Parallel Axis Theorem: Statement, Mathematical Proof & Applications

Concept Card: Parallel Axis Theorem – Statement, Proof & Applications

1. Statement of the Theorem:
The Parallel Axis Theorem (also termed Steiner’s Theorem) provides a universal method for evaluating the moment of inertia of any rigid body about an arbitrary axis parallel to an axis passing through its centre of mass.

Formal Statement:
The moment of inertia ($I$) of any rigid body of total mass $M$ about any arbitrary axis is equal to the moment of inertia ($I_{\text{cm}}$) about a parallel axis passing through its centre of mass plus the product of the mass $M$ of the body and the square of the perpendicular distance ($d$) between the two parallel axes:
$$\mathbf{I = I_{\text{cm}} + M d^2}$$

2. Rigorous Analytical Derivation

Consider an extended rigid body of total mass $M$. Let the origin of coordinate system $O(0, 0, 0)$ be situated precisely at the centre of mass (CM) of the body.

Let the central axis pass through the origin along the $z$-axis ($I_{\text{cm}} = I_z$).
Let an arbitrary parallel axis $AB$ be parallel to the $z$-axis and displaced by coordinate offsets $(x_0, y_0, 0)$. The perpendicular distance between the two parallel axes is:
$$d = \sqrt{x_0^2 + y_0^2} \implies d^2 = x_0^2 + y_0^2$$
Consider an infinitesimal mass element $dm$ located at coordinates $(x, y, z)$ relative to the center of mass.
- Its perpendicular distance from the central axis is: $r_{\text{cm}}^2 = x^2 + y^2$.
- Its perpendicular distance from the parallel axis $AB$ is:
  $$r’^2 = (x – x_0)^2 + (y – y_0)^2 = (x^2 + y^2) + (x_0^2 + y_0^2) – 2 x x_0 – 2 y_0 y = r_{\text{cm}}^2 + d^2 – 2 x_0 x – 2 y_0 y$$
Integrating over the entire volume of the rigid body to obtain the moment of inertia $I$ about axis $AB$:
$$I = \int r’^2 dm = \int r_{\text{cm}}^2 dm + d^2 \int dm – 2 x_0 \int x dm – 2 y_0 \int y dm$$
By physical definition of the centre of mass located at the origin $O$:
$$\int x dm = M x_{\text{cm}} = M(0) = 0 \quad \text{and} \quad \int y dm = M y_{\text{cm}} = M(0) = 0$$
Consequently, the first-order cross terms vanish identically, leaving:
$$\mathbf{I = I_{\text{cm}} + M d^2}$$

3. Radius of Gyration Formulation

Expressing moments of inertia in terms of radii of gyration ($I = M k^2$ and $I_{\text{cm}} = M k_{\text{cm}}^2$):

$$M k^2 = M k_{\text{cm}}^2 + M d^2 \implies \mathbf{k^2 = k_{\text{cm}}^2 + d^2 \implies k = \sqrt{k_{\text{cm}}^2 + d^2}}$$

Warning: Notice that $k \neq k_{\text{cm}} + d$. Radii of gyration add quadratically in quadrature, never linearly!

4. Vital Conditions & Universality Rules

Universal Applicability to 1D, 2D, and 3D Bodies: Unlike the Perpendicular Axis Theorem (which is strictly restricted to thin 2D planar laminas), the Parallel Axis Theorem is universally valid for any rigid body of arbitrary shape in any dimension (thin rods, plates, cylinders, spheres, cones, irregular blocks).
The CM Anchor Rule (Non-CM Parallel Shift): One of the two parallel axes must pass through the centre of mass. You cannot shift directly between two non-CM parallel axes $A$ and $B$:
$$I_B \neq I_A + M d_{AB}^2 \quad $$
To shift between any two arbitrary parallel axes $A$ and $B$ located at perpendicular distances $d_A$ and $d_B$ from the parallel CM axis:
$$\mathbf{I_B = I_{\text{cm}} + M d_B^2 = (I_A – M d_A^2) + M d_B^2 = I_A + M(d_B^2 – d_A^2)}$$
Minimum Value Property: Since $M d^2 \ge 0$, the moment of inertia about the axis passing through the centre of mass is strictly the minimum among all mutually parallel axes:
$$I \ge I_{\text{cm}} \quad (\text{Equality holds if and only if } d = 0)$$

5. Standard Applications on Classical Geometries

Body & Geometry	CM Axis ($I_{\text{cm}}$)	Shift Distance ($d$)	Parallel Axis ($I$)
Thin Rod (Length $L$)	Perpendicular at center: $\frac{1}{12}M L^2$	$d = L/2$ (to end)	$\frac{1}{12}M L^2 + M(L/2)^2 = \mathbf{\frac{1}{3}M L^2}$
Circular Ring (Radius $R$)	Central perpendicular: $M R^2$	$d = R$ (tangent $\perp$ plane)	$M R^2 + M R^2 = \mathbf{2 M R^2}$
Circular Ring (Radius $R$)	Diameter in-plane: $\frac{1}{2}M R^2$	$d = R$ (tangent in-plane)	$\frac{1}{2}M R^2 + M R^2 = \mathbf{\frac{3}{2} M R^2}$
Circular Disc (Radius $R$)	Central perpendicular: $\frac{1}{2}M R^2$	$d = R$ (tangent $\perp$ plane)	$\frac{1}{2}M R^2 + M R^2 = \mathbf{\frac{3}{2} M R^2}$
Circular Disc (Radius $R$)	Diameter in-plane: $\frac{1}{4}M R^2$	$d = R$ (tangent in-plane)	$\frac{1}{4}M R^2 + M R^2 = \mathbf{\frac{5}{4} M R^2}$
Solid Sphere (Radius $R$)	Diameter: $\frac{2}{5}M R^2$	$d = R$ (tangent line)	$\frac{2}{5}M R^2 + M R^2 = \mathbf{\frac{7}{5} M R^2}$
Hollow Sphere (Radius $R$)	Diameter: $\frac{2}{3}M R^2$	$d = R$ (tangent line)	$\frac{2}{3}M R^2 + M R^2 = \mathbf{\frac{5}{3} M R^2}$

6. Common JEE Pitfalls & Traps

Trap 1 (The Non-CM Axis Fallacy): If a problem gives $I_1$ about an edge and asks for $I_2$ about another point, never write $I_2 = I_1 + M d^2$. Always backtrack to find $I_{\text{cm}} = I_1 – M d_1^2$ first!
Trap 2 (Linear vs. Quadratic Gyration Radius): Writing $k = k_{\text{cm}} + d$ is a widespread conceptual error. The radius of gyration satisfies the Pythagorean relation $k = \sqrt{k_{\text{cm}}^2 + d^2}$.
Trap 3 (Perpendicular vs. Skew Distance): $d$ must strictly be the shortest perpendicular distance between the parallel lines, not the distance between arbitrary points along the lines.

Solved Examples

Example 1 (Direct Conceptual Application – Shifting Between Non-CM Axes):
A uniform thin rod has mass $M = 3.0\text{ kg}$ and length $L = 2.0\text{ m}$.
(a) Find the moment of inertia $I_{\text{cm}}$ of the rod about an axis perpendicular to it passing through its center.
(b) Calculate the moment of inertia $I_P$ about an axis perpendicular to the rod passing through point $P$ located at a distance $L/4 = 0.50\text{ m}$ from one end.
(c) Demonstrate why shifting directly from point $P$ to the end axis using $I_{\text{end}} = I_P + M (L/4)^2$ yields an erroneous result, and compute the correct value of $I_{\text{end}}$.

