Concept Card: Perpendicular Axis Theorem – Statement, Proof & Applications
1. Statement of the Theorem:
The Perpendicular Axis Theorem applies strictly to two-dimensional planar bodies of negligible thickness (planar laminas).
Formal Statement:
The moment of inertia ($I_z$) of a plane lamina about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two mutually perpendicular axes ($I_x$ and $I_y$) concurrent with the perpendicular axis and lying within the plane of the lamina:
$$\mathbf{I_z = I_x + I_y}$$
where the lamina lies entirely in the $xy$-plane ($z = 0$), and the three coordinate axes $x$, $y$, and $z$ intersect at a single common origin $O$ and are mutually orthogonal.
2. Rigorous Analytical Derivation
Consider a planar lamina lying entirely in the $xy$-plane ($z = 0$). Let $O(0, 0, 0)$ be the point of concurrence of three mutually perpendicular axes: the $x$-axis and $y$-axis in the plane, and the $z$-axis perpendicular to the plane.
- Consider an infinitesimal mass element $dm$ located at coordinates $(x, y, 0)$ on the lamina.
- The perpendicular distance of the element from the $x$-axis is $|y|$. Thus, the moment of inertia about the $x$-axis is:
$$I_x = \int y^2 dm$$ - The perpendicular distance of the element from the $y$-axis is $|x|$. Thus, the moment of inertia about the $y$-axis is:
$$I_y = \int x^2 dm$$ - The perpendicular distance of the element from the $z$-axis (which passes through the origin $O$ perpendicular to the $xy$-plane) is $r = \sqrt{x^2 + y^2}$. Therefore:
$$I_z = \int r^2 dm = \int (x^2 + y^2) dm$$ - Splitting the integral:
$$I_z = \int x^2 dm + \int y^2 dm = I_y + I_x = \mathbf{I_x + I_y}$$
Q.E.D.
3. Strict 2D Planar Lamina Limitation (Crucial JEE Rule)
The perpendicular axis theorem is strictly invalid for three-dimensional bodies.
- For an extended 3D body with non-zero thickness along the $z$-axis:
$$I_z = \int (x^2 + y^2) dm$$
$$I_x = \int (y^2 + z^2) dm \quad \text{and} \quad I_y = \int (x^2 + z^2) dm$$ - Summing $I_x$ and $I_y$:
$$I_x + I_y = \int (x^2 + y^2 + 2z^2) dm = I_z + 2\int z^2 dm > I_z$$ - Therefore, for a solid sphere, cylinder, cone, or cube, $I_z \neq I_x + I_y$. Attempting to apply $I_z = I_x + I_y$ to any 3D body is a classic conceptual failure in competitive exams!
4. Standard Applications on Classical Planar Laminas
A. Thin Circular Ring (Mass $M$, Radius $R$):
All mass is in the $xy$-plane at distance $R$ from the central $z$-axis, so $I_z = M R^2$.
By continuous circular symmetry in the plane, all diametrical axes are identical: $I_x = I_y = I_d$.
$$I_z = I_x + I_y \implies M R^2 = 2 I_d \implies \mathbf{I_d = \frac{1}{2} M R^2}$$
Radius of gyration: $\mathbf{k_d = \frac{R}{\sqrt{2}}}$.
B. Uniform Circular Disc (Mass $M$, Radius $R$):
From direct integration, the perpendicular central axis has $I_z = \frac{1}{2} M R^2$.
By planar symmetry, every diameter in the disc plane has identical rotational inertia: $I_x = I_y = I_d$.
$$I_z = 2 I_d \implies \frac{1}{2} M R^2 = 2 I_d \implies \mathbf{I_d = \frac{1}{4} M R^2}$$
Radius of gyration: $\mathbf{k_d = \frac{R}{2}}$.
C. Uniform Rectangular Lamina (Mass $M$, Sides $a$ and $b$):
For a rectangular plate centered at the origin with length $a$ along $x$ and width $b$ along $y$:
– Axis through center parallel to width $b$ (along $y$): $I_y = \frac{1}{12} M a^2$.
– Axis through center parallel to length $a$ (along $x$): $I_x = \frac{1}{12} M b^2$.
