In the pulley system shown, the mass of the b | vedclass

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(B) Let the mass of the rod be $M$ and the mass of the ball be $m = 1.2M$. Let $T$ be the tension in the string attached to the rod. The movable pulle

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) Let the mass of the rod be $M$ and the mass of the ball be $m = 1.2M$. Let $T$ be the tension in the string attached to the rod. The movable pulley is supported by two segments of the string, each with tension $T$, so the upward force on the movable pulley is $2T$. This force is transmitted to the ball, so the tension in the string attached to the ball is $2T$.For the ball (mass $m$), the equation of motion is: $m a_1 = 2T - mg$ $(i)$For the rod (mass $M$), the equation of motion is: $M a_2 = Mg - T$ (ii)Since the string length is constant, the constraint relation between accelerations is $a_2 = 2a_1$, or $a_1 = a_2/2$.Substituting $a_1$ into $(i)$: $m(a_2/2) = 2T - mg \Rightarrow T = \frac{m a_2}{4} + \frac{mg}{2}$.Substitute $T$ into (ii): $M a_2 = Mg - (\frac{m a_2}{4} + \frac{mg}{2})$.Rearranging for $a_2$: $a_2(M + m/4) = g(M - m/2)$.$a_2 = g \frac{M - m/2}{M + m/4} = g \frac{1 - (m/M)/2}{1 + (m/M)/4}$.Given $m/M = 1.2$, $a_2 = 10 	imes \frac{1 - 0.6}{1 + 0.3} = 10 	imes \frac{0.4}{1.3} = \frac{4}{1.3} \approx 3.07 \,m/s^2$. The closest option is $3 \,m/s^2$.

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