Rotational Motion Formula Sheet — NEET Physics

Every key Rotational Motion formula, definition and fact for NEET Physics in one place — with common examiner traps and worked examples. Free to read; blurt from memory, then check your gaps.

Syllabus — topics coveredNCERT · 4 sub-topics

Centre of mass of a two-particle system & rigid body
Moment of force, torque, angular momentum & its conservation
Moment of inertia, radius of gyration, theorems of parallel & perpendicular axes
Equilibrium of rigid bodies, equations of rotational motion

Centre of Mass

Two particles

x_{c m} = \frac{m _{1} x _{1} + m _{2} x _{2}}{m _{1} + m _{2}}

m_{1}, m_{2} =

masses,

x_{1}, x_{2} =

their positions; COM lies on the line joining them, closer to the heavier:

m_{1} r_{1} = m_{2} r_{2}

Two masses m1 and m2 with COM between them, m1 r1 = m2 r2 — $m_{1} r_{1} = m_{2} r_{2}$ about the COM.

System / continuous body

r_{c m} = \frac{\sum m _{i} r _{i}}{M} = \frac{1}{M} \int r d m

M = \sum m_{i} =

total mass,

r_{i} =

position of ith particle,

d m =

mass element

Velocity & external force

v_{c m} = \frac{P}{M}; F_{e x t} = M a_{c m}

P =

total linear momentum,

a_{c m} =

COM acceleration; the COM moves as if all mass & all external force acted there

⚡ Shortcut · Removed-mass / cavity trick

Body with a hole

=

(full body)

+

(negative mass filling the hole). Treat the removed part as mass

- m

at its own centroid:

x_{c m} = \frac{M x _{f u ll} - m x _{h o l e}}{M - m}

. Saves a full integral.

COM of common bodies

Body	Centre of mass
Sphere / ring / disc	geometric centre
Cylinder	mid-point of the axis
Solid cone (height $h$ )	on axis, $\frac{h}{4}$ from base
Solid hemisphere ( $R$ )	$\frac{3 R}{8}$ from centre
Triangular lamina	centroid (medians meet)
Semicircular ring	$\frac{2 R}{π}$ from centre
Semicircular disc	$\frac{4 R}{3 π}$ from centre

★ Remember · Key facts

The COM need not lie

inside

the body (ring, horseshoe). Internal forces never shift the COM; only

F_{e x t}

does.

🚫 Examiner Trap · Centre of mass

(1) COM

\neq =

centroid for non-uniform density — the geometric-centre shortcuts assume

uniform

mass. (2) For the cone/hemisphere the COM is

not

at the geometric centre; learn the

\frac{h}{4}

\frac{3 R}{8}

\frac{4 R}{3 π}

values. (3) If

F_{e x t} = 0

the COM keeps moving in a straight line even when the body fragments (a bursting shell's COM continues on the parabola). (4) Use signed coordinates in the removed-mass method — the cavity contributes

- m

Rotational Kinematics & Linear–Rotational Analogy

Angular quantities

▸ $θ$ (rad), $ω = \frac{d θ}{d t}$ (rad/s), $α = \frac{d ω}{d t}$ (rad/ $s^{2}$ ).
▸ $ω$ is directed along the axis (right-hand rule); all points of a rigid body share the same $ω, α$ .

Equations (constant

α

)

Angular SUVAT

ω = ω_{0} + α t; θ = ω_{0} t + \frac{1}{2} α t^{2}; ω^{2} = ω_{0}^{2} + 2 α θ

ω_{0} =

initial angular velocity,

θ =

angle swept; valid only while

α

is constant

Linear links

v = r ω; a_{t} = r α; a_{c} = ω^{2} r = \frac{v ^{2}}{r}

r =

distance of the point from the axis;

a_{t} =

tangential,

a_{c} =

centripetal acceleration

⚡ Shortcut ·

n^{th}

-second angle

Angle turned in the nth second alone:

θ_{n} = ω_{0} + \frac{1}{2} α (2 n - 1)

— exact twin of the linear

S_{n}

. Revolutions

N = θ /2 π

Linear

\leftrightarrow

rotational analogy

Linear	Rotational
$x$	$θ$
$v = d x / d t$	$ω = d θ / d t$
$a$	$α$
mass $m$	moment of inertia $I$
$F = ma$	$τ = I α$
$p = m v$	$L = I ω$
$K = \frac{1}{2} m v^{2}$	$K = \frac{1}{2} I ω^{2}$
$W = F d$	$W = τ θ$
$P = F v$	$P = τ ω$

★ Remember · Use the analogy

Every linear law has a rotational twin — replace

m \to I

F \to τ

v \to ω

p \to L

🚫 Examiner Trap · Rotational kinematics

(1)

θ

MUST be in

radians

for

v = r ω

a_{t} = r α

— convert rev or degrees first (

1 rev = 2 π

). (2)

a_{c} = ω^{2} r

exists even at constant

ω

(uniform rotation is still accelerated);

a_{t} = 0

only when

α = 0

. (3)

ω = 2 π ν = 2 π / T

— don't confuse frequency

ν

(rev/s) with

ω

(rad/s). (4) The angular SUVAT equations fail if

α

is not constant.

Torque & Angular Momentum

Torque (moment of force)

τ = r \times F; τ = r F sin θ = F \times (perp. distance)

r =

position of the application point from the axis,

θ =

angle between

r

and

F

; SI unit N\,m

Position vector r and force F at angle theta; torque = rF sin theta — $τ = r F sin θ$ about O.

Couple

▸Two equal, opposite, non-collinear forces: net force $= 0$ but net torque $= F \times d \neq = 0$ — pure rotation.
▸ $d =$ perpendicular separation of the two lines of action; couple torque is the $same$ about every point.

Angular momentum

L = r \times p = mvr sin θ; L = I ω

(rigid body)

p = m v =

linear momentum,

I =

moment of inertia about the axis; SI unit kg\,

m^{2}

Rotational form of Newton's 2nd law

τ_{e x t} = \frac{d L}{d t}

net external torque equals the rate of change of total angular momentum (about the same point/axis)

Arms out (large I, small omega) vs arms in (small I, large omega): I1 omega1 = I2 omega2 — Arms in $\Rightarrow I ↓, ω ↑$ (so L stays).

Conservation of angular momentum:

τ_{e x t} = 0 \Rightarrow L = I ω = const (I_{1} ω_{1} = I_{2} ω_{2})

🎯 Exam · Where it shows up

Spinning skater pulling arms in, a diver tucking, a collapsing star, a person on a rotating stool — I changes,

ω

adjusts to keep

I ω

fixed.

🚫 Examiner Trap · Torque & angular momentum

(1) Torque depends on the chosen

axis/point

— a force through the axis (

r ∥ F

r = 0

) gives

τ = 0

. (2) Use the perpendicular distance, i.e.

r F sin θ

, not rF. (3) When I falls and

ω

rises,

K = \frac{1}{2} I ω^{2} = \frac{L ^{2}}{2 I}

increases

(muscle work) — L is conserved, KE is not. (4)

L = I ω

holds about a symmetry axis; in general

L

need not be parallel to

ω

Moment of Inertia & Axis Theorems

Definition

I = \sum m_{i} r_{i}^{2} = \int r^{2} d m

r =

perpendicular distance of the mass from the axis; unit kg\,

m^{2}

, dimension

[M L^{2}]

Radius of gyration

I = M k^{2} \Rightarrow k = I / M

k =

distance at which the whole mass

M

would give the same

I

(depends on axis, not on

M

)

I

depends on

▸Mass, its distribution, the position & orientation of the axis.
▸It is the rotational analogue of mass — but $not$ a fixed constant (changes with axis).

Comparative: the two axis theorems

	Parallel axis	Perpendicular axis
Formula	$I = I_{c m} + M d^{2}$	$I_{z} = I_{x} + I_{y}$
Applies to	any rigid body	planar lamina only
Axes	two parallel axes	three mutually $⊥$ axes
Condition	one axis through COM	$z ⊥$ plane; x,y in plane

Parallel axis theorem

I = I_{c m} + M d^{2}

d =

distance between the parallel axis and the COM axis;

I_{c m} = I

about the COM axis

Parallel axis theorem: axis through CM and a parallel axis at distance d, I = I_cm + Md^2 — Parallel-axis theorem.

Perpendicular axis theorem (lamina only)

I_{z} = I_{x} + I_{y}

z is perpendicular to the plane; x,y are two

⊥

in-plane axes through the same point

⚡ Shortcut · Symmetry on a lamina

For a symmetric lamina (disc, ring) the two in-plane diameters are equal:

I_{x} = I_{y} \Rightarrow I_{d iam e t er} = \frac{1}{2} I_{z}

. So disc-diameter

= \frac{1}{2} (\frac{1}{2} M R^{2}) = \frac{1}{4} M R^{2}

instantly.

🚫 Examiner Trap · Moment of inertia & theorems

(1) Parallel-axis uses

I_{c m}

— the COM axis —

not

any parallel axis; d is measured from the COM. (2) Perpendicular-axis theorem is valid for a

flat lamina only

(fails for a sphere/cylinder). (3) You cannot add or subtract k (radius of gyration) like I; combine I first, then take

k = I / M

. (4) I is minimum about an axis through the COM (parallel-axis adds

M d^{2} \geq 0

Moment of Inertia — Standard Bodies

Moment of inertia of ring MR^2, disc MR^2/2, solid sphere 2MR^2/5, rod ML^2/12 — Common bodies and their rotation axes.

Body & axis	$I$
Ring, $⊥$ through centre	$M R^{2}$
Ring, diameter	$\frac{1}{2} M R^{2}$
Ring, tangent ( $⊥$ plane)	$2 M R^{2}$
Ring, tangent (in plane)	$\frac{3}{2} M R^{2}$
Disc, $⊥$ through centre	$\frac{1}{2} M R^{2}$
Disc, diameter	$\frac{1}{4} M R^{2}$
Solid cylinder, own axis	$\frac{1}{2} M R^{2}$
Hollow cylinder, own axis	$M R^{2}$

Body & axis	$I$
Solid sphere, diameter	$\frac{2}{5} M R^{2}$
Hollow sphere, diameter	$\frac{2}{3} M R^{2}$
Rod, $⊥$ at centre	$\frac{1}{12} M L^{2}$
Rod, $⊥$ at one end	$\frac{1}{3} M L^{2}$
Rect. lamina ( $a \times b$ ), $⊥$ centre	$\frac{1}{12} M (a^{2} + b^{2})$
Solid sphere, tangent	$\frac{7}{5} M R^{2}$

⚡ Shortcut ·

k^{2} / R^{2}

at a glance

Hollow

>

solid for the same shape (mass farther out). Ring

\frac{k ^{2}}{R ^{2}} = 1

, disc/cylinder

\frac{1}{2}

, solid sphere

\frac{2}{5}

, hollow sphere

\frac{2}{3}

— these decide the rolling race.

🎯 Exam · Build the rest

Get any other axis from a table value via

I = I_{c m} + M d^{2}

(parallel) or

I_{z} = I_{x} + I_{y}

(lamina). E.g. rod end

= \frac{1}{12} M L^{2} + M (\frac{L}{2})^{2} = \frac{1}{3} M L^{2}

🚫 Examiner Trap · Standard moments of inertia

(1) I uses

R^{2}

L^{2}

— these are

not

proportional to R or L alone. (2) Don't mix up solid vs hollow: solid sphere

\frac{2}{5} M R^{2}

but hollow

\frac{2}{3} M R^{2}

; solid cylinder

\frac{1}{2} M R^{2}

but hollow

M R^{2}

. (3) Disc

=

solid cylinder (both

\frac{1}{2} M R^{2}

, axis along the symmetry axis) — the length doesn't matter. (4) 'Tangent' values already include

M d^{2}

; don't add it again.

Rotational Dynamics & Rolling Motion

Newton's 2nd law for rotation

τ = I α

τ =

net torque about the axis,

I =

moment of inertia,

α =

angular acceleration

Energy, work & power

K_{ro t} = \frac{1}{2} I ω^{2}; W = τ θ; P = τ ω

K_{ro t} =

rotational KE,

W =

work by a constant torque over angle

θ

Rolling without slipping

Rolling wheel: contact point v=0, centre v_cm, top 2v_cm — Contact point at rest; top moves at $2 v_{c m}$ .