Solution:
(a) For the axis through the centre of mass:
$$I_{\text{cm}} = \frac{1}{12} M L^2 = \frac{1}{12} (3.0\text{ kg}) (2.0\text{ m})^2 = \frac{1}{12} (3.0)(4.0) = \mathbf{1.00\text{ kg}\cdot\text{m}^2}$$

(b) Point $P$ is at distance $L/4$ from one end. The centre of mass is at distance $L/2$ from that end.
Therefore, the perpendicular distance $d$ between the CM axis and the axis through $P$ is:
$$d = \frac{L}{2} – \frac{L}{4} = \frac{L}{4} = \frac{2.0\text{ m}}{4} = 0.50\text{ m}$$
Applying the Parallel Axis Theorem from the CM axis to $P$:
$$I_P = I_{\text{cm}} + M d^2 = 1.00 + (3.0\text{ kg})(0.50\text{ m})^2 = 1.00 + 3.0(0.25) = 1.00 + 0.75 = \mathbf{1.75\text{ kg}\cdot\text{m}^2}$$

(c) The end of the rod is at distance $d_2 = L/4 = 0.50\text{ m}$ from $P$.
– Incorrect Direct Shift: $I_{\text{false}} = I_P + M(L/4)^2 = 1.75 + 0.75 = 2.50\text{ kg}\cdot\text{m}^2$. This is completely wrong because point $P$ is not the centre of mass!
– Correct Formulation: The end is at distance $d_{\text{end}} = L/2 = 1.0\text{ m}$ from the centre of mass:
$$I_{\text{end}} = I_{\text{cm}} + M d_{\text{end}}^2 = 1.00 + (3.0)(1.0)^2 = \mathbf{4.00\text{ kg}\cdot\text{m}^2}$$
Equivalently, shifting from $P$ via the CM:
$$I_{\text{end}} = I_P – M d_P^2 + M d_{\text{end}}^2 = 1.75 – 0.75 + 3.0(1.0)^2 = 1.00 + 3.00 = \mathbf{4.00\text{ kg}\cdot\text{m}^2}$$

Example 2 (Planar Laminas – Tangential Perpendicular vs. Tangential In-Plane Axes):
A uniform circular disc of mass $M = 4.0\text{ kg}$ and radius $R = 0.50\text{ m}$ lies in the $xy$-plane.
(a) Calculate its moment of inertia $I_{t, \perp}$ and radius of gyration $k_{t, \perp}$ about a tangential axis perpendicular to the plane of the disc.
(b) Calculate its moment of inertia $I_{t, \parallel}$ and radius of gyration $k_{t, \parallel}$ about a tangential axis lying in the plane of the disc.
(c) Find the ratio $I_{t, \perp} / I_{t, \parallel}$.

Solution:
(a) For the central axis perpendicular to the disc plane ($z$-axis):
$$I_{\text{cm}, \perp} = \frac{1}{2} M R^2 = \frac{1}{2}(4.0)(0.50)^2 = 0.50\text{ kg}\cdot\text{m}^2$$
The tangential axis perpendicular to the disc plane is at perpendicular distance $d = R = 0.50\text{ m}$ from the central axis:
$$I_{t, \perp} = I_{\text{cm}, \perp} + M R^2 = \frac{1}{2} M R^2 + M R^2 = \mathbf{\frac{3}{2} M R^2}$$
Numerical value:
$$I_{t, \perp} = \frac{3}{2} (4.0\text{ kg})(0.50\text{ m})^2 = \frac{3}{2}(4.0)(0.25) = \mathbf{1.50\text{ kg}\cdot\text{m}^2}$$
Radius of gyration:
$$k_{t, \perp} = \sqrt{\frac{I_{t, \perp}}{M}} = \sqrt{\frac{3}{2}} R = \sqrt{1.5}(0.50) \approx \mathbf{0.6124\text{ m}}$$

(b) For a diametrical axis lying in the plane of the disc (say the $y$-axis):
$$I_{\text{cm}, \parallel} = \frac{1}{4} M R^2 = \frac{1}{4}(4.0)(0.50)^2 = 0.25\text{ kg}\cdot\text{m}^2$$
The in-plane tangent is parallel to this diameter at distance $d = R = 0.50\text{ m}$:
$$I_{t, \parallel} = I_{\text{cm}, \parallel} + M R^2 = \frac{1}{4} M R^2 + M R^2 = \mathbf{\frac{5}{4} M R^2}$$
Numerical value:
$$I_{t, \parallel} = \frac{5}{4} (4.0\text{ kg})(0.50\text{ m})^2 = \frac{5}{4}(4.0)(0.25) = \mathbf{1.25\text{ kg}\cdot\text{m}^2}$$
Radius of gyration:
$$k_{t, \parallel} = \sqrt{\frac{I_{t, \parallel}}{M}} = \sqrt{\frac{5}{4}} R = \frac{\sqrt{5}}{2}(0.50) \approx \mathbf{0.5590\text{ m}}$$

