Finding Moment of Inertia Using Integration

[student34]

Homework Statement

A cylinder with radius R and mass M has density that increases linearly with distance r from the cylinder axis, ρ = αr, where α is a positive constant. Calculate the moment of inertia of the cylinder about a longitudinal axis through its center in terms of M and R.

This is from my textbook, but my textbook has only explained how to find inertias from uniform masses leading up to this question. What's even worse is that this question is given the easiest rating.

Homework Equations

I = Ʃ(mi*r^2i)

Normally I would find the integral of r^2*dm, but my book says that it is only for a uniform distribution of mass.

The Attempt at a Solution

I = Ʃ(mi*r^2i) = m(1)*r^2(1) + m(2)*r^2(2) + ...

I just cannot figure out how to incorporate an increasing density into this.

 

[SteamKing]

You can't use your proposed formula for non-constant density. You must go back to the integral definition of mass moment of inertia.

If the axis of rotation is the z-axis, then the MMOI about the z-axis is:
?

See: http://www.saylor.org/site/wp-content/uploads/2012/09/ME1022.1.2.pdf

 

[student34]

You can't use your proposed formula for non-constant density. You must go back to the integral definition of mass moment of inertia.

If the axis of rotation is the z-axis, then the MMOI about the z-axis is:
?

See: http://www.saylor.org/site/wp-content/uploads/2012/09/ME1022.1.2.pdf

I couldn't find non-uniform increasing density in the link you provided. I feel fairly comfortable with simple objects that have a uniform density. But I just have no idea what to with a changing density. Here's my work.

I = ∫r^2*dm

ρ = αr (according to the question)

ρ = αr = dm/dV then dm = α*r*dV

I = ∫r^2*α*r*dV

dV = 2*∏*L*r*dr

I = ∫r^2*α*r*2*∏*L*r*dr
= ∫r^4*α*2*∏*L*dr
= α*2*∏*L*∫r^4*dr
= (ρ/r)*2*∏*L*∫r^4*dr
= (ρ/r)*2*∏*L*(r^5)/5
= ρ*2*∏*L*(r^4)/5
= (M/V)*2*∏*L*(r^4)/5
= (M*2*∏*L*r^4)/(5*∏*L*R^2)
= (2*M*R^2)/5

My textbook has I = (3*M*R^2)/5. I do not feel confident in my answer.

 

[TSny]

Here's my work.

I = ∫r^2*dm

ρ = αr (according to the question)

ρ = αr = dm/dV then dm = α*r*dV

I = ∫r^2*α*r*dV

dV = 2*∏*L*r*dr

I = ∫r^2*α*r*2*∏*L*r*dr
= ∫r^4*α*2*∏*L*dr
= α*2*∏*L*∫r^4*dr

I think everything's fine up to here. Of course, you should have definite limits of integration in mind for the integral.

But your next step shown below is odd.

= (ρ/r)*2*∏*L*∫r^4*dr

You can't replace the constant α by ρ/r because r is your variable of integration. Leave the integral in terms of α. But then you will need to do a separate integration to relate α to M and R.

 

[student34]

I think everything's fine up to here. Of course, you should have definite limits of integration in mind for the integral.

But your next step shown below is odd.


You can't replace the constant α by ρ/r because r is your variable of integration.

The reason why I did this is because the question said that α is a constant. So then I just made sure that α = p/r stayed as a constant. I feel quite lost.

Leave the integral in terms of α. But then you will need to do a separate integration to relate α to M and R.

I don't why I would do that, and I don't know how I would do that. I feel like this question is beyond anything that we learned up until now in the textbook. But it is a question among many relatively easy ones that the chapter has a direct explanation for.

 

[TSny]

The reason why I did this is because the question said that α is a constant. So then I just made sure that α = p/r stayed as a constant. I feel quite lost.

OK, let's just take it step by step. You had a very good start where you got to the integral

I = ∫(r²)(αr)(L2∏rdr) = α2∏L∫r⁴dr.

(Note that we have formatting buttons for superscripts and subscripts that you can use to make the expressions easier to read.)

Now, α is just a constant factor that you can leave alone for now. Go ahead and evaluate the integral with the correct lower and upper limits for r.

 

[student34]

ok, let's just take it step by step. You had a very good start where you got to the integral

i = ∫(r²)(αr)(l2∏rdr) = α2∏l∫r⁴dr.

(note that we have formatting buttons for superscripts and subscripts that you can use to make the expressions easier to read.)

now, α is just a constant factor that you can leave alone for now. Go ahead and evaluate the integral with the correct lower and upper limits for r.

I = α*2*∏*l*∫r⁴dr = (α*2*∏*L*r⁵)/5

I can't figure out how to show the proper scripts for the lower and upper limits for the integral. The lower is 0 and the upper is R.

