### Related Formula vecL = sum left(vecr_i times m vecv_iright)$\vec{L} = \sum \left(\vec{r}_i \times m \vec{v}_i\right)$ vectau_textext = fracdvecLdt$\vec{\tau}_{\text{ext}} = \frac{d\vec{L}}{dt}$ ### Core Logic Since the three masses interact purely through mutual internal gravitational forces, the net external torque acting on the system about any central reference point is zero: vectau_textext = 0 implies vecL_textinitial = vecL_textfinal$\vec{\tau}_{\text{ext}} = 0 \implies \vec{L}_{\text{initial}} = \vec{L}_{\text{final}}$ Let us compute the total angular momentum about the centroid of the equilateral triangle:

From trigonometry, the perpendicular distance from the centroid to the velocity vector along any edge is: r_perp = fraca2sqrt3$r_{\perp} = \frac{a}{2\sqrt{3}}$ The initial angular momentum for one mass about the centroid is L_1 = m V_0 r_perp$L_1 = m V_0 r_{\perp}$. Since all three particles move cyclically in the same direction, their angular momenta reinforce cleanly: L_textnet = 3 cdot left(m V_0 fraca2sqrt3right) = frac32sqrt3 m V_0 a = fracsqrt32 a m V_0$L_{\text{net}} = 3 \cdot \left(m V_0 \frac{a}{2\sqrt{3}}\right) = \frac{3}{2\sqrt{3}} m V_0 a = \frac{\sqrt{3}}{2} a m V_0$ By conservation of angular momentum, this configuration value remains unchanged up to the point of collision. ### Pattern Recognition Mutual internal central forces can never alter the angular momentum of a system. Hence, the solution completely reduces to measuring the static configuration values at t=0$t=0$ about the center of mass symmetry. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion Class 11 Physics: Gravitation

### Core Logic Since density sigma$\sigma$ is independent of the y-coordinate, the vertical center of mass resolves directly by symmetry: mathrmy_mathrmcm = fracmathrmb2$\mathrm{y}_{\mathrm{cm}} = \frac{\mathrm{b}}{2}$

To find the horizontal center of mass, evaluate the continuous mass integral along the x-axis: mathrmx_mathrmcm = fracint_0^mathrma mathrmx \, dmint_0^mathrma mathrmdm = fracint_0^mathrma mathrmx left(fracsigma_0 mathrmxmathrmabright) mathrmb \, dxint_0^mathrma left(fracsigma_0 mathrmxmathrmabright) mathrmb \, dx$\mathrm{x}_{\mathrm{cm}} = \frac{\int_0^{\mathrm{a}} \mathrm{x \, dm}}{\int_0^{\mathrm{a}} \mathrm{dm}} = \frac{\int_0^{\mathrm{a}} \mathrm{x} \left(\frac{\sigma_0 \mathrm{x}}{\mathrm{ab}}\right) \mathrm{b \, dx}}{\int_0^{\mathrm{a}} \left(\frac{\sigma_0 \mathrm{x}}{\mathrm{ab}}\right) \mathrm{b \, dx}}$ mathrmx_mathrmcm = fracint_0^mathrma mathrmx^2 \, mathrmdxint_0^mathrma mathrmx \, mathrmdx = fracleft[ fracmathrmx^33 right]_0^mathrmaleft[ fracmathrmx^22 right]_0^mathrma = fracmathrma^3 / 3mathrma^2 / 2 = frac2mathrma3$\mathrm{x}_{\mathrm{cm}} = \frac{\int_0^{\mathrm{a}} \mathrm{x}^2 \, \mathrm{dx}}{\int_0^{\mathrm{a}} \mathrm{x} \, \mathrm{dx}} = \frac{\left[ \frac{\mathrm{x}^3}{3} \right]_0^{\mathrm{a}}}{\left[ \frac{\mathrm{x}^2}{2} \right]_0^{\mathrm{a}}} = \frac{\mathrm{a}^3 / 3}{\mathrm{a}^2 / 2} = \frac{2\mathrm{a}}{3}$ ### Step 1: Final Position Coordinates The center of mass coordinates are left(frac23 mathrma, fracmathrmb2right)$\left(\frac{2}{3} \mathrm{a}, \frac{\mathrm{b}}{2}\right)$, which matches option (1). ### Pattern Recognition When density varies linearly with position (sigma propto mathrmx$\sigma \propto \mathrm{x}$), the mass distribution shifts outward, moving the center of mass from the geometric midpoint fracmathrma2$\frac{\mathrm{a}}{2}$ to the frac23mathrma$\frac{2}{3}\mathrm{a}$ mark. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion

### Related Formula mathrmI_textdisc = fracmathrmMmathrmR_1^24, quad mathrmI_textring = fracmathrmMmathrmR_2^22, quad mathrmI_textsphere = frac2mathrmMmathrmR_1^25$\mathrm{I}_{\text{disc}} = \frac{\mathrm{M}\mathrm{R}_1^2}{4}, \quad \mathrm{I}_{\text{ring}} = \frac{\mathrm{M}\mathrm{R}_2^2}{2}, \quad \mathrm{I}_{\text{sphere}} = \frac{2\mathrm{M}\mathrm{R}_1^2}{5}$ ### Core Logic Let's list the relevant moment of inertia formulas based on their rotation axes:

From the given problem statements: fracmathrmI_textdiscmathrmI_textring = 2.5 implies fracfracmathrmMmathrmR_1^24fracmathrmMmathrmR_2^22 = frac52 implies fracmathrmR_1^2mathrmR_2^2 = 5$\frac{\mathrm{I}_{\text{disc}}}{\mathrm{I}_{\text{ring}}} = 2.5 \implies \frac{\frac{\mathrm{M}\mathrm{R}_1^2}{4}}{\frac{\mathrm{M}\mathrm{R}_2^2}{2}} = \frac{5}{2} \implies \frac{\mathrm{R}_1^2}{\mathrm{R}_2^2} = 5$ Now, evaluating the second geometric layout ratio: fracmathrmI_textspheremathrmI_textring = mathrmn implies fracfrac2mathrmMmathrmR_1^25fracmathrmMmathrmR_2^22 = mathrmn implies frac4mathrmR_1^25mathrmR_2^2 = mathrmn$\frac{\mathrm{I}_{\text{sphere}}}{\mathrm{I}_{\text{ring}}} = \mathrm{n} \implies \frac{\frac{2\mathrm{M}\mathrm{R}_1^2}{5}}{\frac{\mathrm{M}\mathrm{R}_2^2}{2}} = \mathrm{n} \implies \frac{4\mathrm{R}_1^2}{5\mathrm{R}_2^2} = \mathrm{n}$ Substituting our radius parameter (fracmathrmR_1^2mathrmR_2^2 = 5$\frac{\mathrm{R}_1^2}{\mathrm{R}_2^2} = 5$): mathrmn = frac45 cdot 5 = 4$\mathrm{n} = \frac{4}{5} \cdot 5 = 4$ ### Step 1: Final Value Conclusion The scale value parameter is found to be: mathrmn = 4$\mathrm{n} = 4$ ### Pattern Recognition Be careful with rotation axis descriptions. Disc and ring components rotating along their structural diameter axes use values that are half of their standard perpendicular planar formulas. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion

### Core Logic Since mass scales with the face surface area for discs of identical thickness and material composition: mathrmM = sigma cdot pi mathrmR^2 implies mathrmM propto mathrmR^2$\mathrm{M} = \sigma \cdot \pi \mathrm{R}^2 \implies \mathrm{M} \propto \mathrm{R}^2$

