Two small balls with masses m and 2m are attached to both ends of a rigid rod of length d and negligible mass. If angular momentum of this system is L about an axis (A) passing through its centre of mass and perpendicular to the rod then angular velocity of the system about A is:

### Related Formula L_textrel = m cdot V_textrel cdot r_perp$$L_{\text{rel}} = m \cdot V_{\text{rel}} \cdot r_{\perp}$$ ### Core Logic The relative angular momentum of a particle translating uniformly with respect to another moving reference frame can be found using their relative velocity and the perpendicular distance between their lines of motion. Relative velocity of A with respect to B: V_textrel = V_A - V_B = 72 - 36 = 36 text km/h$$V_{\text{rel}} = V_A - V_B = 72 - 36 = 36 \text{ km/h}$$ ### Step 1: Conversion to SI units V_textrel = 36 times frac518 text m/s = 10 text m/s$$V_{\text{rel}} = 36 \times \frac{5}{18} \text{ m/s} = 10 \text{ m/s}$$ ### Step 2: Final Conclusion Using the perpendicular distance (r_perp = 10 text m$r_{\perp} = 10 \text{ m}$) and mass (m = 10^3 text kg$m = 10^3 \text{ kg}$): L = m cdot V_textrel cdot r_perp$$L = m \cdot V_{\text{rel}} \cdot r_{\perp}$$ L = 1000 times 10 times 10 = 10^5 text kg m^2text/s text (or Jcdottexts)$$L = 1000 \times 10 \times 10 = 10^{5} \text{ kg m}^2\text{/s} \text{ (or J}\cdot\text{s)}$$ ### Pattern Recognition For parallel tracks, the angular momentum of one translating body relative to another is simply m times v_textrelative times texttrack separation$m \times v_{\text{relative}} \times \text{track separation}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Systems of Particles and Rotational Motion Class 11 Physics: Kinematics

### Related Formula Mg - T_2 = Ma$Mg - T_2 = Ma$ T_1 - mg = ma$T_1 - mg = ma$ (T_2 - T_1)r = I alpha = I left(fracarright)$$(T_2 - T_1)r = I \alpha = I \left(\frac{a}{r}\right)$$ ### Core Logic First, evaluate the moment of inertia of the pulley. The pulley consists of: 1. A thin rim of mass M$M$ and radius r$r$. 2. Two rods, each of mass M$M$ and length 2r$2r$ (diameter). Moment of inertia of the rim: I_textrim = Mr^2$I_{\text{rim}} = Mr^2$ Moment of inertia of two rods about the center: I_textrods = 2 times left(fracM(2r)^212right) = 2 times frac4Mr^212 = frac23Mr^2$I_{\text{rods}} = 2 \times \left(\frac{M(2r)^2}{12}\right) = 2 \times \frac{4Mr^2}{12} = \frac{2}{3}Mr^2$ Total I = Mr^2 + frac23Mr^2 = frac53Mr^2$I = Mr^2 + \frac{2}{3}Mr^2 = \frac{5}{3}Mr^2$ ### Step 1: Force Equations From the free body diagrams of the descending block (mass M$M$) and ascending block (mass m$m$): Mg - T_2 = Ma quad text--- (1)$$Mg - T_2 = Ma \quad \text{--- (1)}$$ T_1 - mg = ma quad text--- (2)$$T_1 - mg = ma \quad \text{--- (2)}$$ Torque on the pulley: (T_2 - T_1)r = Ialpha = Ifracar implies T_2 - T_1 = fracIr^2a quad text--- (3)$$(T_2 - T_1)r = I\alpha = I\frac{a}{r} \implies T_2 - T_1 = \frac{I}{r^2}a \quad \text{--- (3)}$$ ### Step 2: Final Conclusion Adding equations (1), (2), and (3): Mg - mg = left(M + m + fracIr^2right)a$$Mg - mg = \left(M + m + \frac{I}{r^2}\right)a$$ Substitute I = frac53Mr^2$I = \frac{5}{3}Mr^2$: (M - m)g = left(M + m + frac5M3right)a$$(M - m)g = \left(M + m + \frac{5M}{3}\right)a$$ (M - m)g = left(frac8M3 + mright)a$$(M - m)g = \left(\frac{8M}{3} + m\right)a$$ a = frac(M-m)gleft[left(frac83right)M + mright]$$a = \frac{(M-m)g}{\left[\left(\frac{8}{3}\right)M + m\right]}$$ ### Pattern Recognition For a real pulley system, the effective mass acting against acceleration incorporates an inertia term: m_texteff = m_1 + m_2 + fracIR^2$m_{\text{eff}} = m_1 + m_2 + \frac{I}{R^2}$. Breaking the complex pulley into fundamental shapes (ring + rods) solves the inertia safely. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Systems of Particles and Rotational Motion Class 11 Physics: Laws of Motion

