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 perpendicular to it. 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 ${\omega }_0$ is minimum, is given by

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

Answer 1

Answer: (B)

Moment of inertia of the system about the axis of rotation (through point $P$) is
$I=m_1{x}^2+m_2(L-x{)}^2$
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 $\frac{1}{2}[2m_{1}x+2m_2(L-x)(-1)]{\omega }_0^2=0$
or $m_{1}x-m_2(L-x)=0(...{\omega }_0\ne 0)$
or $(m_1+m_2)x=m_{2}L$
or $x=\frac{m_{2}L}{m_1+m_2}$