A solid sphere rotates about a vertical axis | vedclass

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(B) Let $T_1$ be the tension in the cord connected to the sphere and $T_2$ be the tension in the cord connected to the block. Let $a$ be the linear ac

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$\sqrt{\frac{20gh}{19}}$

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(B) Let $T_1$ be the tension in the cord connected to the sphere and $T_2$ be the tension in the cord connected to the block. Let $a$ be the linear acceleration of the block.For the solid sphere (moment of inertia $I_s = \frac{2}{5}MR^2$): $T_1 R = I_s \alpha = (\frac{2}{5}MR^2) \frac{a}{R} \implies T_1 = \frac{2}{5}Ma$.For the solid cylinder (moment of inertia $I_c = \frac{1}{2}MR^2$): $(T_2 - T_1) R = I_c \alpha = (\frac{1}{2}MR^2) \frac{a}{R} \implies T_2 - T_1 = \frac{1}{2}Ma$.For the block of mass $M$: $Mg - T_2 = Ma \implies T_2 = M(g - a)$.Substituting $T_1$ and $T_2$ into the cylinder equation: $M(g - a) - \frac{2}{5}Ma = \frac{1}{2}Ma$.$g - a = (\frac{1}{2} + \frac{2}{5})a = \frac{9}{10}a$.$g = a(1 + \frac{9}{10}) = \frac{19}{10}a \implies a = \frac{10g}{19}$.Using $v^2 = u^2 + 2ah$ with $u=0$: $v^2 = 2(\frac{10g}{19})h = \frac{20gh}{19}$.Thus,$v = \sqrt{\frac{20gh}{19}}$.

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