Question

Point masses $m_1$ and $m_2$ are placed at the opposite ends of a rigid rod of length $L$ of 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. $x=\frac{m_{2}L}{m_1+m_2}$
• B. $x=\frac{m_{1}L}{m_1+m_2}$
• C. $x=\frac{m_1}{m_2}L$
• D. $x=\frac{m_2}{m_1}L$

Answer 1

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

Moment of inertia $m_1$ about the axis $I_1=m_1{x}^2$
Moment of inertia of $m_2$ about the axis $I_2=m_2(L-x{)}^2$
K.E. is rotational.
Total K.E. is $E=\frac{1}{2}{I}_1{\omega }_0^2+\frac{1}{2}{I}_2{\omega }_0^2$
$=\frac{1}{2}{\omega }_0^2(m_1{x}^2+m_2(L-x{)}^2)$
Work done is change in K.E.
To minimize $E$, differentiate wrt $x$ and equate to zero.
$m_{1}x-m_2(L-x)=0$
$\Rightarrow x=\frac{m_{2}L}{m_1+m_2}$
Alternatively, work done is minimum when the axis passes through the center of mass.
Center of mass is at $\frac{m_{2}L}{m_1+m_2}$