The pulley shown in figure is made using a thin rim and two rods of length equal to diameter of the rim. The rim and each rod have a mass of M. Two blocks of mass of M and m are attached to two ends of a light string passing over the pulley, which is hinged to rotate freely in vertical plane about its centre. The magnitudes of the acceleration experienced by the blocks is ________ (assume no slipping of string on pulley.)

### 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