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State Board · Class 12 · Physics

Lesson 1: Rotational Dynamics

7 concepts · 56 practice questions with answers · Free to read

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Circular motion: kinematics and force

Link to earlier classes : The motion, force and Newton's laws of classes 9-10 are extended here to circular and rotational motion.

From daily life

When a fan runs, each blade point moves along a circle and changes its direction every instant. So even at constant speed there is an acceleration.

Simple definition

Angular velocity ω = dθ/dt. With period T and frequency n, ω = 2πn = 2π ÷ T. The tangential speed is v = ω r. In uniform circular motion (UCM) the speed is constant and only the direction changes; the acceleration points to the centre and is called centripetal acceleration: a = ω² r = v² ÷ r.

In circular motion the velocity v is tangential and the acceleration a points to the centrevaOrUniform circular motionv = ω ra = ω² r = v² ÷ ra → centre, v → tangentω = 2π n = 2π ÷ TF = m v² ÷ r

In the figure : In the figure the velocity v is along the tangent and the acceleration a points to the centre O.

Rule 1: Angular velocity, period and speed — ω = 2πn = 2π ÷ T and v = ω r. n is in rev/s (Hz), ω is in rad/s.

  • Easy : n = 5 rev/s: ω = 2π × 5 = 31.4 rad/s.
  • Medium : Radius 0.5 m, n = 2 rev/s: v = 2π × 2 × 0.5 = 6.28 m/s.
  • Exam (CET) : r = 0.2 m, n = 3 rev/s: ω = 6π = 18.85 rad/s; a = ω² r = 71.1 m/s².

Rule 2: Centripetal force and centrifugal force — The resultant of the real forces towards the centre is the centripetal force: F = m v² ÷ r = m ω² r. 'Centripetal' tells the direction; it is not a new type of force. Centrifugal force is a pseudo force seen only in the rotating frame, of magnitude m ω² r and directed away from the centre.

  • Easy : m = 0.5 kg, r = 2 m, v = 4 m/s: F = 0.5 × 16 ÷ 2 = 4 N.
  • Medium : m = 2 kg, ω = 3 rad/s, r = 0.5 m: F = 2 × 9 × 0.5 = 9 N.
  • Exam (CET) : A 1000 kg car, v = 20 m/s, r = 100 m: F = 1000 × 400 ÷ 100 = 4000 N.

Rule 3: Angular acceleration and the equations — When the speed changes, α = dω/dt and the tangential acceleration is a_T = α r. For constant α: ω = ω₀ + α t, θ = ω₀ t + ½ α t², ω² = ω₀² + 2 α θ (the same pattern as linear motion).

  • Easy : From rest, α = 2 rad/s² for 5 s: ω = 10 rad/s.
  • Medium : In the same 5 s, θ = ½ × 2 × 25 = 25 rad.
  • Exam (CET) : 25 rad = 25 ÷ 2π = 3.98 revolutions.

Derivation : derivation of a = v² ÷ r

  • Step 1 : In a short time Δt the body turns through Δθ; the velocity turns by Δθ while its length stays v.
  • Step 2 : Change in velocity Δv = v Δθ (for a small angle).
  • Step 3 : Acceleration a = Δv ÷ Δt = v × (Δθ ÷ Δt) = v ω.
  • Step 4 : Putting ω = v ÷ r gives a = v² ÷ r = ω² r, directed to the centre.

CET MCQ (save time) : A motor runs at 600 rpm; r = 0.1 m. v = ? (1) 6.28 m/s (2) 62.8 m/s (3) 0.628 m/s (4) 3.14 m/s → answer 1). Trick : first rpm ÷ 60 ⇒ n = 10 rev/s; v = 2πnr = 6.28.

Remember

ω = 2πn; v = ω r; a = v²/r = ω² r; F = m v²/r. Centrifugal force is a pseudo force, only in the rotating frame.

Common mistake / CET trap

Centripetal force is not an extra force; it is the resultant of forces already present. Do not add it again in a free-body diagram.

Shortcut

Memory clue: "ω = 2π n → v = ω r → a = ω² r (three formulas in one line)".

