- #1
kontejnjer
- 72
- 21
Homework Statement
A homogenous disk with radius [itex]R[/itex] and mass [itex]m[/itex] lies in the xy plane so its center matches the origin [itex]O[/itex]. Point [itex]O'[/itex] is on the z axis at a distance [itex]s[/itex] from point [itex]O[/itex]. Axis [itex]y'[/itex] passes through point [itex]O'[/itex] at an angle [itex]\theta[/itex] with the z axis. Find the moment of inertia around axis [itex]y'[/itex].
Homework Equations
[itex]I_{ij}=\int dm(\delta_{ij}r^2-x_{i}x_{j})[/itex]
The Attempt at a Solution
The new coordinates x',y',z' are connected with the old x,y,z through the equations:
[itex]x'=x[/itex]
[itex]y'=y\text{sin}\theta+(z-s)\text{cos}\theta[/itex]
[itex]z'=-y\text{cos}\theta+(z-s)\text{sin}\theta[/itex]
Since the body has radial symmetry and is 2 dimensional, we can use polar coordinates:
[itex]x=rcos\varphi[/itex]
[itex]y=rsin\varphi[/itex]
and the third coordinate z=0 since its in the xy plane, hence:
[itex]x'=r\text{cos}\varphi[/itex]
[itex]y'=r\text{sin}\varphi \text{sin}\theta-s\:\text{cos}\theta[/itex]
[itex]z'=-r\text{sin}\varphi \text{cos}\theta-s\:\text{sin}\theta[/itex]
The body is homogenous so:
[itex]\sigma=\frac{dm}{dA}=const.\rightarrow dm=\sigma dA[/itex]
[itex]\sigma=\frac{m}{A}=\frac{m}{\pi\:R^{2}}[/itex]
In polar coordinates:
[itex]dA=r\:drd\varphi[/itex]
with limits:
[itex]0\leq r\leq R[/itex]
[itex]0\leq \varphi\leq 2\pi[/itex]
However, I'm not sure whether the "moment of inertia" would in this case mean the [itex]I_{y'\:y'}[/itex] component from the inertia tensor. If that's the case, then:
[itex]I_{y'\:y'}=\int dm(x'^2+z'^2)=\sigma \int \int r\:drd\varphi[(r\text{cos}\varphi)^2+(-r\text{sin}\varphi \text{cos}\theta-s\:\text{sin}\theta)^2][/itex]
which Wolfram evaluates as:
[itex]I_{y'\:y'}=\frac{1}{4} m \left(R^2+R^2 \text{cos}^2\theta+4 s^2 \text{sin}^2\theta\right)[/itex]
Would this be the correct procedure or not?
