Moments and products of Inertia

KBriggs · Hodge Conjecture Offline Joined: 10 Jun 2009 Posts: 71

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.

**Posted:** Sat Nov 07, 2009 9:02 am

Immanuel Bonfils · Riemann Hypothesis Offline Joined: 20 May 2006 Posts: 426

Something is missing in the question; seems the angular velocity is a constant (why should it change ?)...

What for are you needing the inertial tensor? The problem doesn't mention any force acting on the bar.

**Posted:** Tue Nov 17, 2009 11:32 am