Consider a disc of uniform thickness, with a radius of $a$ and a density of $\rho$, rotating with a uniform angular velocity $w$ around its axis of rotation, the $z$-axis. Determine the stress and displacement at each point within the disc is (). (Neglect the gravity of the disc itself.)

Options:
A. $$
\sigma_{r}=\frac{3+2\nu} {8} \rho\omega^{2} a^{2} \left( 1-\frac{r^{2}} {a^{2}} \right), \quad\sigma_{\theta}=-\frac{3+2\nu} {8} \rho\omega^{2} a^{2} \left( 1-\frac{1+3.5\nu} {3+2\nu} \frac{r^{2}} {a^{2}} \right)
$$
$$
u_{r}=\frac{\rho\omega^{2} a^{3} \left( 1-\nu\right)} {8 E} \left[ \left( 3+2\nu\right) \frac{r} {a}-\left( 1+\nu\right) \frac{r^{3}} {a^{3}} \right]
$$
+
B. $$
\sigma_{r}=\frac{3+\nu} {8} \rho\omega^{2} a^{2} \left( 1-\frac{1.5r^{2}} {a^{2}} \right), \quad\sigma_{\theta}=-\frac{3+\nu} {8} \rho\omega^{2} a^{2} \left( 1-\frac{1+3 \nu} {3+\nu} \frac{r} {a^{2}} \right)
$$
$$
u_{r}=\frac{\rho\omega^{2} a^{3} \left( 1-\nu\right)} {8 E} \left[ \left( 3+\nu\right) \frac{2r} {a}-\left( 1+\nu\right) \frac{r^{3}} {a^{3}} \right]
$$
+
C. $$
\sigma_{r}=\frac{4+\nu} {8} \rho\omega^{2} a^{2} \left( 1-\frac{r^{2}} {a^{2}} \right), \quad\sigma_{\theta}=-\frac{4+\nu} {8} \rho\omega^{2} a^{2} \left( 1-\frac{1+4\nu} {4+\nu} \frac{r^{2}} {a^{2}} \right)
$$
$$
u_{r}=\frac{\rho\omega^{2} a^{3} \left( 1-\nu\right)} {8 E} \left[ \left( 4+\nu\right) \frac{r} {a}-\left( 1+\nu\right) \frac{r^{3}} {a^{3}} \right]
$$
+
D. $$
\sigma_{r}=\frac{3+\nu} {8} \rho\omega^{2} a^{3} \left( 1-\frac{r^{2}} {a^{2}} \right), \quad\sigma_{\theta}=-\frac{3+\nu} {8} \rho\omega^{2} a^{2} \left( 1-\frac{1+2\nu} {3+\nu} \frac{r^{2}} {a^{2}} \right)
$$
$$
u_{r}=\frac{\rho\omega^{3} a^{3} \left( 1-\nu\right)} {8 E} \left[ \left( 3+\nu\right) \frac{r} {a}-\left( 1+\nu\right) \frac{r^{3}} {a^{3}} \right]
$$
+
E. $$
\sigma_{r}=\frac{2+\nu} {8} \rho\omega^{2} a^{2} \left( 1-\frac{r^{2}} {a^{2}} \right), \quad\sigma_{\theta}=-\frac{2+\nu} {8} \rho\omega^{2} a^{2} \left( 1-\frac{1+2\nu} {2+\nu} \frac{r^{2}} {a^{2}} \right)
$$
$$
u_{r}=\frac{\rho\omega^{2} a^{3} \left( 1-\nu\right)} {9 E} \left[ \left( 2+\nu\right) \frac{r} {a}-\left( 1+\nu\right) \frac{r^{3}} {a^{3}} \right]
$$
+
F. $$
\sigma_{r}=\frac{3+0.5\nu} {8} \rho\omega^{2} a^{2} \left( 1-\frac{r^{2}} {a^{2}} \right), \quad\sigma_{\theta}=-\frac{3+\nu} {8} \rho\omega^{2} a^{2} \left( 1-\frac{2+3\nu} {3+0.5\nu} \frac{r^{2}} {a^{2}} \right)
$$
$$
u_{r}=\frac{\rho\omega^{2} a^{3} \left( 1-\nu\right)} {8 E} \left[ \left( 3+\nu\right) \frac{r} {a}-\left( 1+1.5\nu\right) \frac{r^{3}} {a^{3}} \right]
$$
+
G. $$
\sigma_{r}=\frac{3+\nu} {8} \rho\omega^{2} a^{2} \left( 1-\frac{r^{2}} {a^{2}} \right), \quad\sigma_{\theta}=-\frac{3+\nu} {8} \rho\omega^{2} a^{2} \left( 1-\frac{1+3 \nu} {3+\nu} \frac{r^{2}} {a^{2}} \right)
$$
$$
u_{r}=\frac{\rho\omega^{2} a^{3} \left( 1-\nu\right)} {8 E} \left[ \left( 3+\nu\right) \frac{r} {a}-\left( 1+\nu\right) \frac{r^{3}} {a^{3}} \right]
$$
+
H. $$
\sigma_{r}=\frac{3+\nu} {7} \rho\omega^{2} a^{2} \left( 1-\frac{r^{2}} {a^{2}} \right), \quad\sigma_{\theta}=-\frac{3+\nu} {9} \rho\omega^{2} a^{2} \left( 1-\frac{1+3\nu} {3+\nu} \frac{r^{2}} {a^{2}} \right)
$$
$$
u_{r}=\frac{\rho\omega^{2} a^{3} \left( 2-\nu\right)} {8 E} \left[ \left( 3+\nu\right) \frac{r} {a}-\left( 1+\nu\right) \frac{r^{3}} {a^{3}} \right]
$$
+
I. $$
\sigma_{r}=\frac{3+\nu} {6} \rho\omega^{2} a^{2} \left( 1-\frac{r^{2}} {a^{2}} \right), \quad\sigma_{\theta}=-\frac{3+\nu} {6} \rho\omega^{2} a^{2} \left( 1-\frac{1+3\nu} {3+\nu} \frac{r^{2}} {a^{2}} \right)
$$
$$
u_{r}=\frac{\rho\omega^{2} a^{3} \left( 1-\nu\right)} {9 E} \left[ \left( 3+\nu\right) \frac{r} {a}-\left( 1+\nu\right) \frac{r^{3}} {a^{3}} \right]
$$
+
J. $$
\sigma_{r}=\frac{3+\nu} {8} \rho\omega^{2} a^{2} \left( 1-\frac{r^{2}} {a^{3}} \right), \quad\sigma_{\theta}=-\frac{3+\nu} {8} \rho\omega a^{2} \left( 1-\frac{1+3\nu} {3+\nu} \frac{r^{2}} {a^{2}} \right)
$$
$$
u_{r}=\frac{\rho\omega^{2} a^{2} \left( 1-\nu\right)} {8 E} \left[ \left( 3+\nu\right) \frac{r} {a}-\left( 1+\nu\right) \frac{r^{3}} {a^{3}} \right]
$$
+
Answer: G