Spot the difference

PointCentripetal forceCentrifugal force
KindResultant of real forcesPseudo force
DirectionTowards the centreAway from the centre
Which frameInertial frameRotating frame

Practice questions

Easy
  1. 1If n = 4 rev/s, what is ω?
  2. 2If T = 0.5 s, what is n?
  3. 3m = 1 kg, v = 6 m/s, r = 3 m: find F.
Medium
  1. 4r = 0.4 m, n = 5 rev/s: find v.
  2. 5ω = 5 rad/s, r = 0.2 m: find a.
  3. 6From rest with α = 4 rad/s², what is ω after 3 s?
Challenging
  1. 7Find θ in the same 3 s.
  2. 8m = 0.1 kg, r = 0.5 m, ω = 4 rad/s: find F.
Show answers (try them yourself first)
  1. 1ω = 2π × 4 = 25.1 rad/s
  2. 2n = 1 ÷ 0.5 = 2 Hz
  3. 3F = 1 × 36 ÷ 3 = 12 N
  4. 4v = 2π × 5 × 0.4 = 12.57 m/s
  5. 5a = 25 × 0.2 = 5 m/s²
  6. 6ω = 4 × 3 = 12 rad/s
  7. 7θ = ½ × 4 × 9 = 18 rad
  8. 8F = 0.1 × 16 × 0.5 = 0.8 N

Horizontal road, well of death and banked road

From daily life

A car taking a turn too fast slips outwards. Friction is what keeps it on the circle, and friction has a limit, so the speed on a turn must be limited.

Simple definition

On a flat circular road friction supplies the centripetal force. The maximum speed is v_max = √(μ r g). On a banked road the horizontal component of N supplies it: tan θ = v² ÷ (r g).

On a banked road the horizontal component of N gives the force towards the centreθNmgN sin θBanked road (no friction)N cos θ = m gN sin θ = m v² ÷ rtan θ = v² ÷ (r g)safe speed: v = √(r g tan θ)← to the centre

In the figure : In the figure N cos θ balances the weight and N sin θ gives the force towards the centre.

Rule 1: Flat road — N = m g and f_s = m v² ÷ r, with f_s ≤ μ N. So v_max = √(μ r g) (independent of mass). The friction between tyre and road is static.

  • Easy : μ = 0.5, r = 20 m: v_max = √(0.5 × 20 × 10) = 10 m/s.
  • Medium : μ = 0.5, r = 45 m: v_max = √225 = 15 m/s.
  • Exam (CET) : v = 72 km/h = 20 m/s, r = 80 m: required μ = v² ÷ (r g) = 400 ÷ 800 = 0.5.

Rule 2: Well of death — A vehicle moves in horizontal circles on a vertical wall: N = m v² ÷ r (centripetal) and friction f_s = m g balances the weight. From f_s ≤ μ N, v_min = √(r g ÷ μ).

  • Easy : r = 8 m, μ = 0.2: v_min = √(8 × 10 ÷ 0.2) = 20 m/s.
  • Medium : r = 2.5 m, μ = 0.25: v_min = √100 = 10 m/s.
  • Exam (CET) : Here the speed has a lower limit; on a flat road it had an upper limit.

Rule 3: Banked road — With no friction: N cos θ = m g, N sin θ = m v² ÷ r, so tan θ = v² ÷ (r g); the safe speed is v_s = √(r g tan θ). With friction: v_max = √[r g (tan θ + μ) ÷ (1 − μ tan θ)] and v_min = √[r g (tan θ − μ) ÷ (1 + μ tan θ)].

  • Easy : θ = 45°, r = 40 m: v_s = √(40 × 10 × 1) = 20 m/s.
  • Medium : v = 10 m/s, r = 50 m: tan θ = 100 ÷ 500 = 0.2, θ = 11.3°.
  • Exam (CET) : r = 40 m, tan θ = 0.75, μ = 0.25: v_s = 17.3 m/s, v_max = 22.2 m/s, v_min = 13.0 m/s.

Derivation : tan θ = v² ÷ (r g) for a banked road

  • Step 1 : No friction. Vertical : N cos θ = m g.
  • Step 2 : Horizontal (towards the centre) : N sin θ = m v² ÷ r.
  • Step 3 : Dividing the second by the first cancels m and N.
  • Step 4 : tan θ = v² ÷ (r g) ⇒ v_s = √(r g tan θ).

CET MCQ (save time) : Banked road r = 100 m, θ = 45°; safe speed? (1) 10 m/s (2) 31.6 m/s (3) 100 m/s (4) 50 m/s → answer 2). Trick : tan 45° = 1 ⇒ v = √(r g) = √1000.

Remember

Flat road: v_max = √(μrg). Well: v_min = √(rg/μ). Banked: tan θ = v²/(rg).

Common mistake / CET trap

The safe speed on a banked road does not depend on mass; and in the v_max formula tan θ + μ goes on top, tan θ − μ in v_min — do not swap the signs.

Shortcut

Memory clue: "flat = upper limit, well = lower limit (which limit where)".

Spot the difference

PointFlat roadWell of death
Centripetal forceFriction f_sNormal reaction N
What balances weightNFriction f_s
Speed limitUpper: √(μrg)Lower: √(rg/μ)

Practice questions

Easy
  1. 1μ = 0.4, r = 25 m: find v_max on a flat road.
  2. 2θ = 45°, r = 90 m: find the safe speed.
  3. 3If v = 20 m/s, r = 50 m, find tan θ.
Medium
  1. 4In a well, r = 5 m, μ = 0.5: find v_min.
  2. 5v = 54 km/h, r = 75 m: find the required μ.
  3. 6If the mass of the car on a banked road is doubled, what happens to v_s?
Challenging
  1. 7r = 20 m, tan θ = 0.5: find v_s.
  2. 8On a flat road, if r becomes four times, by what factor does v_max change?
Show answers (try them yourself first)
  1. 1v = √(0.4 × 25 × 10) = 10 m/s
  2. 2v = √(90 × 10 × 1) = 30 m/s
  3. 3tan θ = 400 ÷ (50 × 10) = 0.8
  4. 4v_min = √(5 × 10 ÷ 0.5) = 10 m/s
  5. 5v = 15 m/s; μ = 225 ÷ (75 × 10) = 0.3
  6. 6Nothing changes (m is not in v_s).
  7. 7v_s = √(20 × 10 × 0.5) = 10 m/s
  8. 82 times (v_max ∝ √r).

Conical pendulum

From daily life

Whirl a stone on a string over your head: the string never becomes fully horizontal; it always sags a little. This is the conical pendulum.

Simple definition

A bob on a string of length L moves in a horizontal circle while the string sweeps a cone. θ is the angle of the string with the vertical. Vertical: T₀ cos θ = m g; horizontal (centripetal): T₀ sin θ = m ω² r. Hence tan θ = ω² r ÷ g = v² ÷ (r g) and r = L sin θ.

Forces in a conical pendulum: string tension T₀ and weight mgθmgT₀rLConical pendulumT₀ cos θ = m gT₀ sin θ = m ω² rT = 2π √(L cos θ ÷ g)r = L sin θ

In the figure : In the figure the vertical component of the tension T₀ balances the weight and the horizontal component gives the force towards the centre.

Rule 1: String tension and angle — From T₀ cos θ = m g, T₀ = m g ÷ cos θ. Tension grows with θ; θ = 90° is impossible because cos 90° = 0 would need infinite tension.

  • Easy : m = 0.2 kg, θ = 60°: T₀ = 0.2 × 10 ÷ 0.5 = 4 N.
  • Medium : m = 0.1 kg, cos θ = 0.8: T₀ = 1 ÷ 0.8 = 1.25 N.
  • Exam (CET) : Why can the string not become horizontal? A horizontal tension cannot balance the weight.

Rule 2: Period and frequency — T = 2π √(L cos θ ÷ g) and n = 1 ÷ T. For given L and g, T depends only on θ, not on the mass.

  • Easy : L = 1 m, θ = 60°: T = 2π √(0.5 ÷ 10) = 1.405 s.
  • Medium : L = 2.5 m, cos θ = 0.8: T = 2π √(2 ÷ 10) = 2.81 s.
  • Exam (CET) : In the same case n = 1 ÷ 2.81 = 0.356 Hz.

Rule 3: Speed and radius — r = L sin θ and v = √(r g tan θ). As θ increases, r, v and the tension increase while T decreases.

  • Easy : L = 2.5 m, sin θ = 0.6: r = 1.5 m.
  • Medium : In that case tan θ = 0.75: v = √(1.5 × 10 × 0.75) = 3.35 m/s.
  • Exam (CET) : A larger θ means a faster bob on a bigger circle.

