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. $x=\frac{m_{2}L}{m_1+m_2}$
• 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 $x=\frac{m_{2}L}{m_1+m_2}$

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