Topic 2.2 Extended G1 – Calculating rotational inertia If a body is made of discrete masses use I = Σm i r i 2 rotational inertia (discrete masses) If a body is made of a continuous distribution of masses, we have I = ∫r 2 dm rotational inertia (continuous masses) This equation is derived in a way analogous to that for the cm. Note: The important thing to remember is that you choose mass elements dm that are located a distance r from the axis of rotation.
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So how are all of the rotational inertias calculated?We won’t derive all of them, but we will derive I for the rod (because it is 1D, and therefore easiest).
Consider a thin rod of mass M and length L with a rotational axis perpendicular to its center.We superimpose a Cartesian cs over the rod, centering it on the axis of rotation.
Now consider the same thin rod of mass M and length L with a rotational axis perpendicular to its end.We superimpose a Cartesian cs over the rod, centering it on the axis of rotation.
Note that the axes of rotation all pass through either the center or the end.What if we have a rotating object like a sphere on the end of a rod, as shown.
axis
The rotational inertia of the rod is Irod,end, but the sphere is not rotating about its center (it is located the length of the rod plus its radius from the axis).This section tells how to find the rotational inertia through any axis parallel to the one passing through the cm of the object.
THE PARALLEL AXIS THEOREMSuppose we know Icm, the rotational inertia of any mass M through it’s center of mass.Then the rotational inertia of that mass through any parallel axis is given by
I = Icm + Mh2
where h is the distance between the new axis and the axis passing through the cm.
PROOF OF THE PARALLEL AXIS THEOREMConsider the irregularly-shaped object of mass M shown below:If we choose an axis passing through the cm, we can find Icm:If we choose a parallel axis passing through some other point located a distance h from the cm, we can find I with respect to the new point.