### Related Formula Velocity of a rolling body from mechanical energy conservation: v = sqrtfrac2gh1 + frack^2R^2$v = \sqrt{\frac{2gh}{1 + \frac{k^2}{R^2}}}$ ### Core Logic Evaluate radius of gyration factor coefficients: 1. For a circular ring: frack^2R^2 = 1$\frac{k^2}{R^2} = 1$ 2. For a solid sphere: frack^2R^2 = frac25$\frac{k^2}{R^2} = \frac{2}{5}$ ### Step 1: Compute Velocity Expressions v_textring = sqrtfrac2gh1 + 1 = sqrtgh$v_{\text{ring}} = \sqrt{\frac{2gh}{1 + 1}} = \sqrt{gh}$ v_textsphere = sqrtfrac2gh1 + frac25 = sqrtfrac10gh7$v_{\text{sphere}} = \sqrt{\frac{2gh}{1 + \frac{2}{5}}} = \sqrt{\frac{10gh}{7}}$ ### Step 2: Calculate the Velocity Ratio fracv_textringv_textsphere = fracsqrtghsqrtfrac10gh7 = sqrtfrac710 = sqrtfrac3.55$\frac{v_{\text{ring}}}{v_{\text{sphere}}} = \frac{\sqrt{gh}}{\sqrt{\frac{10gh}{7}}} = \sqrt{\frac{7}{10}} = \sqrt{\frac{3.5}{5}}$ Matching with the prompt expression format sqrtfracx5$\sqrt{\frac{x}{5}}$ reveals: x = 3.5$x = 3.5$ Rounding to the nearest integer yields 4. ### Pattern Recognition Objects with lower mass concentration near the center (lower frack^2R^2$\frac{k^2}{R^2}$ like the sphere) convert gravitational potential energy into translational kinetic energy more efficiently, rolling faster than hollow equivalents. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion

### Related Formula I_textdisc = frac12 M R^2$I_{\text{disc}} = \frac{1}{2} M R^2$ [cite: 833] I = I_textcm + M d^2 quad text(Parallel Axis Theorem)$I = I_{\text{cm}} + M d^2 \quad \text{(Parallel Axis Theorem)}$ [cite: 848] ### Core Logic Let the original mass density per unit area be sigma$\sigma$. Without any cavity, the moment of inertia is: [cite: 198, 833] I_1 = fracMR^22$I_1 = \frac{MR^2}{2}$ [cite: 833] The mass of the removed small section scales directly with its cut-out area profile: [cite: 198, 834] m = fracMpi R^2 times pi left(fracR3right)^2 = fracM9$m = \frac{M}{\pi R^2} \times \pi \left(\frac{R}{3}\right)^2 = \frac{M}{9}$ [cite: 198, 834, 844] The center of mass of the removed disk sits at a distance d = R - fracR3 = frac2R3$d = R - \frac{R}{3} = \frac{2R}{3}$ away from the primary center O$O$[cite: 198, 205, 848]. Calculating its partial moment of inertia about O$O$ via the parallel axis theorem: [cite: 199, 848] I_2 = fracm r^22 + m d^2 = fracfracM9left(fracR3right)^22 + fracM9left(frac2R3 ight)^2$I_2 = \frac{m r^2}{2} + m d^2 = \frac{\frac{M}{9}\left(\frac{R}{3}\right)^2}{2} + \frac{M}{9}\left(\frac{2R}{3} ight)^2$ [cite: 848] I_2 = fracMR^2162 + frac4MR^281 = fracMR^2 + 8MR^2162 = frac9MR^2162 = fracMR^218$I_2 = \frac{MR^2}{162} + \frac{4MR^2}{81} = \frac{MR^2 + 8MR^2}{162} = \frac{9MR^2}{162} = \frac{MR^2}{18}$ [cite: 848, 850] Subtracting the removed component from the original configuration: [cite: 851] I = I_1 - I_2 = fracMR^22 - fracMR^218 = frac9MR^2 - MR^218 = frac8MR^218 = frac49MR^2$I = I_1 - I_2 = \frac{MR^2}{2} - \frac{MR^2}{18} = \frac{9MR^2 - MR^2}{18} = \frac{8MR^2}{18} = \frac{4}{9}MR^2$ [cite: 851] Matching this with frac4xMR^2$\frac{4}{x}MR^2$, we find x = 9$x = 9$[cite: 200, 208, 851, 853]. ### Pattern Recognition For uniform planar surfaces, mass always scales squarely with linear dimension changes (r rightarrow fracr3 implies m rightarrow fracm9$r \rightarrow \frac{r}{3} \implies m \rightarrow \frac{m}{9}$)[cite: 198, 834, 844]. Always apply the parallel axis theorem to bring the component values to a unified reference point before executing addition or subtraction[cite: 848, 851]. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion

### Related Formula K_textlinear = frac12 m v_textcm^2$K_{\text{linear}} = \frac{1}{2} m v_{\text{cm}}^2$ K_textrotational = frac12 I omega^2$K_{\text{rotational}} = \frac{1}{2} I \omega^2$ ### Core Logic For a solid sphere, the moment of inertia about the center of mass is I = frac25mR^2$I = \frac{2}{5}mR^2$. Since it rolls without slipping, the condition v_textcm = omega R$v_{\text{cm}} = \omega R$ holds. Substituting I$I$ and omega$\omega$ into the ratio: fracK_textlinearK_textrotational = fracfrac12 m v_textcm^2frac12 left(frac25mR^2 ight) left(fracv_textcmR ight)^2$\frac{K_{\text{linear}}}{K_{\text{rotational}}} = \frac{\frac{1}{2} m v_{\text{cm}}^2}{\frac{1}{2} \left(\frac{2}{5}mR^2 ight) \left(\frac{v_{\text{cm}}}{R} ight)^2}$ fracK_textlinearK_textrotational = frac1frac25 = frac52$\frac{K_{\text{linear}}}{K_{\text{rotational}}} = \frac{1}{\frac{2}{5}} = \frac{5}{2}$ ### Pattern Recognition The ratio of translational to rotational kinetic energy for any rolling body is given by fracmR^2I_textcm$\frac{mR^2}{I_{\text{cm}}}$. For a solid sphere, this becomes frac12/5 = frac52$\frac{1}{2/5} = \frac{5}{2}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion

### Related Formula t = sqrtfrac2ella_textcm$t = \sqrt{\frac{2\ell}{a_{\text{cm}}}}$ a_textcm = fracg sin theta1 + fracI_textcmMR^2$a_{\text{cm}} = \frac{g \sin \theta}{1 + \frac{I_{\text{cm}}}{MR^2}}$ ### Core Logic For a solid sphere: I_textsolid = frac25MR^2 implies a_1 = fracgsintheta1 + 2/5 = frac57gsintheta$I_{\text{solid}} = \frac{2}{5}MR^2 \implies a_1 = \frac{g\sin\theta}{1 + 2/5} = \frac{5}{7}g\sin\theta$. For a hollow sphere: I_texthollow = frac23MR^2 implies a_2 = fracgsintheta1 + 2/3 = frac35gsintheta$I_{\text{hollow}} = \frac{2}{3}MR^2 \implies a_2 = \frac{g\sin\theta}{1 + 2/3} = \frac{3}{5}g\sin\theta$. Comparing accelerations: a_1 > a_2$a_1 > a_2$ Since acceleration of the solid sphere is greater, it takes less time to descend the incline: t_1 < t_2$t_1 < t_2$

### Pattern Recognition Smaller moment of inertia mass distribution distribution (more concentrated at the center) yields larger acceleration down an incline, meaning a quicker descent. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion