Rotational Motion Formula Sheet — JEE Main Physics

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

Syllabus — topics coveredNTA · 14 sub-topics

Centre of mass of a two-particle system
Centre of mass of a rigid body
Basic concepts of rotational motion
Moment of a force
Torque
Angular momentum
Conservation of angular momentum
Moment of inertia
Radius of gyration
Values of moments of inertia for simple geometrical objects
Parallel and perpendicular axes theorems
Equilibrium of rigid bodies
Rigid body rotation equations
Comparison of linear and rotational motions

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 JEE Main?

This Rotational Motion formula sheet covers all the high-yield Physics formulas, definitions and theorems you need for JEE Main, across Centre of mass of a two-particle system, Centre of mass of a rigid body, Basic concepts of rotational motion, Moment of a force, Torque — each shown with the key result and, where useful, a worked example.

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How should I revise Rotational Motion formulas?

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