Inertia of a particle

Question

The inertia of a ring is $I$. The ring is on the point $B$ of a rod AB of length 2a, having negligible inertia. The inertia of the particle at A is $I+4Ma^2$ where $M$ is the mass of the ring.

However I use a different approach to calculate inertia of particle at A.

Calculating the inertia of particle at the midpoint of AB = $I+Ma^2$

Shifting the calculated inertia at the point A =$I+Ma^2+Ma^2=I+2Ma^2$

Why does the discrepancy arises?

That isnt making sense where is ring and what is the point? – Archis Welankar May 30 '17 at 13:37 — Archis Welankar, May 30 '17 at 13:37

score 0 · Answer 1 · answered May 30 '17 at 12:46

0

In taking moment of inertia we always take moment of inertia about midpoint and then use parallel axis theorem so you cannot use this theorem about one end of the rod.Hoping that its MI of a cylinder with radius $r$ and length $2a $

answered May 30 '17 at 12:46

Archis Welankar

15,910

Please see the edit. – mathnoob123 May 30 '17 at 12:54

score 0 · Answer 2 · answered May 30 '17 at 13:10

You are not using the Parallel Axes Theorem correctly. If $m$ is the mass of the body and the centroid of the body is at $G$ then the moment of inertia of the body about an axis through $A$ at a distance $d$ from a parallel axis through $G$ is given by $$I_A=I_G+md^2$$

So the first calculation is correct and the second one is wrong.

Inertia of a particle

2 Answers2