– Central axis perpendicular to the plate ($z$-axis):
$$\mathbf{I_z = I_x + I_y = \frac{1}{12} M (a^2 + b^2)}$$
D. Uniform Square Lamina (Mass $M$, Side $a$):
Here $a = b$, so $I_x = I_y = \frac{1}{12} M a^2$, and $I_z = \frac{1}{6} M a^2$.
In-Plane Rotational Invariance Theorem: For any square lamina, any two mutually perpendicular in-plane axes passing through the center (such as the two diagonals $d_1$ and $d_2$) must satisfy $I_{d_1} + I_{d_2} = I_z = \frac{1}{6} M a^2$. By fourfold rotational symmetry, $I_{d_1} = I_{d_2}$, which gives:
$$\mathbf{I_{\text{diagonal}} = \frac{1}{12} M a^2}$$
In fact, the moment of inertia of a uniform square plate about any line in its plane passing through its center is identically $\mathbf{\frac{1}{12} M a^2}$!
5. Common JEE Pitfalls & Traps
- Trap 1 (The 3D Cylinder / Sphere Blunder): Applying $I_z = I_x + I_y$ to a cylinder or sphere. Remember: The theorem is valid only for 2D flat sheets of negligible thickness.
- Trap 2 (Non-Concurrent Axes): The three axes must strictly intersect at a single common point $O$. You cannot add $I_x$ through the center and $I_y$ through an edge to get $I_z$!
- Trap 3 (Non-Orthogonal In-Plane Axes): The two in-plane axes $x$ and $y$ must be mutually perpendicular (angle between them must be exactly $90^\circ$).
Solved Examples
Example 1 (Direct Application – Ring & Disc Diametrical Properties):
A uniform circular disc of mass $M = 2.0\text{ kg}$ and radius $R = 0.60\text{ m}$ lies in the $xy$-plane centered at origin $O$.
(a) Use the perpendicular axis theorem to derive the moment of inertia $I_d$ of the disc about any diameter.
(b) Calculate the radius of gyration $k_d$ about a diameter.
(c) If a constant torque $\tau = 0.90\text{ N}\cdot\text{m}$ is applied to spin the disc about its diameter from rest, calculate the angular velocity $\omega$ acquired after $t = 4.0\text{ s}$.
Solution:
(a) The central perpendicular axis ($z$-axis) has moment of inertia $I_z = \frac{1}{2} M R^2$.
Choosing any two orthogonal diameters along the $x$ and $y$ axes: by rotational symmetry of the circle, $I_x = I_y = I_d$.
Applying the Perpendicular Axis Theorem:
$$I_z = I_x + I_y = 2 I_d \implies I_d = \frac{1}{2} I_z = \frac{1}{4} M R^2$$
Numerical value:
$$I_d = \frac{1}{4} (2.0\text{ kg}) (0.60\text{ m})^2 = \frac{1}{4} (2.0)(0.36) = \mathbf{0.180\text{ kg}\cdot\text{m}^2}$$
(b) Radius of gyration about a diameter:
$$k_d = \sqrt{\frac{I_d}{M}} = \sqrt{\frac{\frac{1}{4}M R^2}{M}} = \frac{R}{2} = \frac{0.60\text{ m}}{2} = \mathbf{0.30\text{ m}}$$
(c) Angular acceleration about the diametrical axis:
$$\alpha = \frac{\tau}{I_d} = \frac{0.90\text{ N}\cdot\text{m}}{0.180\text{ kg}\cdot\text{m}^2} = 5.0\text{ rad/s}^2$$
Angular velocity after $4.0\text{ s}$:
$$\omega = \alpha t = (5.0\text{ rad/s}^2)(4.0\text{ s}) = \mathbf{20.0\text{ rad/s}}$$
Example 2 (In-Plane Invariance of Square Plates):
A uniform square plate of mass $M = 3.0\text{ kg}$ and side length $a = 1.0\text{ m}$ lies in the $xy$-plane with its center at the origin $O$.
(a) Find the moment of inertia $I_x$ and $I_y$ about axes parallel to its edges passing through its center.
(b) Find the moment of inertia $I_z$ about the axis perpendicular to the plate passing through its center.
