Inertia of a beam with added masses

[cmcd]

Homework Statement

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

Homework Equations

I'm not sure what equations to use.

The Attempt at a Solution

And I have no solution.

 

[SteamKing]

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

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

 

[cmcd]

Thanks SteamKing for the reply.
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)

 

[SteamKing]

Thanks SteamKing for the reply.
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)

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?

 

[rezeew]

I would approach this problem by first understanding the concept of moment of inertia. Moment of inertia is a measure of an object's resistance to rotational motion. It is dependent on the mass and distribution of mass of the object. In this case, the beam has a moment of inertia of (mL^2)/12 about its centre.

Now, when additional masses are added at one end and at the centre of the beam, the overall moment of inertia will change. This can be calculated by using the parallel axis theorem, which states that the moment of inertia of an object can be calculated by adding the moment of inertia of the object about its own center of mass to the product of the mass of the object and the square of the distance between the center of mass and the new axis.

Applying this theorem to the given problem, we can calculate the moment of inertia about the center of the beam as follows:

I = (mL^2)/12 + (m/2)((L/2)^2) + (m/2)(L/2)^2

= (mL^2)/12 + (mL^2)/8 + (mL^2)/8

= (mL^2)/12 + (mL^2)/4

= (mL^2)/3

Therefore, the correct answer is (d) - (1/3)mL^2. This shows that the moment of inertia of the beam with added masses is larger than the moment of inertia of the original beam, indicating that it will have a greater resistance to rotational motion. This is consistent with the concept of inertia, which states that an object will resist changes in its state of motion.

In conclusion, as a scientist, I would use the parallel axis theorem and the concept of moment of inertia to solve this problem and determine the correct answer.