Worked example 10.1: Equilibrium of two rods

Question: Suppose that two uniform rods (of negligible thickness) are welded together at right-angles, as shown in the diagram below. Let the first rod be of mass $m_1=5.2 {\rm kg}$ and length $l_1=1.3 {\rm m}$. Let the second rod be of mass $m_2=3.4 {\rm kg}$ and length $l_2=0.7 {\rm m}$. Suppose that the system is suspended from a pivot point located at the free end of the first rod, and then allowed to reach a stable equilibrium state. What angle $\theta$ does the first rod subtend with the downward vertical in this state?

Answer: Let us adopt a coordinate system in which the $x$-axis runs parallel to the second rod, whereas the $y$-axis runs parallel to the first. Let the origin of our coordinate system correspond to the pivot point. The centre of mass of the first rod is situated at its mid-point, whose coordinates are

$$(x_1, y_1) = (0, l_1/2).$$

Likewise, the centre of mass of the second rod is situated at its mid-point, whose coordinates are

$$(x_2, y_2) = (l_2/2, l_1).$$

It follows that the coordinates of the centre of mass of the whole system are given by

$$x_{cm} = \frac{m_1 x_1+ m_2 x_2}{m_1+m_2} = \frac{1}{2} \f... ...{m_1+m_2}= \frac{3.4\times 0.7}{2\times 8.6} = 0.138 {\rm m},$$

and

$$y_{cm} = \frac{m_1 y_1+ m_2 y_2}{m_1+m_2} = \frac{m_1 l_1... ... = \frac{5.2\times 1.3/2+3.4\times 1.3}{8.6} = 0.907 {\rm m}.$$

The angle $\theta$ subtended between the line joining the pivot point and the overall centre of mass, and the first rod is simply

$$\theta = \tan^{-1}\left(\frac{x_{cm}}{y_{cm}}\right) = \tan^{-1}0.152= 8.65^\circ.$$

When the system reaches a stable equilibrium state then its centre of mass is aligned directly below the pivot point. This implies that the first rod subtends an angle $\theta=8.65^\circ$ with the downward vertical.