Question 12.59: A small, 200-g collar D can slide on portion AB of a rod whi...

A small, 200-g collar D can slide on portion AB of a rod which is bent as shown. Knowing that the rod rotates about the vertical AC at a constant rate and that α = 30° and r = 600 mm, determine the range of values of the speed ν for which the collar will not slide on the rod if the coefficient of static friction between the rod and the collar is 0.30.

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Case 1: ν=νmin\underline{\nu=\nu_{\min }}, impending motion downward

+Fx=max:NWsin30=mν2rcos30orN=mgsin30+ν2rcos30+ΣFy=may:FWcos30=mν2rsin30orF=mgcos30ν2rsin30\begin{aligned}+\nearrow F_{x}=m a_{x}: \quad N-W \sin 30^{\circ} & =m \frac{\nu^{2}}{r} \cos 30^{\circ} \\\text{or}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad N & =m\left\lgroup g \sin 30^{\circ}+\frac{\nu^{2}}{r} \cos 30^{\circ}\right\rgroup \\+\nwarrow \Sigma F_{y}=m a_{y}: \quad F-W \cos 30^{\circ} & =-m \frac{\nu^{2}}{r} \sin 30^{\circ} \\\text{or}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad F & =m\left\lgroup g \cos 30^{\circ}-\frac{\nu^{2}}{r} \sin 30^{\circ}\right\rgroup\end{aligned}

Now F=μsN\quad\quad\quad\quad F=\mu_{s} N

Then mgcos30ν2rsin30=μs×mgsin30+ν2rcos30\quad m\left\lgroup g \cos 30^{\circ}-\frac{\nu^{2}}{r} \sin 30^{\circ}\right\rgroup=\mu_{s} \times m\left\lgroup g \sin 30^{\circ}+\frac{\nu^{2}}{r} \cos 30^{\circ}\right\rgroup

or 

ν2=gr1μstan30μs+tan30=(9.81 m/s2)(0.6 m)10.3tan300.3+tan30\begin{aligned}\nu^{2} & =g r \frac{1-\mu_{s} \tan 30^{\circ}}{\mu_{s}+\tan 30^{\circ}} \\& =\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(0.6 \mathrm{~m}) \frac{1-0.3 \tan 30^{\circ}}{0.3+\tan 30^{\circ}}\end{aligned}

or νmin=2.36 m/s\quad\quad\quad\nu_{\min }=2.36 \mathrm{~m} / \mathrm{s}

Case 2: ν=νmax\quad \underline{\nu=\nu_{\max }}, impending motion upward

+ΣFx=max:NWsin30=mν2rcos30orN=mgsin30+ν2rcos30+ΣFy=may:F+Wcos30=mν2rsin30orF=mgcos30+ν2rsin30\begin{aligned}+\nearrow \Sigma F_{x}=m a_{x}: \quad N-W \sin 30^{\circ} & =m \frac{\nu^{2}}{r} \cos 30^{\circ} \\\text{or}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad N & =m\left\lgroup g \sin 30^{\circ}+\frac{\nu^{2}}{r} \cos 30^{\circ}\right\rgroup \\+\searrow \Sigma F_{y}=m a_{y}: \quad F+W \cos 30^{\circ} & =m \frac{\nu^{2}}{r} \sin 30^{\circ} \\\text{or}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad F & =m\left\lgroup-g \cos 30^{\circ}+\frac{\nu^{2}}{r} \sin 30^{\circ}\right\rgroup\end{aligned}

Now F=μsN\quad\quad\quad\quad F=\mu_{s} N

Then mgcos30+ν2rsin30 =μs×mgsin30+ν2rcos30\quad m\left\lgroup-g \cos 30^{\circ}+\frac{\nu^{2}}{r} \sin 30^{\circ}\right\rgroup  =\mu_{s} \times m\left\lgroup g \sin 30^{\circ}+\frac{\nu^{2}}{r} \cos 30^{\circ}\right\rgroup

or

ν2=gr1+μstan30tan30μs=(9.81 m/s2)(0.6 m)1+0.3tan30tan300.3\begin{aligned}\nu^{2} & =g r \frac{1+\mu_{s} \tan 30^{\circ}}{\tan 30^{\circ}-\mu_{s}} \\& =\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(0.6 \mathrm{~m}) \frac{1+0.3 \tan 30^{\circ}}{\tan 30^{\circ}-0.3}\end{aligned}

or νmax=4.99 m/s\quad\quad\quad\nu_{\max }=4.99 \mathrm{~m} / \mathrm{s}

For the collar not to slide 2.36 m/s<ν<4.99 m/s\quad\quad 2.36\mathrm{\ m}/\mathrm{s} \lt \nu \lt 4.99\mathrm{\ m}/\mathrm{s}\blacktriangleleft

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