Solved Problem on Dynamics

A rod with length L is leaning at an angle θ relative to the vertical. In the rod, there is a ring that can slide without friction. The system is rotating around a vertical axis through the lower end with constant angular velocity. Determine angular velocity that must be carried out by the rod so that the ring remains at rest in the midpoint of the rod.

Problem data:

Length of rod: L;
Angle of leaning of the rod: θ;
Assuming the acceleration due to gravity: g.

Problem diagram:

Forces that act in the ring (Figure 1):

\( {\vec F}_{g} \): gravitational force on the ring;
\( \vec{N} \): normal reaction force of the rod over the ring.

Figure 1

Solution

The leaning angle of the rod is θ, as the sum of the internal angles of a triangle is equal to 180°, the angle α between the rod and the horizontal will be (Figure 2-A).

\[ \alpha +\theta +90°=180°\Rightarrow \alpha =90°-\theta \]

The angle between the rod and the normal reaction force is equal to 90°

\[ \alpha +\beta =90°\Rightarrow 90°-\theta +\beta =90° \Rightarrow \beta =\theta \]

the angle between the normal reaction and the horizontal is also θ (Figure 2-B).

Figure 2

We draw the forces on a coordinate system xy (Figure 3). As the ring revolves around the vertical axis, we write Newton's Second Law for a body in a circular motion

\[ \begin{gather} \bbox[#99CCFF,10px] {{\vec{F}}_{cp}=m{\vec{a}}_{cp}} \tag{I} \end{gather} \]

The component of the normal reaction \( \vec N \) in the x direction will be

\[ \begin{gather} N_{x}=N\cos \theta \tag{II} \end{gather} \]

the only force acting in the x direction is the component of the normal reaction

\[ \begin{gather} F_{cp}=N_{x}\\[5pt] F_{cp}=N\cos \theta \tag{III} \end{gather} \]

Figure 3

The centripetal acceleration is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {a_{cp}=\frac{v^{2}}{r}} \tag{IV} \end{gather} \]

substituting expressions (III) and (IV) into expression (I)

\[ \begin{gather} N\cos \theta =m\frac{v^{2}}{r} \tag{V} \end{gather} \]

The tangential speed is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {v=\omega r} \tag{VI} \end{gather} \]

substituting the expression (VI) into expression (IV)

\[ \begin{gather} N\cos \theta =m\frac{(\omega r)^{2}}{r}\\[5pt] N\cos \theta=m\frac{\omega ^{2}r^{2}}{r}\\[5pt] N\cos \theta =m\omega ^{2}r \tag{VII} \end{gather} \]

The component of the normal reaction force \( \vec N \) in the y direction will be

\[ \begin{gather} N_{y}=N\sin \theta \tag{VIII} \end{gather} \]

We want the ring to remain at rest, so, the resultant forces in this direction should be equal to zero

\[ \begin{gather} N_{y}=F_{g} \tag{IX} \end{gather} \]

The gravitational force is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {F_{g}=mg} \tag{X} \end{gather} \]

substituting expressions (VIII) and (X) into expression (IX)

\[ \begin{gather} N\sin \theta =mg \tag{XI} \end{gather} \]

Dividing the expression (XI) by expression (VII)

\[ \begin{gather} \frac{\cancel{N}\sin \theta}{\cancel{N}\cos \theta}=\frac{\cancel{m}g}{\cancel{m}\omega^{2}r} \end{gather} \]

From the Trigonometry \( \operatorname{tg}\theta =\frac{\sin \theta}{\cos \theta} \)

\[ \begin{gather} \tan \theta =\frac{g}{\omega ^{2}r}\\[5pt] \omega^{2}=\frac{g}{\tan \theta r} \tag{XII} \end{gather} \]

While the rod rotates around a vertical axis the ring, fixed in the midpoint \( \left(\frac{L}{2}\right) \), describes a circumference of radius r in a horizontal plane (Figure 4). By the figure, the sine of the angle θ will be

\[ \begin{gather} \sin \theta =\frac{\text{opposite leg}}{\text{hypotenuse}}=\frac{r}{\dfrac{L}{2}}\\[5pt] r=\frac{L}{2}\sin \theta \tag{XIII} \end{gather} \]

Figure 4

substituting the expression (XIII) into expression (XII)

\[ \begin{gather} \omega ^{2}=\frac{g}{\tan \theta\dfrac{L}{2}\sin \theta }\\[5pt] \omega^{2}=\frac{2g}{L\tan \theta \sin \theta} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {\omega =\sqrt{\frac{2g}{L\tan \theta\sin \theta }}} \end{gather} \]

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