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Figure 10-33 shows an arrangement of 15 identical disks
Chapter 10, Problem 40(choose chapter or problem)
Figure 10-33 shows an arrangement of 15 identical disks that have been glued together in a rod-like shape of length L = 1.0000 m and (total) mass \(M\) = 100.0 mg. The disk 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?
Questions & Answers
QUESTION:
Figure 10-33 shows an arrangement of 15 identical disks that have been glued together in a rod-like shape of length L = 1.0000 m and (total) mass \(M\) = 100.0 mg. The disk 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?
Step 1 of 4
We are asked to find the moment of Inertia of the arrangement and the percentage error if we consider the formula given in table 10.2e for the shape if considered to be a uniform rod.
Concept to be used:
Parallel axis theorem:
Suppose we want to find the rotational inertia I of a body of mass M about a given axis. Let h be the perpendicular distance between the given axis and the axis through the center of mass (remember these two axes must be parallel).
Then the rotational inertia I about the given axis is \(I=I_{\text {com }}+m h^{2}\) (parallel-axis theorem).
Here, the center of mass is a uniform disc whose moment of inertia is given as,
\(I_{\text {com }}=\frac{1}{2} m R^{2}\)
Given data:
\(\begin{array}{l}\text{Length of arrangement: } L=1.00 \mathrm{~m}\\ \text{Mass of arrangement: } m=100 \mathrm{mg}\\ \text{Radius of each disc:}\\ R=\frac{\frac{L}{15}}{2}=\frac{L}{30}=0.033 \mathrm{~m}\\ \text{Mass of each disc:}\\ m=\frac{M}{15}=\frac{100 \mathrm{mg}}{15}=6.66 \mathrm{mg}\end{array}\)