### Related Formula 1. Torque (tau$\tau$) produced by a tangential pulling force: tau = F cdot R$\tau = F \cdot R$ 2. Newton's second law for rotation: tau = I cdot alpha$\tau = I \cdot \alpha$ ### Core Logic We are given: - Radius of the wheel R = 0.2 \ mathrmm$R = 0.2 \ \mathrm{m}$ - Applied force F = 10 \ mathrmN$F = 10 \ \mathrm{N}$ - Angular acceleration alpha = 2 \ mathrmrad/s^2$\alpha = 2 \ \mathrm{rad/s^2}$ ### Step 1: Calculate torque and moment of inertia First, find the torque tau$\tau$: tau = F cdot R = 10 \ mathrmN times 0.2 \ mathrmm = 2 \ mathrmN cdot m$\tau = F \cdot R = 10 \ \mathrm{N} \times 0.2 \ \mathrm{m} = 2 \ \mathrm{N \cdot m}$ Next, calculate the moment of inertia I$I$ using tau = Ialpha$\tau = I\alpha$: I = fractaualpha = frac2 \ mathrmN cdot m2 \ mathrmrad/s^2 = 1 \ mathrmkg cdot m^2$I = \frac{\tau}{\alpha} = \frac{2 \ \mathrm{N \cdot m}}{2 \ \mathrm{rad/s^2}} = 1 \ \mathrm{kg \cdot m^2}$ Thus, the moment of inertia is 1 mathrm~kgcdot m^2$1 \mathrm{~kg\cdot m^2}$. ### Pattern Recognition Sees: Pulley/wheel torque with basic rotational dynamics. Trap: Attempting to integrate gravity (g = 10text m/s^2$g = 10\text{ m/s}^2$) into mass equations. Gravity is a redundant distractor here because the pulling tension force is explicitly defined! Shortcut: Directly compute torque as tau = F R$\tau = F R$ and divide by the angular acceleration alpha$\alpha$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion

### Related Formula 1. Moment of inertia of a circular ring of mass M$M$ and radius R$R$ about a diametrical axis in its plane: I_textdia = frac12 M R^2$I_{\text{dia}} = \frac{1}{2} M R^2$ 2. Parallel Axis Theorem: I = I_textcm + M d^2$I = I_{\text{cm}} + M d^2$ where d$d$ is the perpendicular distance between the center of mass axis and the parallel axis. ### Core Logic The axis of rotation is tangential and lies in the plane of the ring. Thus, the distance from the center of mass axis (which is also diametrical and in-plane) is equal to the radius R$R$. Using the parallel axis theorem: I_texttangent = I_textdia + M R^2 = frac12 M R^2 + M R^2 = frac32 M R^2$I_{\text{tangent}} = I_{\text{dia}} + M R^2 = \frac{1}{2} M R^2 + M R^2 = \frac{3}{2} M R^2$ ### Step 1: Express in terms of diameter r$r$ The question specifies the diameter of the ring is r$r$. Therefore, the radius R$R$ is: R = fracr2$R = \frac{r}{2}$ Substitute R = fracr2$R = \frac{r}{2}$ into the moment of inertia formula: I_texttangent = frac32 M left(fracr2right)^2 = frac38 M r^2$I_{\text{tangent}} = \frac{3}{2} M \left(\frac{r}{2}\right)^2 = \frac{3}{8} M r^2$ Thus, the moment of inertia is frac38 M r^2$\frac{3}{8} M r^2$. ### Pattern Recognition Sees: Circular ring tangential axis in-plane. Trap: Directly using radius r$r$ instead of diameter r$r$. Read variables carefully: often r$r$ is radius, but here it is explicitly given as diameter! Shortcut: Standard in-plane tangent is frac32MR^2$\frac{3}{2}MR^2$. Substitute R = fracr2$R = \frac{r}{2}$ to get frac38Mr^2$\frac{3}{8}Mr^2$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion

### Related Formula For a continuous system modeled as discrete point masses located at their respective centers of mass: x_textcom = fracm_1x_1 + m_2x_2m_1 + m_2$x_{\text{com}} = \frac{m_1x_1 + m_2x_2}{m_1 + m_2}$ y_textcom = fracm_1y_1 + m_2y_2m_1 + m_2$y_{\text{com}} = \frac{m_1y_1 + m_2y_2}{m_1 + m_2}$ ### Core Logic Let the uniform linear mass density of the rod be lambda$\lambda$. - Total length is 5L$5L$. - One segment of length 2L$2L$ lies on the x-axis. Its mass is 2m = lambda(2L)$2m = \lambda(2L)$ and its center of mass is at (L, 0)$(L, 0)$. - The remaining segment of length 3L$3L$ lies on the y-axis. Its mass is 3m = lambda(3L)$3m = \lambda(3L)$ and its center of mass is at (0, 1.5L)$(0, 1.5L)$. ### Step 1: Calculate COM Coordinates Find the coordinates of the system's center of mass: x_textcom = frac2m(L) + 3m(0)2m + 3m = frac2L5 = 0.4L$x_{\text{com}} = \frac{2m(L) + 3m(0)}{2m + 3m} = \frac{2L}{5} = 0.4L$ y_textcom = frac2m(0) + 3m(1.5L)2m + 3m = frac4.5L5 = 0.9L$y_{\text{com}} = \frac{2m(0) + 3m(1.5L)}{2m + 3m} = \frac{4.5L}{5} = 0.9L$ Given L = 10 mathrm~cm$L = 10 \mathrm{~cm}$: x_textcom = 0.4 times 10 = 4 mathrm~cm$x_{\text{com}} = 0.4 \times 10 = 4 \mathrm{~cm}$ y_textcom = 0.9 times 10 = 9 mathrm~cm$y_{\text{com}} = 0.9 \times 10 = 9 \mathrm{~cm}$ ### Step 2: Vector Form Expressing in vector notation: vecr_textcom = 4hatmathbfi + 9hatmathbfj$\vec{r}_{\text{com}} = 4\hat{\mathbf{i}} + 9\hat{\mathbf{j}}$ ### Pattern Recognition Sees: L-shaped rod formed by bending a total length L_texttotal$L_{\text{total}}$. Shortcut: Treat each arm as a point mass at its geometric midpoint. For segments of ratio 2:3$2:3$, the COM divides the distance between their midpoints in the inverse ratio 3:2$3:2$ closer to the heavier segment on the y-axis. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion

### Related Formula Parallel Axis Theorem: I_textaxis = I_textcom + M R^2$I_{\text{axis}} = I_{\text{com}} + M R^2$ Standard Moments of Inertia about center of mass: - Disc about diameter: I_textdisc,dia = fracMR^24$I_{\text{disc,dia}} = \frac{MR^2}{4}$ - Solid sphere: I_textsphere = frac25MR^2$I_{\text{sphere}} = \frac{2}{5}MR^2$ - Spherical shell: I_textshell = frac23MR^2$I_{\text{shell}} = \frac{2}{3}MR^2$ ### Core Logic The axis of rotation PQ passes through the center of the top disc (A) along its diameter. - Top disc (A): I_A = fracMR^24$I_A = \frac{MR^2}{4}$ - Bottom-left solid sphere (B): Center lies at distance R$R$ from the axis PQ. I_B = I_textcom + M R^2 = frac25MR^2 + MR^2 = frac75MR^2$I_B = I_{\text{com}} + M R^2 = \frac{2}{5}MR^2 + MR^2 = \frac{7}{5}MR^2$ - Bottom-right spherical shell (C): Center lies at distance R$R$ from the axis PQ. I_C = I_textcom + M R^2 = frac23MR^2 + MR^2 = frac53MR^2$I_C = I_{\text{com}} + M R^2 = \frac{2}{3}MR^2 + MR^2 = \frac{5}{3}MR^2$ ### Step 1: Calculate Total System Moment of Inertia Sum the contributions: I_textPQ = I_A + I_B + I_C$I_{\text{PQ}} = I_A + I_B + I_C$ I_textPQ = fracMR^24 + frac75MR^2 + frac53MR^2$I_{\text{PQ}} = \frac{MR^2}{4} + \frac{7}{5}MR^2 + \frac{5}{3}MR^2$ To add the fractions, find a common denominator (60$60$): I_textPQ = left( frac15 + 84 + 10060 right) MR^2 = frac19960 MR^2$I_{\text{PQ}} = \left( \frac{15 + 84 + 100}{60} \right) MR^2 = \frac{199}{60} MR^2$ ### Step 2: Express in terms of standard Disc Moment We are given I = fracMR^24 implies MR^2 = 4I$I = \frac{MR^2}{4} \implies MR^2 = 4I$. Substitute this in the expression: I_textPQ = frac19960 (4I) = frac19915 I$I_{\text{PQ}} = \frac{199}{60} (4I) = \frac{199}{15} I$ Comparing with I_textPQ = fracx15 I$I_{\text{PQ}} = \frac{x}{15} I$ yields x = 199$x = 199$. ### Pattern Recognition Sees: Composite body consisting of three standard symmetric shapes about a tangent/offset axis. Shortcut: Sum the central inertia terms and the offset terms separately. Offset masses are only B and C, so the offset sum is 2MR^2$2MR^2$. The central sum is (1/4 + 2/5 + 2/3)MR^2$(1/4 + 2/5 + 2/3)MR^2$. Adding these directly yields the combined fractional factor of 199/60$199/60$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion

### 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