- 1). Multiply half the mass (M) by the sum of the inner and outer radii of the disk if the cutout is in the center of the disk. Thus, the moment of inertia equals 1/2 (M) (R_out + R_in), where R_out is the radius of the outer ring and R_in is the radius of the inner ring. Given a mass of 50 kg and inner/outer radii of 2 meters and 3 meters, respectively, calculate 1/2 (50) (2 + 5) to get a moment of inertia of 175 kg/m^2.
- 2). Measure the distance (D) from the center of the disk to the center of the cutout if the latter is not in the center of the disk. Square it and then multiply it by 2 to get A. Given a distance of 3 meters, calculate 2 (3^2) to get 18 meters.
- 3). Square the radius (r) of the hole and add it to A to get B. Given a radius of 5 meters, calculate 5^2 + 18 to get 43 meters.
- 4). Divide the square of the hole radius by the square of the disk radius (R), and multiply the result by B to get C. Given a hole radius of 5 meters and a disk radius of 13 meters, calculate 5^2 / 13^2 to get .1479. Multiply the result by 43 to get 6.3597.
- 5). Subtract C from the square of the disk radius to get D. Given the example, square 13 to get 169, and then subtract 6.3597 from it to get 162.6403.
- 6). Multiply D by half of the disk's mass to determine the moment of inertia. Given a mass of 20 kg, multiply half of it (10) by 162.6403 to get a moment of inertia of 1626.403 kg/m^2. The general formula for steps 2 through 6 is:
1/2 (M) [ R^2 - r^2/R^2 (r^2 + 2d^2) ]
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