Question 17

1. Angular Momentum Conservation about pivot $C$: Initially, $\vec{L}_i = \vec{r}_{P/C} \times m\vec{v}_1$. Given $P$ is at $45^\circ$ below the horizontal relative to center $O$, the coordinates relative to $C$ are $(R \cos 45^\circ, -(R + R \sin 45^\circ))$. $\vec{L}_i = [R \cos 45^\circ \hat{i} - R(1 + \sin 45^\circ) \hat{j}] \times [-mv_1 \hat{i}] = -mv_1 R(1 + \sin 45^\circ) \hat{k}$. Magnitude $L_i = 0.02 \times 100 \times 0.2(1 + 0.707) = 0.6828 \text{ kg m}^2\text{s}^{-1}$.

Final particle angular momentum: $\vec{L}_{f,p} = [R \cos 45^\circ \hat{i} - R(1 + \sin 45^\circ) \hat{j}] \times [-mv_2 \hat{j}] = -mv_2 R \cos 45^\circ \hat{k}$. Magnitude $L_{f,p} = 0.02 \times 90 \times 0.2 \times 0.707 = 0.2545 \text{ kg m}^2\text{s}^{-1}$.

Angular momentum of disk $L_d = L_i - L_{f,p} = 0.6828 - 0.2545 = 0.4283 \text{ kg m}^2\text{s}^{-1}$.

2. Angular Velocity calculation: $I_C = I_{cm} + MR^2 = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2 = 1.5 \times 1 \times (0.2)^2 = 0.06 \text{ kg m}^2$. $\omega = \frac{L_d}{I_C} = \frac{0.4283}{0.06} \approx 7.138 \text{ rad/s}$.

3. Height Change of Center $O$: From energy conservation, $\frac{1}{2} I_C \omega^2 = Mg \Delta h$. $\Delta h = \frac{0.5 \times 0.06 \times (7.138)^2}{1 \times 10} = \frac{1.528}{10} \approx 0.153 \text{ m}$. Rounding to two decimal places, we get $0.15 \text{ m}$.