Question

2. Two point masses m and m2 are separated by a massless rod of length L. (a) Write an expression for the moment of inertia I about an axis perpendicular to the rod and passing through it a distance x from mass mi. (b) Calculate dl/dx and show that I is at a minimum when the axis passes through the center of mass of the system.

Answer 1

(a) I = m₁ x² + m₂ x (L - x)² Answer

(b) dI/dx = 2m₁ x - 2m₂ (L - x )

Now, for minimising moment of Inertia, dI/sx = 0

therefore,

2m₁ x - 2m₂ (L - x ) = 0

x = m₂ L / (m₁ + m₂ ) , which is the position of center of mass, measured from m₁

Hence proved that, moment of inertia is minimum, when axis passes through the center of mass