# Inertia of a beam with added masses

1. Apr 16, 2014

### cmcd

1. The problem statement, all variables and given/known data

A beam of mass m and length L with a moment of inertia (mL^2)/12 carries additional masses of m/2 at one end and at its centre. The moment of inertia about its centre is now...

(a) - (5/24)mL^2
(b) - 0
(c) - (1/12)mL^2
(d) - (1/3)mL^2
(e) - (17/96)L^2

2. Relevant equations

I'm not sure what equations to use.

3. The attempt at a solution

And I have no solution.

2. Apr 16, 2014

### SteamKing

Staff Emeritus
Well, what formulas have you studied? Most people will list something.

Can you at least figure out what the center of mass will be?

3. Apr 16, 2014

### cmcd

I'm revising for exams in two weeks or so and I can't remember doing any questions like this, although we did some theory on centres of gravity and inertia. They were really proofs though. No I don't understand how to get the centre of mass. Although I could give it a go;

Is it 1/8* L from the centre of the beam to the end of the beam with the additional mass?

→ m * g * (1/2) * y = m * g * (3/2) * x

where x is the distance of the centre of gravity from the centre of the beam and y is the distance of the centre of gravity from end of the beam with the additional mass.

→ y = 3x; y + x = L * (1/2);

→ L * (1/2) - x = 3x

→ x = L * (1/8)

4. Apr 16, 2014

### SteamKing

Staff Emeritus
Well, the proofs should tell you something. I mean, you were trying to prove that the c.o.m. of
a given shape was related to its dimensions, right? Ditto for the moment of inertia.
Was one of the proofs called the 'parallel axis theorem' by any chance?

In any event, your calculations for the c.o.m. are wrong. You want to calculate the
first moment of all the masses about a common origin, and then divide this moment by the total mass.
You do know what a moment is, don't you?