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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. In the above question if the rod has a mass 1 kg, then its initial angular acceleration

\[\begin {array} {1 1} (a)\;\frac{7g}{11} & \quad (b)\;\frac{8g}{11} \\ (c)\;\frac{9g}{11} & \quad  (d)\;\frac{10 g}{11} \end {array}\]

1 Answer

$(m_1g-m_2g)l/2=\bigg(m_1(l/2)^2+m_2(l/2)^2+\frac{ml^2}{12}\bigg) \alpha$
$3g\;l/2= \bigg[\large\frac{5l^2}{4}+\frac{2l^2}{4}+\frac{l^2}{12}\bigg]$
$3gl/2=\large\frac{22 l^2}{12}\alpha$
$\qquad\alpha=\large\frac{9g}{11 l}$
answered Nov 27, 2013 by meena.p
edited Jun 21, 2014 by lmohan717

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