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

Options

A. `x = (m_(2))/(m_(1)) ""L`
B. `x = (m_(2) L)/(m_(1) + m_(2))`
C. `x = (m_(1) L)/(m_(1) + m_(2))`
D. `x = (m_(1))/(m_(2)) "" L`

Answer

B

Explanation

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 = (1)/(2) I omega_(0)^(2) = (1)/(2) [m_1 x^(2) + m_2 (L - x)^(2) ] omega_(0)^(2)`
For W to be minimum , `(dW)/(dx) = 0`
i.e., `(1)/(2) [2 m_1 x + 2m_2 (L- x) (- 1)] 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) L` or ` x = (m_(2) L)/(m_1 + m_2)`