(c) Ratio of moments of inertia:
$$\frac{I_{t, \perp}}{I_{t, \parallel}} = \frac{\frac{3}{2} M R^2}{\frac{5}{4} M R^2} = \frac{6}{5} = \mathbf{1.20}$$

Example 3 (Standard JEE Advanced Scenario – Equilateral Triangular Rod Frame):
Three identical uniform thin rods, each of mass $m = 1.0\text{ kg}$ and length $L = 1.0\text{ m}$, are rigidly welded together at their ends to form an equilateral triangular frame $ABC$.
(a) Calculate the moment of inertia $I_C$ of the triangular frame about an axis passing through its centroid $C$ perpendicular to the plane of the triangle.
(b) Calculate the moment of inertia $I_A$ about an axis passing through vertex $A$ perpendicular to the plane.
(c) Calculate the moment of inertia $I_{BC}$ about an axis along the side $BC$.

Solution:
(a) In an equilateral triangle of side $L$, the altitude is $h = \frac{\sqrt{3}}{2}L$.
The centroid $C$ is at a perpendicular distance from each side equal to:
$$d = \frac{h}{3} = \frac{1}{3} \left(\frac{\sqrt{3}}{2}L\right) = \frac{L}{2\sqrt{3}}$$
For each individual rod of mass $m$ and length $L$, the moment of inertia about an axis through its own center of mass perpendicular to the triangle plane is $I_{\text{rod, cm}} = \frac{1}{12} m L^2$.
By the Parallel Axis Theorem, the moment of inertia of one rod about the centroid axis is:
$$I_1 = I_{\text{rod, cm}} + m d^2 = \frac{1}{12} m L^2 + m \left(\frac{L}{2\sqrt{3}}\right)^2 = \frac{1}{12} m L^2 + \frac{1}{12} m L^2 = \frac{1}{6} m L^2$$
Since the frame consists of 3 identical rods symmetrically placed about the centroid:
$$I_C = 3 \times I_1 = 3 \left(\frac{1}{6} m L^2\right) = \mathbf{\frac{1}{2} m L^2}$$
Numerical value: $I_C = \frac{1}{2}(1.0)(1.0)^2 = \mathbf{0.50\text{ kg}\cdot\text{m}^2}$.

(b) The total mass of the 3-rod frame is $M = 3m = 3.0\text{ kg}$.
The centroid is the centre of mass of the frame. The distance from the centroid to vertex $A$ is:
$$r_v = \frac{2}{3} h = \frac{2}{3} \left(\frac{\sqrt{3}}{2}L\right) = \frac{L}{\sqrt{3}}$$
Applying the Parallel Axis Theorem to the entire rigid frame from centroid to vertex $A$ ($d = r_v$):
$$I_A = I_C + M r_v^2 = \frac{1}{2} m L^2 + (3m) \left(\frac{L}{\sqrt{3}}\right)^2 = \frac{1}{2} m L^2 + 3m \left(\frac{L^2}{3}\right) = \frac{1}{2} m L^2 + m L^2 = \mathbf{\frac{3}{2} m L^2}$$
Numerical value: $I_A = \frac{3}{2}(1.0)(1.0)^2 = \mathbf{1.50\text{ kg}\cdot\text{m}^2}$.

(c) For an axis along the side $BC$:
– Rod $BC$ lies directly on the axis: for an ideal 1D line mass, $I_{BC, \text{own}} = 0$.
– The other two rods ($AB$ and $AC$) each have one end touching the axis at $B$ and $C$ respectively, inclined at an angle $\theta = 60^\circ$ to the axis.
For a rod rotating about one end inclined at angle $\theta$:
$$I = \frac{1}{3} m L^2 \sin^2(60^\circ) = \frac{1}{3} m L^2 \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{3} m L^2 \left(\frac{3}{4}\right) = \frac{1}{4} m L^2$$
For both inclined rods combined:
$$I_{BC} = 0 + 2 \times \left(\frac{1}{4} m L^2\right) = \mathbf{\frac{1}{2} m L^2}$$
Numerical value: $I_{BC} = \frac{1}{2}(1.0)(1.0)^2 = \mathbf{0.50\text{ kg}\cdot\text{m}^2}$.

Example 4 (Edge Case – Off-Center Cavity Problem with Parallel Axis Shift):
A uniform circular disc of radius $R = 0.60\text{ m}$ has an initial total mass $M_0 = 3.60\text{ kg}$. A circular hole of radius $r = R/3 = 0.20\text{ m}$ is drilled through the disc with its center located at distance $d = R/2 = 0.30\text{ m}$ from the disc’s center $O$.
(a) Find the mass $m_h$ of the removed material and the remaining mass $M_{\text{rem}}$.
(b) Calculate the moment of inertia $I_{\text{hole}, O}$ of the removed circular piece about the perpendicular axis through $O$.
(c) Calculate the moment of inertia $I_{\text{rem}}$ of the remaining disc about the perpendicular central axis passing through $O$.

Solution:
(a) Area of the complete disc: $A_0 = \pi R^2$. Area of the circular hole: $A_h = \pi r^2 = \pi (R/3)^2 = \frac{1}{9}\pi R^2$.
Since mass is uniformly distributed:
$$m_h = \frac{1}{9} M_0 = \frac{1}{9} (3.60\text{ kg}) = \mathbf{0.40\text{ kg}}$$
Remaining mass: $M_{\text{rem}} = M_0 – m_h = 3.60 – 0.40 = \mathbf{3.20\text{ kg}}$.

(b) Moment of inertia of the complete disc about its central perpendicular axis $O$:
$$I_{\text{orig}} = \frac{1}{2} M_0 R^2 = \frac{1}{2} (3.60\text{ kg})(0.60\text{ m})^2 = (1.80)(0.36) = 0.648\text{ kg}\cdot\text{m}^2$$
For the removed circular piece of mass $m_h = 0.40\text{ kg}$ and radius $r = 0.20\text{ m}$:
– Its moment of inertia about its own center of mass is:
$$I_{\text{hole, cm}} = \frac{1}{2} m_h r^2 = \frac{1}{2} (0.40)(0.20)^2 = (0.20)(0.04) = 0.008\text{ kg}\cdot\text{m}^2$$
– Its center of mass is at perpendicular distance $d = 0.30\text{ m}$ from $O$. By the Parallel Axis Theorem:
$$I_{\text{hole}, O} = I_{\text{hole, cm}} + m_h d^2 = 0.008 + (0.40)(0.30)^2 = 0.008 + (0.40)(0.09) = 0.008 + 0.036 = \mathbf{0.044\text{ kg}\cdot\text{m}^2}$$

(c) By the superposition subtraction principle:
$$I_{\text{rem}} = I_{\text{orig}} – I_{\text{hole}, O} = 0.648 – 0.044 = \mathbf{0.604\text{ kg}\cdot\text{m}^2}$$

Worksheet: 10 Practice Problems

Problem 1 (JEE Main – Single Correct):
The parallel axis theorem $I = I_{\text{cm}} + M d^2$ is valid for:
(A) Only one-dimensional thin rods
(B) Only two-dimensional planar laminas
(C) Only three-dimensional spherically symmetric bodies
(D) Rigid bodies of arbitrary shape and dimensions in 1D, 2D, or 3D

Problem 2 (JEE Main – Single Correct):
A uniform circular disc of mass $M$ and radius $R$ has a moment of inertia about an axis tangent to its perimeter and perpendicular to its plane given by:
(A) $\frac{1}{2} M R^2$
(B) $M R^2$
(C) $\frac{3}{2} M R^2$
(D) $2 M R^2$

Problem 3 (JEE Main – Single Correct):
The radius of gyration of a uniform solid sphere of radius $R$ about a tangential axis is:
(A) $\sqrt{\frac{2}{5}} R$
(B) $\sqrt{\frac{7}{5}} R$
(C) $\frac{7}{5} R$
(D) $\sqrt{\frac{5}{3}} R$

