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A light rod of length 1 m is pivoted at its centre and two masses 5 kg and 2 kg are hung from the ends as shown. The critical angular acceleration of the rod if it is assumed to be massless is?

\[\begin {array} {1 1} (a)\;\frac{6g}{7} & \quad (b)\;\frac{5g}{4} \\ (c)\;\frac{3g}{2} & \quad  (d)\;\frac{2g}{3} \end {array}\]

1 Answer

Since the rod is massless , the tensions in the 2 strings will be the same.
Writing torque equation :
$5 g \times l/2- 2g \times l/2=I \alpha$
$\qquad= [m_1 \times (l/2) ^2 +m_2 (l/2)^2] \alpha$
$3g l/2=\bigg[\large\frac{5l^2}{4} +\frac{2l^2}{4} \bigg] \alpha$
$3g l /2=\large\frac{7l^2}{4} \alpha$
$\alpha = \large\frac{6g}{7}$
answered Nov 27, 2013 by meena.p

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