question_answer

A uniform rod of length \['l'\] which is free to rotate about horizontal axis passing through one of its end is released. Find acceleration of free end of the rod just after release.

A)  \[\frac{3g}{2}\]                               

B)  \[\frac{2g}{3}\]

C)  \[\frac{g}{2}\]                                  

D)  \[\frac{g}{3}\]

Correct Answer: A

Solution :

\[\tau =\]force \[\times \]shortest distance\[=I\alpha \] \[mg\frac{\ell }{2}=\frac{m{{\ell }^{2}}}{3}\alpha \] \[\alpha =\frac{3g}{2\ell }\] \[{{a}_{T}}=\alpha \,\,r=\frac{3g}{2\ell }\times \ell =\frac{3g}{2}\] \[{{a}_{c}}={{w}^{2}}r=0\] \[\therefore \overrightarrow{a}=\overrightarrow{{{a}_{T}}}+\overrightarrow{{{a}_{c}}}\] \[\therefore a=\frac{3g}{2}\]