mathrmM_1 = mathrmM_0, quad mathrmM_2 = sigma pi (2mathrmR_1)^2 = 4mathrmM_0$\mathrm{M}_1 = \mathrm{M}_0, \quad \mathrm{M}_2 = \sigma \pi (2\mathrm{R}_1)^2 = 4\mathrm{M}_0$ The moment of inertia formula for a disc is: mathrmI = frac12mathrmMmathrmR^2 implies mathrmI propto mathrmMmathrmR^2 propto mathrmR^4$mathrm{I} = \frac{1}{2}\mathrm{M}\mathrm{R}^2 \implies \mathrm{I} \propto \mathrm{M}\mathrm{R}^2 \propto \mathrm{R}^4$ Calculating the ratio for the given radii configuration: fracmathrmI_1mathrmI_2 = fracmathrmM_1 mathrmR_1^2mathrmM_2 mathrmR_2^2 = fracmathrmM_0 cdot mathrmR_1^24mathrmM_0 cdot (2mathrmR_1)^2 = frac116$\frac{\mathrm{I}_1}{\mathrm{I}_2} = \frac{\mathrm{M}_1 \mathrm{R}_1^2}{\mathrm{M}_2 \mathrm{R}_2^2} = \frac{\mathrm{M}_0 \cdot \mathrm{R}_1^2}{4\mathrm{M}_0 \cdot (2\mathrm{R}_1)^2} = \frac{1}{16}$ ### Step 1: Value Convergence Comparing this fraction to 1/mathrmx$1/\mathrm{x}$ yields: mathrmx = 16$\mathrm{x} = 16$ ### Pattern Recognition For 2D uniform laminar objects, scaling the radius changes both the mass factor (by mathrmR^2$\mathrm{R}^2$) and the distribution distance (by mathrmR^2$\mathrm{R}^2$), resulting in an overall mathrmR^4$\mathrm{R}^4$ dependency rule. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion

### Related Formula Torque equation about the instantaneous center of zero velocity (bottom contact point P$P$): tau_P = I_P alpha$\tau_P = I_P \alpha$ For a solid sphere, the moment of inertia about the center is I_c = frac25MR^2$I_c = \frac{2}{5}MR^2$. By the parallel axis theorem: I_P = I_c + MR^2 = frac75MR^2$I_P = I_c + MR^2 = \frac{7}{5}MR^2$ ### Core Logic Since the sphere rolls without slipping, we can conveniently write the torque equation about the lowest point of contact P$P$ because static friction passes through this point and exerts zero torque. - Distance from point P$P$ to the top highest point is 2R$2R$. - Tangential force F = 49mathrm~N$F = 49\mathrm{~N}$. - Mass of solid sphere, M = 20mathrm~kg$M = 20\mathrm{~kg}$. tau_P = F times 2R$\tau_P = F \times 2R$ Substitute tau_P$\tau_P$ and I_P$I_P$ into the torque equation: F times 2R = left(frac75MR^2right) alpha$F \times 2R = \left(\frac{7}{5}MR^2\right) \alpha$ ### Step 1: Solving for Linear Acceleration For pure rolling, the acceleration of the center of mass a$a$ is related to angular acceleration alpha$\alpha$ by a = Ralpha$a = R\alpha$: 2F R = frac75MR^2 left(fracaRright)$2F R = \frac{7}{5}MR^2 \left(\frac{a}{R}\right)$ 2F = frac75 M a implies a = frac10F7M$2F = \frac{7}{5} M a \implies a = \frac{10F}{7M}$ Substitute the numerical values (F = 49mathrm~N$F = 49\mathrm{~N}$ and M = 20mathrm~kg$M = 20\mathrm{~kg}$): a = frac10 times 497 times 20 = frac490140 = 3.5mathrm~m/s^2$a = \frac{10 \times 49}{7 \times 20} = \frac{490}{140} = 3.5\mathrm{~m/s}^2$ ### Step 2: Analysis of Friction Force Direction Let's write force equations to verify consistency: F + f = M a$F + f = M a$ 49 + f = 20 times 3.5 = 70 implies f = 21mathrm~N$49 + f = 20 \times 3.5 = 70 \implies f = 21\mathrm{~N}$ Since f$f$ is positive, static friction acts in the forward direction. Rolling without slipping is fully maintained since the required static friction coefficient is well within realistic limits. ### Pattern Recognition Calculating torque about the bottom contact point is a powerful shortcut for rolling-without-slipping questions! It completely bypasses having to guess or set up equations for the friction direction. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion