Moment of inertia of a solid cone about its axis For a solid circular cone of mass m and base radius r, what is the moment of inertia about its own vertical (symmetry) axis?

Introduction / Context:
Moments of inertia about symmetry axes are essential in rotational dynamics and vibration analysis. For common solids (disc, cylinder, sphere, cone), standard formulas save derivation time during exams and design work.

Given Data / Assumptions:

• Homogeneous solid cone.
• Mass = m; base radius = r.
• Axis of rotation is the cone’s vertical (symmetry) axis through the apex and center of base.

Concept / Approach:
The cone can be treated as a stack of differential disks of radius varying linearly from 0 at the tip to r at the base. Integrating each disk’s moment of inertia (1/2 * dm * ρ^2) over the height yields the standard result I_axis = (3/10) * m * r^2.

Step-by-Step Solution (outline):
Let radius vary with height: ρ = (r/h) * x.For a thin disk: dI = (1/2) * ρ^2 * dm.Relate dm to density and geometry; integrate from x = 0 to h.Result simplifies to I = 3/10 * m * r^2.

Verification / Alternative check:
Compare with solid cylinder I = 1/2 * m * r^2: since a cone concentrates less mass near the rim than a cylinder of the same r, its I must be less than 0.5 * m * r^2. The value 0.3 * m * r^2 satisfies this expectation.

Why Other Options Are Wrong:

• 3mr^2/5, 2mr^2/5, 4mr^2/5 are larger than the cylinder’s 1/2 mr^2, which is inconsistent for a cone.

Common Pitfalls:

• Confusing the cone’s axial I with its I about a centroidal diameter, which has a different value.

Final Answer:
3mr2/10