(c) Prove that the moment of inertia about either of its diagonals is equal to $I_x$, and calculate its numerical value.
(d) Prove that the moment of inertia about ANY line in the plane of the plate passing through $O$ inclined at an arbitrary angle $\theta$ is constant.
Solution:
(a) Consider strips of length $a$ parallel to the $x$-axis. Each strip is at distance $y$ from the $x$-axis.
$$I_x = \frac{1}{12} M a^2 = \frac{1}{12} (3.0\text{ kg})(1.0\text{ m})^2 = \mathbf{0.250\text{ kg}\cdot\text{m}^2}$$
By symmetry between $x$ and $y$:
$$I_y = \frac{1}{12} M a^2 = \mathbf{0.250\text{ kg}\cdot\text{m}^2}$$
(b) Applying the Perpendicular Axis Theorem:
$$I_z = I_x + I_y = \frac{1}{12} M a^2 + \frac{1}{12} M a^2 = \frac{1}{6} M a^2 = \frac{1}{6}(3.0)(1.0)^2 = \mathbf{0.500\text{ kg}\cdot\text{m}^2}$$
(c) Let the two diagonals be $D_1$ and $D_2$. In a square, diagonals are mutually perpendicular and intersect at the center $O$.
By the Perpendicular Axis Theorem:
$$I_{D_1} + I_{D_2} = I_z = \frac{1}{6} M a^2$$
By fourfold reflection symmetry, $I_{D_1} = I_{D_2} = I_{\text{diag}}$:
$$2 I_{\text{diag}} = \frac{1}{6} M a^2 \implies I_{\text{diag}} = \frac{1}{12} M a^2 = \mathbf{0.250\text{ kg}\cdot\text{m}^2} = I_x$$
(d) For an axis in the plane making an angle $\theta$ with the $x$-axis, the perpendicular distance of any point $(x, y)$ from this line is $u = -x\sin\theta + y\cos\theta$.
$$I_\theta = \int u^2 dm = \int (y\cos\theta – x\sin\theta)^2 dm = \cos^2\theta \int y^2 dm + \sin^2\theta \int x^2 dm – 2\sin\theta\cos\theta \int x y\,dm$$
By reflection symmetry across both axes, $\int x y\,dm = 0$.
$$I_\theta = I_x \cos^2\theta + I_y \sin^2\theta$$
Since $I_x = I_y = \frac{1}{12} M a^2$:
$$I_\theta = \left(\frac{1}{12} M a^2\right)(\cos^2\theta + \sin^2\theta) = \mathbf{\frac{1}{12} M a^2 = \text{constant!}}$$
Thus, the moment of inertia of a square plate about ANY central in-plane axis is completely invariant with respect to rotation.
Example 3 (Standard JEE Advanced Scenario – Elliptical Plate & Combined Theorems):
A uniform flat elliptical plate of mass $M = 2.0\text{ kg}$ is bounded by $\frac{x^2}{a^2} + \frac{y^2}{b^2} \le 1$ with semi-major axis $a = 0.60\text{ m}$ along the $x$-axis and semi-minor axis $b = 0.40\text{ m}$ along the $y$-axis.
It is given that $I_x = \frac{1}{4} M b^2$ and $I_y = \frac{1}{4} M a^2$.
(a) Determine the moment of inertia $I_z$ about the central axis perpendicular to the plate.
(b) Calculate the eccentricity $e$ of the ellipse and the coordinate position of its focus $F_1(ae, 0)$.
(c) Using the Parallel Axis Theorem, calculate the moment of inertia $I_{F_1}$ about an axis perpendicular to the plate passing through focus $F_1$.
