Question

In the adjacent figure a uniform rod of length $L$ and mass $m$ is kept at rest in horizontal position on an elevated edge. The value of $x$ (consider the figure-3.120) is such that the rod will have maximum angular acceleration $\alpha$ as soon as it set free.

Answer 1

$x = \frac{L}{2} \left(1 - \frac{1}{\sqrt{3}}\right)$

Answer 2

The angular acceleration $\alpha$ of the rod about the pivot point $O$ is given by $\alpha = \frac{\tau}{I}$, where $\tau$ is the torque and $I$ is the moment of inertia about $O$. The weight of the rod is $mg$, acting at the center of mass $G$. The distance of $G$ from the left end is $L/2$. The pivot $O$ is at a distance $x$ from the left end. The figure shows that the pivot $O$ is to the left of the center of mass $G$, which implies $x < L/2$. The distance of the center of mass $G$ from the pivot $O$ is $d = L/2 - x$. The torque about $O$ is $\tau = mg \cdot d = mg(L/2 - x)$. The moment of inertia of the rod about $O$ is given by the parallel axis theorem: $I = I_{cm} + md^2 = \frac{mL^2}{12} + m(L/2 - x)^2$. Thus, the angular acceleration is: $\alpha = \frac{mg(L/2 - x)}{\frac{mL^2}{12} + m(L/2 - x)^2}$ Let $y = L/2 - x$. Since $x < L/2$, $y > 0$. The expression for $\alpha$ becomes: $\alpha = \frac{mgy}{\frac{mL^2}{12} + my^2} = \frac{gy}{\frac{L^2}{12} + y^2}$ To find the value of $y$ that maximizes $\alpha$, we differentiate $\alpha$ with respect to $y$ and set the derivative to zero: $\frac{d\alpha}{dy} = g \frac{\left(\frac{L^2}{12} + y^2\right)(1) - y(2y)}{\left(\frac{L^2}{12} + y^2\right)^2} = g \frac{\frac{L^2}{12} - y^2}{\left(\frac{L^2}{12} + y^2\right)^2}$ Setting $\frac{d\alpha}{dy} = 0$, we get $\frac{L^2}{12} - y^2 = 0$, which implies $y^2 = \frac{L^2}{12}$. Since $y > 0$, we have $y = \sqrt{\frac{L^2}{12}} = \frac{L}{2\sqrt{3}}$. This value of $y$ corresponds to a maximum as the derivative changes from positive to negative. Now we substitute back $y = L/2 - x$: $\frac{L}{2} - x = \frac{L}{2\sqrt{3}}$ Solving for $x$: $x = \frac{L}{2} - \frac{L}{2\sqrt{3}} = \frac{L}{2} \left(1 - \frac{1}{\sqrt{3}}\right)$ This value of $x$ is less than $L/2$, which is consistent with the figure. We can rationalize the term $\frac{1}{\sqrt{3}}$ as $\frac{\sqrt{3}}{3}$: $x = \frac{L}{2} \left(1 - \frac{\sqrt{3}}{3}\right) = \frac{L}{2} \left(\frac{3 - \sqrt{3}}{3}\right) = \frac{L(3 - \sqrt{3})}{6}$