Derivation : period of the conical pendulum

  • Step 1 : T₀ cos θ = m g and T₀ sin θ = m ω² r; dividing gives tan θ = ω² r ÷ g.
  • Step 2 : Put r = L sin θ : sin θ ÷ cos θ = ω² L sin θ ÷ g.
  • Step 3 : Cancel sin θ : ω² = g ÷ (L cos θ).
  • Step 4 : T = 2π ÷ ω ⇒ T = 2π √(L cos θ ÷ g).

CET MCQ (save time) : Conical pendulum L = 2.5 m, θ = 60°; T = ? (1) 3.14 s (2) 2.22 s (3) 4.44 s (4) 1.57 s → answer 2). Trick : cos 60° = 0.5 ⇒ L cos θ = 1.25; T = 2π√(0.125) = 2.22.

Remember

T₀ cos θ = mg; T₀ sin θ = mω²r; T = 2π√(L cos θ/g); the string is never horizontal.

Common mistake / CET trap

Do not use the simple-pendulum T = 2π√(L/g) here; it is L cos θ. And r is L sin θ, not L.

Shortcut

Memory clue: "vertical = cos, horizontal = sin (the two components of T₀)".

Spot the difference

PointSimple pendulumConical pendulum
MotionOscillation in a vertical planeHorizontal circle
Period2π√(L/g)2π√(L cos θ/g)
Centripetal forceNot applicableT₀ sin θ

Practice questions

Easy
  1. 1m = 0.5 kg, θ = 60°: find T₀.
  2. 2L = 1.6 m, cos θ = 0.5: find T.
  3. 3L = 1 m, sin θ = 0.6: find r.
Medium
  1. 4For L = 0.9 m, cos θ = 0.8, find the period.
  2. 5tan θ = 0.75, r = 0.6 m: find v.
  3. 6When θ increases, does T increase or decrease?
Challenging
  1. 7L = 1 m, θ = 30°: find T₀ ÷ (m g). (cos 30° = 0.866)
  2. 8What happens to the period if the mass is doubled?
Show answers (try them yourself first)
  1. 1T₀ = 5 ÷ 0.5 = 10 N
  2. 2T = 2π √(0.8 ÷ 10) = 1.78 s
  3. 3r = 1 × 0.6 = 0.6 m
  4. 4T = 2π √(0.72 ÷ 10) = 1.69 s
  5. 5v = √(0.6 × 10 × 0.75) = 2.12 m/s
  6. 6It decreases (cos θ decreases).
  7. 71 ÷ 0.866 = 1.155
  8. 8Nothing changes (m is not in the formula).

Vertical circular motion

From daily life

In a roller-coaster loop or a stone whirled in a vertical circle, a minimum speed is needed at the top so that it does not fall; the string is most stretched at the bottom.

Simple definition

For motion controlled only by gravity, kinetic and potential energy change into each other and the speed is not constant. At the top A: m g + T_A = m v_A² ÷ r; at the bottom B: T_B − m g = m v_B² ÷ r. Energy conservation gives v_B² = v_A² + 4 r g.

Vertical circle: A (top), B (bottom), C and D (string horizontal)ABCDmgT_BVertical circle (string)v_A(min) = √(r g)v_B(min) = √(5 r g)v_C(min) = √(3 r g)T_B − T_A = 6 m gRod: v ≈ 0 at A is fine

In the figure : In the figure at A both weight and tension point down (to the centre); at B tension points up and weight down.

Live figure

Vertical circular motion

In the app you can move this figure with your finger.

Open in the app

Rule 1: Mass tied to a string — For the string to stay taut, T_A ≥ 0 at the top. Minimum speeds: v_A = √(r g), v_B = √(5 r g), and at the horizontal position v_C = √(3 r g). Also T_B − T_A = 6 m g; at minimum speed T_A = 0 and T_B = 6 m g.

  • Easy : r = 0.9 m: v_A(min) = √9 = 3 m/s.
  • Medium : For the same r, v_B(min) = √(5 × 9) = 6.71 m/s.
  • Exam (CET) : m = 0.2 kg: at minimum speed T_B = 6 × 0.2 × 10 = 12 N.

Rule 2: Mass tied to a rod — A rigid rod needs no tension at the top, so the minimum speed at A is nearly zero (v_A ≈ 0). Then v_B(min) = √(4 r g) = 2√(r g). Still T_B − T_A = 6 m g.

  • Easy : r = 0.5 m: v_B(min) for a rod = √(4 × 0.5 × 10) = 4.47 m/s.
  • Medium : For the same r with a string, v_B(min) = √25 = 5 m/s (more).
  • Exam (CET) : Why is a smaller speed enough for a rod? Because a rod can push as well as pull.

Rule 3: Convex bridge — At the top of a convex bridge N = m g − m v² ÷ r. As the speed grows N falls; contact is lost at v_max = √(r g) (N = 0). At the bottom of a dip N = m g + m v² ÷ r.

  • Easy : r = 40 m: v_max = √(40 × 10) = 20 m/s.
  • Medium : m = 1000 kg, v = 10 m/s, r = 40 m: N = 10000 − 1000 × 100 ÷ 40 = 7500 N.
  • Exam (CET) : In a dip at the same speed N = 10000 + 2500 = 12500 N.

Derivation : minimum speeds in the vertical circle

  • Step 1 : At the top A : m g + T_A = m v_A² ÷ r. For a taut string T_A ≥ 0.
  • Step 2 : At the minimum speed T_A = 0 ⇒ v_A = √(r g).
  • Step 3 : Energy conservation A → B (height drop 2r) : ½ m v_B² = ½ m v_A² + m g (2r).
  • Step 4 : v_B² = v_A² + 4 r g = 5 r g ⇒ v_B = √(5 r g).
  • Step 5 : At B T_B − m g = m v_B² ÷ r; from these T_B − T_A = 6 m g.

CET MCQ (save time) : A stone on a string in a vertical circle of r = 1.6 m; minimum speed at the top? (1) 2 m/s (2) 4 m/s (3) 16 m/s (4) 8 m/s → answer 2). Trick : √(r g) = √16 directly.

Remember

String: v_A = √(rg), v_B = √(5rg), v_C = √(3rg); T_B − T_A = 6mg. Rod: v_A ≈ 0, v_B = 2√(rg).

Common mistake / CET trap

v_B(min) = √(5rg) is only for a string; for a rod it is √(4rg). T_A = 0 holds only at the minimum speed; at higher speeds the tension is not zero.

Shortcut

Memory clue: "one, five, three — the √(rg), √(5rg), √(3rg) of A, B, C (for a string)".

Spot the difference

PointStringRod
Minimum speed at A√(rg)≈ 0
Minimum speed at B√(5rg)2√(rg)
T_B − T_A6mg6mg

Practice questions

Easy
  1. 1r = 1.6 m: find v_A(min) for a string.
  2. 2r = 0.4 m, string: find v_B(min).
  3. 3m = 0.5 kg: find T_B at minimum speed.
Medium
  1. 4Maximum speed to stay in contact on a convex bridge with r = 90 m?
  2. 5m = 800 kg, v = 10 m/s, r = 50 m: find N at the top.
  3. 6r = 0.9 m: find v_C(min) at the horizontal position.
Challenging
  1. 7Use T_B − T_A = 6 m g to find the difference for m = 0.1 kg.
  2. 8r = 0.8 m, rod: find v_B(min).
Show answers (try them yourself first)
  1. 1v = √(1.6 × 10) = 4 m/s
  2. 2v = √(5 × 0.4 × 10) = 4.47 m/s
  3. 3T_B = 6 × 0.5 × 10 = 30 N
  4. 4v = √(90 × 10) = 30 m/s
  5. 5N = 8000 − 800 × 100 ÷ 50 = 6400 N
  6. 6v_C = √(3 × 0.9 × 10) = 5.20 m/s
  7. 76 × 0.1 × 10 = 6 N
  8. 8v = √(4 × 0.8 × 10) = 5.66 m/s

Moment of inertia and the axis theorems

From daily life

In linear motion mass m measures inertia; in rotation it also matters how far the mass is from the axis. So two wheels of equal mass can need very different effort to spin.

Simple definition

The moment of inertia is I = Σ mᵢ rᵢ², unit kg m². The rotational kinetic energy is KE = ½ I ω². The radius of gyration K is defined by I = M K², so K = √(I ÷ M).

Parallel axes : rod about its centre and about its endI_C = M L² ÷ 12I = M L² ÷ 3h = L ÷ 2I = I_C + M h²

In the figure : The rod about its centre and about its end : I = I_C + M h² (h = L ÷ 2).

Rule 1: I of common shapes (own axis) — Ring M R²; disc / solid cylinder ½ M R²; hollow cylinder M R²; solid sphere ⅖ M R²; hollow sphere ⅔ M R²; rod about its centre M L² ÷ 12, about an end M L² ÷ 3. For equal M and R the ring has the largest I (all mass far from the axis).

  • Easy : M = 2 kg, R = 0.5 m: ring I = 0.5 kg m²; disc I = 0.25 kg m².
  • Medium : For the same M, R a solid sphere has I = ⅖ × 2 × 0.25 = 0.2 kg m².
  • Exam (CET) : Disc K = R ÷ √2 = 0.5 ÷ 1.414 = 0.354 m; ring K = R = 0.5 m.

Rule 2: Parallel and perpendicular axes theorems — Parallel axes: I = I_C + M h² (I_C is about the parallel axis through the centre of mass; h is the distance between the axes). Perpendicular axes (flat thin objects only): I_z = I_x + I_y. About a diameter, a disc has ¼ M R² and a ring ½ M R².

  • Easy : Rod M = 3 kg, L = 1.2 m: I_C = 3 × 1.44 ÷ 12 = 0.36 kg m².
  • Medium : About an end: h = 0.6 m, I = 0.36 + 3 × 0.36 = 1.44 kg m² = M L² ÷ 3.
  • Exam (CET) : Disc M = 2 kg, R = 0.5 m about a tangent in its plane: I = ¼ M R² + M R² = 0.125 + 0.5 = 0.625 kg m².

Rule 3: Rotational kinetic energy — KE = ½ I ω² = ½ I (2πn)². I plays the role of m and ω the role of v.

  • Easy : I = 0.8 kg m², ω = 50 rad/s: KE = ½ × 0.8 × 2500 = 1000 J.
  • Medium : Disc M = 10 kg, R = 0.4 m: I = ½ × 10 × 0.16 = 0.8 kg m².
  • Exam (CET) : Doubling ω makes the KE four times (KE ∝ ω²).

Derivation : I = ½ M R² of a disc

  • Step 1 : A disc is a set of thin rings; take one of radius r and width dr.
  • Step 2 : Its mass dm = (M ÷ π R²) × 2π r dr = (2M ÷ R²) r dr.
  • Step 3 : I = ∫ r² dm = (2M ÷ R²) ∫₀ᴿ r³ dr = (2M ÷ R²) × R⁴ ÷ 4.
  • Step 4 : I = ½ M R².

CET MCQ (save time) : Solid sphere M = 5 kg, R = 0.2 m; I about a tangent = ? (1) 0.08 (2) 0.28 (3) 0.20 (4) 0.48 kg m² → answer 2). Trick : about a tangent I = (⅖ + 1) M R² = 7/5 M R².

Remember

Ring M R²; disc ½ M R²; solid sphere ⅖ M R²; hollow sphere ⅔ M R²; parallel axes I = I_C + M h².

Common mistake / CET trap

The perpendicular-axes theorem does not apply to thick (3-D) objects. And in the parallel-axes theorem one axis must pass through the centre of mass.

Shortcut

Memory clue: "sphere: two-fifths (solid), two-thirds (hollow) (⅖ and ⅔)".

Spot the difference

ShapeAxisIUnit / dimensions
RingOwn axisM R²kg m²; [M L²]
Disc / solid cylinderOwn axis½ M R²kg m²; [M L²]
Solid / hollow sphereDiameter⅖ M R² / ⅔ M R²kg m²; [M L²]

Practice questions

Easy
  1. 1M = 4 kg, R = 0.5 m: find I of a ring.
  2. 2The disc of the same M, R: find I.
  3. 3I = 8 kg m², M = 2 kg: find K.
Medium
  1. 4Rod M = 2 kg, L = 3 m: find I about an end.
  2. 5I = 2 kg m², ω = 10 rad/s: find KE.
  3. 6Solid sphere M = 5 kg, R = 0.2 m: find I.
Challenging
  1. 7Ring M = 2 kg, R = 0.5 m: find I about a diameter.
  2. 8Disc M = 2 kg, R = 0.5 m: find I about an axis through the edge, perpendicular to the plane. (I_C = ½ M R², h = R)
Show answers (try them yourself first)
  1. 1I = 4 × 0.25 = 1 kg m²
  2. 2I = ½ × 1 = 0.5 kg m²
  3. 3K = √(8 ÷ 2) = 2 m
  4. 4I = 2 × 9 ÷ 3 = 6 kg m²
  5. 5KE = ½ × 2 × 100 = 100 J
  6. 6I = ⅖ × 5 × 0.04 = 0.08 kg m²
  7. 7I = ½ × 2 × 0.25 = 0.25 kg m²
  8. 8I = 0.25 + 2 × 0.25 = 0.75 kg m²

Angular momentum, torque and its conservation

From daily life

An ice skater spins slowly with arms out and spins fast when the arms are pulled in. Nobody pushes her, yet the spin changes.

Simple definition

Angular momentum is L = I ω (or L = r p sin θ), unit kg m² s⁻¹. Torque is τ = I α = dL/dt, unit N m. If the external torque is zero, L stays constant: I₁ ω₁ = I₂ ω₂, so when I decreases ω (and n) increases.

Skater : arms out (large I) and arms in (small I)arms out : small ωarms in : large ωI ω = constant

In the figure : Pulling the arms in lowers I and raises ω; L = I ω stays the same.

Rule 1: L = I ω and τ = I α — I is like m, ω like v, L like p and τ like F. Power is P = τ ω.

  • Easy : I = 0.4 kg m², ω = 10 rad/s: L = 4 kg m² s⁻¹.
  • Medium : I = 2 kg m², τ = 6 N m: α = 3 rad/s²; after 4 s from rest ω = 12 rad/s.
  • Exam (CET) : Then KE = ½ × 2 × 144 = 144 J; with τ = 5 N m, ω = 20 rad/s the power is 100 W.

Rule 2: Conservation of angular momentum — If τ_ext = 0, L = I ω = 2π I n is constant. Hence n₂ = n₁ I₁ ÷ I₂. Skaters, dancers and divers all use this principle.

  • Easy : I₁ = 3 kg m², n₁ = 1 rev/s; after pulling the arms in, I₂ = 1: n₂ = 3 rev/s.
  • Medium : ω₁ = 12 rad/s, I₁ = 0.5; a second disc is added so the total I = 1.0: ω = 0.5 × 12 ÷ 1.0 = 6 rad/s.
  • Exam (CET) : A planet's angular momentum about the Sun is constant; it moves faster when closer.

Rule 3: L as r × p — For a particle L = r p sin θ, with θ the angle between r and p. In circular motion θ = 90°, so L = m v r = m r² ω.

  • Easy : m = 0.2 kg, v = 5 m/s, r = 2 m: L = 0.2 × 5 × 2 = 2 kg m² s⁻¹.
  • Medium : For the same particle ω = v ÷ r = 2.5 rad/s and I = m r² = 0.8; L = I ω = 2 ✓.
  • Exam (CET) : The unit of torque is N m but it is not energy (J) — they are different quantities.

Derivation : conservation of L and the kinetic energy

  • Step 1 : L = I ω ⇒ dL ÷ dt = I α = τ (I constant).
  • Step 2 : τ_ext = 0 ⇒ dL ÷ dt = 0 ⇒ I₁ ω₁ = I₂ ω₂.
  • Step 3 : KE = ½ I ω² = L² ÷ (2 I) (since ω = L ÷ I).
  • Step 4 : With L fixed, halving I doubles the KE — the skater does work pulling the arms in.

CET MCQ (save time) : A skater's I halves (τ = 0); the KE becomes? (1) same (2) double (3) half (4) four times → answer 2). Trick : KE = L² ÷ 2I ⇒ I ÷ 2 ⇒ KE × 2.

Remember

L = Iω; τ = Iα = dL/dt; τ = 0 ⇒ L constant ⇒ n₂ = n₁ I₁/I₂; P = τω.

Common mistake / CET trap

Even though L is constant, KE is not; the skater's KE increases when the arms go in (work is done).

Shortcut

Memory clue: "less I, more ω — same L (the skater's principle)".

Spot the difference

PointLinearRotationalUnit / dimensions (rotational)
InertiamIkg m²; [M L²]
Momentump = m vL = I ωkg m² s⁻¹; [M L² T⁻¹]
Force / torqueF = m aτ = I αN m; [M L² T⁻²]