Solution:
(a) By the Perpendicular Axis Theorem for the flat lamina in the $xy$-plane:
$$I_z = I_x + I_y = \frac{1}{4} M b^2 + \frac{1}{4} M a^2 = \frac{1}{4} M (a^2 + b^2)$$
Numerical value:
$$I_z = \frac{1}{4} (2.0\text{ kg}) \left[(0.60)^2 + (0.40)^2\right] = \frac{1}{2} (0.36 + 0.16) = \frac{1}{2}(0.52) = \mathbf{0.260\text{ kg}\cdot\text{m}^2}$$
(b) Eccentricity $e$ of the ellipse:
$$e = \sqrt{1 – \frac{b^2}{a^2}} = \sqrt{1 – \frac{0.16}{0.36}} = \sqrt{1 – \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3} \approx 0.7454$$
The distance of focus $F_1$ from origin $O$ is $d = a e$. Therefore:
$$d^2 = a^2 e^2 = a^2 – b^2 = (0.60)^2 – (0.40)^2 = 0.36 – 0.16 = \mathbf{0.20\text{ m}^2}$$
(c) The axis through focus $F_1$ is parallel to the central $z$-axis at perpendicular distance $d = \sqrt{0.20}\text{ m}$.
Applying the Parallel Axis Theorem:
$$I_{F_1} = I_z + M d^2 = I_z + M(a^2 – b^2)$$
Substitute algebraic expressions:
$$I_{F_1} = \frac{1}{4} M (a^2 + b^2) + M(a^2 – b^2) = M\left[\frac{5}{4}a^2 – \\frac{3}{4}b^2\right]$$
Numerical value:
$$I_{F_1} = 0.260\text{ kg}\cdot\text{m}^2 + (2.0\text{ kg})(0.20\text{ m}^2) = 0.260 + 0.400 = \mathbf{0.660\text{ kg}\cdot\text{m}^2}$$
Example 4 (Edge Case – Right-Angled Isosceles Triangular Lamina):
A uniform thin flat plate of mass $M$ forms a right-angled isosceles triangle with legs of length $a$ aligned along the $+x$ and $+y$ axes ($0 \le x \le a$, $0 \le y \le a – x$).
(a) Derive the moment of inertia $I_x$ and $I_y$ about the legs of the triangle.
(b) Find the moment of inertia $I_z$ about the axis perpendicular to the plate passing through the right-angle vertex $(0, 0)$.
(c) Derive the moment of inertia $I_{\text{hyp}}$ about the hypotenuse.
Solution:
(a) Area of the triangle $A = \frac{1}{2} a^2$. Surface density $\sigma = \frac{M}{A} = \frac{2M}{a^2}$.
For $I_x$, consider a horizontal strip of length $(a – y)$ and thickness $dy$ at height $y$:
$$dm = \sigma (a – y)\,dy$$
$$I_x = \int_0^a y^2 dm = \sigma \int_0^a y^2 (a – y)\,dy = \sigma \left[ a\frac{y^3}{3} – \frac{y^4}{4} \right]_0^a = \sigma \left( \frac{a^4}{3} – \frac{a^4}{4} \right) = \sigma \frac{a^4}{12}$$
Substitute $\sigma = \frac{2M}{a^2}$:
$$I_x = \left(\frac{2M}{a^2}\right) \left(\frac{a^4}{12}\right) = \mathbf{\frac{1}{6} M a^2}$$
By exact symmetry between $x$ and $y$ legs:
$$I_y = \mathbf{\frac{1}{6} M a^2}$$
(b) Applying the Perpendicular Axis Theorem at the right-angle vertex $(0, 0)$:
$$I_z = I_x + I_y = \frac{1}{6} M a^2 + \frac{1}{6} M a^2 = \mathbf{\frac{1}{3} M a^2}$$
(c) The hypotenuse is the line $x + y = a$. The perpendicular distance of any point $(x, y)$ from the hypotenuse is $u = \frac{a – (x + y)}{\sqrt{2}}$.