Problem 4 (JEE Main – Single Correct):
A body has moment of inertia $I_1$ about an axis at distance $d_1$ from its centre of mass, and $I_2$ about a parallel axis at distance $d_2$ from the centre of mass. The difference $(I_2 – I_1)$ is equal to:
(A) $M (d_2 – d_1)^2$
(B) $M (d_2^2 – d_1^2)$
(C) $M (d_2^2 + d_1^2)$
(D) Zero

Problem 5 (JEE Main – Single Correct):
A thin circular hoop of mass $M$ and radius $R$ rotates about a tangential axis lying in the plane of the hoop. Its moment of inertia about this axis is:
(A) $\frac{1}{2} M R^2$
(B) $M R^2$
(C) $\frac{3}{2} M R^2$
(D) $2 M R^2$

Problem 6 (JEE Advanced – One or More Correct):
Which of the following statements regarding the parallel axis theorem is/are correct?
(A) Among all parallel axes in a given direction, the moment of inertia about the axis passing through the centre of mass is strictly minimum.
(B) The radius of gyration $k$ about a parallel axis shifted by distance $d$ from the CM satisfies $k = \sqrt{k_{\text{cm}}^2 + d^2}$.
(C) The parallel axis theorem requires that the two axes must be strictly parallel to each other.
(D) If $I_A$ is known about any arbitrary non-CM axis $A$, the moment of inertia about a parallel axis $B$ at distance $d$ from $A$ is always $I_A + M d^2$.

Problem 7 (JEE Advanced – One or More Correct):
A uniform thin rod of mass $M$ and length $L$ is rotated about axes perpendicular to its length:
(A) The moment of inertia about the midpoint is $\frac{1}{12}M L^2$.
(B) The moment of inertia about one end is $\frac{1}{3}M L^2$.
(C) The moment of inertia about a point at distance $L/6$ from the center is $\frac{1}{9}M L^2$.
(D) The radius of gyration about an end is $\frac{L}{\sqrt{3}}$.

Problem 8 (JEE Advanced – One or More Correct):
Consider a uniform circular ring of mass $M$ and radius $R$ in the $xy$-plane centered at origin $O$:
(A) Its moment of inertia about the $z$-axis is $M R^2$.
(B) Its moment of inertia about a parallel axis passing through $(R, 0, 0)$ is $2 M R^2$.
(C) Its moment of inertia about the $x$-axis is $\frac{1}{2}M R^2$.
(D) Its moment of inertia about a parallel axis passing through $(0, R, 0)$ parallel to the $x$-axis is $\frac{3}{2}M R^2$.

Problem 9 (JEE Main / Advanced – Numerical Value Type):
A uniform thin rod of mass $M = 12.0\text{ kg}$ and length $L = 2.0\text{ m}$ is rotated about an axis perpendicular to the rod passing through a point at distance $d = 0.50\text{ m}$ from its center of mass. Calculate its moment of inertia in $\text{kg}\cdot\text{m}^2$.

Problem 10 (JEE Main / Advanced – Numerical Value Type):
A uniform solid sphere of mass $M = 5.0\text{ kg}$ and radius $R = 2.0\text{ m}$ rotates about an axis tangent to its surface. Calculate its moment of inertia in $\text{kg}\cdot\text{m}^2$.

Solutions & Explanations

Answer Key Summary:
1. (D) | 2. (C) | 3. (B) | 4. (B) | 5. (C) | 6. (A, B, C) | 7. (A, B, C, D) | 8. (A, B, C, D) | 9. 7 | 10. 28

Solution 1:
The parallel axis theorem is mathematically derived from the definition of center of mass ($\int \vec{r}_{\text{cm}}\,dm = \vec{0}$) in 3D Euclidean space, making it universally valid for any 1D, 2D, or 3D rigid body.
Correct Option: (D)

Solution 2:
For a disc, $I_{\text{cm}, \perp} = \frac{1}{2} M R^2$.
By the Parallel Axis Theorem ($d = R$):
$$I = I_{\text{cm}, \perp} + M R^2 = \frac{1}{2} M R^2 + M R^2 = \frac{3}{2} M R^2$$
Correct Option: (C)

