Rotational Motion: theory | Physics Circle

01

CONCEPT

Moment of Inertia, Radius of Gyration & Standard Bodies

In one line
Moment of inertia is the rotational inertia of a body: it tells how hard it is to start, stop or change its rotation about an axis.

RIGID BODY

7.1.1 Rotation about a fixed axis

Rigid body: the distance between any two particles never changes.
In rotation about a fixed axis every particle moves in a circle centred on the axis.
All particles have the same θ, ω and α.
Their linear speeds differ: v = rω.

MOMENT OF INERTIA

7.1.2 I = Σ mr²

I = mr²

r = perpendicular distance from the axis

I = m₁r₁² + m₂r₂² + m₃r₃² + …

continuous body: I = ∫ r² dm

SI unit kg m²; dimensions [ML²T⁰].
Plays the role in rotation that mass plays in translation.
Depends on the mass, its distribution and the position of the axis.
Does not depend on ω, α, torque or angular momentum.
It is neither a vector nor a scalar (it changes with the axis): a tensor.
Same M and R: a hollow body has a larger I than a solid one (mass lies farther out).

TWO POINT MASSES

7.1.3 About their centre of mass

Axis through the CM, ⊥ to the line joining them

r₁ = m₂ rm₁ + m₂ · r₂ = m₁ rm₁ + m₂

I = μr², μ = m₁m₂m₁ + m₂

μ is the reduced mass: μ < m₁ and μ < m₂.
I about the CM is the least of all parallel axes.
Used for diatomic molecules (H₂, HCl).

RADIUS OF GYRATION

7.1.4 I = MK²

I = MK² ⇒ K = √IM

K is the distance from the axis at which the whole mass can be put to give the same I.
For n equal particles K is the root-mean-square distance:

K = √r₁² + r₂² + … + r_n²n

Depends on the shape, the mass distribution and the axis; not on the total mass.
SI unit metre, dimensions [L].
Disc about its axis: K = R√2.

STANDARD BODIES

7.1.5 Learn these by heart

ring

disc

rod

sphere

cylinder

plate

Body	Axis	I
Ring	⊥ plane, centre	MR²
Ring	diameter	12MR²
Disc	⊥ plane, centre	12MR²
Disc	diameter	14MR²
Annular disc	⊥ plane, centre	M2(R₁² + R₂²)
Annular disc	diameter	M4(R₁² + R₂²)
Solid cylinder	own axis	12MR²
Thin hollow cylinder	own axis	MR²
Thick hollow cylinder	own axis	M2(R₁² + R₂²)
Solid cylinder	⊥ axis through centre	M(L²12 + R²4)
Solid sphere	diameter	25MR²
Spherical shell	diameter	23MR²
Thin rod	⊥, through centre	ML²12
Thin rod	⊥, through one end	ML²3
Rectangular plate	⊥ plane, centre	M(l² + b²)12
Square plate	⊥ plane, centre	Ma²6

Same M and R: I_ring : I_disc = 2 : 1; K_ring : K_disc = √2 : 1.
Same M and R: I_shell > I_{solid sphere}.
Order of I (same M, R, about a diameter or own axis): ring = hollow cylinder > shell > disc = solid cylinder > solid sphere.

Practise PYQs on this topicMoment of Inertia & Radius of Gyration · 204 questions with solutions

02

CONCEPT

Theorems of Parallel & Perpendicular Axes

In one line
Know I about one axis; these two theorems give I about any parallel axis or about the axis ⊥ to a plane body.

PARALLEL AXES

7.2.1 Shifting the axis sideways

Two parallel axes, d apart

I = I_cm + Md²

I_cm: about the parallel axis through the centre of mass.
d: distance between the two axes.
Works for every body (1D, 2D or 3D).
I is least about the axis through the CM.
Used for I about a tangent of a ring, disc or sphere.

rod, axis at x from the centre: I = M(L²12 + x²)

rod, one end: ML²12 + M(L2)² = ML²3

disc, tangent ⊥ plane: 12MR² + MR² = 32MR²

solid sphere, tangent: 25MR² + MR² = 75MR²

square plate, corner (⊥ plane): Ma²6 + Ma²2 = 23Ma²

PERPENDICULAR AXES

7.2.2 Only for plane laminas

x and y in the plane, z ⊥ to it

I_z = I_x + I_y

Only for thin plane bodies: rings, discs, plates.
x and y lie in the plane; z is ⊥ to it through their meeting point.
Used to find I about a diameter of a ring or disc.

disc: I_x = I_y = I_D ⇒ 12MR² = 2I_D ⇒ I_D = 14MR²

ring: I_D = 12MR²

For a symmetric plane body (disc, square plate) I is the same about every in-plane axis through the centre, perpendicular or not.
Square plate: every in-plane axis through the centre (sides or diagonals) has I = Ma²12.

TANGENTS

7.2.3 Using both theorems

Body	tangent in plane	tangent ⊥ plane
Ring	32MR²	2MR²
Disc	54MR²	32MR²
Annular disc	M4(5R₂² + R₁²)	M2(3R₂² + R₁²)

Body	Axis	I
Solid sphere	tangent	75MR²
Spherical shell	tangent	53MR²
Solid cylinder	generator (tangent ∥ axis)	32MR²
Thin hollow cylinder	generator	2MR²
Solid cylinder	diameter of an end face	M(L²3 + R²4)

Find I about the CM axis first, then shift with Md².
Never use the perpendicular axes theorem for a sphere or a cylinder.

Practise PYQs on this topicMoment of Inertia & Radius of Gyration · 204 questions with solutions

03

CONCEPT

Systems of Bodies & Cut-outs

In one line
Moments of inertia about the same axis simply add; a hole is a negative piece.

ADDING PIECES

7.3.1 I of a system

I_system = I₁ + I₂ + I₃ + …

Take every piece about the same axis.
A piece whose centre is off the axis: I_cm + Md² (parallel axes).
Rods lying along the axis add nothing.

RODS AND FRAMES

7.3.2 Common results

System (rods of mass M, length L)	Axis	I
two masses m at the ends of a rod	centre, ⊥ rod	(M + 6m)L²12
rods along x, y, z from the origin	z-axis	2ML²3
H of three rods	one side	4ML²3
square frame of four rods	centre, ⊥ plane	4ML²3
square frame of four rods	one side	5ML²3

RINGS, DISCS AND SPHERES

7.3.3 Pieces off the axis

Ring or disc touching the axis in its plane: tangent value.
Two spheres (M, R) with centres 2a apart, axis through the middle: 2(25MR² + Ma²).
Seven touching rings, axis through the centre ring: MR² + 6(MR² + 4MR²) = 31MR².
Seven touching discs: 12MR² + 6(12MR² + 4MR²) = 552MR².

A on the axis; B and C touch it

CUT-OUTS

7.3.4 Remove a part

Hole of radius R2; its rim passes through O

I_remaining = I_whole − I_removed

Mass of the removed piece ∝ its area (plate) or volume (solid).
Take the removed piece about the same axis (use parallel axes).
All three moments of inertia must be about one axis.

Hole of radius R2 whose rim passes through O: mass M4, centre R2 from O.

I_hole = 12·M4·R²4 + M4·R²4 = 332MR²

I_remaining = 1632MR² − 332MR² = 1332MR²

disc (9M, R) less a concentric disc (M, R3): I = 409MR²

An arc cut from a ring: the rest is still at distance R ⇒ I = (remaining mass) × R².
A coaxial cylinder carved out: find its mass from the volume fraction first.

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Bodies of variable density

Split the body into thin rings (disc) or thin strips (rod).
Disc with mass per unit area σ(r): dm = σ · 2πr dr, then I = ∫r² dm.
Rod with mass per unit length λ(x): dm = λ dx, then I = ∫x² dm about the end.

σ = kr² (disc): I = 23MR²

More mass far from the axis ⇒ I larger than for a uniform body.

Hollow cone and cube from a sphere

Thin hollow cone about its axis: I = 12MR² (R = radius of the open end).
Solid cone about its axis: I = 310MR².
Cube of side a about an axis through its centre ⊥ to a face: I = Ma²6.

Practise PYQs on this topicMoment of Inertia & Radius of Gyration · 204 questions with solutions

04

CONCEPT

Bending, Recasting & Scaling

In one line
When a body is bent, melted or rescaled, write its mass in terms of its size first; then I follows from the standard formula.

LOOPS FROM ONE WIRE

7.4.1 Mass ∝ radius

M = λ(2πR) ∝ R ⇒ I = MR² ∝ R³

λ = mass per unit length (same wire).
I₁I₂ = (R₁R₂)³: ratio 27 ⇒ radii 3 : 1; ratio 8 ⇒ 2 : 1.

ROD BENT INTO A RING

7.4.2 Length becomes the circumference

2πR = L ⇒ R = L2π

ring ⊥ axis: I = ML²4π² · diameter: I = ML²8π²

I_{rod, centre}I_{ring, diameter} = 2π²3

Wire bent into a semicircle: πR = L, so R = Lπ.
Semicircular wire of mass M: I = MR² about the ⊥ axis through the centre and 12MR² about the diameter joining its ends.
Coil of length L, density λ, tangent in its plane: I = 32(λL)(L2π)² = 3λL³8π².

DISCS AND PLATES

7.4.3 Same material and thickness

M = ρπR²t ∝ R² ⇒ I ∝ R⁴

Radii R and 3R ⇒ I in the ratio 1 : 81.
Same mass and thickness, different densities: R² ∝ 1d ⇒ I ∝ 1d.
To get the largest I from two metals, put the denser one outside.

SPHERES

7.4.4 Same mass or same density

Condition	Radius	I
same mass	R changes	I ∝ R²
same density	M ∝ R³	I ∝ R⁵
same mass, densities ρ	R ∝ ρ⁻¹³	I ∝ ρ⁻²³

RECASTING

7.4.5 Melt and remould

Mass and volume stay the same; only the shape changes.
Write the new size from the volume, then use the new formula.

sphere → n equal spheres: radius Rn^1/3, each I′ = In^5/3

n = 27: each small sphere has I′ = I ÷ 243.

disc → sphere: πR²t = 43πr³

Disc of thickness R6 recast into a sphere: r = R2, I_sphere = I_disc5.
Sphere recast into a disc with equal I about its edge: 32Mr² = 25MR² ⇒ r = 2R√15.

RODS

7.4.6 Cut and rescale

Rod cut into two halves: each has I = 18 of the original about its own centre (mass halves, L² quarters).
Same rod, double the length (same λ): I ∝ L³.

Practise PYQs on this topicMoment of Inertia & Radius of Gyration · 204 questions with solutions

05

CONCEPT

Torque, Couple & Equilibrium

In one line
Torque is the turning effect of a force about an axis; it does in rotation what force does in translation.

TORQUE

7.5.1 τ = r × F

Only F sin θ (⊥ to r) turns the body

τ = r × F

τ = rF sin θ = F × d, d = r sin θ

d = perpendicular distance of the line of action from the axis (lever arm).
Torque is an axial vector, ⊥ to both r and F (right-hand rule).
Maximum when F ⊥ r: τ = rF.
Zero when F is along r (θ = 0° or 180°) or passes through the axis.
Unit N m; dimensions [ML²T⁻²] (same as work, but not energy).
Large r ⇒ large torque: door handle far from the hinges, long spanner, long pump handle.

TORQUE IN COMPONENTS

7.5.2 Cartesian form

τ_x = yF_z − zF_y
τ_y = zF_x − xF_z
τ_z = xF_y − yF_x

Torque about a point A: use r − r_A in place of r.
Several forces: τ = τ₁ + τ₂ + τ₃ + … (vector sum).

ROTATION ↔ TRANSLATION

7.5.3 The analogy

Rotation	Translation
τ = Iα	F = ma
τ = dLdt	F = dpdt
W = ∫τ dθ	W = ∫F ds
P = τω	P = Fv
I	m
L = Iω	p = mv

COUPLE

7.5.4 Equal, opposite, parallel forces

Two forces F, d apart

moment of a couple τ = F × d

Net force is zero ⇒ no translation; it produces pure rotation.
Its moment is the same about every point: independent of the axis.
A couple cannot be balanced by a single force, only by another couple.
Torque: one external force (the axis gives the reaction). Couple: both forces external.

twisting a wire: τ = Cθ, W = 12Cθ²

C = torsional constant (couple per unit twist).

EQUILIBRIUM

7.5.5 Balance of a rigid body

ΣF = 0 and Στ = 0

Beam balance or see-saw

F₁l₁ = F₂l₂

Take torques about the support or about an unknown force (it then drops out).
The weight of a uniform rod acts at its middle.
Rod on two knife edges d apart, CM at x from A: N_A = W(d − x)d.

Forces	ΣF	Στ	Motion
equal, opposite, same line	0	0	rest
couple	0	≠ 0	rotation
unequal, same line	≠ 0	0	translation
unequal, different lines	≠ 0	≠ 0	both (rolling)

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Toppling

A body tilted slightly comes back if the vertical line through its centre of gravity still falls inside its base.
If that line falls outside the base, the torque of the weight turns it over.
Low centre of gravity and a wide base ⇒ more stable.

Ladder against a wall

Forces: weight at the middle, normal reaction and friction at the floor, normal reaction at the (smooth) wall.
Use ΣF_x = 0, ΣF_y = 0 and Στ = 0 about the foot of the ladder.

smooth wall, rough floor: μ_min = 12 tan θ

θ = angle of the ladder with the floor.

Practise PYQs on this topicTorque, Couple & Equilibrium of Rigid Bodies · 51 questions with solutions

06

CONCEPT

Rotational Kinematics & τ = Iα

In one line
Angle, angular velocity and angular acceleration describe rotation; τ = Iα tells how a torque changes it.

ANGULAR DISPLACEMENT

7.6.1 θ

θ = arc length sradius r

Unit radian; dimensionless.
2π rad = 360° = one revolution.
Small angular displacements are axial vectors: s = θ × r.
Every particle of a rigid body turns through the same θ.

ANGULAR VELOCITY

7.6.2 ω

ω = dθdt · ω_av = θ₂ − θ₁t₂ − t₁

ω = 2πT = 2πn

v = ω × r (v = rω)

Unit rad/s; dimensions [T⁻¹], same as frequency.
Direction along the axis (right-hand screw rule).
n rpm ⇒ ω = 2πn60 rad/s.
ω is the same for all particles; v ∝ r.

ANGULAR ACCELERATION

7.6.3 α

α = dωdt = d²θdt² · a_t = α × r

Unit rad/s²; dimensions [T⁻²].
α = 0 ⇒ uniform rotation.
Axial vector along the change in ω.

EQUATIONS OF ROTATION

7.6.4 Constant α

Translation	Rotation
v = u + at	ω = ω₀ + αt
s = ut + 12at²	θ = ω₀t + 12αt²
v² = u² + 2as	ω² = ω₀² + 2αθ
s_n = u + a2(2n − 1)	θ_n = ω₀ + α2(2n − 1)
s = u + v2t	θ = ω₀ + ω2t

α not constant: use ω = dθdt, α = dωdt = ωdωdθ and integrate.
ω = α − βt stops at t = αβ after turning θ = α²2β.

τ = Iα

7.6.5 Newton’s second law for rotation

τ = Iα

τ = ΔLΔt = I(ω − ω₀)t

Tangential force F at the rim: τ = FR.
To bring ω to zero in time t: τ = Iωt; to reverse it: τ = 2Iωt.
Same torque: α ∝ 1I, so a disc speeds up faster than a ring of the same M and R.

STRING WOUND ON A BODY

7.6.6 Falling mass and rotating drum

Mass m hangs from a string wound on a body (I, R)

a = mgR²I + mR² · T = mgII + mR²

after falling h: v² = 2mghR²I + mR² · ω² = 2mghI + mR²

String pulled with a force F instead: α = FR ÷ I.
No slipping: a = Rα and v = Rω.
Disc of mass m with a bob of mass m: ω = 2R√gh3.

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Two masses over a pulley with inertia

a = (m₁ − m₂)gR²(m₁ + m₂)R² + I

Tensions on the two sides are different: (T₁ − T₂)R = Iα.
A massless pulley (I = 0) gives the usual Atwood result.

Rod released from a horizontal position

Rod pivoted at one end, released horizontal: α = 3g2L.
Acceleration of the free end = 3g2, more than g.

Practise PYQs on this topicRotational Kinematics & Dynamics · 83 questions with solutions

07

CONCEPT

Angular Momentum

In one line
Angular momentum is the moment of linear momentum about a point or an axis.

PARTICLE

7.7.1 L = r × p

L is ⊥ to the plane of r and p

L = r × p

L = rp sin θ = p × d, d = r sin θ

Axial vector, ⊥ to the plane of rotation.
Unit kg m² s⁻¹ = J s; dimensions [ML²T⁻¹] (same as Planck’s constant h).
Maximum when p ⊥ r: L = mvr.
Zero when p is along r.
Only the transverse part of p counts.

L_x = yp_z − zp_y
L_y = zp_x − xp_z
L_z = xp_y − yp_x

STRAIGHT-LINE MOTION

7.7.2 L need not be zero

A particle moving in a straight line has L = mv d about any point off the line.
d is the same everywhere on the line ⇒ L is constant along it.
L = 0 only about points on the line itself.
Projectile at its highest point: L = m(v cos θ)H about the point of projection.

CIRCULAR MOTION AND RIGID BODIES

7.7.3 L = Iω

L = mvr = mr²ω = Iω

For a body rotating about a fixed axis L = Iω, along the axis.
Uniform circular motion: L is constant in magnitude and direction.
Earth about its axis: L = 25MR² · 2πT.
Bohr model: L = nh2π.
Fixed axis: L about every point on the axis is the same.

TORQUE AND ANGULAR MOMENTUM

7.7.4 Rate of change

τ = dLdt

With L = Iω and I constant: τ = Idωdt = Iα.
Central force (along r) ⇒ τ = 0 ⇒ L constant.
L of a system = vector sum of L of its parts.

ANGULAR IMPULSE

7.7.5 Large torque for a short time

J = ∫τ dt = τ_av Δt = ΔL

Area under the τ–t graph = change in angular momentum.
Average torque of brakes: τ = Iωt.

Practise PYQs on this topicAngular Momentum & its Conservation · 98 questions with solutions

08

CONCEPT

Conservation of Angular Momentum

In one line
If the external torque is zero, the total angular momentum of a system stays constant.

THE LAW

7.8.1 τ = 0 ⇒ L constant

τ_ext = dLdt = 0 ⇒ L = constant

I₁ω₁ = I₂ω₂ ⇒ ω ∝ 1I

Internal forces cannot change L.
I decreases ⇒ ω increases, and the reverse.
Kinetic energy is not conserved: work is done by internal forces.

K = L²2I ⇒ K₂K₁ = I₁I₂

CHANGING I

7.8.2 Examples

Folding the arms lowers I, so ω rises

Dancer, skater or circus artist pulls in the arms ⇒ spins faster.
Man on a turntable folds his arms (I → 75%) ⇒ ω and K rise by 33.3%.
A diver tucks in to somersault, then stretches out before entering water.
A planet moves faster near the sun (equal areas in equal times).
Mass on a string pulled through a hole: mvr constant ⇒ K ∝ 1r².

sphere shrinks (mass fixed): I ∝ R² ⇒ T ∝ R²

Earth shrinks to half its radius ⇒ day = 24 ÷ 4 = 6 h.
Sphere cools to radius R ÷ n ⇒ ω becomes n²ω.

BODIES JOINED

7.8.3 Discs put together, masses added

ω = I₁ω₁ + I₂ω₂I₁ + I₂

K = (I₁ω₁ + I₂ω₂)²2(I₁ + I₂)

loss of KE = I₁I₂2(I₁ + I₂)(ω₁ − ω₂)²

A disc at rest placed gently on a rotating one: ω₂ = 0 in the formulas.
Identical ring placed on a ring: ω → ω2; half the KE is lost.
A child sitting on a rotating platform or masses sliding out on rods: I rises, ω falls, L stays.
Particles moving out along rods: add mr² for each new position.

STRIKING A PIVOTED ROD

7.8.4 L about the pivot

A particle that hits and sticks to a pivoted rod: L about the pivot is conserved (the pivot force has no torque about it).
Sum L = mvd of every particle before the impact, with signs for the sense of turning.

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Collision of a particle with a pivoted rod

The pivot exerts a force, so linear momentum is not conserved.
That force has no torque about the pivot, so L about the pivot is conserved.

m v d = (I_rod + md²)ω

Rod free in space (no pivot): conserve both p and L about the centre of mass.

Practise PYQs on this topicAngular Momentum & its Conservation · 98 questions with solutions

09

CONCEPT

Rotational Kinetic Energy, Work & Power

In one line
A rotating body has kinetic energy even when it stays in place: K = ½Iω².

ROTATIONAL KE

7.9.1 Three forms

K = 12Iω² = 12Lω = L²2I

L = √(2IK)

Rotational	Translational
K = 12Iω²	K = 12mv²
K = 12Lω	K = 12pv
K = L²2I	K = p²2m

Same	Rule
I	K ∝ ω² (ω up 20% ⇒ K up 44%)
L	K ∝ 1I
K	L ∝ √I
ω	K ∝ I

ω = 1 rad/s ⇒ I = 2K numerically.
Frequency f: ω = 2πf, so K = 2π²If².

WORK

7.9.2 Done by a torque

W = ∫τ dθ (constant τ: W = τθ)

W = ΔK = 12I(ω₂² − ω₁²)

Work–energy theorem for rotation.
Torque to produce K in time t: find ω from K, then α = ω ÷ t.

POWER

7.9.3 Rate of doing work

P = dWdt = τω = Iαω

Analogue of P = Fv.
Instantaneous ω = PIα.

ROD PIVOTED AT ONE END

7.9.4 Energy method

Use mgh_cm = 12Iω² with I = mL²3 and h_cm = change in height of the centre.

turned θ from the upright position: ω = √6gL sin θ2

upright → falls flat: ω = √3gL

swings with maximum ω: centre rises h = L²ω²6g

α at angle θ from the vertical = 3g sin θ2L

COMPOUND PENDULUM

7.9.5 Rigid body swinging about a pivot

T = 2π√Lg, L = l² + K²l

l = distance of the CM from the pivot; K = radius of gyration about the CM.

Body (pivot on the rim)	L	T
Ring, ⊥ plane	2R	2π√2Rg
Ring, in its plane	32R	2π√3R2g
Disc, ⊥ plane	32R	2π√3R2g
Disc, in its plane	54R	2π√5R4g
Spherical shell	53R	2π√5R3g
Solid sphere	75R	2π√7R5g

Practise PYQs on this topicRotational Kinetic Energy, Work & Power · 56 questions with solutions

10

CONCEPT

Rolling Motion

In one line
Rolling = translation of the centre + rotation about the centre; without slipping the contact point is at rest.

SLIDING, SPINNING, ROLLING

7.10.1 Three kinds of motion

Only rolling has both v and ω

Motion	Kinetic energy	Example
sliding (ω = 0)	12mv²	block on ice; F = 0
spinning (v = 0)	12Iω²	fan blades
rolling	12mv² + 12Iω²	wheel on a road

ROLLING WITHOUT SLIPPING

7.10.2 Contact point at rest

v_cm = Rω

Speeds of points on the rim

Top point: 2v. Centre: v. Contact point P: 0.
Ends of the horizontal diameter: √2 v at 45°.
All rim points have the same speed relative to the centre (Rω).
Pure rotation about the contact point P at that instant (instantaneous axis).
Friction does no work in pure rolling; no energy is lost.
In half a turn the contact point moves πR forward and 2R up: displacement R√(π² + 4).

I_P = I + MR²

KINETIC ENERGY

7.10.3 Split between translation and rotation

K = 12mv² + 12Iω² = 12mv²(1 + β), β = K²R²

K = 12I_Pω²

K_RK_T = β · K_RK = β1 + β · K_TK = 11 + β

Body	β	K	K_T : K	K_R : K
ring, hollow cylinder	1	mv²	1 : 2	1 : 2
disc, solid cylinder	12	34mv²	2 : 3	1 : 3
solid sphere	25	710mv²	5 : 7	2 : 7
hollow sphere	23	56mv²	3 : 5	2 : 5

K_R = K_T only for a ring (β = 1).
K_R = 50% of K_T ⇒ β = ½: disc or solid cylinder.
Same m and v: total K is largest for a ring, least for a solid sphere.
Work needed to stop a rolling body = its total K.

Practise PYQs on this topicRolling Motion · 96 questions with solutions

11

CONCEPT

Rolling on an Inclined Plane

In one line
Rolling down an incline, the lost potential energy is shared between translation and rotation, so the body is slower than a sliding one.