Rolling condition & KE

v_{c m} = R ω; K = \frac{1}{2} m v_{c m}^{2} + \frac{1}{2} I ω^{2} = \frac{1}{2} m v_{c m}^{2} (1 + \frac{k ^{2}}{R ^{2}})

v_{c m} =

speed of the centre,

k =

radius of gyration; total KE

=

translational

+

rotational

Rolling down a rough incline (

θ

)

a = \frac{g sin θ}{1 + k ^{2} / R ^{2}}; v = \frac{2 g h}{1 + k ^{2} / R ^{2}}

θ =

incline angle,

h =

vertical drop; v is the speed at the bottom (independent of M,R)

Comparative: rolling bodies down an incline

Body	$k^{2} / R^{2}$	$a / (g sin θ)$
Solid sphere	$\frac{2}{5}$	$\frac{5}{7}$ (fastest)
Disc / cylinder	$\frac{1}{2}$	$\frac{2}{3}$
Hollow sphere	$\frac{2}{3}$	$\frac{3}{5}$
Ring / hoop	$1$	$\frac{1}{2}$ (slowest)

Race down the incline

▸Smaller $k^{2} / R^{2} \Rightarrow$ larger a: $solid sphere fastest$ , ring slowest (independent of M,R).
▸Order is the same for final speed and for arrival time.

🎯 Exam · Pure-rolling condition

Needs

μ_{s} \geq \frac{tan θ}{1 + R ^{2} / k ^{2}}

(for a disc,

μ_{s} \geq \frac{1}{3} tan θ

). Static friction acts up the plane and does

no work

— mechanical energy is conserved.

🚫 Examiner Trap · Rotational dynamics & rolling

(1) In rolling, friction is

static

and does no work — don't deduct 'friction loss' from energy. (2) Total KE has BOTH terms

\frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2}

; using only

\frac{1}{2} m v^{2}

is the classic slip. (3) The contact point is instantaneously at rest (

v = 0

), the top moves at

2 v_{c m}

. (4)

a = \frac{g sin θ}{1 + k ^{2} / R ^{2}}

independent

of M and R — heavier/bigger bodies don't win. (5) If

μ_{s}

is too small the body rolls

with slipping

and these formulas fail.

Equilibrium of Rigid Bodies & Centre of Gravity

Conditions for equilibrium

\sum F = 0 (translational) and \sum τ = 0 (rotational)

both are needed & independent — a couple has

\sum F = 0

but

\sum τ \neq = 0

;

\sum τ = 0

may be taken about ANY point

Ladder against a smooth wall on a rough floor: weight W, wall normal N_w, floor normal N_f, friction f — Ladder: take torques about the foot to find $N_{w}$ .

Centre of gravity: The point where the total gravitational torque is zero. It coincides with the COM when

g

is uniform over the body.

Comparative: COM vs centre of gravity

	Centre of mass	Centre of gravity
Defined by	mass distribution	weight (torque of $g$ )
Depends on $g$ ?	no	yes
Uniform $g$	coincide	coincide
Huge body / varying $g$	fixed point	shifts toward stronger $g$

Problem strategy

▸Take torques about a point where an unknown force acts — it drops out.
▸Lever / see-saw balance: $F_{1} d_{1} = F_{2} d_{2}$ about the pivot.
▸Solve $\sum F_{x} = 0, \sum F_{y} = 0, \sum τ = 0$ together.

⚡ Shortcut · Pick the smart pivot

Take torques about the point where the most unknowns act (e.g. the ladder's foot, a hinge). Each force through the pivot has zero moment, collapsing 3 unknowns to 1 equation.

🎯 Exam · Toppling vs sliding

A body

topples

when the vertical line through its COM falls outside the base; it

slides

first if friction is too small. Compare the two thresholds.

🚫 Examiner Trap · Equilibrium

(1) BOTH

\sum F = 0

AND

\sum τ = 0

are required —

\sum F = 0

alone allows a spinning couple. (2) For full equilibrium

\sum τ = 0

about

every

point, so pick the most convenient one. (3) A smooth (frictionless) wall exerts only a

normal

force — no friction term there. (4) Stable vs unstable depends on whether the COM

rises

or falls on a small displacement.

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Frequently Asked Questions

What are the most important Rotational Motion formulas for NEET?

This Rotational Motion formula sheet covers all the high-yield Physics formulas, definitions and facts you need for NEET, across Centre of mass of a two-particle system & rigid body, Moment of force, torque, angular momentum & its conservation, Moment of inertia, radius of gyration, theorems of parallel & perpendicular axes, Equilibrium of rigid bodies, equations of rotational motion — each shown with the key result and, where useful, a worked example.

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How should I revise Rotational Motion for NEET?

Blurt the Rotational Motion key points from memory, then check against this sheet to find your gaps — and practise a few previous-year questions on the chapter to make sure you can apply them under time pressure.

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