Moments and products of inertia

[KBriggs]

Homework Statement

Hey all

I have a physics problem that I need some help setting up the problem:

A bar of length 2L rotates with angular velocity $$\omega$$, which points in the +y direction. The bar makes an angle $$\alpha$$ with the y axis, with its centre corresponding to the origin. The bar has linear mass density $$\rho(r) = \frac {b}{L^2}r^2$$, where r is the distance along the rod measured from the origin. At the instant that the bar is in the xy plane, find the angular velocity.

Now, I am having trouble computing the moments of inertia. Using the inertia tensor notation, and since $$\omega = (0,\omega,0)$$, it is clear that only the $$I_{xy}, I_{yy}, I_{zy}$$ components will appear in the solution. For those who might not know,

$$I_{xy} = - \int{xydm}$$
$$I_{yy} = \int{(x^2 + z^2)dm}$$
$$I_{zy} = - \int{zydm}$$

Now I know that $$dm = \rho(r)dr$$ but I have no idea how to get $$\rho(r)dr$$

The class has as yet not covered double integrals, so there must be a way to reduce these three to a single variable integration. Note that z=0 throughout the problem.

Any help setting up would be appreciated, and if anything in unclear I'll try to make it clearer.

Homework Equations

$$I_{xy} = - \int{xydm}$$
$$I_{yy} = \int{(x^2 + z^2)dm}$$
$$I_{zy} = - \int{zydm}$$
$$\rho(r) = \frac {b}{L^2}r^2$$

The Attempt at a Solution

I first tried rotating the coordinate axes so that the x' acis corresponded to the length of the rod, but I ended up getting different answer when I integrated directly to find L, and when I used the inertia tensor notation. So I am really stuck right at the beginning of the problem. Some help setting up the integral I need to solve to find the moments and prducts of inertia would be enough for me to solve the rest on my own.

We have not covered double integrals yet, so there must be a way to reduce all of the above to singel variable problems.

Any help is appreciated :)

EDIT: update. I think I have solved for the first product of inertia:

We have that $$\tan\alpha=\frac{x}{y} => y = \frac{x}{\tan\alpha}$$
Also, $$r=\frac{x}{\sin\alpha} => dr = \frac{dx}{\sin\alpha}$$
$$\rho(r) = \frac{b}{L^2}(x^2+\frac{x^2}{\tan\alpha})$$
Putting this together, we have that

$$I_{xy} = -\int_{-L\sin\alpha}^{L\sin\alpha}(\frac{x^2}{\tan\alpha})\frac{b}{L^2}(x^2+\frac{x^2}{\tan\alpha})\frac{dx}{\sin\alpha}$$

And simplifying and solving, we get

$$I_{xy} = \frac{2bL^3}{5}(\sin^3\alpha\cos\alpha + \sin^2\alpha\cos^2\alpha)$$

Can someone check this and tell me if I am even close to the right track?

 

[tiny-tim]

Hey KBriggs!

(have an alpha: α and an omega: ω )

Now, I am having trouble computing the moments of inertia. Using the inertia tensor notation, and since $$\omega = (0,\omega,0)$$, it is clear that only the $$I_{xy}, I_{yy}, I_{zy}$$ components will appear in the solution.

Sorry, but you're doing this completely the wrong way.

Let's start again …

the moment of inertia tensor has to be calculated relative to the principal axes …

doing it along the (sort-of-arbitrary ) x y and z axes won't work.

Those axes, of course, are along the rod (call that r) and any two perpendicular axes perpendicular to the rod (call that s) … so you need I_rr I_rs etc.

Then, relative to those axes, ω is not (0,ω,0), it's (cosα,sinα,0).

Carry on from there. ​

 

[KBriggs]

No, it will still work for arbitrary axes, the only difference is that if you use the principle axes, all of the products of inertia will be 0 and the matrix will be a diagonal matrix.

I am fairly sure that I got the answer (I got the same answer using the tensor as i did by integrating $$\int(rxv)dm$$ directly.

What I did was say that
$$y=\frac{x}{\tan\alpha}$$
$$r=\frac{x}{\sin\alpha}$$
$$dr = \frac{dx}{\sin\alpha}$$
$$dm=\frac{b}{L^2\sin\alpha}(x^2+\frac{x^2}{\tan^2\alpha})dx$$

Then all of the products and moments of inertia mentioned before are integartions over x only, and can be integrated from $$-Lsin\alpha$$ to $$Lsin\alpha$$.


I end up with:

$$L=<\frac{2bL^3w}{5}(\sin^3\alpha\cos\alpha + \sin\alpha\cos^3\alpha), \frac{2bL^3w}{5}(\sin^4\alpha + \sin^2\alpha\cos^2\alpha), 0>$$

I don't suppose anyone actually wants to verify that answer for me? ^_^