Pulley of Atwood machine has mass m₁and is free to rotate about its geometrical center without friction. Centre of mass of the pulley is at a distance R/2 from geometrical centre, where R is the radius of the pulley. There is no slipping between the string and pulley. The ratio of maximum acceleration to minimum acceleration of block of mass m₁ is (Take m₁R² as moment of inertia of pulley about axis of rotation $\frac{m_{1}}{m_{2}} = 3)$

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Detailed Solution

For block m₁, $m_{1} g - T_{1} = m_{1} a$

For block m₂, $T_{2} - m_{2} g = m_{2} a$

No slipping between string and pulley, $a = αR$

For maximum acceleration, the COM of pulley should be on right side of geometrical center,

$T_{1} R + \frac{m_{1} gR}{2} - T_{2} R = (m_{1} R^{2}) α$

After solving we get, $a_{\max} = \frac{g}{2}$

For minimum acceleration, the COM of pulley should be on left side of geometrical center,

$T_{1} R - \frac{m_{1} gR}{2} - T_{2} R = (m_{1} R^{2}) α$

After solving we get, $a_{\min} = \frac{g}{14}$

Ratio, $\frac{a_{\max}}{a_{\min}} = \frac{7}{1}$

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