Question

Point masses $m_1$ and $m_2$ are placed at the opposite ends of a rigid rod of length $L$, and negligible mass. The rod is to be set rotating about an axis
The Position of point $P$ on this rod through which the axis should pass so that the work required to set the rod rotating with angular velocity $\omega$ is minimum, is given by

• A. $x=\frac{m_2}{m_1}L$
• B. $x=\frac{m_{2}L}{m_1+m_2}$
• C. $x=\frac{m_1}{m_2}L$
• D. $x=\frac{m_{1}L}{m_1+m_2}$

Answer 1

The correct option is B $x=\frac{m_{2}L}{m_1+m_2}$

$I=m_1{x}^2+m_2(L-x{)}^2$
By work energy theorem,
Work done to set the rod rotating with angular velocity ${\omega }_0=$ Increase in rotational kinetic energy
$W=\frac{1}{2}I{\omega }_0^2=\frac{1}{2}[m_1{x}^2+m_2(L-x{)}^2]{\omega }_0^2$
For $W$ to be minimum, $\frac{dW}{dx}=0$
i.e. $\frac{1}{2}[2m_{1}x+2m_2(L-x\left)\right(-1\left)\right]{\omega }_0^2=0$
or $m_{1}x-m_2(L-x)=0(\because {\omega }_0\ne 0)$
or $(m_1+m_2)x=m_{2}Lorx=\frac{m_{2}L}{m_1+m_2}$