Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

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 .
Can you answer this question?

1 Answer

0 votes
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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App