A uniform rod AB of length ell and mass M is | vedclass

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(A) Let the rod have length $\ell$ and mass $M$. The center of mass of the uniform rod is at its midpoint,at a distance $\ell/2$ from end $A$.Taking t

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$m, \frac{1}{x}$

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(A) Let the rod have length $\ell$ and mass $M$. The center of mass of the uniform rod is at its midpoint,at a distance $\ell/2$ from end $A$.Taking the torque about the point of suspension:The torque due to mass $m$ at end $A$ is $	au_1 = m \cdot g \cdot x$ (counter-clockwise).The torque due to the weight of the rod $Mg$ acting at its center is $	au_2 = M \cdot g \cdot (\frac{\ell}{2} - x)$ (clockwise).For the rod to be horizontal,the net torque must be zero:$m \cdot g \cdot x = M \cdot g \cdot (\frac{\ell}{2} - x)$$m \cdot x = M \cdot \frac{\ell}{2} - M \cdot x$$m = (M \cdot \frac{\ell}{2}) \cdot \frac{1}{x} - M$This equation is of the form $y = k \cdot X + c$,where $y = m$,$X = \frac{1}{x}$,$k = M \cdot \frac{\ell}{2}$,and $c = -M$.Thus,plotting $m$ versus $\frac{1}{x}$ will yield a straight line.

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