### Related Formula For a solid cylinder: Parallel to its length passing through center: I = fracM'R^22$I = \frac{M'R^{2}}{2}$ Perpendicular to length through center: I = fracM'R^24 + fracM'L^212$I = \frac{M'R^{2}}{4} + \frac{M'L^{2}}{12}$ Parallel Axis Theorem: I = I_textcm + M'd^2$I = I_{\text{cm}} + M'd^{2}$ ### Core Logic The square loop is formed by four identical solid cylinders, each of mass M' = fracM4$M' = \frac{M}{4}$. The given axis passes through the mid-points of two opposite cylinders. This means two cylinders have the axis passing perpendicularly through their centers, and the other two cylinders are parallel to the axis at a distance of L/2$L/2$. ### Step 1: Moment of Inertia for cylinders bisected perpendicularly For the two cylinders perpendicular to the axis of rotation: I_1 = 2 times left(fracM'R^24 + fracM'L^212right)$$I_{1} = 2 \times \left(\frac{M'R^{2}}{4} + \frac{M'L^{2}}{12}\right)$$ ### Step 2: Moment of Inertia for cylinders parallel to axis For the two cylinders parallel to the axis, distance d = L/2$d = L/2$. Apply the parallel axis theorem: I_2 = 2 times left[fracM'R^22 + M'left(fracL2right)^2right]$$I_{2} = 2 \times \left[\frac{M'R^{2}}{2} + M'\left(\frac{L}{2}\right)^{2}\right]$$ ### Step 3: Total Moment of Inertia I_textnet = I_1 + I_2 = 2left(fracM'R^24 + fracM'L^212right) + 2left(fracM'R^22 + fracM'L^24right)$$I_{\text{net}} = I_{1} + I_{2} = 2\left(\frac{M'R^{2}}{4} + \frac{M'L^{2}}{12}\right) + 2\left(\frac{M'R^{2}}{2} + \frac{M'L^{2}}{4}\right)$$ I_textnet = fracM'R^22 + fracM'L^26 + M'R^2 + fracM'L^22$$I_{\text{net}} = \frac{M'R^{2}}{2} + \frac{M'L^{2}}{6} + M'R^{2} + \frac{M'L^{2}}{2}$$ I_textnet = frac3M'R^22 + frac4M'L^26 = frac3M'R^22 + frac2M'L^23$$I_{\text{net}} = \frac{3M'R^{2}}{2} + \frac{4M'L^{2}}{6} = \frac{3M'R^{2}}{2} + \frac{2M'L^{2}}{3}$$ ### Step 4: Substitute Total Mass Substitute M' = M/4$M' = M/4$: I = frac32left(fracM4right)R^2 + frac23left(fracM4right)L^2$$I = \frac{3}{2}\left(\frac{M}{4}\right)R^{2} + \frac{2}{3}\left(\frac{M}{4}\right)L^{2}$$ I = frac38MR^2 + frac16ML^2$$I = \frac{3}{8}MR^{2} + \frac{1}{6}ML^{2}$$ ### Pattern Recognition Sees: "square loop of cylinders" + "mass M of entire loop" → Always remember M_i = M/4$M_{i} = M/4$. Calculate individual moment of inertia carefully considering whether the cylinder is oriented parallel or perpendicular. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Systems of Particles and Rotational Motion