I don't quite get the picture here. The exact moment of inertia, for rotation about an axis that is perpendicular to the axis of the glued-together rod, will depend upon the diameter of the discs as well as the length of the rod. I don't get the significance of the number of discs, since they are glued together into one road.
The formula I = (1/2) m L^2 applies to a rod that is very narrow compared to its length, and rotated about its center, about an axis PERPENDICULAR to the axis of the rod
The figure below
shows an arrangement of 15 identical disks that have been glued together in a rod-like shape of length L = 1.4500 m and (total) mass M = 135.000 g. The arrangement can rotate about a perpendicular axis through its central disk at point O.
(a) What is the rotational inertia of the arrangement about that axis?
(b) If we approximated the arrangement as being a uniform rod of mass M and length L, what percentage error would we make in using the formula in Table 10-2e to calculate the rotational inertia?
%
the formula given is I=1/12 ML^2
2 answers
Use parallel axis theorem for each individual particle
1 answer