question_answer

A thin rod of length \[l\]and mass m is suspended from one of its ends. It is set into oscillation about a horizontal axis which is passing through the suspension point. Its angular speed is\[\omega \]while passing through its mean position. How high will the center of mass rise from the lowest position?

A) \[\frac{{{\omega }^{2}}{{l}^{2}}}{2g}\]                   

B) \[\frac{{{\omega }^{2}}{{l}^{2}}}{3g}\]

C)  \[\frac{{{\omega }^{2}}{{l}^{2}}}{g}\]                    

D)  \[\frac{{{\omega }^{2}}{{l}^{2}}}{6g}\]

Correct Answer: D

Solution :

 Applying energy conservation, we get \[mgh=\frac{I{{\omega }^{2}}}{2}=\frac{m{{l}^{2}}}{3}\times \frac{{{\omega }^{2}}}{2}\Rightarrow \frac{{{\omega }^{2}}{{l}^{2}}}{6g}\]