The moment of inertia of a thin rod of mass m and length L about an axis perpendicualar to it and passing through the point at a distance L/3 from the center equals

Answer 1

Hello Tawfiq,

This equation has to parts. We have the MOI about the center axis, which is the circular cross-section.

I_Center = 1 / 12 * m * L²; m is mass, L is length of rod

The parallel axis theorem (PAT) states that there is now an additive term for the total MOI. This new term is a known formula for a rod, given as: m * d²; where d is the distance from the new axis to the center,

d = L / 3

Combine both terms and you get

I = 1 / 12 * m * L² + m * L² / 9

Simplifying the equation, we get:

I = 7 / 36 * m * L²

Hope this helps!

-Andrew Evans

Answer 2

So here we can apply the parallel axis theorem, which relates a center of mass moment of inertia to one off-center as follows: I_new = I_center + M*(r)^2, where r is the distance moved. About the center, the moment of inertia for a perpendicular axis would have been ML^2/12. However it is a distance L/3 from the center, so I_new = I_center + M*(L/3)^2 = ML^2/12 + ML^2/9 = 7ML^2/36.

This answer is complete if the initial I_center was given to your or available in an appendix. Let me know if that was the case, otherwise I can also explain how I_center is derived from first principles.