A massless rod of length L connects three iron masses If the

A massless rod of length L connects three iron masses. If the first mass (at x = 0) has a mass of 1.90 kg and the second mass (at x = L) has a mass of 8.77 kg and the third mass (located at the center of mass of the system) has a mass of 148 kg, what is the moment of inertia I for a free rotation about the center of mass?

Solution

m1 = 1.90 Kg

x1 = 0 m

m2 = 8.77 Kg

x2 = L

m3 = 148 Kg

let the position of centre of mass is Xcm

Xcm = (m1 * x1 + m2 * L)/(m1 + m2)

Xcm = (1.90 * 0 + 8.77 * L)/(1.90 + 8.77)

Xcm = 0.822 * L

moment of inertia = m1 * (0.822 * L)^2 + m2 * (L - 0.822 * L)^2

moment of inertia = 1.90 * (0.822 * L)^2 + 8.77 * (L - 0.822 * L)^2

moment of inertia = 1.562 * L^2

the moment of inertia is 1.562 * L^2