A rod of length l and mass m lies on a fixed | vedclass

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(C) Let $a_{cm}$ be the acceleration of the centre of mass of the rod and $\alpha$ be the angular acceleration about the centre of mass. The accelerat

Vedclass · Accepted Answer

$4$

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(C) Let $a_{cm}$ be the acceleration of the centre of mass of the rod and $\alpha$ be the angular acceleration about the centre of mass. The acceleration of the end of the rod where the string is attached is $a = a_{cm} + \alpha (l/2)$.From the equation of motion for the block $m_1$: $m_1 g - T = m_1 a$.For the rod,the force equation is $T = m a_{cm}$ and the torque equation about the centre of mass is $T(l/2) = I \alpha = (ml^2/12) \alpha$,which gives $T = (m l / 6) \alpha$,so $a_{cm} = (l/6) \alpha$.Substituting $\alpha = 6 a_{cm} / l$ into the expression for $a$: $a = a_{cm} + (l/2)(6 a_{cm} / l) = a_{cm} + 3 a_{cm} = 4 a_{cm}$.Substituting $T = m a_{cm}$ and $a = 4 a_{cm}$ into the block's equation: $m_1 g - m a_{cm} = m_1 (4 a_{cm}) \implies m_1 g = (4 m_1 + m) a_{cm}$.Thus,$a_{cm} = \frac{m_1}{4 m_1 + m} g = \frac{g}{4 + (m/m_1)}$.To maximize $a_{cm}$,we need to minimize the denominator $4 + (m/m_1)$. As $m_1 	o \infty$,$m/m_1 	o 0$,so $a_{cm, max} = g/4$. Therefore,$n = 4$.

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