Physics - Rotation Question with Solution | TestHub

Let the mass of the uniform rod be $M$ and its length be $L$. The center of mass of the rod is at a distance $L/2$ from end A.

The rod is suspended at point P, which is at a distance $x$ from A. For the rod to remain horizontal in equilibrium, the net torque about the suspension point P must be zero.

The torque due to the suspended mass $m$ at A is $mgx$ in the anti-clockwise direction.

The torque due to the weight of the rod is $Mg\left(\frac{L}{2}-x\right)$ in the clockwise direction.

Equating the two torques for rotational equilibrium:

$mgx=Mg\left(\frac{L}{2}-x\right)$

$mx=M\left(\frac{L}{2}\right)-Mx$

Dividing the entire equation by $x$:

$m=\frac{ML}{2}\left(\frac{1}{x}\right)-M$

This equation is of the form $y=kz+c$, where $y=m$, $z=\frac{1}{x}$, $k=\frac{ML}{2}$, and $c=-M$.

Therefore, plotting $m$ against $\frac{1}{x}$ will give a straight line.

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