Moment Of Inertia

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Four spheres of diameter 2a and mass M are placed with their centres on the four corners of a square of side b. The moment of inertia of the system about one side of the square is:

RankGuru · Accepted Answer

1. **Moment of Inertia of a Single Sphere**: For a solid sphere about its diameter, $I_{CM} = \frac{2}{5}Ma^2$.
2. **Spheres on the axis**: Two spheres have their centers directly on the side of the square being used as the axis. For these, $I_1 = 2 	imes \left(\frac{2}{5}Ma^2ight) = \frac{4}{5}Ma^2$.
3. **Spheres at distance b**: The other two spheres are at a perpendicular distance $b$ from the axis. Using the **Parallel Axis Theorem**: $I_{sphere} = I_{CM} + Mb^2 = \frac{2}{5}Ma^2 + Mb^2$.
4. **Total for these two spheres**: $I_2 = 2 	imes \left(\frac{2}{5}Ma^2 + Mb^2ight) = \frac{4}{5}Ma^2 + 2Mb^2$.
5. **System Total**: $I_{total} = I_1 + I_2 = \frac{4}{5}Ma^2 + \frac{4}{5}Ma^2 + 2Mb^2 = \frac{8}{5}Ma^2 + 2Mb^2$.

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