Amount of Energy Loss in Off Center Disk Particle Collision

To find the amount of energy lost during the collision, we analyze the angular momentum of the system about the fixed pivot point $C$ before and after the collision, followed by calculating the initial and final kinetic energies.

1. System Parameters and Coordinates

• Mass of the disk: $M = 1 \text{ kg}$
• Radius of the disk: $R = 0.2 \text{ m}$
• Mass of the particle: $m = 20 \text{ g} = 0.02 \text{ kg}$
• Initial velocity of the particle: $\vec{v}_i = -100 \hat{i} \text{ ms}^{-1}$
• Final velocity of the particle: $\vec{v}_f = -90 \hat{j} \text{ ms}^{-1}$

Taking the pivot point $C$ as the origin $(0,0)$ in the $XY$ plane:

• The center $O$ of the disk is located at $(0, -R)$.
• Point $P$ on the circumference makes an angle of $45^\circ$ with the vertical line passing through $C$ and $O$.
• Position vector of point $P$ relative to $C$: $\vec{r}_P = R \sin 45^\circ \hat{i} - \left(R + R \cos 45^\circ\right) \hat{j} = \frac{R}{\sqrt{2}} \hat{i} - R\left(1 + \frac{1}{\sqrt{2}}\right) \hat{j}$

Substituting $R = 0.2 \text{ m}$: $\vec{r}_P = \frac{0.2}{\sqrt{2}} \hat{i} - 0.2\left(1 + \frac{1}{\sqrt{2}}\right) \hat{j}$

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2. Moment of Inertia of the Disk

By the parallel axis theorem, the moment of inertia of the uniform circular disk about the pivot point $C$ is: $I_C = I_{\text{CM}} + M R^2 = \frac{1}{2} M R^2 + M R^2 = \frac{3}{2} M R^2$

Substituting the values: $I_C = \frac{3}{2} (1) (0.2)^2 = 0.06 \text{ kg m}^2$

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3. Conservation of Angular Momentum about Pivot $C$

Since external forces during the short collision interval act at the pivot point $C$, the angular momentum about $C$ is conserved:

Initial Angular Momentum ($\vec{L}_i$):

$\vec{L}_i = \vec{r}_P \times (m \vec{v}_i)$ $\vec{L}_i = \left[ \frac{R}{\sqrt{2}} \hat{i} - R\left(1 + \frac{1}{\sqrt{2}}\right) \hat{j} \right] \times \left( -m v_1 \hat{i} \right)$ $\vec{L}_i = -m v_1 R\left(1 + \frac{1}{\sqrt{2}}\right) \hat{k}$

Substituting $m = 0.02 \text{ kg}$, $v_1 = 100 \text{ ms}^{-1}$, and $R = 0.2 \text{ m}$: $L_{i, z} = -(0.02)(100)(0.2)\left(1 + \frac{1}{\sqrt{2}}\right) = -0.4\left(1 + \frac{1}{\sqrt{2}}\right) \text{ kg m}^2\text{s}^{-1}$

Final Angular Momentum ($\vec{L}_f$):

$\vec{L}_f = I_C \omega \hat{k} + \vec{r}_P \times (m \vec{v}_f)$ $\vec{r}_P \times (m \vec{v}_f) = \left[ \frac{R}{\sqrt{2}} \hat{i} - R\left(1 + \frac{1}{\sqrt{2}}\right) \hat{j} \right] \times \left( -m v_2 \hat{j} \right) = -m v_2 \frac{R}{\sqrt{2}} \hat{k}$

Substituting $v_2 = 90 \text{ ms}^{-1}$: $L_{p, f, z} = -(0.02)(90) \frac{0.2}{\sqrt{2}} = -\frac{0.36}{\sqrt{2}} \text{ kg m}^2\text{s}^{-1}$

Equating initial and final angular momentum along the $z$-axis: $I_C \omega + L_{p, f, z} = L_{i, z}$ $0.06 \omega = -0.4\left(1 + \frac{1}{\sqrt{2}}\right) - \left(-\frac{0.36}{\sqrt{2}}\right)$ $0.06 \omega = -0.4 - \frac{0.04}{\sqrt{2}} \approx -0.4 - 0.028284 = -0.428284 \text{ kg m}^2\text{s}^{-1}$ $\omega = \frac{-0.428284}{0.06} \approx -7.138 \text{ rad s}^{-1}$

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4. Calculation of Kinetic Energy Loss

• Initial Kinetic Energy ($K_i$): $K_i = \frac{1}{2} m v_1^2 = \frac{1}{2} (0.02) (100)^2 = 100 \text{ J}$

• Final Kinetic Energy ($K_f$): $K_{\text{particle}, f} = \frac{1}{2} m v_2^2 = \frac{1}{2} (0.02) (90)^2 = 81 \text{ J}$ $K_{\text{disk}, f} = \frac{1}{2} I_C \omega^2 = \frac{1}{2} (0.06) (-7.138)^2 \approx 1.5285 \text{ J}$ $K_f = 81 + 1.5285 = 82.5285 \text{ J}$

• Energy Loss ($\Delta E$): $\Delta E = K_i - K_f = 100 - 82.5285 = 17.4715 \text{ J}$

Thus, the amount of energy loss in the collision is approximately $17.47 \text{ J}$ (or within the range $17$ to $18 \text{ J}$).