Question 12.58: A small, 300-g collar D can slide on portion AB of a rod whi...

First note \quad \nu_{D}=r \dot{\theta}_{A B C}

+\uparrow \Sigma F_{y}=0: \quad N \sin 40^{\circ}-W=0

or \quad\quad\quad\quad  N=\frac{m g}{\sin 40^{\circ}}

\overset{+}{\longrightarrow } \Sigma F_{n}=m a_{n}: \quad N \cos 40^{\circ}=m \frac{\nu_{D}^{2}}{r}

or \quad\frac{m g}{\sin 40^{\circ}} \cos 40^{\circ}=m \frac{\left(r \dot{\theta}_{A B C}\right)^{2}}{r}

or 

\begin{aligned}r & =\frac{g}{\dot{\theta}_{A B C}^{2}} \frac{1}{\tan 40^{\circ}} \\& =\frac{9.81 \mathrm{~m} / \mathrm{s}^{2}}{(5 \mathrm{\ rad} / \mathrm{s})^{2}} \frac{1}{\tan 40^{\circ}} \\& =0.468 \mathrm{~m}\end{aligned}

or \quad\quad\quad\quad r=468 \mathrm{~mm}\blacktriangleleft