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) such that the rod will have maximum angular acceleration $\alpha$ (as it is released from position shown)

• A. $x$ is equal to $\frac{L}{2\sqrt{3}}$
• B. $\alpha$ is equal to $\frac{g\sqrt{3}}{2L}$
• C. $\alpha$ is equal to $\frac{g\sqrt{3}}{L}$
• D. $x$ is equal to $\frac{L}{\sqrt{3}}$

Answer 1

Answer: (A), (C)

$\alpha$ is equal to $\frac{g\sqrt{3}}{L}$

Apply $\tau =I\alpha$ about O
$mgx=(\frac{m{L}^2}{12}+m{x}^2)\alpha$
(using parallel-axis theorem)
Her $\alpha$ can be written as
$\alpha =\frac{mgx}{(\frac{m{L}^2}{12}+m{x}^2)}$,


For $\alpha$ to be maximum $\frac{d\alpha }{dx}=0$
$\frac{(\frac{m{L}^2}{12}+m{x}^2)mg-mgx\left(2mx\right)}{{(\frac{m{L}^2}{12}+m{x}^2)}^2}=0$
$x^2=\frac{L^2}{12}$
which gives $x=\frac{L}{2\sqrt{3}}$
On substituting we get
$\alpha =\frac{\sqrt{3}g}{L}$