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Mechanics

Parallel-Axis Theorem

The parallel-axis theorem states that the moment of inertia of a rigid body about any arbitrary axis is equal to the moment of inertia about a parallel axis passing through its centre of mass plus the product of the mass of the body and the square of the perpendicular distance between the two axes. It applies to any rigid body regardless of its geometry or planar symmetry.

UP TGT ScienceUP PGT PhysicsStudy Notes + Practice

Scope and prerequisites

TGT core

Direct numerical calculations using I = I<sub>cm</sub> + Md<sup>2</sup> for standard uniform bodies such as rods, rings, discs, and spheres about parallel axes.

PGT extension

Advanced applications involving composite bodies, continuous mass distributions with variable linear or areal density, and tensor-based rotational dynamics.

Prerequisites

Definition of Moment of Inertia and Radius of Gyration
Properties and coordinate determination of Centre of Mass
Integration techniques and vector summation in rigid body dynamics

Statement of Parallel-Axis Theorem

The theorem relates the moment of inertia I about an axis to the moment of inertia I_cm about a parallel axis passing through the centre of mass.
Mathematical expression: $I = I_{cm} + Md^{2}$, where M is the total mass of the body and d is the perpendicular distance between the two parallel axes.
Both axes must remain strictly parallel to each other for the theorem to hold true.

Condition of Applicability

One of the two axes of rotation must necessarily pass through the centre of mass of the rigid body.
The theorem is valid for both two-dimensional (planar) and three-dimensional (solid) rigid bodies of any arbitrary shape.
It holds for continuous mass distributions as well as discrete systems of particles.

Physical Interpretation

The moment of inertia is minimum about an axis passing through the centre of mass compared to any other parallel axis.
Shifting the axis away from the centre of mass always increases the moment of inertia by the additive term Md².
The term Md² represents the rotational inertia contributed by treating the entire mass concentrated at the centre of mass combined with the displaced axis.

Significance in Rotational Dynamics

Simplifies complex moment of inertia calculations by reducing them to standard centre of mass values.
Essential for analyzing compound pendulums and bodies rotating about a fixed hinge or pivot that does not coincide with the centre of mass.
Used alongside the perpendicular-axis theorem to determine moments of inertia for complex three-dimensional geometries.

Derivations

Derivation of Parallel-Axis Theorem for a Continuous Body

Conditions: Rigid body of mass M, axis passing through CM, and a parallel axis at distance d.

Let an axis pass through the centre of mass C along the z-axis. Consider a mass element dm located at position vector r = (x, y, z) relative to C.
The moment of inertia about the axis passing through the centre of mass is given by $I_{cm} = \int (x^2 + y^2) dm$.
Consider a parallel axis at a perpendicular distance d along the x-axis. The distance squared of the mass element dm from this new axis is given by ((x + d)^2 + y^2).
The moment of inertia about the new axis is $I = \int ((x + d)^2 + y^2) dm = \int (x^2 + y^2 + 2dx + d^2) dm$.
Expanding the integral: $I = \int (x^2 + y^2) dm + 2d \int x dm + d^2 \int dm$.
By definition of the centre of mass relative to the origin at CM, the term \int x dm = 0. Also, \int dm = M and \int (x^2 + y^2) d$m = I_{cm}$.
Substituting these yields the final relation: $I = I_{cm} + Md^2$.

Result: $I = I_{cm} + Md^2$

Formula sheet

Formula / law	Equation	Variables and units	Conditions
Parallel-Axis Theorem	$$I = I_{cm} + M d^2$$	I = Moment of inertia about new axis, $I_{cm} = Moment of inertia about parallel axis through CM$, M = Total mass, d = Perpendicular distance kg m^2	One axis must pass through the centre of mass and both axes must be parallel.
Moment of Inertia of Uniform Rod (about CM)	$$I_{cm} = \frac{1}{12} M L^2$$	M = Mass of rod, L = Length of rod kg m^2	Axis perpendicular to rod passing through its midpoint.
Moment of Inertia of Uniform Rod (about End)	$$I = \frac{1}{3} M L^2$$	M = Mass of rod, L = Length of rod kg m^2	Axis perpendicular to rod passing through one end (derived using $d = L/2)$.
Moment of Inertia of Circular Ring (about Tangent in Plane)	$$I = \frac{3}{2} M R^2$$	M = Mass of ring, R = Radius of ring kg m^2	Axis in the plane of the ring and tangent to its circumference ($I_{cm} = 1/2 MR^2$, d = R).
Moment of Inertia of Circular Disc (about Tangent in Plane)	$$I = \frac{5}{4} M R^2$$	M = Mass of disc, R = Radius of disc kg m^2	Axis in the plane of the disc and tangent to its edge ($I_{cm} = 1/4 MR^2$, d = R).
Moment of Inertia of Circular Disc (about Tangent Perpendicular to Plane)	$$I = \frac{3}{2} M R^2$$	M = Mass of disc, R = Radius of disc kg m^2	Axis perpendicular to the plane of the disc passing through its edge ($I_{cm} = 1/2 MR^2$, d = R).
Moment of Inertia of Solid Sphere (about Tangent)	$$I = \frac{7}{5} M R^2$$	M = Mass of sphere, R = Radius of sphere kg m^2	Axis tangent to the surface of the solid sphere ($I_{cm} = 2/5 MR^2$, d = R).
Moment of Inertia of Hollow Sphere (about Tangent)	$$I = \frac{5}{3} M R^2$$	$M = Mass of solid/thin spherical shell$, R = Radius kg m^2	Axis tangent to the surface of the thin spherical shell ($I_{cm} = 2/3 MR^2$, d = R).

Graphs and experiments

Variation of Moment of Inertia with Axis Shift (Graph)

Draw: A parabolic curve plotting Moment of Inertia (I) on the y-axis against perpendicular distance (d) on the x-axis, with the minimum at I_{cm} when d = 0.

Interpret: Shows that the moment of inertia increases quadratically with distance d from the centre of mass axis according to $I = I_{cm} + Md^2$.

Exam use: Interpreting graphical dependence of rotational inertia on axis position in objective questions.

Torsional Pendulum Experiment

Draw: Experimental setup showing a rigid body suspended by a wire, rotated horizontally to measure time period T.

Interpret: Allows experimental verification of moment of inertia and verification of parallel-axis theorem by adding symmetrical masses at known distances d.

Exam use: Understanding laboratory verification methods for rotational mechanics parameters.

Compound Pendulum Oscillation Setup

Draw: A rigid bar pivoted about an off-centre knife edge oscillating under gravity.

Interpret: Time period $T = 2*\pi*\sqrt(I / (M g l))$, where I is evaluated using the parallel-axis theorem about the pivot point ($I = I_{cm} + Ml^2)$.

Exam use: Solving numericals involving compound pendulum frequency and equivalent length.

Solved numericals

TGT core

Find the moment of inertia of a uniform thin rod of mass 2 kg and length 1.0 m about an axis passing through one of its ends and perpendicular to its length.

Identify the formula for the moment of inertia of a rod about its centre of mass: $I_{cm} = (1/12) M L^2$.
Calculate $I_{cm} = (1/12) * 2 kg * (1.0 m)^2 = 2 / 12 = 1/6 kg m^2$.
Identify the distance d from the centre of mass to the end of the rod: $d = L / 2 = 0.5 m$.
Apply the parallel-axis theorem: $I = I_{cm} + M d^2 = (1/6) + 2 * (0.5)^2 = (1/6) + 2 * (0.25) = (1/6) + 0.5 = 2/3 kg m^2$.

Answer: 0.67 kg m^2

TGT core

Calculate the moment of inertia of a circular disc of mass 4 kg and radius 0.5 m about a tangent lying in the plane of the disc.

The moment of inertia of a disc about its diameter (passing through CM in its plane) is $I_{cm} = (1/4) M R^2$.
Calculate $I_{cm} = (1/4) * 4 kg * (0.5 m)^2 = 1 * 0.25 = 0.25 kg m^2$.
The distance from the diameter to the tangent in the plane is equal to the radius: d = R = 0.5 m.
Apply the parallel-axis theorem: $I = I_{cm} + M d^2 = 0.25 + 4 * (0.5)^2 = 0.25 + 4 * 0.25 = 0.25 + 1.0 = 1.25 kg m^2$.

Answer: 1.25 kg m^2

PGT extension

A uniform square plate has mass M and side length a. Find its moment of inertia about an axis passing through one of its corners and perpendicular to the plane of the plate.

The moment of inertia of a square plate about an axis perpendicular to its plane through the centre of mass is given by $I_{cm} = (1/6) M a^2 (derived using perpendicular-axis theorem with I_{xx} = I_{yy} = (1/12) M a^2)$.
The perpendicular distance d from the centre of the square to any corner is the half-diagonal: $d = \sqrt((a/2)^2 + (a/2)^2) = (a * \sqrt(2)) / 2 = a / \sqrt(2)$.
Square the distance: $d^2 = a^2 / 2$.
Apply the parallel-axis theorem: $I = I_{cm} + M d^2 = (1/6) M a^2 + M (a^2 / 2) = M a^2 * (1/6 + 1/2) = M a^2 * (1/6 + 3/6) = (4/6) M a^2 = (2/3) M a^2$.

Answer: (2/3) M a^2

Assumptions and limitations

The theorem assumes the body is rigid, meaning the relative positions of mass elements do not change under rotation.
It is strictly required that the two axes under consideration are mutually parallel.
One of the axes must pass precisely through the centre of mass; the theorem cannot directly transfer inertia between two arbitrary non-CM parallel axes without an intermediate step via the CM.

Common misconceptions

Common error: Applying $I = I_{cm} + Md^2 between any two parallel axes$, even if neither passes through the centre of mass.

Correct: One of the axes must pass through the centre of mass of the body.

The integral term \int x dm is only guaranteed to vanish when the origin of the coordinate system is located at the centre of mass.

Common error: Assuming the parallel-axis theorem only applies to two-dimensional planar objects.

Correct: The theorem applies equally to all 1D, 2D, and 3D rigid bodies.

The derivation relies on spatial coordinates (x, y, z) and holds true for any continuous or discrete three-dimensional mass distribution.

Common error: Confusing the perpendicular distance d with the total distance along the body's surface.

Correct: d must always be the perpendicular distance measured orthogonal to the direction of the parallel axes.

The shift in the moment of inertia depends strictly on the perpendicular separation between the lines of the axes of rotation.

Quick revision

Parallel-axis theorem formula: $I = I_{cm} + Md^2$, where one axis passes through the centre of mass.
Parallel-Axis Theorem: $I = I_{cm} + M d^2$
Moment of Inertia of Uniform Rod (about CM): $I_{cm} = \frac{1}{12} M L^2$

Must remember

The parallel-axis theorem is expressed mathematically as $I = I_{cm} + M d^2$, where one axis must pass through the centre of mass.
The moment of inertia is always minimum about an axis passing through the centre of mass compared to any other parallel axis.
For a uniform rod of mass $M$ and length $L$, the moment of inertia about an axis through its end is $\frac{1}{3} M L^2$.
For a circular disc, the moment of inertia about a tangent in its plane is $\frac{5}{4} M R^2$, and about a tangent perpendicular to its plane is $\frac{3}{2} M R^2$.
For a solid sphere, the moment of inertia about a tangent to its surface is $\frac{7}{5} M R^2$.
The term $Md^2$ represents the rotational inertia contributed by treating the entire mass concentrated at the centre of mass combined with the displaced axis.

Glossary

Parallel-Axis TheoremA theorem relating the moment of inertia about any axis to that about a parallel axis passing through the centre of mass: $I = I_{cm} + M d^2$.

Centre of Mass ($I_{cm}$)The moment of inertia of a rigid body about an axis passing through its centre of mass, representing the minimum possible rotational inertia for a given set of parallel axes.

Perpendicular Distance ($d$)The shortest perpendicular distance measured between the two parallel axes of rotation.

Rotational Inertia ($I$)A scalar quantity that determines the torque needed for a desired angular acceleration about a rotational axis, measured in $\text{kg m}^2$.

Comparison of Moments of Inertia for Standard Objects about CM and Tangential Axes

Body	About CM ($I_{cm}$)	About Tangent (Parallel)
Uniform Rod	$\frac{1}{12} M L^2$ (midpoint)	$\frac{1}{3} M L^2$ (end)
Circular Ring	$\frac{1}{2} M R^2$ (perpendicular to plane)	$\frac{3}{2} M R^2$ (in plane)
Circular Disc	$\frac{1}{2} M R^2$ (perpendicular to plane)	$\frac{3}{2} M R^2$ (perpendicular to plane, edge)
Solid Sphere	$\frac{2}{5} M R^2$ (diameter)	$\frac{7}{5} M R^2$ (surface tangent)

Practice Quiz

Choose an option to reveal the explanation.

1. What is the mathematical expression for the parallel-axis theorem?

Explanation: The parallel-axis theorem states that $I = I_{cm} + M d^2$, where $I_{cm}$ is the moment of inertia about the centre of mass, $M$ is total mass, and $d$ is perpendicular distance.

2. Which of the following is a mandatory condition for applying the parallel-axis theorem?

Explanation: One of the two parallel axes must necessarily pass through the centre of mass of the rigid body for the theorem to hold true.

3. The moment of inertia of a uniform rod of mass $M$ and length $L$ about an axis passing through its end and perpendicular to its length is:

Explanation: Using $I = I_{cm} + M d^2$ where $I_{cm} = \frac{1}{12} M L^2$ and $d = L/2$, we get $I = \frac{1}{12} M L^2 + M (L/2)^2 = \frac{1}{3} M L^2$.

4. What is the moment of inertia of a circular ring of mass $M$ and radius $R$ about a tangent in the plane of the ring?

Explanation: For a ring, $I_{cm}$ in its plane (about diameter) is $\frac{1}{2} M R^2$. Using $d = R$, $I = \frac{1}{2} M R^2 + M R^2 = \frac{3}{2} M R^2$.

5. For which types of bodies is the parallel-axis theorem applicable?

Explanation: The theorem is valid for both two-dimensional and three-dimensional rigid bodies of any arbitrary shape, as well as continuous and discrete mass distributions.

6. How does the moment of inertia about an axis through the centre of mass compare to any other parallel axis?

Explanation: The moment of inertia is minimum about an axis passing through the centre of mass compared to any other parallel axis.

7. What is the moment of inertia of a solid sphere of mass $M$ and radius $R$ about a tangent touching its surface?

Explanation: For a solid sphere, $I_{cm} = \frac{2}{5} M R^2$. Using $d = R$, $I = \frac{2}{5} M R^2 + M R^2 = \frac{7}{5} M R^2$.

8. What is the moment of inertia of a circular disc of mass $M$ and radius $R$ about a tangent perpendicular to its plane?

Explanation: For a disc, $I_{cm}$ perpendicular to its plane is $\frac{1}{2} M R^2$. Using $d = R$, $I = \frac{1}{2} M R^2 + M R^2 = \frac{3}{2} M R^2$.

9. What does the additive term $M d^2$ physically represent in the parallel-axis theorem?

Explanation: The term $M d^2$ represents the rotational inertia contributed by treating the entire mass concentrated at the centre of mass combined with the displaced axis.

10. Why is the integral term $\int x \, dm$ equal to zero in the derivation of the parallel-axis theorem?

Explanation: The integral term $\int x \, dm$ represents the position of the centre of mass relative to the origin, which is identically zero when the origin is chosen at the centre of mass.

11. What is the moment of inertia of a hollow sphere (thin spherical shell) of mass $M$ and radius $R$ about a tangent to its surface?

Explanation: For a thin spherical shell, $I_{cm} = \frac{2}{3} M R^2$. Adding $M R^2$ gives $I = \frac{5}{3} M R^2$.

12. What is the correct definition of the distance $d$ in the parallel-axis theorem formula?

Explanation: $d$ must always be the perpendicular distance measured orthogonal to the direction of the parallel axes.

13. What is the moment of inertia of a circular disc of mass $M$ and radius $R$ about a tangent in its plane?

Explanation: For a disc about a diameter in its plane, $I_{cm} = \frac{1}{4} M R^2$. Using $d = R$, $I = \frac{1}{4} M R^2 + M R^2 = \frac{5}{4} M R^2$.

14. Which theorem is often used in conjunction with the parallel-axis theorem to determine moments of inertia for complex 3D geometries?

Explanation: The parallel-axis theorem is frequently used alongside the perpendicular-axis theorem to determine moments of inertia for complex three-dimensional geometries.

15. In advanced rotational dynamics (PGT extension), how is the parallel-axis theorem generalized for continuous mass distributions with variable density?

Explanation: Advanced applications involve integrating mass distributions with variable density and extending tensor-based rotational dynamics using weighted spatial coordinates.

PYQ Patterns

These are pattern-based practice questions. Authentic year labels are only used when verified from a source.

1. The moment of inertia of a uniform rod of mass $M$ and length $L$ about an axis passing through its centre and perpendicular to its length is $\frac{1}{12} M L^2$. What is its moment of inertia about a parallel axis passing through one of its ends?

Explanation: Using parallel-axis theorem: $I = I_{cm} + M d^2 = \frac{1}{12} M L^2 + M (L/2)^2 = \frac{1}{12} M L^2 + \frac{1}{4} M L^2 = \frac{1}{3} M L^2$.

2. If the moment of inertia of a circular disc about an axis passing through its centre and perpendicular to its plane is $\frac{1}{2} M R^2$, its moment of inertia about a tangent in its plane is:

Explanation: First find $I_{cm}$ about a diameter in the plane: $\frac{1}{4} M R^2$. Then apply parallel-axis theorem with $d = R$: $I = \frac{1}{4} M R^2 + M R^2 = \frac{5}{4} M R^2$.

3. A solid sphere of mass $M$ and radius $R$ has a moment of inertia $\frac{2}{5} M R^2$ about its diameter. What is its moment of inertia about a tangent touching its surface?

Explanation: Using $I = I_{cm} + M d^2$ with $I_{cm} = \frac{2}{5} M R^2$ and $d = R$, we get $I = \frac{2}{5} M R^2 + M R^2 = \frac{7}{5} M R^2$.

4. Which of the following conditions is absolutely necessary for the validity of the parallel-axis theorem $I = I_{cm} + M d^2$?

Explanation: Both axes must remain strictly parallel to each other for the parallel-axis theorem to hold true.

5. For a thin spherical shell (hollow sphere) of mass $M$ and radius $R$, the moment of inertia about a diameter is $\frac{2}{3} M R^2$. What is its moment of inertia about a tangent parallel to the diameter?

Explanation: Applying the parallel-axis theorem: $I = \frac{2}{3} M R^2 + M R^2 = \frac{5}{3} M R^2$.

6. What is the ratio of the moment of inertia of a uniform rod about an axis through its end to that about an axis through its centre perpendicular to the rod?

Explanation: End axis moment of inertia is $\frac{1}{3} M L^2$; centre axis moment of inertia is $\frac{1}{12} M L^2$. Ratio = $(\frac{1}{3}) / (\frac{1}{12}) = 4 : 1$.

Timed Topic Test

Attempt these questions under a ten-minute limit using a separate timer.

1. If the moment of inertia of a rigid body about an axis through its centre of mass is $I_0$, what is the moment of inertia about a parallel axis at a distance $2d$ from the centre of mass?

Explanation: Using $I = I_{cm} + M (2d)^2 = I_0 + 4 M d^2$.

2. A square lamina of mass $M$ and side $a$ has an axis passing through its centre and perpendicular to its plane. What is its moment of inertia about an axis along one of its edges?

Explanation: For a square lamina, $I_{cm}$ (perpendicular to plane) = $\frac{1}{6} M a^2$. The distance from centre to edge is $d = a/2$. Using parallel-axis theorem: $I = \frac{1}{6} M a^2 + M(a/2)^2 = \frac{1}{6} M a^2 + \frac{1}{4} M a^2 = \frac{5}{12} M a^2$ ... wait, let's recalculate: $\frac{2}{3} M a^2$ is for edge axis in plane, but perpendicular to plane edge axis is $\frac{1}{6} + \frac{1}{4} = \frac{5}{12}$. Wait, for a square lamina about an edge in its plane, $I = \frac{1}{3} M a^2$. Let's check option B: $\frac{2}{3} M a^2$ is incorrect. Let's use standard formula: for axis along edge in its plane, $I = \frac{1}{3} M a^2$.

3. Which misconception about the parallel-axis theorem is invalid?

Explanation: Using the theorem where one axis passes through the CM and both are parallel is the correct condition, hence not a misconception.

4. What is the moment of inertia of a circular ring of mass $M$ and radius $R$ about a tangent perpendicular to the plane of the ring?

Explanation: For a ring, $I_{cm}$ perpendicular to plane is $M R^2$. Using $d = R$, $I = M R^2 + M R^2 = 2 M R^2$.

5. Two parallel axes are separated by a distance $d$. Axis 1 passes through the centre of mass. Axis 2 gives a moment of inertia twice that of Axis 1. What is the distance $d$ in terms of the radius of gyration $k$ about the centre of mass?

Explanation: Given $I_2 = 2 I_{cm} \implies I_{cm} + M d^2 = 2 I_{cm} \implies M d^2 = I_{cm} = M k^2 \implies d = k$.

6. Why does the parallel-axis theorem fail if neither of the two axes passes through the centre of mass?

Explanation: The cross-term involving $\int x \, dm$ only vanishes when the reference axis passes through the centre of mass.

7. What is the moment of inertia of a uniform rod of mass $M$ and length $L$ about an axis parallel to the rod at a distance $d$ from it?

Explanation: The moment of inertia of a thin rod about an axis along its length (passing through its CM) is approximately zero (treating it as a line mass). Thus, about a parallel axis at distance $d$, $I = 0 + M d^2 = M d^2$.

8. In compound pendulum experiments, why is the parallel-axis theorem essential?

Explanation: It is essential for analyzing compound pendulums and bodies rotating about a fixed hinge or pivot that does not coincide with the centre of mass.

9. If a solid cylinder of mass $M$ and radius $R$ has a moment of inertia about its central geometric axis equal to $\frac{1}{2} M R^2$, what is its moment of inertia about a tangent parallel to its central axis?

Explanation: For a solid cylinder, $I_{cm} = \frac{1}{2} M R^2$ about its geometric axis. Using $d = R$, $I = \frac{1}{2} M R^2 + M R^2 = \frac{3}{2} M R^2$.

10. What is the physical significance of the statement that moment of inertia is minimum about the centre of mass?

Explanation: Since $I = I_{cm} + M d^2$ and $M d^2 \ge 0$, $I_{cm}$ is the lowest possible value, meaning minimum rotational inertia and minimum torque required for angular acceleration.