Laws of MotionHard

Question

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 center. The magnitudes of the acceleration experienced by the blocks is $\_\_\_\_$ (assume no slipping of string on pulley.)

Options

A.$\frac{(M - m)g}{\left\lbrack \left( \frac{13}{6} \right)M + m \right\rbrack}$
B.$\frac{(M - m)g}{M + m}$
C.$\frac{(M - m)g}{\left\lbrack \left( \frac{8}{3} \right)M + m \right\rbrack}$
D.$\frac{(M - m)g}{2M + m}$

Solution

$$\begin{array}{r} Mg - T_{2} = Ma\#(1) \\ {\text{ }T}_{1} - mg = ma\#(2) \\ \left( {\text{ }T}_{2} - T_{1} \right)r = I\frac{a}{r}\#(3) \\ \ (1) + (2) + (3)\#(3) \\ \ (M - m)g = \left( M + m + \frac{I}{r^{2}} \right)a\#(3) \end{array}$$

Here $I = Mr^{2} + \frac{M \times (2r)^{2}}{12} \times 2$

$$\begin{matrix} & \ = \left( 1 + \frac{2}{3} \right){Mr}^{2} \\ & \ = \frac{5}{3}{Mr}^{2} \end{matrix}$$

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