### 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

### Related Formula The definition of angular momentum vector vecL$\vec{L}$ relative to the origin is: vecL = vecr times vecp = m(vecr times vecv) $\vec{L} = \vec{r} \times \vec{p} = m(\vec{r} \times \vec{v}) $ In scalar form for a horizontal speed component at a maximum altitude H$H$: L = m v_x H$L = m v_{x} H$ ### Core Logic At maximum height, the vertical speed component drops to zero, and the projectile travels entirely horizontally along the curve profile as shown in

: v_x = v_0 cos 45^circ = fracv_0sqrt2$v_{x} = v_{0} \cos 45^{\circ} = \frac{v_{0}}{\sqrt{2}}$ H = fracv_0^2 sin^2 45^circ2g = fracv_0^24g$H = \frac{v_{0}^{2} \sin^{2} 45^{\circ}}{2g} = \frac{v_{0}^{2}}{4g}$ ### Step 1: Calculating Magnitude and Vector Direction Evaluate the horizontal vector cross components: L = m left(fracv_0sqrt2 ight) left(fracv_0^24g ight) = fracmv_0^34sqrt2g$L = m \left(\frac{v_{0}}{\sqrt{2}} ight) \left(\frac{v_{0}^{2}}{4g} ight) = \frac{mv_{0}^{3}}{4\sqrt{2}g}$ Using the right-hand rule, vecr$\vec{r}$ points into quadrant-1 while velocity points towards +hati$+\hat{i}$. Therefore, vecr times vecv$\vec{r} \times \vec{v}$ tracks clockwise, yielding a negative hatk$\hat{k}$ orientation (along the negative z-axis). ### Pattern Recognition At peak height, always map vecL$\vec{L}$ using m cdot v_textpeak cdot y_textmax$m \cdot v_{\text{peak}} \cdot y_{\text{\max}}$. This scalar form simplifies the calculation significantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion

### Related Formula The linear acceleration a$a$ for pure rolling motion down an incline profile is: a = fracgsintheta1 + fracImr^2$a = \frac{g\sin\theta}{1 + \frac{I}{mr^{2}}}$ ### Core Logic For a uniform solid cylinder, the moment of inertia around its central axis is: I = frac12mr^2 implies fracImr^2 = frac12$I = \frac{1}{2}mr^{2} \implies \frac{I}{mr^{2}} = \frac{1}{2}$ ### Step 1: Calculating Acceleration Substitute theta = 45^circ$\theta = 45^{\circ}$ and the cylinder inertial factor into the formula : a = fracgsin 45^circ1 + frac12 = fracfracgsqrt2frac32$a = \frac{g\sin 45^{\circ}}{1 + \frac{1}{2}} = \frac{\frac{g}{\sqrt{2}}}{\frac{3}{2}}$ a = frac2g3sqrt2 = fracsqrt2g3$a = \frac{2g}{3\sqrt{2}} = \frac{\sqrt{2}g}{3}$ ### Pattern Recognition Solid cylinder rolls with acceleration matching frac23 gsintheta$\frac{2}{3} g\sin\theta$. Since sin 45^circ = frac1sqrt2$\sin 45^{\circ} = \frac{1}{\sqrt{2}}$, this simplifies directly to fracsqrt2g3$\frac{\sqrt{2}g}{3}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion