Moment of Inertia of a Disc: Is it paradoxical?

[Aurelius120]

Homework Statement

Calculate the Moment of Inertia of Disc about axis passing through its center and perpendicular to plane

Relevant Equations

$dI=\int{x^2 dm }$

In deriving the MOI of a ring about its center perpendicular to plane, our teacher said think of ring as made up of $dm$ units each at $R$ distance from the axis therefore the MOI becomes: $$I=R^2\int{dm}=MR^2$$

If disc is considered to be made up of rods of $dx$ thickness, in this manner

then by the former logic its MOI should be$$I=R^2\frac{\int{dm}}{3}$$ However the correct method of derivation using unequal rods or concentric rings shows that it is $$I=\frac{MR^2}{2}$$ Doesn't it seem paradoxical??

 

[PeroK]

Doesn't it seem paradoxical??

More wrong than paradoxical, I would say.

 

[Orodruin]

As you can clearly see from your image, your assumption of fixed width rods does not correspond to a constant density disc.

 

[Aurelius120]

@Oroduin constant density should mean $$dm=\sigma R dx$$. Area of strip being $Rdx$

 

[haruspex]

@Oroduin constant density should mean $$dm=\sigma R dx$$. Area of strip being $Rdx$

If you take sample areas of your diagram, the rods (constant thickness) will occupy a larger fraction of the area for samples close to the disc centre than in samples nearer the edge. That does not represent constant density.
Instead of rods, consider the triangular slices.

 

[BvU]

constant density should mean $$dm=\sigma R dx$$. Area of strip being $Rdx$

And the integral is from $x=?$ to $x= ?$

$\$

 

[kuruman]

In deriving the MOI of a ring about its center perpendicular to plane, our teacher said think of ring as made up of $dm$ units each at $R$ distance from the axis therefore the MOI becomes: $$I=R^2\int{dm}=MR^2$$

If disc is considered to be made up of rods of $dx$ thickness, in this manner
then by the former logic its MOI should be$$I=R^2\frac{\int{dm}}{3}$$ However the correct method of derivation using unequal rods or concentric rings shows that it is $$I=\frac{MR^2}{2}$$ Doesn't it seem paradoxical??

Paradox resolved. Your teacher said "ring" but you read which you misapplied to "disc".

 

[haruspex]

Paradox resolved. Your teacher said "ring" but you read "disc".

As I read post #1, the ring case is only being used as an example of how to find MoIs. The actual task is to do the same for a disc.

 

[BvU]

Paradox resolved. Your teacher said "ring" but you read which you misapplied to "disc".

I wonder what the OP finds for the area of a circle

$\$

 

[kuruman]

As I read post #1, the ring case is only being used as an example of how to find MoIs. The actual task is to do the same for a disc.

I stand corrected.

 

[Aurelius120]

If you take sample areas of your diagram, the rods (constant thickness) will occupy a larger fraction of the area for samples close to the disc centre than in samples nearer the edge. That does not represent constant density.
Instead of rods, consider the triangular slices.

But as the thickness approaches zero shouldn't the area of triangular and rectangular slices both become lines? And a circle is made of lines

If I take a large bunch of wires and glue them to form a circle won't its MOI be $\frac{MR^2}{3}$

 

[haruspex]

But as the thickness approaches zero shouldn't the area of triangular and rectangular slices both become lines? And a circle is made of lines

If I take a large bunch of wires and glue them to form a circle won't its MOI be $\frac{MR^2}{3}$

Suppose you were to stick on thin rectangular strips radially to cover the circle. In the centre, they would stack up, forming a cone. No matter how thin you make the strips this won't change.

 

[PeroK]

But as the thickness approaches zero shouldn't the area of triangular and rectangular slices both become lines? And a circle is made of lines

If I take a large bunch of wires and glue them to form a circle won't its MOI be $\frac{MR^2}{3}$

I can't make any sense of either of these statements. The thing you are missing is this: you can view a disc as a set of circles of increasing radius. And you can use this idea to calculate the area or MOI of a disc. I'm surprised you haven't seen this technique before. It applies across area, volume and MOI calculations for many shapes and solids.

In any case, you need a variable ($r$, say) to represent the variable radius of the circles in this case.

 

[Orodruin]

But as the thickness approaches zero shouldn't the area of triangular and rectangular slices both become lines? And a circle is made of lines

If I take a large bunch of wires and glue them to form a circle won't its MOI be $\frac{MR^2}{3}$

Simply put: No.

If you try this then the disc will be denser in the middle. What you need if you want to go that way is the MoI of infinitesimal circle segments.

 

[kuruman]

But as the thickness approaches zero shouldn't the area of triangular and rectangular slices both become lines? And a circle is made of lines

If I take a large bunch of wires and glue them to form a circle won't its MOI be $\frac{MR^2}{3}$

I think you have a conceptual difficulty here about how the idea behind assembling typical mass elements $dm$. The simple truth is that, if you expect to assemble a mass of uniform density starting from a typical element, you should be able to reverse the process and disassemble the mass to get that typical element. Let's see how that works here.

You propose to disassemble the disc into wedges (like a pizza) of radius $R$ which, when subdivided into many pieces results in a typical piece that looks like a stick. Start with 4 pieces. A typical piece is a wedge that has arc length $S_4=\frac{2\pi R}{4}.$ Cut each piece in half and you end up with a wedge that has arc length $S_8=\frac{2\pi R}{8}.$ Keep halving to get 1024 pieces and you will end up with a wedge has arc length $S_{\text{1024}}=\frac{2\pi R}{1024}.$

You can see from this that no matter how many times you halve the wedges, you will end up with wedge mass elements that have finite arc length $dS=Rd\theta$ at one side and taper to a point at the other side. You will never end up with stick mass elements that have uniform thickness $dx$ from one end to the other. It follows from the simple truth that you cannot assemble a disc from sticks of uniform mass density because you cannot subdivide it into sticks of uniform mass density.

However, if you had a pizza made in a rectangular tray, then a stick and not a wedge would be the appropriate shape to reassemble it. The impossibility of turning a rectangular shape into a circle is also known as squaring the circle.

 

[Aurelius120]

Thanks