Question Stem for Question Nos. 17 and 18

A uniform circular disk of radius $$0.2\,\mathrm{m}$$ and mass $$1\,\mathrm{kg}$$ 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 $$20\,\mathrm{g}$$, travelling along negative $$x$$ direction in the $$XY$$ plane with speed $$100\,\mathrm{ms^{-1}}$$, hits the circumference of the disk at a point $$P$$. After collision the particle moves along negative $$y$$ direction at a speed of $$90\,\mathrm{ms^{-1}}$$.

[Given: the acceleration due to gravity $$(g)=-10\hat{j}\,\mathrm{ms^{-2}}$$]

Amount of energy loss (in J) in the collision is:

Angular momentum about the pivot point $$C$$ is conserved during the collision, and the mechanical energy loss is the difference between initial and final kinetic energies of the system.

Using coordinates relative to center $$O(0,0)$$:

$$C = (0, R) = (0, 0.2)$$

$$P = (R\sin 45^\circ, -R\cos 45^\circ) = \left(0.2\frac{1}{\sqrt{2}}, -0.2\frac{1}{\sqrt{2}}\right)$$

Finding the position vector from $$C$$ to $$P$$:

$$\vec{r}_{P/C} = \vec{r}_P - \vec{r}_C = 0.2\frac{1}{\sqrt{2}}\hat{i} - \left(0.2 + 0.2\frac{1}{\sqrt{2}}\right)\hat{j}$$

Finding the initial and final angular momentum of the particle about $$C$$:

$$\vec{L}_{i,p} = \vec{r}_{P/C} \times (m\vec{u}) = \left[0.2\frac{1}{\sqrt{2}}\hat{i} - \left(0.2 + 0.2\frac{1}{\sqrt{2}}\right)\hat{j}\right] \times (-0.02 \times 100\hat{i})$$

$$\vec{L}_{i,p} = -\left(0.2 + 0.2\frac{1}{\sqrt{2}}\right)(2)\hat{k} = -0.4\left(1 + \frac{1}{\sqrt{2}}\right)\hat{k}$$

$$\vec{L}_{f,p} = \vec{r}_{P/C} \times (m\vec{v}) = \left[0.2\frac{1}{\sqrt{2}}\hat{i} - \left(0.2 + 0.2\frac{1}{\sqrt{2}}\right)\hat{j}\right] \times (-0.02 \times 90\hat{j})$$

$$\vec{L}_{f,p} = \left(0.2\frac{1}{\sqrt{2}}\right)(-1.8)\hat{k} = -\frac{0.36}{\sqrt{2}}\hat{k}$$

Using parallel axis theorem for the disk about pivot $$C$$:

$$I_C = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}(1)(0.2)^2 = 0.06\text{ kg}\cdot\text{m}^2$$

From conservation of angular momentum about $$C$$:

$$L_{i,p} = L_{f,p} + I_C\omega$$

$$-0.4 - \frac{0.4}{\sqrt{2}} = -\frac{0.36}{\sqrt{2}} + 0.06\omega$$

$$0.06\omega = -0.4 - \frac{0.04}{\sqrt{2}} = -0.4 - 0.02828 = -0.42828 \implies \omega \approx -7.138\text{ rad}\cdot\text{s}^{-1}$$

Finding kinetic energies:

$$K_i = \frac{1}{2}mu^2 = \frac{1}{2}(0.02)(100)^2 = 100\text{ J}$$

$$K_f = \frac{1}{2}mv^2 + \frac{1}{2}I_C\omega^2 = \frac{1}{2}(0.02)(90)^2 + \frac{1}{2}(0.06)(-7.138)^2 = 81 + 1.53 = 82.53\text{ J}$$

Finding energy loss: $$\Delta K = K_i - K_f = 100 - 82.53 = 17.47\text{ J}$$