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(A) Since there are no external horizontal forces acting on the system,the center of mass of the system remains at its initial horizontal position.Let

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$1.5$

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(A) Since there are no external horizontal forces acting on the system,the center of mass of the system remains at its initial horizontal position.Let the length of the rod be $L = 5 \ m$. The center of mass of the two equal masses $m$ is at the midpoint of the rod,i.e.,at a distance $L/2 = 2.5 \ m$ from either end.Initially,the rod is vertical,so the center of mass is at a horizontal distance of $0$ from the vertical axis passing through the midpoint.When the rod makes an angle $	heta = 37^o$ with the vertical,the horizontal distance of the center of mass from the initial vertical axis remains $0$.Let $x_1$ be the horizontal displacement of the lower mass and $x_2$ be the horizontal displacement of the upper mass from the initial vertical axis.Due to the symmetry of the system,the center of mass is always at the midpoint. The horizontal position of the midpoint is $x_{cm} = (x_1 + x_2) / 2 = 0$.This implies $x_1 = -x_2$. The distance between the two masses is $L = 5 \ m$. The horizontal projection of the rod is $L \sin(37^o) = 5 \cdot (3/5) = 3 \ m$.Thus,$|x_1| + |x_2| = 3 \ m$. Since $|x_1| = |x_2|$,we have $2|x_1| = 3 \ m$,which gives $|x_1| = 1.5 \ m$.

Two particles of mass $m$ each are fixed at the opposite ends of a massless rod of length $5 \ m$ 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. (in $m$)

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