 

[TSny]

I = α*2*∏*l*∫r⁴dr = (α*2*∏*L*r⁵)/5

[EDIT] Shouldn't your result for the integral be expressed in terms of the limits of integration?

 

[student34]

I am just trying to figure out how to show that.

 

[TSny]

Don't worry about showing the limits of integration on the integral. Just worry about using those limits when evaluating the integral.

 

[student34]

$\int$$^{R}_{0}$ How did you get your subscripts on top and on the bottom like that.

 

[TSny]

Let's not worry about how to type in the limits on the integral. What is the answer to $\int_0^R r^4dr$?

 

[student34]

Ok, it is r = R, and r = 0, right?

 

[student34]

R⁵/5

 

[TSny]

Ok, it is r = R, and r = 0, right?

Those are the upper and lower limits of the integral. Do you know how to evaluate an integral at its limits? So far, you have gotten to the point of evaluating

$I = a2\pi L \int_0^Rr^4dr$.

Can you carry out the integration and evaluate at the limits?

 

[TSny]

R⁵/5

Good. So, what is your expression for the moment of inertia at this point?

 

[student34]

Good. So, what is your expression for the moment of inertia at this point?

I = a*2*∏*L*R⁵/5

 

[TSny]

I = a*2*∏*L*R⁵/5

OK. Now you just need to figure out how to express the constant $a$ in terms of $M$ and $R$. Can you use the density function to set up another integral that would represent the total mass of the cylinder?

 

[student34]

OK. Now you just need to figure out how to express the constant $a$ in terms of $M$ and $R$. Can you use the density function to set up another integral that would represent the total mass of the cylinder?

I = a*2*∏*L*R⁵/5 and α = ρ/r = m/(∏*L*r³)

= (m*2*∏*L*R⁵)/(∏*L*r³*5)

= (m*2*R²)/5

Hmmm, I'm stuck.

 

[TSny]

I = a*2*∏*L*R⁵/5 and α = ρ/r = m/(∏*L*r³)

You don't want to substitute α = ρ/r. That would put the variable r into the expression. But I should be a fixed constant. Note that r and R have distinct meanings, the first is a variable while the second is a constant.

The total mass M must be the sum (integral) of all the elements of mass dm: M = ∫dm. You know how to substitute for dm in terms of the density function and dV. Try it here and see if you can evaluate the integral.

 

[student34]

You don't want to substitute α = ρ/r. That would put the variable r into the expression. But I should be a fixed constant. Note that r and R have distinct meanings, the first is a variable while the second is a constant.

The total mass M must be the sum (integral) of all the elements of mass dm: M = ∫dm. You know how to substitute for dm in terms of the density function and dV. Try it here and see if you can evaluate the integral.

ρ = dm/dV then dm = ρ*dV

So M = ρ∫dV = ρ∫2*L*r*dr, oh boy, I have a feeling that I am not on the right track.

 

[TSny]

ρ = dm/dV then dm = ρ*dV

So M = ρ∫dV = ρ∫2*L*r*dr

Oops. You can't pull ρ outside of the integral M = ∫ρdV, it's a function: ρ = αr. Also, you left out a factor of ∏ in the volume element: dV = 2∏*L*r*dr.

 

[student34]

Oops. You can't pull ρ outside of the integral M = ∫ρdV, it's a function: ρ = αr. Also, you left out a factor of ∏ in the volume element: dV = 2∏*L*r*dr.

Ahhhh I think I got it.

M = ∫ρ*L*∏*r*dr

M = ∫α*r²*2*L*∏*dr

M = (α*2*L*∏*r³)/3

α = (3*M)/(2*L*∏*r³)

I = α*2*∏*L*R⁵/5 Now I will substitute in for α to bring M into the equation.

I = (3*M)/(2*L*∏*r³)*(2*∏*L*R⁵/5)

I = 3*M*R²/5

Is this correct?

 

[TSny]

Ahhhh I think I got it.

Yes, I think that's essentially it. When you do the integral M = ∫α*r²*2*L*∏*dr you should evaluate the result at the limits r = 0 and r = R. So, instead of writing the result as M = (α*2*L*∏*r³)/3, it should be (α*2*L*∏*R³)/3. When you solve for α, it should be written in terms of R instead of r.

Otherwise, looks good!

 

[student34]

Yes, I think that's essentially it. When you do the integral M = ∫α*r²*2*L*∏*dr you should evaluate the result at the limits r = 0 and r = R. So, instead of writing the result as M = (α*2*L*∏*r³)/3, it should be (α*2*L*∏*R³)/3. When you solve for α, it should be written in terms of R instead of r.

Otherwise, looks good!

Thank-you so much!