$$I_{\text{hyp}} = \int u^2 dm = \frac{1}{2} \int (a – x – y)^2 dm$$
Using coordinate transformations or standard triangular plate integrals, the moment of inertia about the base of an isosceles triangle of altitude $h = \frac{a}{\sqrt{2}}$ is:
$$I_{\text{base}} = \frac{1}{6} M h^2 = \frac{1}{6} M \left(\frac{a}{\sqrt{2}}\right)^2 = \frac{1}{6} M \left(\frac{a^2}{2}\right) = \mathbf{\frac{1}{12} M a^2}$$
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
The perpendicular axis theorem $I_z = I_x + I_y$ is strictly valid for:
(A) Solid cylinders
(B) Solid spheres
(C) Two-dimensional planar laminas
(D) Any three-dimensional rigid body
Problem 2 (JEE Main – Single Correct):
A circular wire loop of mass $M$ and radius $R$ lies in the $xy$-plane. Its moment of inertia about the $x$-axis (a diameter) is:
(A) $M R^2$
(B) $\frac{1}{2} M R^2$
(C) $\frac{1}{4} M R^2$
(D) $2 M R^2$
Problem 3 (JEE Main – Single Correct):
A uniform circular disc of mass $M$ and radius $R$ has a moment of inertia $I$ about an axis in its plane touching its edge tangentially. The value of $I$ is:
(A) $\frac{1}{4} M R^2$
(B) $\frac{1}{2} M R^2$
(C) $\frac{5}{4} M R^2$
(D) $\frac{3}{2} M R^2$
Problem 4 (JEE Main – Single Correct):
A uniform square plate of mass $M$ and side $a$ has a moment of inertia about an axis passing through its center perpendicular to its plane given by:
(A) $\frac{1}{12} M a^2$
(B) $\frac{1}{6} M a^2$
(C) $\frac{1}{3} M a^2$
(D) $\frac{2}{3} M a^2$
Problem 5 (JEE Main – Single Correct):
A uniform rectangular sheet of mass $M$ has dimensions $6.0\text{ m} \times 4.0\text{ m}$. The ratio of its moment of inertia about an axis passing through its center parallel to its length to that parallel to its breadth is:
(A) $4 : 9$
(B) $9 : 4$
(C) $2 : 3$
(D) $3 : 2$
Problem 6 (JEE Advanced – One or More Correct):
Which of the following statements regarding the perpendicular axis theorem is/are correct?
(A) The three axes $x$, $y$, and $z$ must intersect at a common origin.
(B) The axes $x$ and $y$ must be mutually perpendicular and lie within the plane of the lamina.
(C) For a 3D solid sphere, $I_x + I_y = I_z + 2\int z^2 dm > I_z$.
(D) The theorem can be directly used to find the moment of inertia of a flat circular disc about its diameter from its central perpendicular inertia.
Problem 7 (JEE Advanced – One or More Correct):
For a uniform square lamina of side $a$ and mass $M$ lying in the $xy$-plane with center at the origin:
(A) The moment of inertia about the diagonal $y = x$ is $\frac{1}{12} M a^2$.
(B) The moment of inertia about the diagonal $y = -x$ is $\frac{1}{12} M a^2$.
(C) The moment of inertia about any arbitrary line in the $xy$-plane passing through the origin is $\frac{1}{12} M a^2$.
(D) The moment of inertia about an axis perpendicular to the plate passing through one corner is $\frac{2}{3} M a^2$.
Problem 8 (JEE Advanced – One or More Correct):
A thin uniform circular ring and a uniform circular disc have identical mass $M$ and radius $R$ lying in the $xy$-plane centered at $O$:
(A) $I_{\text{ring}, z} = 2 I_{\text{ring}, x}$.
(B) $I_{\text{disc}, z} = 2 I_{\text{disc}, x}$.
(C) $I_{\text{ring}, x} = I_{\text{disc}, z}$.
(D) $I_{\text{ring}, x} = 2 I_{\text{disc}, x}$.
Problem 9 (JEE Main / Advanced – Numerical Value Type):
A uniform rectangular plate of mass $M = 6.0\text{ kg}$ has length $a = 4.0\text{ m}$ and width $b = 2.0\text{ m}$. Calculate its moment of inertia in $\text{kg}\cdot\text{m}^2$ about an axis perpendicular to the plate passing through its geometric center.
Problem 10 (JEE Main / Advanced – Numerical Value Type):
A uniform square plate has mass $M = 12.0\text{ kg}$ and side length $a = 2.0\text{ m}$. Calculate the moment of inertia of the plate (in $\text{kg}\cdot\text{m}^2$) about any of its diagonals.
Solutions & Explanations
Answer Key Summary:
1. (C) | 2. (B) | 3. (C) | 4. (B) | 5. (A) | 6. (A, B, C, D) | 7. (A, B, C, D) | 8. (A, B, C, D) | 9. 10 | 10. 4
Solution 1:
The perpendicular axis theorem is derived under the condition that all mass elements have $z = 0$, which holds strictly only for 2D planar laminas.
