question_answer

A thin and uniform rod of mass M and length L is held vertical on floor with large friction. The rod is released from rest so that it falls by rotating about is contact-point with the floor without slipping. Which of the following statement(s) is/are correct. When the rod makes an angle \[60{}^\circ \] with vertical?  [g is the acceleration due to gravity]

A) The normal reaction force from the floor on the mg rod will be \[\frac{mg}{16}\]

B) The angular acceleration of the rod will be \[\frac{2g}{L}\]

C) The angular speed of the rod will be \[\sqrt{\frac{3g}{2L}}\]

D) The radial acceleration of the rod's center of mass will be  \[\frac{3g}{4}\]

Correct Answer: A , C , D

Solution :

By E.C., \[\frac{1}{2}\text{I}{{w}^{2}}=+mg\frac{\ell }{4}\]

\[\Rightarrow \]   \[\frac{1}{2}\frac{m\ell }{3}{{\omega }^{2}}=\text{mg}\frac{\ell }{4}\]

\[\Rightarrow \]   \[\omega =\frac{m{{\ell }^{2}}}{3}{{\omega }^{2}}=mg\frac{\ell }{{{4}^{2}}}\]

\[\Rightarrow \]   \[\omega \,=\,\sqrt{\frac{3g}{2l}}\]

\[{{a}_{c}}=\,{{\omega }^{2}}\text{R}=\frac{3g}{2l}\times \frac{l}{2}=\frac{3g}{4}\]

\[\alpha =\frac{\tau }{\text{I}}\]

\[=\frac{mg\frac{l}{2}\text{sin}60{}^\circ }{\frac{m{{l}^{2}}}{3}}=\frac{3\sqrt{3g}}{4l}\]

\[a=\alpha \frac{l}{2}\text{sin}60{}^\circ +{{\omega }^{2}}\frac{l}{2}\text{cos}e60{}^\circ \]

\[(R\alpha )\]

\[=\,\frac{9g}{16}+\frac{6g}{16}=\frac{15g}{16}\]

\[\therefore \]      \[mg-\text{N}=ma\]

\[\Rightarrow \]   \[\text{N}\,=\,mg-\frac{15mg}{16}\]

\[\text{N}=\frac{\text{Mg}}{\text{16}}\]

Solution :

Same as above

Solution :

Same as above