question_answer

A uniform rectangular parallelopiped of mass m having sides \[l\], \[2l\] and \[4l\] is placed in turn on each of its three sides on a horizontal surface. What is the potential energy of the parallelopiped in each of these positions? Which position will be the most stable?

Answer:

                When the parallelopiped is placed on its small, middle and large side, its centre of gravity will lie at heights \[2l,\]\[l\]and \[l/2\]respectively above the horizontal surface. If \[{{U}_{1}},{{U}_{2}}\]and \[{{U}_{3}}\]are the potential energies in the three Cases, then \[{{U}_{1}}=mg\times 2l=2mgl\] \[{{U}_{2}}=mg\times l=mgl\] \[{{U}_{3}}=mg\times \frac{1}{2}=\frac{1}{2}mgl\] When the parallelepiped rests on its large side, its potential energy is minimum. So this position is the position of stable equilibrium.