Correct Option: (C)
Solution 2:
For a thin ring, $I_z = M R^2$. By planar symmetry, $I_x = I_y = I_d$.
$I_z = 2 I_d \implies I_d = \frac{1}{2} M R^2$.
Correct Option: (B)
Solution 3:
For a disc, diameter in-plane has $I_d = \frac{1}{4} M R^2$.
By the Parallel Axis Theorem, for an in-plane tangent at distance $d = R$:
$$I = I_d + M R^2 = \frac{1}{4} M R^2 + M R^2 = \frac{5}{4} M R^2$$
Correct Option: (C)
Solution 4:
For a square plate, $I_x = I_y = \frac{1}{12} M a^2$.
By the Perpendicular Axis Theorem:
$$I_z = I_x + I_y = \frac{1}{12} M a^2 + \frac{1}{12} M a^2 = \frac{1}{6} M a^2$$
Correct Option: (B)
Solution 5:
Let length $a = 6.0\text{ m}$ along $x$ and width $b = 4.0\text{ m}$ along $y$.
– Axis parallel to length (parallel to $x$): $I_x = \frac{1}{12} M b^2$.
– Axis parallel to breadth (parallel to $y$): $I_y = \frac{1}{12} M a^2$.
$$\text{Ratio} = \frac{I_x}{I_y} = \frac{b^2}{a^2} = \frac{4.0^2}{6.0^2} = \frac{16}{36} = \frac{4}{9}$$
Correct Option: (A)
Solution 6:
– (A) True: Coordinate setup requires concurrent orthogonal axes.
– (B) True: Standard condition of the theorem.
– (C) True: Mathematical proof showing why 3D bodies fail the theorem.
– (D) True: $2 I_d = I_z \implies I_d = I_z / 2$.
Correct Options: (A, B, C, D)
Solution 7:
– (A, B) True: $I_{\text{diag}} = \frac{1}{12} M a^2$.
– (C) True: $I_\theta = I_x \cos^2\theta + I_y \sin^2\theta = \frac{1}{12} M a^2$.
– (D) True: Distance from center to corner is $d = a/\sqrt{2} \implies I_{\text{corner}} = I_z + M d^2 = \frac{1}{6} M a^2 + M(a^2/2) = \frac{4}{6} M a^2 = \frac{2}{3} M a^2$.
All options are correct.
Correct Options: (A, B, C, D)
Solution 8:
– (A) True: $I_{\text{ring}, z} = M R^2 = 2(\frac{1}{2} M R^2) = 2 I_{\text{ring}, x}$.
– (B) True: $I_{\text{disc}, z} = \frac{1}{2} M R^2) True: $I_{\text{disc}, z} = \frac{1}{2} M R^2 = 2(\frac{1}{4} M R^2) = 2 I_{\text{disc}, x}$.
– (C) True: $I_{\text{ring}, x} = \frac{1}{2} M R^2$ and $I_{\text{disc}, z} = \frac{1}{2} M R^2$.
– (D) True: $I_{\text{ring}, x} = \frac{1}{2} M R^2 = 2(\frac{1}{4} M R^2) = 2 I_{\text{disc}, x}$.
All options are correct.
Correct Options: (A, B, C, D)
Solution 9:
$$I_z = \frac{1}{12} M (a^2 + b^2) = \frac{1}{12} (6.0\text{ kg}) (4.0^2 + 2.0^2) = \frac{1}{12} (6.0) (16.0 + 4.0) = \frac{1}{12} (6.0)(20.0) = \mathbf{10.0\text{ kg}\cdot\text{m}^2}$$
Correct Answer: 10
Solution 10:
For any square plate about its diagonal:
$$I_{\text{diag}} = \frac{1}{12} M a^2 = \frac{1}{12} (12.0\text{ kg}) (2.0\text{ m})^2 = \frac{1}{12} (12.0)(4.0) = \mathbf{4.0\text{ kg}\cdot\text{m}^2}$$
Correct Answer: 4