A fly wheel having mass 3 text kg$3 \text{ kg}$ and radius 5 text m$5 \text{ m}$ is free to rotate about a horizontal axis. A string having negligible mass is wound around the wheel and the loose end of the string is connected to a 3 text kg$3 \text{ kg}$ mass. The mass is kept at rest initially and released. Kinetic energy of the wheel when the mass descends by 3 text m$3 \text{ m}$ is ____ J. (g = 10 text m/s^2$g = 10 \text{ m/s}^{2}$ )

### 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

### Related Formula r_textcm = fracm_1r_1 + m_2r_2m_1 + m_2$$r_{\text{cm}} = \frac{m_{1}r_{1} + m_{2}r_{2}}{m_{1} + m_{2}}$$ I_textcm = mu d^2 = left(fracm_1m_2m_1 + m_2right)d^2$$I_{\text{cm}} = \mu d^{2} = \left(\frac{m_{1}m_{2}}{m_{1} + m_{2}}\right)d^{2}$$ L = Iomega$L = I\omega$ ### Step 1: Calculate Moment of Inertia about COM Let mass m$m$ be at the origin. Position of 2m$2m$ is d$d$. X_textcm = fracm(0) + 2m(d)m + 2m = frac2d3$$X_{\text{cm}} = \frac{m(0) + 2m(d)}{m + 2m} = \frac{2d}{3}$$ Distance of mass m$m$ from COM is frac2d3$\frac{2d}{3}$. Distance of mass 2m$2m$ from COM is d - frac2d3 = fracd3$d - \frac{2d}{3} = \frac{d}{3}$. I = mleft(frac2d3right)^2 + 2mleft(fracd3right)^2$$I = m\left(\frac{2d}{3}\right)^{2} + 2m\left(\frac{d}{3}\right)^{2}$$ I = frac4md^29 + frac2md^29 = frac6md^29 = frac2md^23$$I = \frac{4md^{2}}{9} + \frac{2md^{2}}{9} = \frac{6md^{2}}{9} = \frac{2md^{2}}{3}$$ ### Step 2: Calculate Angular Velocity L = Iomega$L = I\omega$ omega = fracLI$$\omega = \frac{L}{I}$$ omega = fracLfrac2md^23 = frac3L2md^2$$\omega = \frac{L}{\frac{2md^{2}}{3}} = \frac{3L}{2md^{2}}$$ ### Pattern Recognition Sees: "two point masses" + "rotation about COM" → Quickly use reduced mass mu$\mu$ moment of inertia shortcut: I_textcm = mu d^2 = (fracm cdot 2m3m)d^2 = frac23md^2$I_{\text{cm}} = \mu d^{2} = (\frac{m \cdot 2m}{3m})d^2 = \frac{2}{3}md^2$ to save time. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Systems of Particles and Rotational Motion

### Related Formula vecv = fracdvecrdt$$\vec{v} = \frac{d\vec{r}}{dt}$$ vecp = mvecv$$\vec{p} = m\vec{v}$$ veca = fracdvecvdt$$\vec{a} = \frac{d\vec{v}}{dt}$$ vecL = vecr times vecp$$\vec{L} = \vec{r} \times \vec{p}$$ ### Core Logic Under a sign change of coordinates (parity transformation or spatial inversion) vecr rightarrow -vecr$\vec{r} \rightarrow -\vec{r}$. Then, velocity vecv = fracdvecrdt rightarrow -vecv$\vec{v} = \frac{d\vec{r}}{dt} \rightarrow -\vec{v}$. Linear momentum vecp = mvecv rightarrow -vecp$\vec{p} = m\vec{v} \rightarrow -\vec{p}$. Acceleration veca = fracdvecvdt rightarrow -veca$\vec{a} = \frac{d\vec{v}}{dt} \rightarrow -\vec{a}$. Angular momentum vecL = vecr times vecp$\vec{L} = \vec{r} \times \vec{p}$. Under the transformation: vecL' = (-vecr) times (-vecp) = vecr times vecp = vecL$\vec{L}' = (-\vec{r}) \times (-\vec{p}) = \vec{r} \times \vec{p} = \vec{L}$. ### Step 1: Final Conclusion Since both position and momentum vectors change sign, their cross product (angular momentum) retains its original sign. It does not flip. ### Pattern Recognition Angular momentum is a pseudovector (or axial vector). True vectors (polar vectors) flip signs under spatial inversion, but pseudovectors (which are cross products of two polar vectors) do not. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Systems of Particles and Rotational Motion