### Related Formula Velocity of a point on a rolling wheel at angular position theta$\theta$ from the lowest point: v = 2 v_textcm sinleft(fractheta2right)$v = 2 v_{\text{cm}} \sin\left(\frac{\theta}{2}\right)$ ### Core Logic At the highest point, theta = 180^circ$\theta = 180^\circ$, so: v_texttop = 2v_textcm = 8text m/s implies v_textcm = 4text m/s$v_{\text{top}} = 2v_{\text{cm}} = 8\text{ m/s} \implies v_{\text{cm}} = 4\text{ m/s}$ For a particle on the rim at the same horizontal level as the center, the angle from the lowest point is theta = 90^circ$\theta = 90^\circ$. ### Step 1: Compute Speed at Mid-Height Substituting theta = 90^circ$\theta = 90^\circ$ into our velocity relation: v_textmid = 2v_textcm sin(45^circ) = 2 times 4 times frac1sqrt2 = 4sqrt2text m/s$v_{\text{mid}} = 2v_{\text{cm}} \sin(45^\circ) = 2 \times 4 \times \frac{1}{\sqrt{2}} = 4\sqrt{2}\text{ m/s}$

### Pattern Recognition The contact point with the ground is the Instantaneous Center of Rotation (ICR). Distance to top point is 2R$2R$, distance to mid-level point is sqrtR^2+R^2 = sqrt2R$\sqrt{R^2+R^2} = \sqrt{2}R$. Velocity scales linearly with distance from ICR. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Rotational Motion

### Related Formula Conservation of Angular Momentum (since no external torque acts): I_1 omega_1 = I_2 omega_2$I_1 \omega_1 = I_2 \omega_2$ For a solid sphere, moment of inertia is: I = frac25MR^2$I = \frac{2}{5}MR^2$ Mass scales with volume: M propto R^3$M \propto R^3$ ### Core Logic When the radius reduces to R_2 = fracR2$R_2 = \frac{R}{2}$, the mass scales cubically: M_2 = M_1 left(fracR/2Rright)^3 = fracM_18$M_2 = M_1 \left(\frac{R/2}{R}\right)^3 = \frac{M_1}{8}$ Now, compute the new moment of inertia I_2$I_2$: I_2 = frac25 M_2 R_2^2 = frac25 left(fracM_18right) left(fracR2right)^2 = frac25 M_1 R^2 times frac132 = fracI_132$I_2 = \frac{2}{5} M_2 R_2^2 = \frac{2}{5} \left(\frac{M_1}{8}\right) \left(\frac{R}{2}\right)^2 = \frac{2}{5} M_1 R^2 \times \frac{1}{32} = \frac{I_1}{32}$ ### Step 1: Compute Final Angular Velocity Using conservation of angular momentum: I_1 omega_1 = left(fracI_132right) omega_2 implies omega_2 = 32 omega_1$I_1 \omega_1 = \left(\frac{I_1}{32}\right) \omega_2 \implies \omega_2 = 32 \omega_1$ Hence, the value of x$x$ is **32**. ### Pattern Recognition Since inertia of a solid sphere scales with M R^2$M R^2$ and M propto R^3$M \propto R^3$, the net moment of inertia scales with R^5$R^5$. Shrinking the radius by half (1/2$1/2$) cuts down inertia by a factor of (1/2)^5 = 1/32$(1/2)^5 = 1/32$. Velocity must scale up by 32 to conserve momentum. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Rotational Motion

### Related Formula Fundamental mathematical definition of torque: vectau = vecr times vecF $ vec{\tau} = \vec{r} \times \vec{F} $ Rotational analogue of Newton's second law: vectau = fracdvecLdt = Ivecalpha $ vec{\tau} = \frac{dvec{L}}{dt} = Ivec{\alpha} $ Linear momentum relations: vecL = vecr times vecp implies vectau = fracddt(vecr times vecp) $ vec{L} = \vec{r} \times \vec{p} implies \vec{\tau} = \frac{d}{dt}(\vec{r} \times \vec{p}) $ ### Core Logic Let's check each expression sequentially: * **A.** vectau=vecItimesvecL$\vec{\tau}=\vec{I}\timesvec{L}$ is dimensionally incorrect (Moment of inertia I$I$ is primarily treated as a tensor or scalar placeholder, not crossed directly like this). * **B.** vectau=fracdvecLdt = fracddt(vecrtimesvecp)$\vec{\tau}=\frac{dvec{L}}{dt} = \frac{d}{dt}(\vec{r}\timesvec{p})$ is fundamentally correct. * **C.** vectau=vecrtimesvecF = vecrtimesfracdvecpdt$\vec{\tau}=\vec{r}\timesvec{F} = \vec{r}\times\frac{dvec{p}}{dt}$ is correct since vecF = fracdvecpdt$\vec{F} = \frac{dvec{p}}{dt}$. * **D.** vectau=Ivecalpha$\vec{\tau}=Ivec{\alpha}$ is the standard scalar component/fixed axis formulation. * **E.** vectau=vecrtimesvecF$\vec{\tau}=\vec{r}\timesvec{F}$ is the true physical vector definition. Thus, statements B, C, D, and E are universally correct representations. ### Pattern Recognition Torque can be represented either through geometric structural parameters (position and force cross products) or via kinematic response properties (rate of change of angular momentum or product of rotational inertia and acceleration). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion

### Related Formula Acceleration vector equation via secondary derivation: veca = fracd^2vecrdt^2 $\vec{a} = \frac{d^2\vec{r}}{dt^2} $ Force equation: vecF = mveca $ vec{F} = mvec{a} $ ### Core Logic Given position tracking trace: vecr = a(haticosomega t + hatjsinomega t) $ vec{r} = a(\hat{i}cosomega t + \hat{j}sinomega t) $ Velocity vector vecv$\vec{v}$: vecv = fracdvecrdt = aomega(-hatisinomega t + hatjcosomega t) $ vec{v} = \frac{dvec{r}}{dt} = aomega(-\hat{i}sinomega t + \hat{j}cosomega t) $ ### Step 1: Differentiate to find Acceleration veca = fracdvecvdt = aomega^2(-haticosomega t - hatjsinomega t) $ vec{a} = \frac{dvec{v}}{dt} = aomega^2(-\hat{i}cosomega t - \hat{j}sinomega t) $ veca = -omega^2 left[ a(haticosomega t + hatjsinomega t) ight] = -omega^2vecr $ vec{a} = -\omega^2 \left[ a(\hat{i}cosomega t + \hat{j}sinomega t) ight] = -\omega^2\vec{r} $ ### Step 2: Establish Force Direction vecF = mveca = -momega^2vecr$ vec{F} = mvec{a} = -momega^2\vec{r}$ The minus sign indicates the net centripetal pulling force aligns explicitly **opposite to the direction of vecr$\vec{r}$**. ### Pattern Recognition The expression describes a standard uniform circular motion profile. In circular configurations, acceleration and centripetal forces point radially inward, directly opposing the outbound position tracker vector. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinematics Class 11 Physics: System of Particles and Rotational Motion

### 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