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Two particles of mass $m$ each are fixed at the opposite ends of a massless rod of length $5m$ which is oriented vertically on a smooth horizontal surface and released. Find the displacement of the lower mass on the ground when the rod makes an angle of $37^o$ with the vertical. ........ $m$
$1.5$
$2$
$2.5$
$3.5$
Solution
The rod is not pinned on the vertical in a point about it rotates and thus because of the symmetry of the system (equal masses at both ends) it needs to rotate about a point situated half the distance from each end. It is just like when you set loose the rod in middle air. Because of the symmetry it has, it will start rotating about its middle point (half distance from each end). See the figure below.
Therefore as can be seen from the figure the displacement on the horizontal of the lower mass is
$x=\left(\frac{l}{2}\right) \cdot \cos (37)=2.5 \cdot \sin (37)=1.5 m$