Rotational Motion

Concept:Angular momentum about pivot $C$ is conserved during the instantaneous collision because gravity is non‑impulsive and the reaction at $C$ exerts zero torque about $C$. After the collision, the disk’s rotational kinetic energy is fully converted into gravitational potential energy as its centre rises.Explanation:The disk of mass $M = 1\ \mathrm{kg}$ and radius $R = 0.2\ \mathrm{m}$ is pivoted at its top point $C$.The particle of mass $m = 20\ \mathrm{g} = 0.02\ \mathrm{kg}$ hits at point $P$ on the circumference, making an angle $45^\circ$ with the vertical (in the fourth quadrant).Before collision, particle velocity $\vec{v}_1 = -100\hat{\imath}\ \mathrm{m/s}$; after collision, $\vec{v}_2 = -90\hat{\jmath}\ \mathrm{m/s}$.Position of centre $O$ relative to $C$: $\vec{r}_{O/C} = -R\hat{\jmath} = -0.2\hat{\jmath}$.Position of $P$ relative to $O$: $\vec{r}_{P/O} = \frac{R}{\sqrt{2}}\hat{\imath} - \frac{R}{\sqrt{2}}\hat{\jmath} = \frac{0.2}{\sqrt{2}}(\hat{\imath} - \hat{\jmath})$.Therefore $\vec{r}_{P/C} = \vec{r}_{P/O} - \vec{r}_{C/O} = \frac{0.2}{\sqrt{2}}\hat{\imath} - \left(0.2 + \frac{0.2}{\sqrt{2}}\right)\hat{\jmath}$.Initial angular momentum about $C$ (only particle):$$\vec{L}_{C,\text{initial}} = \vec{r}_{P/C} \times (m\vec{v}_1) = \left[\frac{0.2}{\sqrt{2}}\hat{\imath} - \left(0.2+\frac{0.2}{\sqrt{2}}\right)\hat{\jmath}\right] \times (-2\hat{\imath}) = -\left(0.4 + \frac{0.4}{\sqrt{2}}\right)\hat{k} .$$Final angular momentum about $C$ (particle + disk rotating with angular speed $\omega$):$$\vec{L}_{C,\text{final}} = \left[\vec{r}_{P/C} \times (m\vec{v}_2)\right] + I_C\,\omega\hat{k} ,$$where $\vec{r}_{P/C} \times (m\vec{v}_2) = \left[\frac{0.2}{\sqrt{2}}\hat{\imath} - \left(0.2+\frac{0.2}{\sqrt{2}}\right)\hat{\jmath}\right] \times (-1.8\hat{\jmath}) = -\frac{0.36}{\sqrt{2}}\hat{k}$.Moment of inertia of the disk about $C$ (parallel axis theorem):$$I_C = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2 = \frac{3}{2}(1)(0.2)^2 = 0.06\ \mathrm{kg\,m^2}.$$Conservation of angular momentum ($\vec{L}_{C,\text{initial}} = \vec{L}_{C,\text{final}}$) gives:$$-\left(0.4 + \frac{0.4}{\sqrt{2}}\right)\hat{k} = -\frac{0.36}{\sqrt{2}}\hat{k} + 0.06\omega\hat{k} \Rightarrow 0.4 + \frac{0.04}{\sqrt{2}} = 0.06\omega .$$Using $\sqrt{2} \approx 1.414$:$$0.4 + 0.0283 = 0.4283 = 0.06\omega \Rightarrow \omega = \frac{0.4283}{0.06} \approx 7.138\ \mathrm{rad\,s^{-1}}.$$Rotational kinetic energy of the disk just after collision:$$K_{\text{disk}} = \frac{1}{2} I_C \omega^2 = \frac{1}{2} \times 0.06 \times (7.138)^2 \approx 1.5285\ \mathrm{J}.$$As the disk swings up, its kinetic energy converts to gravitational potential energy. Let $\Delta h$ be the maximum vertical rise of the centre $O$:$$K_{\text{disk}} = Mg\,\Delta h \Rightarrow \Delta h = \frac{1.5285}{1 \times 10} = 0.15285\ \mathrm{m} \approx 0.15\ \mathrm{m}.$$Answer:$0.15\ \mathrm{m}$ (option A: $0.14$ is close but the correct value is $0.15$).