Moment of intertia of thin disc

[BradP]

Homework Statement

This is not a homework question but it could be. I am trying to understand why the moment of inertia of a thin disc about its axis of rotation is 1/2 MR².

Homework Equations

The rotational moment of inertia of a system of points is ^{}\summr².

The Attempt at a Solution

However, if you just took one point at radius r and divided its mass into as many points as it took to make a virtual ring with radius r, the result of ^{}\summr² is mr², right? So how is this the equation for a solid disc and not a thin ring?

 

[rock.freak667]

Consider a mass at a distance 'r' on the disc.

The total moment of intertia of the disc is ∫r² dm = ∫ r² σ dA where σ is the mass per unit area.

Now dA is just the circumference of that ring * width of the ring (dr)

So if the radius of the ring is 'r' what is the circumference?

 

[BradP]

Consider a mass at a distance 'r' on the disc.

The total moment of intertia of the disc is ∫r² dm = ∫ r² σ dA where σ is the mass per unit area.

Now dA is just the circumference of that ring * width of the ring (dr)

So if the radius of the ring is 'r' what is the circumference?

So the integral becomes \sigma∫r²(2*pi*r)*dr.

(I cannot type any more. Whenever I enter a latex symbol it shows up as some random unrelated symbol.)

 

[rock.freak667]

So the integral becomes \sigma∫r²(2*pi*r)*dr.

(I cannot type any more. Whenever I enter a latex symbol it shows up as some random unrelated symbol.)

Yes and the 'r' varies from 0 to R.

 

[BradP]

Okay, I see how that works. The sigma*pi*r^2 turns into M. But I still do not see why the concept that I illustrated does not also create 1/2*M*R^2. Suppose you start with a mass M at a distance R from the center. Then you divide that mass into 100 pieces and you equally distribute the pieces around a point in a ring, all at a distance R. The moment of inertia of each piece would be 1/100*1/2*M*R^2. Applying ^{}\summr² yields 100*1/100*1/2*M*R^2, which equals 1/2*M*R^2. Isn't that right? How could it be the same?

 

[rock.freak667]

Okay, I see how that works. The sigma*pi*r^2 turns into M. But I still do not see why the concept that I illustrated does not also create 1/2*M*R^2. Suppose you start with a mass M at a distance R from the center. Then you divide that mass into 100 pieces and you equally distribute the pieces around a point in a ring, all at a distance R. The moment of inertia of each piece would be 1/100*1/2*M*R^2. Applying ^{}\summr² yields 100*1/100*1/2*M*R^2, which equals 1/2*M*R^2. Isn't that right? How could it be the same?

Because you are dividing up the mass into equal pieces at the the same radius. So when you sum them up you would get that.

Usually to get the moment of inertia you would start with a small delta element and then form an expression for the inertia of that and then integrate over the entire mass.

 

[BradP]

Because you are dividing up the mass into equal pieces at the the same radius. So when you sum them up you would get that.

Yeah, that is what I did. My point is that it created a ring with mass M. So why doesn't the formula work?

Usually to get the moment of inertia you would start with a small delta element and then form an expression for the inertia of that and then integrate over the entire mass.

I used a delta element of 1/100th of the total mass, and I found the expression of the inertia of that as 1/100*M*R^2. I then integrated over the entire ring, from 0 to 2*pi.

 

[rock.freak667]

I used a delta element of 1/100th of the total mass, and I found the expression of the inertia of that as 1/100*M*R^2. I then integrated over the entire ring, from 0 to 2*pi.

It is best that you use a delta element of mass 'dm' but, with your way, if you take an element of mass 0.01M at a distance of R; then you've divided the mass into 100 pieces and you placed them all at a distance 'R'. Which is not the case in a disc. All pieces are not equidistant from the center.

 

[BradP]

I know. This is the formula I derived for a ring, not a disc. My question is why I ended up with the same formula as that of a disc. The derivation seems sound to me.

 

[rock.freak667]

I know. This is the formula I derived for a ring, not a disc. My question is why I ended up with the same formula as that of a disc. The derivation seems sound to me.

you have I_disc = 0.5mr² and I_ring =mr².

the only time you ended up with mr² is initially when you took individual elements at equal distances of 'r' and summed them up. That forms a ring and not a disc.

You should not end up with the inertia of the ring when you do it for a disc.

But in your OP, you took one point and subdivided that, which would give you mr² being the inertia of a point mass.

 

[BradP]

you have I_disc = 0.5mr² and I_ring =mr².

the only time you ended up with mr² is initially when you took individual elements at equal distances of 'r' and summed them up. That forms a ring and not a disc.

You should not end up with the inertia of the ring when you do it for a disc.

Oh wow, I thought I was getting the same thing. I don't know how I overlooked the (1/2), sorry.

But in your OP, you took one point and subdivided that, which would give you mr² being the inertia of a point mass.

Well, the point was that I formed a ring by subdividing the point mass, but that doesn't matter anymore.