Question

Point masses m1 m2 are placed at opposite ends

Answer 1

(a) $x = \frac{m_2L}{m_1+m_2}$

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 $E = \frac{1}{2} I_1w_0^2 + \frac{1}{2}I_2w_0^2 = \frac{1}{2} w_0^2(m_1x^2 + m_2(L-x)^2)$

Work done is change in kinetic energy.

To minimize E, differentiate wrt x and equate to zero.

$m_1x - m_2(L-x) = 0\\\\x = \frac{m_2L}{m_1+m_2}$

Alternatively, work done is minimum when the axis passes through the center of mass.

Center of mass is at $\frac{m_2L}{m_1 + m_2}$