Moment of inertia of a solid sphere about a tangential axis Find the area moment of inertia (second moment of mass) of a solid sphere of mass m and radius r about an axis tangent to its surface.
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A2 m r^2 / 3
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B2 m r^2 / 5
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C7 m r^2 / 3
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D7 m r^2 / 5
Answer
Correct Answer: 7 m r^2 / 5
Explanation
Introduction / Context:Moments of inertia for standard solids are key in rotational dynamics. The tangent-axis value can be obtained using the parallel-axis theorem from the centroidal value.
Given Data / Assumptions:
- Solid (uniform) sphere of mass m and radius r.
- Known centroidal MOI of a solid sphere: I_center = 2 m r^2 / 5 about any diameter.
- Axis sought is tangent to the sphere (parallel to a diameter and shifted by r).
Concept / Approach:Apply the parallel-axis theorem: I_tangent = I_center + m d^2, where d is the perpendicular distance between axes. Here, d = r.
Step-by-Step Solution:
Start with I_center = 2 m r^2 / 5.Distance from center to tangent axis: d = r.Use parallel-axis theorem: I_tangent = I_center + m r^2.Compute: I_tangent = (2/5) m r^2 + 1 * m r^2 = (2/5 + 5/5) m r^2 = 7 m r^2 / 5.Verification / Alternative check:Dimensional check: units are kg·m^2. Value exceeds the centroidal MOI, as expected when shifting the axis away from the center.
Why Other Options Are Wrong:
- 2 m r^2 / 5: centroidal value only, not tangent.
- 2 m r^2 / 3 and 7 m r^2 / 3: incorrect coefficients; no standard derivation yields these.
Common Pitfalls:Forgetting to add m r^2 when moving to a tangent axis; mixing up solid versus hollow sphere formulas.
Final Answer:7 m r^2 / 5