Question
In the arrangement shown in figure, 'a' represent the magnitude of acceleration of small blocks A and B while 'T' is the tension in the massless string passing over the frictionless and massless pulley. The sum of the masses of blocks A and B is constant. For this system, a linear relationship can be obtained between
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Solution
[Key Concept: In an Atwood machine with fixed total mass, eliminating the mass difference links T linearly to a2.]
$a = \frac{\left( m_{A} - m_{B} \right)g}{\left( m_{A} + m_{B} \right)}\ T = \frac{2{\text{ }m}_{A}m_{B}g}{\left( m_{A} + m_{B} \right)}$
$$\begin{matrix} & a^{2} = \frac{g^{2}}{\left( m_{A} + m_{B} \right)^{2}} \times \left( m_{A}^{2} + m_{B}^{2} - 2m_{A}m_{B} \right) \\ & a^{2} = \frac{\left( m_{A}^{2} + m_{B}^{2} \right)g^{2}}{\left( m_{A} + m_{B} \right)^{2}} - \frac{2m_{A}m_{B}(g)(g)}{\left( m_{A} + m_{B} \right)\left( m_{A} + m_{B} \right)} \end{matrix}$$
$$\begin{matrix} & a^{2} = \frac{\left\lbrack \left( m_{A}^{2} + m_{B}^{2} \right) - 2m_{A}m_{B} + 2m_{A}m_{B} \right\rbrack g^{2}}{\left( m_{A} + m_{B} \right)^{2}} - \frac{Tg}{\left( m_{A} + m_{B} \right)} \\ & a^{2} = \frac{\left\lbrack \left( m_{A}^{2} + m_{B}^{2} \right) - 2m_{A}m_{B} \right\rbrack g^{2}}{\left( m_{A} + m_{B} \right)^{2}} - \frac{Tg}{\left( m_{A} + m_{B} \right)} \\ & a^{2} = \left\lbrack 1 - \frac{2m_{A}m_{B}}{\left( m_{A} + m_{B} \right)^{2}} \right\rbrack g^{2} - \frac{Tg}{\left( m_{A} + m_{B} \right)} \\ & a^{2} = g^{2} - \frac{2m_{A}m_{B}g^{2}}{\left( m_{A} + m_{B} \right)^{2}} - \frac{Tg}{\left( m_{A} + m_{B} \right)} \\ & a^{2} = g^{2} - \left\lbrack \frac{2g}{\left( m_{A} + m_{B} \right)^{2}} \right\rbrack T \\ & Y = C - mx \end{matrix}$$
Linear relation between $a^{2}$ & T
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