Moment of Inertia

Circle moment of inertia

Find the principal moments of inertia of a circle of radius $R$ and mass $m$. Use the integration over the circle length. (Hint: first define the infinitesimal moment of inertia).

The inertia moment is a measure of how a body that has certain dimensions, opposes to changes in rotation motion, exactly the same as a mass inertia opposes to changes in linear motion. For a mass m situated at distance r from the center of rotation, its inertia moment is defined as $I=mr^2$. The inertia moment of a body that has measurable dimensions, is the sum of all constituent particles inertia. We define the principal moments of inertia when the body rotates around the principal axes of rotation (where $L=(r \times p) || v$).

In general $I=mr^2$     so $dI=r^2*dm$  for all inertia moments.

We choose a cartesian system, having its origin in the center of the circle as in the figure.

For the rotation around x axis that goes through the center of circle we write:

$I_{xx}=∫_L y^2 dm$    where $y=R*\sin⁡ θ$ from polar coordinates   and $dm=m/L*dl=m/2πR*(R*dθ)$

Thus the principal moment of inertia for rotation around x is

$I_{xx}=m/2πR*∫_0^2π (R*\sin⁡ θ)^2 (R*dθ)=(m/2πR)*R^3  (∫_0^{2π} \sin^2 ⁡θdθ )=[(mR^2)/2π]*π=(mR^2)/2$

For the rotation of the circle around y axis that goes through the origin:

$I_{yy}=∫_L x^2 dm$    where $x=R*\cos⁡ θ$    and $dm=m/L*dl=m/2πR*(R*dθ)$

Thus

$I_{yy}=m/2πR*∫_0^2π (R*\cos⁡ θ)^2 (R*dθ)=(m/2πR)*R^3 (∫_0^{2π} \cos^2⁡ θ )=[(mR^2)/2π]*π=(mR^2)/2$

For the rotation around the z axis we write:

$I_{zz}=∫_L z^2 dm$    where $z^2=R^2=x^2+y^2$

$I_{zz}=∫_L (x^2+y^2 )dm=∫_L x^2 dm+∫_L y^2 dm=I_{xx}+I_{yy}=(mR^2)/2+(mR^2)/2=mR^2$

The central axes theorem states that for all bodies that are placed in the plane (xy), the third axis (z) moment of inertia is equal to the sum of the x and y moments of inertia. This can be easily demonstrated from the above equation for $I_{zz}$

Continue reading on the rotational inertia…