12.8 \mathrm{~m} / \mathrm{s}^2\end{aligned}
A free-body diagram and kinetic diagram for the system are shown in Fig. 2. The inertial terms on your kinetic diagram include a couple \bar{I}_E \alpha (since gear E is in centroidal rotation), a couple \bar{I}_{O B} \alpha, and two vector components m_{O B}\left(\bar{a}_{O B}\right)_n and m_{O B}\left(\bar{a}_{O B}\right)_t at the mass center of OB.
Preliminary Calculations: The magnitudes of the weights are
\begin{aligned}W_E &=m_E g=(4 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^2\right)=39.2 \mathrm{~N} \\W_{O B} &=m_{O B} g=(3 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^2\right)=29.4 \mathrm{~N}\end{aligned}
Since you know the accelerations, you can compute the magnitudes of the components and couples on your kinetic diagram, as
\begin{aligned}\bar{I}_E \alpha=m_E \bar{k}_E^2 \alpha &=(4 \mathrm{~kg})(0.085 \mathrm{~m})^2\left(40 \mathrm{rad} / \mathrm{s}^2\right)=1.156 \mathrm{~N} \cdot \mathrm{m} \\m_{O B}\left(\bar{a}_{O B}\right)_t &=(3 \mathrm{~kg})\left(8 \mathrm{~m} / \mathrm{s}^2\right)=24.0 \mathrm{~N} \\m_{O B}\left(\bar{a}_{O B}\right)_n &=(3 \mathrm{~kg})\left(12.8 \mathrm{~m} / \mathrm{s}^2\right)=38.4 \mathrm{~N} \\\bar{I}_{O B} \alpha=\left(\frac{1}{12} m_{O B} L^2\right) \alpha &=\frac{1}{12}(3 \mathrm{~kg})(0.400 \mathrm{~m})^2\left(40 \mathrm{rad} / \mathrm{s}^2\right)=1.600 \mathrm{~N} \cdot \mathrm{m}\end{aligned}
Equations of Motion. Setting the system of the external forces shown in your free-body diagram equal to the inertia terms in your kinetic diagram, you obtain the following equations, which you can solve as
\begin{aligned}+\circlearrowleft \Sigma M_O=\dot{H}_O & \\F(0.120 \mathrm{~m}) &=\bar{I}_E \alpha + m_{O B}\left(\bar{a}_{O B}\right)_t(0.200 \mathrm{~m}) + \bar{I}_{O B} \alpha \\F(0.120 \mathrm{~m}) &=1.156 \mathrm{~N} \cdot \mathrm{m} + (24.0 \mathrm{~N})(0.200 \mathrm{~m}) + 1.600 \mathrm{~N} \cdot \mathrm{m}\end{aligned}
F=63.0 \mathrm{~N} \quad \mathrm{~F}=63.0 \mathrm{~N} \downarrow
\stackrel{+}{\rightarrow} \Sigma F_x=\Sigma m \bar{a}_x: \quad R_x=m_{O B}\left(\bar{a}_{O B}\right)_t
R_x=24.0 \mathrm{~N} \quad \mathbf{R}_x=24.0 \mathrm{~N} \rightarrow
\begin{aligned}+\uparrow \Sigma F_y=\Sigma m \bar{a}_y: \quad R_y – F – W_E – W_{O B} &=m_{O B}\left(\bar{a}_{O B}\right)_n \\R_y-63.0 \mathrm{~N} – 39.2 \mathrm{~N}-29.4 \mathrm{~N} &=38.4 \mathrm{~N}\end{aligned}
R_y=170.0 \mathrm{~N} \quad \mathbf{R}_y=170.0 \mathrm{~N} \uparrow
REFLECT and THINK: When you drew your kinetic diagram, you put your inertia terms at the center of mass for the gear and for the rod. Alternatively, you could have found the center of mass for the system and put the vectors \bar{I}_{A O B} \alpha, m_{A O B} \bar{a}_x and m_{A O B} \bar{a}_y on the diagram. Finally, you could have found an overall I_O for the combined gear and rod and used Eq. 16.8 to solve for force F.
\Sigma M_O=I_O \alpha (16.8)