Does dm/dV = m/V for a cylinder?

[Science Student]

When finding the moment of inertia, I was taught that the procedure is to use a constant P for uniform density to equal a derivative of mass over a derivative of volume, dm/dV. And we know that P also equals all of the mass over all of the volume. This seems to mean that the ratio dm/dV equals the ratio m/V. But dV for a cylinder equals 2(pi)*r*h*dr which does not quite equal the volume (pi)(r^2)*h.

 

Or do ratios of infinitesimals to real values work differently than real ratios to real ratios? If so, how?

[swansont]

dm/dV will in general, not simply be m/V. You can see this in the cylindrical geometry you have given, and it's also true of a sphere. Both have the characteristic of depending on r in a nonlinear way, i.e. a shell of thickness dr has a volume that depends on r in some way. A Cartesian coordinate system integral (a cube or rectangular prism) is the exception rather than the rule.

[studiot]

 

dm/dV will in general, not simply be m/V.

 

For an isotropic, homogeneous body, surely they will be.

 

dm/dV is just the limit as dV tends to zero and produces the scalar point density finction.

 

Science Student the problem lies in the fact you have differentiated with respect to the wrong variable.However, congratulations for using you scientific acumen to recognise that something was awry.

 

Presumabbly this floows on from your previous question and I will post a more detailed reply in the same vein later.

Edited January 28, 2014 by studiot

[Endercreeper01]

If the relationship between mass and volume is linear, [latex]\frac{dM}{dV}[/latex] is just [latex]\frac{M}{V}[/latex]. This is because if [latex]M[/latex] linearly depends on [latex]V[/latex], [latex]M=\rho V[/latex]. [latex]\frac{dM}{dV}[/latex] would get you the same value as [latex]\frac{M}{V}[/latex], as [latex]\frac{dM}{dV}[/latex] is just the limit as [latex]\frac{M}{V}[/latex] approaches 0.

[studiot]

 

is just the limit as approaches 0.

 

 

I don't think you quite mean this.

[Science Student]

 

For an isotropic, homogeneous body, surely they will be.

 

dm/dV is just the limit as dV tends to zero and produces the scalar point density finction.

 

Science Student the problem lies in the fact you have differentiated with respect to the wrong variable.However, congratulations for using you scientific acumen to recognise that something was awry.

 

Presumabbly this floows on from your previous question and I will post a more detailed reply in the same vein later.

Okay thanks, I am still unsure about all of this.

[studiot]

 

Okay thanks, I am still unsure about all of this.

 

 

OK I owe you a post that has been a long time in the making.

 

It is taking quite a bit to produce this last one but now I know you are still interested in moment of inertia I will complete it.

[studiot]

I have tried to tailor this development to include answers to your questions.

First a bit of background that may help later.

There are many sorts of moments and it is easy to get them mixed up, unfortunately some textbooks do this.

 

The First Moment from a point, P to an axis X-X is the product of some desired quantity, Q, located at P and the shortest or perpendicular distance to the axis.

 

 

First Moment = Q . d

 

This sounds obvious or pompous, depending upon your point of view, but there is more.

 

In Physics the quantity may be a force, an area, a mass, a pressure or other things.

In some disciplines e.g. Statistics or Economics non-physical quantities are employed.

 

We usually don’t bother to include the ‘First’, just calling this moment the moment of the quantity concerned or even just the moment in the case of force.

 

However we find it is useful to define a second type of moment that is called – yes –

The Second Moment.

This is defined to be the product of Q and the square of the distance d

 

Second Moment = Q . d²

 

The moment of inertia is one of these second moments and is found when we place mass M_q at P. That is P is a mass point.

 

MOI = M_q . d²,

 

MOI has the units ML² or Kg (metres)²

 

Another second moment is called the second moment of area.

This is found when an area A is located at P.

Thus Second moment of area = A d²

 

Since this is an area times a distance times a distance it has units L⁴ or (metres)⁴

 

The problem is that many texts, particularly in the field of strength of materials, wrongly call this the moment of inertia so beware.

Here is an example of what you might see. The column labelled Moment of Inertia in the table really refers to the second moment of area of a plane section.

Note the formula for the “moment of inertia” in the column and the equation at the bottom where you might use this – in bending (flexure) calculations in statics.

 

 

A body or plane figure (called a lamina) is made up of lots of points P.

The relevant first or second moment of the whole body is the sum of all the individual moments of the points P.

 

I mentioned that you need to differentiate and integrate with respect to the appropriate variable.

You also need to choose you elemental region appropriately to sum over the whole body. What works for the calculation of one particular quantity may not work for a different quantity.

 

To show this difference I will work through the calculation of volume and MOI of a cylinder of radius a. This will also show why the length is apparently missing from the MOI.

 

To calculate the volume we can consider an elemental disk of thickness dx and sum all the disks from x=0 to x=L (where L is the length) by integrating between these limits.

 

 

This approach works just fine for the volume and therefore for the mass which we can obtain by multiplying the volume by the density.

But when we come to consider moment of inertia there is a problem.

 

In the sketch consider two small elements of face area of the disk, A1 and A2, as shown, such that A1 is distance r1 from X-X and A2 is distance r2.

 

For the volume calculation each elemental area generates a strip (along the length of the cylinder) of the same volume and therefore the same mass. That is the volume does not depend upon r.

 

Vol of strip generated by (A1) = Vol of strip generated by (A2)

Mass of strip generated by (A1) = Mass of strip generated by (A2)

 

So all we have to do is sum all the elements A to get the overall face area, which we know is

 

[math]{A_{face}} = \pi {a^2}[/math]

 

However the moment of inertia depends upon r (as r^2) so the contributions from A1 and its strip will be different from the contribution from A2 and its strip.

Just integrating along X-X , as we did to find the volume, will not take this into account.

 

So we must choose a more appropriate elemental quantity and variable to integrate.

 

Noting that all the material at distance r from X-X has the same moment of inertia and that this material forms a hollow shell cylinder

 

 

Note the units of the MOI. This is the version you should use for calculations involving the rotational dynamics of a solid cylinder about its own axis

[Science Student]

Thank-you very much, you are very generous with your time and knowledge.