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(C) Let the radius of the rim be $r$. The moment of inertia $I$ of the pulley consists of the rim and two rods.$I = I_{	ext{rim}} + 2 	imes I_{	ext

Vedclass · Accepted Answer

$\frac{( M - m ) g }{\left[\left(\frac{8}{3}\right) M + m \right]}$

Vedclass · Answer

(C) Let the radius of the rim be $r$. The moment of inertia $I$ of the pulley consists of the rim and two rods.$I = I_{	ext{rim}} + 2 	imes I_{	ext{rod}}$$I = Mr^2 + 2 	imes \left( \frac{M(2r)^2}{12} ight) = Mr^2 + 2 	imes \left( \frac{4Mr^2}{12} ight) = Mr^2 + \frac{2}{3}Mr^2 = \frac{5}{3}Mr^2$.Let $a$ be the acceleration of the blocks and $T_1, T_2$ be the tensions in the string.The equations of motion are:$Mg - T_2 = Ma$ ... $(1)$$T_1 - mg = ma$ ... $(2)$$(T_2 - T_1)r = I \alpha = I \left( \frac{a}{r} ight) \implies T_2 - T_1 = \frac{I}{r^2} a = \frac{5}{3}Ma$ ... $(3)$Adding $(1)$, $(2)$, and $(3)$:$(M - m)g = (M + m + \frac{5}{3}M)a$$(M - m)g = (\frac{8}{3}M + m)a$$a = \frac{(M - m)g}{\frac{8}{3}M + m}$.

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