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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 $$(g)=-10\hat{j}\,\mathrm{ms^{-2}}$$]
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:
Correct Answer: 0.15
Conservation of mechanical energy governs the pure rotation of the disk about its pivot point $$C$$ after the collision, where kinetic energy is entirely converted into gravitational potential energy at the highest point of swing.
$$I_C = \frac{3}{2}MR^2 = 0.06\text{ kg}\cdot\text{m}^2$$
rom conservation of mechanical energy post-collision:
$$E_i = E_f$$
$$\frac{1}{2}I_C\omega^2 = Mg\Delta h_{\text{cm}}$$
Substituting values to find the maximum height change $$\Delta h$$ of center $$O$$:
$$\frac{1}{2}(0.06)(7.138)^2 = (1)(10)\Delta h$$
$$0.03 \times 50.95 = 10\Delta h$$
$$1.5285 = 10\Delta h \implies \Delta h = 0.15285\text{ m} \approx 0.15\text{ m}$$
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