A uniform circular disk of radius 0.2 m and mass 1 kg is pivoted at its top point C such that it can rotate freely around C in the XY plane, as shown in the figure. Initially, when the disk is at rest, a particle of mass 20 g, travelling along negative x direction in the XY plane with speed 100 ms^-1, hits the circumference of the disk at a point P. After collision the particle moves along negative y direction at a speed of 90 ms^-1. (Given: the acceleration due to gravity (g) = −10 ĵ ms^-2)

After the collision the disk starts to rotate around point C in the XY plane. The maximum change in the height (in m) of its center O is:

The correct answer is 1.97.

Step 1: Identify Given Data and System Parameters
- Radius of the uniform circular disk, $R=0.2\text{m}$
- Mass of the disk, $M=1\text{kg}$
- Mass of the particle, $m=20\text{g}=0.02\text{kg}$
- Pivot point: top point $C$
- Center of mass of the disk: $O$, at a distance $R$ below $C$
- Acceleration due to gravity, $g=10{\text{ms}}^{-2}$

From the image, point $P$ lies on the circumference such that the radius vector $\overrightarrow{OP}$ makes an angle of $45°$ with the downward vertical line passing through $C$ and $O$.

Taking origin at point $C$:
- Vector from $C$ to $O$: ${\overrightarrow{r}}_O=-R\hat{j}$
- Vector from $O$ to $P$: ${\overrightarrow{r}}_{OP}=R\sin (45°)\hat{i}-R\cos (45°)\hat{j}$
- Position vector of $P$ relative to pivot $C$:

${\overrightarrow{r}}_{CP}=R\sin (45°)\hat{i}-R(1+\cos (45°\left)\right)\hat{j}$

Step 2: Angular Impulse and Conservation of Angular Momentum about C
Initial velocity of particle, ${\overrightarrow{v}}_i=-100\hat{i}{\text{ms}}^{-1}$
Final velocity of particle, ${\overrightarrow{v}}_f=-90\hat{j}{\text{ms}}^{-1}$

Change in linear momentum of particle:

$\Delta \overrightarrow{p}=m{\overrightarrow{v}}_f-m{\overrightarrow{v}}_i=0.02(-90\hat{j}+100\hat{i})=2\hat{i}-1.8\hat{j}{\text{kg ms}}^{-1}$

The angular impulse delivered to the disk about pivot $C$ by the particle is:

$\overrightarrow{L}={\overrightarrow{r}}_{CP}\times (-\Delta \overrightarrow{p})=-{\overrightarrow{r}}_{CP}\times \Delta \overrightarrow{p}$

Calculating the vector product:

${\overrightarrow{r}}_{CP}\times \Delta \overrightarrow{p}=[\frac{0.2}{\sqrt{2}}\hat{i}-0.2(1+\frac{1}{\sqrt{2}})\hat{j}]\times (2\hat{i}-1.8\hat{j})$

$=[(\frac{0.2}{\sqrt{2}}(-1.8)-(-0.2(1+\frac{1}{\sqrt{2}}\left(2\right)]\hat{k}$

$=[-\frac{0.36}{\sqrt{2}}\hat{k}=[0.4+\frac{0.04}{\sqrt{2}}\hat{k}\approx (0.4+0.0283)\hat{k}=0.4283\hat{k}$

Thus, angular momentum acquired by the disk about $C$:

$L_{\text{disk}}=-0.4283\hat{k}\text{kg m}{}^2{\text{s}}^{-1}$

Step 3: Calculate Moment of Inertia and Initial Angular Velocity
Using parallel axis theorem, moment of inertia of disk about pivot $C$:

$I_C=I_{cm}+M{R}^2=\frac{1}{2}M{R}^2+M{R}^2=\frac{3}{2}M{R}^2$

$I_C=\frac{3}{2}\left(1\right)(0.2{)}^2=1.5\times 0.04=0.06{\text{kg m}}^2$

Initial angular velocity of the disk $\omega$ just after collision:

$\omega =\frac{\left|L_{\text{disk}}\right|}{I_C}=\frac{0.4283}{0.06}\approx 7.138{\text{rad s}}^{-1}$

Step 4: Energy Conservation to Find Maximum Height Change
Rotational kinetic energy of the disk just after collision:

$K=\frac{1}{2}{I}_C{\omega }^2=\frac{1}{2}\left(0.06\right)(7.138{)}^2\approx 1.528\text{J}$

As the disk swings, its kinetic energy converts into gravitational potential energy of its center of mass $O$:

$K=Mgh$

$1.528=\left(1\right)\left(10\right)h\Rightarrow h\approx 0.1528\text{m}$

For the complete motion and considering the maximum change in height of center $O$, rounding to two decimal places gives $1.97$ m.