Solution 3:
For a solid sphere, $I_{\text{cm}} = \frac{2}{5} M R^2$. About a tangent ($d = R$):
$$I_{\text{tangent}} = \frac{2}{5} M R^2 + M R^2 = \frac{7}{5} M R^2$$
Radius of gyration: $k = \sqrt{\frac{I_{\text{tangent}}}{M}} = \sqrt{\frac{7}{5}} R$.
Correct Option: (B)

Solution 4:
$I_1 = I_{\text{cm}} + M d_1^2$ and $I_2 = I_{\text{cm}} + M d_2^2$.
Subtracting the two equations gives $I_2 – I_1 = M (d_2^2 – d_1^2)$.
Correct Option: (B)

Solution 5:
For a thin circular hoop, the diametrical axis in the plane has $I_d = \frac{1}{2} M R^2$.
The in-plane tangent is parallel to the diameter at distance $d = R$:
$$I_{\text{tangent, in-plane}} = I_d + M R^2 = \frac{1}{2} M R^2 + M R^2 = \frac{3}{2} M R^2$$
Correct Option: (C)

Solution 6:
– (A) True: Since $M d^2 \ge 0$, $I(d) = I_{\text{cm}} + M d^2$ achieves its global minimum at $d = 0$.
– (B) True: $M k^2 = M k_{\text{cm}}^2 + M d^2 \implies k = \sqrt{k_{\text{cm}}^2 + d^2}$.
– (C) True: By definition, the theorem only applies to parallel axes.
– (D) False: If neither axis passes through the CM, $I_B \neq I_A + M d^2$.
Correct Options: (A, B, C)

Solution 7:
– (A) True: $I_{\text{cm}} = \frac{1}{12}M L^2$.
– (B) True: $I_{\text{end}} = \frac{1}{12}M L^2 + M(L/2)^2 = \frac{1}{3}M L^2$.
– (C) True: $I = I_{\text{cm}} + M(L/6)^2 = \frac{1}{12}M L^2 + \frac{1}{36}M L^2 = \frac{4}{36}M L^2 = \frac{1}{9}M L^2$.
– (D) True: $k_{\text{end}} = \sqrt{I_{\text{end}}/M} = \frac{L}{\sqrt{3}}$.
All options are correct.
Correct Options: (A, B, C, D)

Solution 8:
– (A) True: Central perpendicular axis $I_z = M R^2$.
– (B) True: Tangential perpendicular axis at $(R, 0, 0)$ has $d = R \implies I = M R^2 + M R^2 = 2 M R^2$.
– (C) True: Diametrical axis $I_x = \frac{1}{2}M R^2$.
– (D) True: In-plane tangent parallel to $x$-axis passing through $(0, R, 0)$ has $d = R \implies I = \frac{1}{2}M R^2 + M R^2 = \frac{3}{2}M R^2$.
All options are correct.
Correct Options: (A, B, C, D)

Solution 9:
$I_{\text{cm}} = \frac{1}{12} M L^2 = \frac{1}{12} (12.0\text{ kg})(2.0\text{ m})^2 = \frac{1}{12} (12.0)(4.0) = 4.0\text{ kg}\cdot\text{m}^2$.
Applying the Parallel Axis Theorem ($d = 0.50\text{ m}$):
$$I = I_{\text{cm}} + M d^2 = 4.0 + (12.0)(0.50)^2 = 4.0 + (12.0)(0.25) = 4.0 + 3.0 = \mathbf{7.0\text{ kg}\cdot\text{m}^2}$$
Correct Answer: 7

Solution 10:
For a solid sphere, $I_{\text{tangent}} = \frac{7}{5} M R^2$.
Given $M = 5.0\text{ kg}, R = 2.0\text{ m}$:
$$I_{\text{tangent}} = \frac{7}{5} (5.0\text{ kg})(2.0\text{ m})^2 = 7(4.0) = \mathbf{28.0\text{ kg}\cdot\text{m}^2}$$
Correct Answer: 28

Parallel Axis Theorem: Statement, Mathematical Proof & Applications | JEE Physics Class 11