ENERGY

7.11.1 Height h, from rest

Rolling without slipping down an incline

mgh = 12mv²(1 + β)

v = √2gh1 + β

a = g sin θ1 + β

t = 1sin θ√2h(1 + β)g

β = K²R² depends only on the shape, not on mass or radius.
Smaller β ⇒ larger v and a, shorter time.
v does not depend on θ; a and t do.

WHICH BODY WINS?

7.11.2 Same incline, released together

Body	β	v	a
ring, hollow cylinder	1	√(gh)	12g sin θ
disc, solid cylinder	12	√4gh3	23g sin θ
solid sphere	25	√10gh7	57g sin θ
hollow sphere	23	√6gh5	35g sin θ

Order of arrival: solid sphere, then disc or solid cylinder, then hollow sphere, then ring.
A solid body beats a hollow one of the same shape.

ROLLING, SLIDING, FALLING

7.11.3 Compare from the same height

Motion	v	a	t
rolling	√2gh1 + β	g sin θ1 + β	1sin θ√2h(1 + β)g
sliding (smooth)	√(2gh)	g sin θ	1sin θ√2hg
free fall	√(2gh)	g	√2hg

v_rollv_slide = 1√(1 + β)

Solid sphere: v_roll : v_slide = √5 : √7; a_roll : a_slide = 5 : 7.
Ring: v_roll = v_slide√2. Disc: v_roll = v_slide√23.

ROLLING UP

7.11.4 Energy in reverse

h = v²(1 + β)2g · distance up the ramp s = hsin θ

rolling into a spring: 12mv²(1 + β) = 12kx²

Solid sphere at 2 m/s up a 30° ramp: h = 0.28 m, s = 0.56 m.
A body cannot roll on an incline without friction.

JEE · NEET

Corner

Friction in rolling

No friction ⇒ the body only slides; it cannot roll on an incline.
Static friction acts up the incline and does no work in pure rolling.

f = mg sin θ · β1 + β

μ ≥ β1 + β tan θ, β = K²R²

μ needed: ring > hollow sphere > disc > solid sphere.
Ring: μ ≥ ½ tan θ; solid sphere: μ ≥ (27) tan θ.

Slipping to pure rolling

A body pushed on a rough floor first slides; kinetic friction slows v and raises ω.
Pure rolling starts when v = Rω; after that friction becomes zero on level ground.
Angular momentum about the contact point is conserved during slipping: e.g. a spinning ring placed on the floor rolls with v = v₀2 (from ω₀R = v₀).

Practise PYQs on this topicRolling Motion · 96 questions with solutions

EXAM-DAY CARD

Just see this before the exam

I = Σmr² = MK²

moment of inertia

I = I_cm + Md²

parallel axes

τ = r × F = Iα

torque

L = Iω, τ = dLdt

angular momentum

K = L²2I

rotational KE

a = g sin θ1 + β

rolling on incline

FORMULA SHEET

Whole chapter on one page

Moment of inertia

I = Σmr² = MK²

ring MR², disc ½MR²

sphere 25MR², shell 23MR²

rod ML²12 (centre), ML²3 (end)

Theorems & scaling

I = I_cm + Md²

I_z = I_x + I_y (laminas)

same wire: I ∝ R³

disc, same t: I ∝ R⁴

Torque & kinematics

τ = rF sin θ = Iα

ω = ω₀ + αt, ω² = ω₀² + 2αθ

θ = ω₀t + ½αt²

v = rω, a_t = rα

Angular momentum

L = r × p = mvr = Iω

τ = dLdt, J = ΔL

I₁ω₁ = I₂ω₂

ω = I₁ω₁ + I₂ω₂I₁ + I₂

Energy & power

K = ½Iω² = L²2I

W = τθ = ΔK

P = τω

T = 2π√Lg, L = l² + K²l

Rolling

β = K²R²; v = Rω, top 2v

K = ½mv²(1 + β)

v² = 2gh1 + β

a = g sin θ1 + β

TOP MISTAKES

Avoid these = easy marks

MISTAKE ✗

Perpendicular axes theorem for a sphere

Parallel axes from a non-CM axis

KE conserved when a disc is added

Rolling: K = ½mv² only

CORRECT ✓

Only for plane laminas

I = I_cm + Md², shift from the CM

Only L is conserved; KE is lost

K = ½mv²(1 + β)