Confusion over dV of a cylinder.

[Science Student]

I know that the volume of a cylinder is V = (pi)*L*r^2.

 

But, when it comes to dV of cylinder, my book seems to imply two different answers. On one page of my textbook dV = dx*dy*dz, where x, y and z are the three dimensions; okay that makes sense. But then when it explains how to find the moment of inertia of a solid cylinder, it suggests using dV = 2*(pi)*L*r*dr to put the differential and integration variables in the same terms, which also makes sense. But now we are using a derivative of only one dimension while r and L are its actual values.

 

So the explanation of dV has all 3 dimensions differentiated. But in the problem, they only differentiate one of the dimensions. Does anyone know how dV can mean two different things?

Edited January 19, 2014 by Science Student

[swansont]

Does the moment of inertia of a cylinder (about its axis) depend on L?

[studiot]

 

Does the moment of inertia of a cylinder (about its axis) depend on L?

 

 

Since the OP asked about a solid cylinder, the answer is yes it does.

 

However, ScienceStudent, you need to be a good deal clearer about what you mean. We would not normally develop the moment of inertia in terms of volume.

 

The axial moment of inertia is Wr²/2g so it depends upon length inasmuch as we add up the contributions from all the disks along the length, but tell us more about what you book says for further comment.

[swansont]

 

Since the OP asked about a solid cylinder, the answer is yes it does.

 

 

 

Um, no, it doesn't. Moment of inertia of a cylinder of mass m depends on r. Not L.

[studiot]

 

Um, no, it doesn't. Moment of inertia of a cylinder of mass m depends on r. Not L.

 

 

But if you increase L, do you not increase m?

Edited January 19, 2014 by studiot

[swansont]

But if you increase L, do you not increase m?

it depends on how the problem is phrased, and we don't know that. But given the answer we were given, it's likely it was phrased as an object with a fixed mass.

[studiot]

 

it depends on how the problem is phrased, and we don't know that. But given the answer we were given, it's likely it was phrased as an object with a fixed mass.

 

 

The OP said nothing about the mass being fixed or otherwise.

 

Why would the mass be fixed?

That implies that either the radius or the density decreases as the length increases. that makes no sense.

 

The fact remains that it takes an engine twice the effort to run up a shaft of twice the length, but the same material and diameter.

 

I think the important point to make to the OP is that there is one property that varies with the volume, but without regard to shape and we call that the mass.

For any shape whatsoever the mass = volume times the density.

 

But the moment of inertia contains information about the distribution of that mass and depends not only on the volume but the shape of that volume and therefore the distribution of the mass in it.

 

I think it is legitimate to assume constant density for that exercise.

 

I did also say in post#3 that we really need more detail.

[swansont]

The OP said nothing about the mass being fixed or otherwise.

 

Why would the mass be fixed?

That implies that either the radius or the density decreases as the length increases. that makes no sense.

 

You're right, it says nothing. But we know what answer was given: one that did not vary with length.

 

There are obvious situations where density can differ: thermal expansion and using different materials immediately come to mind.

[Science Student]

Sorry for the confusion, I really just want to know what dV actually means for cylinder with uniform density and fixed dimensions. I think that the textbook wants us to assume the values for the dimension are fixed but equal L, r, P,V just to get a general formula for the inertia of any cylinder. However, the book does want r to vary when integrating the infinitesimally thin shells of the cylinder as they need to change in circumference.

[swansont]

Ah, so it's not even varying the volume, it's the integral to calculate it. IOW, the mass is fixed, and L is not a variable of integration - it's also a constant for the moment of inertia case

[Science Student]

Ah, so it's not even varying the volume, it's the integral to calculate it. IOW, the mass is fixed, and L is not a variable of integration - it's also a constant for the moment of inertia case

Yeah, this the first semester of first year physics. It seems to be an introduction to finding inertia by using integration.

[studiot]

Integration is often most useful as summation, adding in the contributions of very small parts of a whole.

 

That is what you are describing in this case. It is better not to think of dV as a variation but as a small part of the whole.

[Endercreeper01]

Sorry for the confusion, I really just want to know what dV actually means for cylinder with uniform density and fixed dimensions.

[latex]\partial V[/latex] is just an infinitesimal volume. It can mean different things because [latex]V[/latex] is a function of multiple variables.

Let's say that [latex]V[/latex] is a function of some variable [latex]n[/latex]. To find [latex]\partial V[/latex], we can imagine [latex]\frac{\partial V}{\partial n}[/latex] is a ratio between two infinitesimals, [latex]\partial V[/latex] and [latex]\partial n[/latex]. Because of this, we just multiply the partial derivative by [latex]\partial n[/latex] to get [latex]\partial V[/latex], i.e [latex]\partial V=\frac{\partial V}{\partial n}\partial n[/latex]

This can be applied to the different variables V depends on, such as x, y, and z.

For example, let's say that

[latex]V(x, y, z)= x^2 + y^2 +z - xyz^2[/latex]

In this case, we can take the partial derivative with respect to 3 variables: x, y, and z.

We will find [latex]\partial V[/latex] with respect to x.

We find that

[latex]\frac{\partial V}{\partial x}=2x+y^2-yz^2[/latex]

and so

[latex]\frac{\partial V}{\partial x} \partial x=(2x+y^2-yz^2) \ dx[/latex]

For cylinders, [latex]V[/latex] is a function of [latex]z[/latex] and [latex]r[/latex].

The volume of a cylinder is given by

[latex]V=\pi r^2 z[/latex]

And so we have 2 values of [latex]\partial V[/latex],

[latex]\partial V=\frac{\partial V}{\partial r} \partial r=2\pi r z dr[/latex]

and

[latex]\partial V= \frac{\partial V}{\partial z}\partial z=\pi r^2 dz[/latex]

In a cylinder with uniform density, [latex]dV[/latex] is just [latex]dV=\frac{1}{\rho }dM[/latex]

Edited January 20, 2014 by Endercreeper01