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.

The Blue Check Mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.

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

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

Then \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 

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

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

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

Then \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

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

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

12.59.
12.59..

Related Answered Questions