Questions: Perpendicular Axis Theorem

Question 1

A student applies the perpendicular axis theorem to a solid cylinder of radius R and height h. They note I_z = ½MR² and use symmetry (I_x = I_y) to conclude I_x = ¼MR². Is this correct?

• A. Yes — the perpendicular axis theorem applies to any axially symmetric object
• B. No — the theorem requires the object to be planar (all mass in a single plane), and a cylinder with height h has mass distributed at nonzero z-values, violating this requirement
• C. The result ¼MR² is approximately correct for thin cylinders where h << R
• D. No — the symmetry argument I_x = I_y fails for cylinders; symmetry only applies to disks

Answer: (B)

The perpendicular axis theorem applies ONLY to planar objects — those with all mass in the z = 0 plane. The proof uses the fact that for a mass element at (x, y, 0), r_z² = x²+y² = r_y²+r_x² exactly. For a cylinder with height h, mass elements at (x, y, z) with z ≠ 0 have r_x = √(y²+z²) ≠ y, so I_x + I_y ≠ I_z. The correct moment for a cylinder about a diameter includes an h² term and exceeds ¼MR².

Question 2

For a uniform thin disk of mass M and radius R, what is the moment of inertia about a diameter, and how does the perpendicular axis theorem derive it efficiently?

• A. I = ½MR², derived by direct integration of r² dm over the disk
• B. I = ¼MR², derived by noting I_z = ½MR² and using I_z = I_x + I_y with I_x = I_y by rotational symmetry, giving I_x = ¼MR²
• C. I = MR², derived using the parallel axis theorem to shift from center of mass to the rim
• D. I = ⅓MR², derived from the general formula for a planar object rotating about an in-plane axis

Answer: (B)

The perpendicular axis theorem at its most elegant: I_z = ½MR² for the symmetry axis (from direct integration). The disk has rotational symmetry, so any two perpendicular diameters are equivalent: I_x = I_y. The theorem gives I_z = I_x + I_y = 2I_x, so I_x = ¼MR². This avoids integrating over a tilted axis — much harder. Symmetry plus the theorem converts a difficult integral into simple algebra.

Question 3

The perpendicular axis theorem states I_z = I_x + I_y and applies to any object as long as the z-axis is perpendicular to the plane containing x and y.

• T. True
• F. False

Answer: False

The theorem applies only to planar objects — those with all mass at z = 0. For a 3D object, a mass element at (x, y, z) has r_x = √(y²+z²) and r_y = √(x²+z²), so I_x + I_y = Σm(y²+z²) + Σm(x²+z²) = Σm(x²+y²) + 2Σmz² = I_z + 2Σmz². The sum equals I_z only when z = 0 for all mass elements. Applying the theorem to solid 3D objects (cylinders, spheres) produces wrong answers.

Question 4

For a uniform thin disk, the moment of inertia is the same about any diameter, equal to ¼MR².

• T. True
• F. False

Answer: True

By the rotational symmetry of the disk, all diameters are geometrically equivalent — the disk looks the same from every direction in its plane. So I_x = I_y for any pair of perpendicular diameters. The perpendicular axis theorem gives I_z = I_x + I_y = 2I_x, so I_x = ¼MR². This value is the same for any diameter, because the disk's circular symmetry ensures all diameters are equivalent axes.

Question 5

Why does the perpendicular axis theorem require the object to be planar, and what goes wrong when you try to apply it to a three-dimensional object?

Model answer: The proof rests on r_z² = r_x² + r_y², which holds only when z = 0 for every mass element. For a planar object: r_z = √(x²+y²), r_x = y (since z=0), r_y = x, so r_z² = x²+y² = r_y²+r_x². For a 3D object with a mass element at (x, y, z): r_x = √(y²+z²) and r_y = √(x²+z²), so I_x + I_y = Σm(x²+y²) + 2Σmz² = I_z + 2Σmz² ≠ I_z unless z = 0 everywhere.

This is the most common error with this theorem. The planarity condition is not a technicality — it is what makes the 2D Pythagorean theorem applicable to distances from axes. A cylinder, sphere, or any solid 3D object fails this condition. Always verify that the object is a flat plate, disk, ring, or thin shell in a single plane before applying the theorem. The parallel axis theorem, which shifts rather than rotates the axis, has no such planarity restriction.