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A uniform rod of mass 20 kg and length 1.6 m is pivoted at its end and swings in the vertical plane. Angular acceleration of the rod just after the rod is released from rest in the horizontal position

\[\begin {array} {1 1} (a)\;\frac{15g}{16} & \quad (b)\;\frac{17\;g}{16} \\ (c)\;\frac{16\;g}{15} & \quad  (d)\;\frac{g}{15} \end {array}\]

1 Answer

Torque of the weight $mg= I \alpha$
$mg \times \large\frac{l}{2}=\large\frac{ml^2}{3} \alpha$
$\alpha= \large\frac{3g}{2l}$
$\qquad= \large\frac{3 \times g}{2 \times 1.6}$
$\qquad= \large\frac{30\;g}{2 \times 16}$
$\qquad= \large\frac{15\;g}{16}$
answered Nov 27, 2013 by meena.p

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