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., \[\frac{1}{2}\text{I}{{w}^{2}}=+mg\frac{\ell }{4}\] |
\[\Rightarrow \] \[\frac{1}{2}\frac{m\ell }{3}{{\omega }^{2}}=\text{mg}\frac{\ell }{4}\] |
\[\Rightarrow \] \[\omega =\frac{m{{\ell }^{2}}}{3}{{\omega }^{2}}=mg\frac{\ell }{{{4}^{2}}}\] |
\[\Rightarrow \] \[\omega \,=\,\sqrt{\frac{3g}{2l}}\] |
\[{{a}_{c}}=\,{{\omega }^{2}}\text{R}=\frac{3g}{2l}\times \frac{l}{2}=\frac{3g}{4}\] |
\[\alpha =\frac{\tau }{\text{I}}\] |
\[=\frac{mg\frac{l}{2}\text{sin}60{}^\circ }{\frac{m{{l}^{2}}}{3}}=\frac{3\sqrt{3g}}{4l}\] |
\[a=\alpha \frac{l}{2}\text{sin}60{}^\circ +{{\omega }^{2}}\frac{l}{2}\text{cos}e60{}^\circ \] |
\[(R\alpha )\] |
\[=\,\frac{9g}{16}+\frac{6g}{16}=\frac{15g}{16}\] |
\[\therefore \] \[mg-\text{N}=ma\] |
\[\Rightarrow \] \[\text{N}\,=\,mg-\frac{15mg}{16}\] |
\[\text{N}=\frac{\text{Mg}}{\text{16}}\] |
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