# Homework Help: Ranking Rotational Inertia Task

1. Oct 20, 2014

### B3NR4Y

1. The problem statement, all variables and given/known data
I'm given the following four structures with their rotational axes marked, I have to rank them from largest to smallest rotational inertias, I have trouble with that apparently.

2. Relevant equations
No equations are really necessary, just my book says the more mass there is on one side of the rotation axis, the higher the rotational inertia.

3. The attempt at a solution
I gave the following four answers
a>b>d=c (Before I read the chapter, just using intuition)
a>b>d>c
a>b>c>d (again before I read the chapter, at this point I realized I was completely off)
c=d>b>a (I said c and d are equal because the mass they both separate is about equal, and b is greater than a because it seems it would be "harder" to rotate about that axis then a, a would rotate freely, b would have a bit of trouble, c and d would balance)

2. Oct 21, 2014

### haruspex

You can compare a with c and b with d easily by using the parallel axis theorem (but it should also be reasonably intuitive). Comparing a with b and c with d is somewhat harder. I don't find it intuitively obvious. One trick you can use is to cancel out equivalent contributors. E.g., the horizontal member of b cancels one vertical member of a.

3. Oct 21, 2014

### Orodruin

Staff Emeritus
It is not a matter of how much is on each side, it is a matter of how far away from the axis of rotation the mass is. Haruspex' suggestion of cancelling equivalent contributions is a good one (I assume all bars are to be considered as having the same length and mass).

4. Oct 21, 2014

### B3NR4Y

Yeah I read my professor's lecture notes for today (these exercises are prelecture so he knows where we stand, but they're graded on accuracy so I try to get right) and my task here made more sense. I got b>a>d>c, which I think is right. However I do not know the parallel axis theorem, or I have never given it a name. What is that?

5. Oct 21, 2014

### haruspex

Yes, I get that answer too.
For present purposes, the parallel axis theorem tells you that the moment of inertia about an axis through the mass centre (as in c and d) is less than about any other axis parallel to it. This tells you a > c and b > d. It remains to compare a and d. For those, you can cancel the horizontal of d with the two lower halves of the verticals in a, and the horizontal of a with one vertical of d. For what's left, it should be evident that a > d.