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## Q. 2.6

A bar is hanging in equilibrium in the position shown in Figure 2.14(a). Determine the potential energy of the bar in terms of $\theta$ the counterclockwise angular position of the bar from its equilibrium position when (a) the datum is taken to be the horizontal plane at the bottom of the bar when in equilibrium, (b) the datum is taken as the horizontal plane through the mass center when the bar is in equilibrium, and (c) the datum is taken to be the horizontal plane through the pin support. ## Verified Solution

(a) As the bar swings through an angle $\theta$, as illustrated in Figure 2.14(b), the mass center is a distance

$h=\frac{L}{2} +\frac{L}{2} \left(1-\cos \theta \right)$         (a)

and has a potential energy with respect to the datum of

$V=mg\frac{L}{2} \left(2-\cos \theta \right)$           (b)

(b) Using a horizontal plane through G as a datum, we have

$V=mg\frac{L}{2} \left(1-\cos \theta \right)$          (c)

(c) Using a horizontal plane through $\omicron$ as a datum, we have

$V= -mg\frac{L}{2}\cos \theta$          (d) 