Question 16.SP.7: A 6 × 8 in. rectangular plate weighing 60 lb is suspended fr......

a. Angular Acceleration. We observe that as the plate rotates about point A, its mass center G describes a circle of radius \bar{r} with center at A.
Since the plate is released from rest (ω = 0), the normal component of the acceleration of G is zero. The magnitude of the acceleration \overline{ \text{a} } of the mass center G is thus \bar{a}=\bar{r} \alpha. We draw the diagram shown to express that the external forces are equivalent to the effective forces:
+\downarrow \Sigma M_A=\Sigma\left(M_A\right)_{\text{eff}}: \quad W \bar{x}=(m \bar{a}) \bar{r}  +  \bar{I} \alpha
Since \bar{a}=\bar{r} \alpha, we have

W \bar{x}=m(\bar{r} \alpha) \bar{r}+\bar{I} \alpha \quad \alpha=\frac{W \bar{x}}{\frac{W}{\text{g}} \bar{r}^2  +  \bar{I}}              (1)

The centroidal moment of inertia of the plate is

\begin{aligned}\bar{I}=\frac{m}{12}\left(a^2+b^2\right) & =\frac{60\text{ lb}}{12\left(32.2\text{ ft}/ s ^2\right)}\left[\left(\frac{8}{12}\text{ ft}\right)^2+\left(\frac{6}{12}\text{ ft} \right)^2\right] \\& =0.1078\text{ lb}\cdot \text{ ft} \cdot s ^2\end{aligned}

Substituting this value of \bar{I} together with W = 60 lb, \bar{r}=\frac{5}{12}\text{ ft}\text {, and } \bar{x}=\frac{4}{12} \text{ ft}
into Eq. (1), we obtain

\alpha=+46.4\text{ rad}/ s ^2 \quad \alpha =46.4 \text{ rad} / s ^2 \downarrow

b. Reaction at A. Using the computed value of a, we determine the magnitude of the vector m \overline{ \text{a} } attached at G.

m \bar{a}=m \bar{r} \alpha=\frac{60\text{ lb}}{32.2\text{ ft}/ s ^2}\left(\frac{5}{12}\text{ ft}\right)\left(46.4\text{ rad}/ s ^2\right)=36.0\text{ lb}

Showing this result on the diagram, we write the equations of motion
\begin{aligned}\stackrel{+}{\rightarrow} \Sigma F_x=\Sigma\left(F_x\right)_{\text{eff}}: A_x &=-\frac{3}{5}(36\text{ lb}) \\& =-21.6\text{ lb}& A _x=21.6\text{ lb}\leftarrow\end{aligned}
\begin{aligned}+\uparrow \Sigma F_y=\Sigma\left(F_y\right)_{\text{eff}}: \quad A_y&-60\text{ lb}=-\frac{4}{5}(36\text{ lb}) \\A_y&=+31.2\text{ lb}\quad &A _y=31.2\text{ lb}\uparrow\end{aligned}
The couple \bar{I} \alpha is not involved in the last two equations; nevertheless, it should be indicated on the diagram.