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A uniform rod of length l is free to rotate in a vertical plane about a fixed horizontal axis through 0. The rod is allowed to rotate from its unstable vertical position. Then the angular velocity of the rod when it has turned through an angle $\theta$ is


\[\begin {array} {1 1} (a)\;\sqrt {\frac{3g}{l}}. \sin \theta/2 & \quad (b)\;\sqrt {\frac{6g}{l}}. \sin \theta/2 \\ (c)\;\sqrt {\frac{3g}{l}}. \cos \theta/2 & \quad  (d)\;\sqrt {\frac{6g}{l}}. \cos \theta/2 \end {array}\]
I dint understand how did you write that height mgl/2(1-cos@) plzz explain by diagram .
I also not understand it

1 Answer

Since the reaction force at the height is not doing any work, total mechanical energy is conserved.
$\Delta PE= mg l/2 (1- \cos \theta)$
$\Delta K  = \large\frac{1}{2} I_0. w^2$
$\qquad=\large\frac{1}{2} \frac{ml^2}{3}.w^2$
$\large\frac{1}{2} \large\frac{ml^2}{3}$$ w^2= mgl$$\;.\ \sin ^2 \theta/2$
$w= \sqrt {\large\frac{6g}{l}} \sin \theta/2$
answered Nov 27, 2013 by meena.p
edited Jun 21, 2014 by lmohan717

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