MHT-CET Physics · Rotational Dynamics

Parallel and Perpendicular Axis Theorems

Two theorems carry a known moment of inertia to a new axis: shift it parallel by d and add Md²; for a flat body, the axis perpendicular to the plane has the sum of the two in-plane ones.

All Rotational Dynamics notes MHT-CET Physics strategy

Why this matters

25 PYQs, twelve HARD — the HARDEST page in the chapter. Two in three of the HARD ones are a composite body (seven discs, a square of rods, spheres on a rod, a disc with a hole) added up piece by piece with the parallel-axis theorem. Two further questions ask where a centre of mass lies.

Concept 1 of 4: The Parallel-Axis Theorem

The centre of mass is the axis of least resistance. Move the axis a distance d away, parallel to itself, and every bit of mass is on average d further out — the moment of inertia grows by exactly Md². The theorem only works FROM the centre-of-mass axis.

Definition

$I = I_{\text{cm}} + Md^2$ , $d$ measured from the centre of mass.
Rod about an end: $\dfrac{ML^2}{12} + M\left(\dfrac{L}{2}\right)^2 = \dfrac{ML^2}{3}$ .
Disc, perpendicular axis through the rim: $\dfrac{3}{2}MR^2$ ; tangent in its plane: $\dfrac{5}{4}MR^2$ .
Ring, tangent in its plane: $\dfrac{3}{2}MR^2$ ; solid sphere, tangent: $\dfrac{7}{5}MR^2$ .
Square plate, perpendicular axis at a corner: $\dfrac{Ma^2}{6} + M\dfrac{a^2}{2} = \dfrac{2}{3}Ma^2$ .
The largest I for a family of parallel axes is at the point FURTHEST from the centre of mass.

Parallel-axis theorem

I = I_{\text{cm}} + Md^2

Worked example

A disc of mass M and radius R. Moment of inertia about a perpendicular axis through a point halfway from the centre to the rim?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 10th May Shift 2 · Q17Moderate

Example 1 · Rotational Dynamics · Parallel and Perpendicular Axis Theorems

The moment of inertia of a uniform square plate about an axis perpendicular to its plane and passing through the centre is

\frac{Ma^2}{6}

, where '

M

' is the mass and '

a

' is the side of square plate. Moment of inertia of this plate about an axis perpendicular to its plane and passing through one of its corners is

Shifting from an axis that is not through the centre of mass

From an edge to a point midway you cannot add

M(\frac{R}{2})^2

. Go back to the centre-of-mass axis first, then out to the new one.

Drill 9 more on the parallel-axis theorem

Concept 2 of 4: The Perpendicular-Axis Theorem

For a flat body, the distance of a point from the axis perpendicular to the plane satisfies r² = x² + y², so its moment of inertia is the sum of those about two perpendicular axes lying in the plane and crossing it. It is how a disc's diameter value, MR²/4, comes from its axis value MR²/2.

Definition

Plane lamina only: $I_z = I_x + I_y$ , with $x$ and $y$ in the plane, meeting on the $z$ axis.
Ring: $I_z = MR^2$ ⇒ diameter $\dfrac{MR^2}{2}$ . Disc: $\dfrac{MR^2}{2}$ ⇒ diameter $\dfrac{MR^2}{4}$ .
Square plate: by symmetry every in-plane axis through the centre has the same I, so $I_z = 2I_{\text{in-plane}}$ — diagonal and midline alike.
Rods along x, y and z from the origin, each $\dfrac{ML^2}{3}$ about an end: about the z axis only the x and y rods count.

Perpendicular-axis theorem (lamina)

I_z = I_x + I_y

Worked example

A square plate has

I = \dfrac{Ma^2}{6}

about the perpendicular axis through its centre. About a diagonal?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 15th May Shift 1 · Q16Moderate

Example 2 · Rotational Dynamics · Parallel and Perpendicular Axis Theorems

Three thin rods, each of mass

M

and length

L

are placed along X, Y and Z axes which are mutually perpendicular. One end of each rod is at origin. M.I. of the system about Z axis is

Using it on a three-dimensional body

I_z = I_x + I_y

needs every bit of mass to lie in the x–y plane. For a sphere or a cylinder it gives nonsense; use the standard results instead.

Drill 2 more on the perpendicular-axis theorem

Concept 3 of 4: Composite Bodies: Add the Parts, Subtract the Holes

Moments of inertia about the SAME axis simply add. Break the body into pieces whose own I you know, shift each to the common axis with Md², and add. A hole is a piece with negative mass: compute the whole, then subtract the part cut away.

Definition

Four rods welded into a square, axis through its centre: each $\dfrac{ML^2}{12} + M\left(\dfrac{L}{2}\right)^2 = \dfrac{ML^2}{3}$ ; total $\dfrac{4ML^2}{3}$ .
Seven touching discs in a hexagon: centre $\dfrac{MR^2}{2}$ , each outer one $\dfrac{MR^2}{2} + M(2R)^2$ ; total $\dfrac{55}{2}MR^2$ .
Disc with a hole of diameter $R$ touching the centre: removed mass $\dfrac{M}{4}$ , its I about the centre $\dfrac{3}{32}MR^2$ ; left $\dfrac{13}{32}MR^2$ .
Spheres in a row, axis through one centre: each adds $\dfrac{2}{5}MR^2 + Md^2$ , $d$ = its centre's distance.
Point masses: just $\sum mr^2$ — three at an equilateral triangle's vertices, axis through one vertex parallel to the opposite side: $2m\left(\dfrac{\sqrt{3}}{2}L\right)^2 = \dfrac{3}{2}mL^2$ .

Adding about one axis

I = \sum_{\text{parts}} \left(I_{\text{cm},i} + m_i d_i^2\right) \;-\; I_{\text{removed}}

Worked example

A disc (mass M, radius R) carries a small particle of mass m on its rim. I about the disc's axis?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 11th May Shift 2 · Q14Hard

Example 3 · Rotational Dynamics · Parallel and Perpendicular Axis Theorems

From a disc of mass 'M' and radius 'R', a circular hole of diameter 'R' is cut whose rim passes through the centre. The moment of inertia of the remaining part of the disc about perpendicular axis passing through the centre is

Forgetting a sphere's own moment of inertia

A sphere on the axis still has

\frac{2}{5}MR^2

about it; a sphere off the axis has that PLUS

Md^2

. Treating either as a point mass drops a term and misses every option.

Drill 9 more on composite bodies: add the parts, subtract the holes

Concept 4 of 4: Where the Centre of Mass Lies

The centre of mass is the mass-weighted average position. It sits closer to the heavier body, and for equal masses arranged symmetrically it sits at the centre of the symmetry.

Definition

$x_{\text{cm}} = \dfrac{m_1x_1 + m_2x_2 + \cdots}{m_1 + m_2 + \cdots}$ .
Two masses a distance $d$ apart: $\dfrac{m_2}{m_1 + m_2}d$ from $m_1$ .
Equal masses at the vertices of an equilateral triangle: at the centroid, where the medians meet.

Centre of mass

x_{\text{cm}} = \frac{\sum m_ix_i}{\sum m_i}

Worked example

3 kg at x = 0 and 1 kg at x = 8 m. Centre of mass?

Practice this conceptself-check · 1 quick reps

The same idea in a real exam question:

MHT-CET · 2021 · May Shift 1 · Q6Easy

Example 4 · Rotational Dynamics · Parallel and Perpendicular Axis Theorems

The spheres of masses 2 kg and 4 kg are situated at the opposite ends of a wooden bar of length 9 m. Where does the center of mass of the system lie?

Measuring from the wrong end

The centre of mass is nearer the HEAVIER mass. 2 kg and 4 kg on a 9 m bar: 6 m from the 2 kg, which is 3 m from the 4 kg — the options offer both ends.

Drill 1 more on where the centre of mass lies

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (4)

The Parallel-Axis Theorem
Parallel-axis theorem
$I = I_{\text{cm}} + Md^2$
The Perpendicular-Axis Theorem
Perpendicular-axis theorem (lamina)
$I_z = I_x + I_y$
Composite Bodies: Add the Parts, Subtract the Holes
Adding about one axis
$I = \sum_{\text{parts}} \left(I_{\text{cm},i} + m_i d_i^2\right) \;-\; I_{\text{removed}}$
Where the Centre of Mass Lies
Centre of mass
$x_{\text{cm}} = \frac{\sum m_ix_i}{\sum m_i}$

Watch out for (4)

Shifting from an axis that is not through the centre of mass
→ The Parallel-Axis Theorem
Using it on a three-dimensional body
→ The Perpendicular-Axis Theorem
Forgetting a sphere's own moment of inertia
→ Composite Bodies: Add the Parts, Subtract the Holes
Measuring from the wrong end
→ Where the Centre of Mass Lies

Test yourself on Rotational Dynamics

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