Question

Two point mass $50\text{g}$ and $100\text{g}$ are attached to the ends of a rod of length $2\text{m}$ and of negligible mass. The rod which was initially at rest is set into rotation about an axis perpendicular to the length of the rod. Find the position of axis of rotation from $50\text{g}$ about which the work required to set the rod into rotation with an uniform angular velocity of $2\text{rad/s}$, is minimum.

• A. $\frac{2}{3}\text{m}$
• B. $\frac{4}{3}\text{m}$
• C. $1\text{m}$
• D. $\frac{1}{3}\text{m}$

Answer 1

The correct option is B $\frac{4}{3}\text{m}$

Let the position of the axis of rotation from $50\text{g}$ be ${}^{\prime }{x}^{\prime }$. About this axis of rotation, moment of inertia of system remains constant.
From work-energy theorem,
$W=\frac{1}{2}I{w}^2$


Here, moment of inertia of the system is
$I=m_1{x}^2+m_2(l-x{)}^2$
$=0.05(x{)}^2+0.1(2-x{)}^2$
$=0.05x^2+0.1[2^2+x^2-2(2\left)x\right)]$
$I=0.15x^2+0.4-0.4x$

Now, for the work to be minimum, moment of inertia should be minimum
$\therefore$ For $I$ to be minimum, $\frac{dI}{dx}=0$
$\frac{d}{dx}(0.15x^2+0.4-0.4x)=0$
$\Rightarrow 0.3x-0.4=0$
$\Rightarrow x=\frac{4}{3}\text{m}$