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_1x^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_1x^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} $