Moment of inertia around the z-axis

[cherry]

Homework Statement

Determine the moment of inertia about the z axis of the assembly which consists of the 1.5-kg rod CD and the 7-kg disk.

Relevant Equations

I_zz=int(r_z^2)dm

I am not getting the right answer and I don't know where I am going wrong. Help would be appreciated, thanks!

(My apologies if my equations are not formatted correctly, I am still having trouble using LaTex on here)

My work:
\mathrm{I}_{zz}=[\frac{1}{12}(1.5)(0.2)^{2})+1.5(0.1sin(26.56))^{2}]+[\frac{1}{2}(7)(0.1)^{2}+7(0.2sin(26.56))^{2}]=0.09898{kg}\cdot{m^2}
Correct answer:
\mathrm{I}_{zz}=0.0915{kg}\cdot{m^2}

 

[Lnewqban]

Are you considering the location of the rod's center of mass?

 

[erobz]

I think it would help you ( for starters) if you tried to derive the mass moment of inertia of a thin-rod about a vertical axis passing through its center at some angle $\theta$ from horizontal using the definition:

$$ I = \int_V \rho r^2 ~ dV $$

It's not easy to tell where you are going wrong without steps, thought process, and working equations in Latex... So lets just start from the beginning.

 

[erobz]

Homework Statement: Determine the moment of inertia about the z axis of the assembly which consists of the 1.5-kg rod CD and the 7-kg disk.
Relevant Equations: I_zz=int(r_z^2)dm

View attachment 359342

I am not getting the right answer and I don't know where I am going wrong. Help would be appreciated, thanks!

(My apologies if my equations are not formatted correctly, I am still having trouble using LaTex on here)

My work:
$$ \mathrm{I}_{zz}=[\frac{1}{12}(1.5)(0.2)^{2}) +1.5(0.1\sin(26.56))^{2}] +[\frac{1}{2}(7)(0.1)^{2}+7(0.2\sin(26.56))^{2}]=0.09898{kg}\cdot{m^2}$$


Correct answer:
\mathrm{I}_{zz}=0.0915{kg}\cdot{m^2}

Fixed your Latex.

You enter that with the following code:

$$ \mathrm{I}_{zz}=[\frac{1}{12}(1.5)(0.2)^{2}) +1.5(0.1\sin(26.56))^{2}] +[\frac{1}{2}(7)(0.1)^{2}+7(0.2\sin(26.56))^{2}]=0.09898{kg}\cdot{m^2}$$

See: LaTeX Guide

 

[cherry]

I think it would help you ( for starters) if you tried to derive the mass moment of inertia of a thin-rod about a vertical axis passing through its center at some angle $\theta$ from horizontal using the definition:

$$ I = \int_V \rho r^2 ~ dV $$

View attachment 359386

It's not easy to tell where you are going wrong without steps, thought process, and working equations in Latex... So lets just start from the beginning.

The center of mass of a thin rod is at L/2.
My derivation for the mass-moment of inertia of a thin rod about a vertical axis based on the diagram provided is the following (x is along the axis of the rod):
$$ \rho=\frac{m}{L} \longrightarrow dm=\frac{m}{L}dx $$
$$ I=\int_V{\rho r^2}dV $$
$$ I=\int_{0}^{L}\frac{m} {L}dx\cdot(xcos\theta)^2 $$
$$ =\frac{m}{L}cos^2\theta\int_{0}^{L}x^2dx $$
$$ =\frac{m}{L}cos^2\theta(\frac{1}{3}L^3) $$
$$ I=\frac{1}{3}mL^2cos^2\theta $$

Based on the question, theta ends up being:
$$ \\\theta=90-tan^{-1}(\frac{0.1}{0.2}) $$
$$ =90-26.565 $$
$$ \theta=63.43 $$

Please let me know if I'm on the right track! Thank you.

 

[erobz]

The center of mass of a thin rod is at L/2.
My derivation for the mass-moment of inertia of a thin rod about a vertical axis based on the diagram provided is the following (x is along the axis of the rod):
$$ \rho=\frac{m}{L} \longrightarrow dm=\frac{m}{L}dx $$
$$ I=\int_V{\rho r^2}dV $$
$$ I=\int_{0}^{L}\frac{m} {L}dx\cdot(xcos\theta)^2 $$
$$ =\frac{m}{L}cos^2\theta\int_{0}^{L}x^2dx $$
$$ =\frac{m}{L}cos^2\theta(\frac{1}{3}L^3) $$
$$ I=\frac{1}{3}mL^2cos^2\theta $$

Based on the question, theta ends up being:
$$ \\\theta=90-tan^{-1}(\frac{0.1}{0.2}) $$
$$ =90-26.565 $$
$$ \theta=63.43 $$

Please let me know if I'm on the right track! Thank you.

I was asking for a derivation about a central axis. Also, your element $dV$ doesn't seem to make sense. Neither do your limits of integration.

Just say the cross-sectional area of the element is $A$, and the uniform density of the bar is $\rho$, and they are both constant with respect to $r$. Try again for the axis shown in the diagram.

Actually lets just take it piece by piece. What is $dV$ ( the volume of the differential element in red) in terms of $dr$? Remember, $r$ is perpendicular to the axis of rotation. Just write me the integral (without limits), with the constants factored out.

 

[Lnewqban]

The center of mass of a thin rod is at L/2.

That is correct respect to the lenght of the rod.
But it is not in a horizontal position.

You would need to use the distances between the centers of mass of both objects and the common axis of rotation.

 

[erobz]

So, $\rho$ isn't a linear density in the equation. $dm = \rho dV$, so that was confusing for me. I also see that you have transformed to $x$ along the length of the bar instead of keeping with $r$. Ok.

Your main issue here is that you have the limits of integration incorrect (as @Lnewqban is pointing out) for the axis in the diagram. You are unwittingly computing $I$ about an axis passing through the end of the bar. Which is what you have in your problem, but not what I asked.