Posted by **Jake** on Thursday, March 4, 2010 at 11:40pm.

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

- Physics. -
**drwls**, Friday, March 5, 2010 at 7:56am
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

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