(a) [tex]x = \frac{m_2L}{m_1+m_2}[/tex]
Explanation:
Given:
Moment of Inertia of m₁ about the axis, I₁ = m₁x²
Moment of Inertia of m₂ about the axis. I₂ = m₂ (L - x)²
Kinetic energy is rotational.
Total kinetic energy is [tex]E = \frac{1}{2} I_1w_0^2 + \frac{1}{2}I_2w_0^2 = \frac{1}{2} w_0^2(m_1x^2 + m_2(L-x)^2)[/tex]
Work done is change in kinetic energy.
To minimize E, differentiate wrt x and equate to zero.
[tex]m_1x - m_2(L-x) = 0\\\\x = \frac{m_2L}{m_1+m_2}[/tex]
Alternatively, work done is minimum when the axis passes through the center of mass.
Center of mass is at [tex]\frac{m_2L}{m_1 + m_2}[/tex]