Practice questions

Easy
  1. 1I = 0.5 kg m², ω = 8 rad/s: find L.
  2. 2I = 4 kg m², τ = 20 N m: find α.
  3. 3τ = 10 N m, ω = 5 rad/s: find P.
Medium
  1. 4I₁ = 4, n₁ = 0.5 rev/s; I₂ = 1: find n₂.
  2. 5m = 2 kg, v = 3 m/s, r = 0.5 m: find L.
  3. 6I = 3 kg m², α = 2 rad/s²: find ω after 5 s from rest.
Challenging
  1. 7A disc with I = 2 and ω = 6 has a second stationary disc of I = 1 dropped on it. Find the new ω.
  2. 8A skater halves her I by pulling in the arms. What happens to n?
Show answers (try them yourself first)
  1. 1L = 0.5 × 8 = 4 kg m² s⁻¹
  2. 2α = 20 ÷ 4 = 5 rad/s²
  3. 3P = 10 × 5 = 50 W
  4. 4n₂ = 0.5 × 4 ÷ 1 = 2 rev/s
  5. 5L = 2 × 3 × 0.5 = 3 kg m² s⁻¹
  6. 6ω = 2 × 5 = 10 rad/s
  7. 7ω = 2 × 6 ÷ 3 = 4 rad/s
  8. 8n doubles.

Rolling motion

From daily life

Race a ring, a disc and a sphere down the same slope and the sphere wins and the ring comes last, whatever their mass or radius.

Simple definition

In pure rolling linear motion of the centre of mass and rotation about the axis go on together. The total kinetic energy is KE = ½ M v² (1 + K² ÷ R²). Down a slope v = √[2 g h ÷ (1 + K² ÷ R²)] and a = g sin θ ÷ (1 + K² ÷ R²).

Rolling down a slope : sphere first, ring lastspherediscringa = g sin θ ÷ (1 + K² ÷ R²)θ

In the figure : A sphere, disc and ring released on one slope : the one with the smallest K² ÷ R² (sphere) arrives first.

Live figure

Rolling motion

In the app you can move this figure with your finger.

Open in the app

Rule 1: Values of K² ÷ R² — Ring / hollow cylinder 1; disc / solid cylinder ½; solid sphere ⅖; hollow sphere ⅔. The smaller K² ÷ R², the larger the acceleration down a slope.

  • Easy : Ring: a = g sin θ ÷ 2; with sin θ = 0.5, a = 2.5 m/s².
  • Medium : Disc: a = 10 × 0.5 ÷ 1.5 = 3.33 m/s²; solid sphere: a = 5 ÷ 1.4 = 3.57 m/s².
  • Exam (CET) : Order: sphere > disc > ring; a frictionless sliding block with g sin θ = 5 m/s² is fastest of all.

Rule 2: Division of energy — Translational KE : rotational KE = 1 : K² ÷ R²; the translational share of the total is 1 ÷ (1 + K² ÷ R²). For a solid sphere the ratio is 5 : 2, i.e. 71.4 % translational.

  • Easy : Ring: 50 % translational, 50 % rotational.
  • Medium : Hollow sphere: 1 : ⅔ = 3 : 2, i.e. 60 % translational, 40 % rotational.
  • Exam (CET) : Solid sphere M = 2 kg, v = 3 m/s: total KE = ½ × 2 × 9 × 1.4 = 12.6 J (translational 9 J).

Rule 3: Speed at the bottom of a slope — Energy conservation gives v = √[2 g h ÷ (1 + K² ÷ R²)], independent of M and R. Pure rolling needs static friction, which does no work, so no energy is lost.

  • Easy : Solid sphere, h = 0.7 m: v = √(14 ÷ 1.4) = √10 = 3.16 m/s.
  • Medium : For the same h a ring: v = √(2gh ÷ 2) = √(g h) = 2.65 m/s.
  • Exam (CET) : A frictionless sliding block: v = √(2gh) = 3.74 m/s (the largest).

Derivation : speed while rolling down a slope

  • Step 1 : Energy conservation : M g h = ½ M v² + ½ I ω².
  • Step 2 : I = M K² and for pure rolling ω = v ÷ R.
  • Step 3 : ½ I ω² = ½ M v² (K² ÷ R²) ⇒ M g h = ½ M v² (1 + K² ÷ R²).
  • Step 4 : v = √[2 g h ÷ (1 + K² ÷ R²)]; M and R do not appear.

CET MCQ (save time) : Solid sphere and ring on the same slope : a_sphere ÷ a_ring = ? (1) 7 : 10 (2) 10 : 7 (3) 1 : 1 (4) 5 : 7 → answer 2). Trick : a ∝ 1 ÷ (1 + K²/R²) ⇒ (1/1.4) ÷ (1/2) = 10/7.

Remember

KE = ½Mv²(1 + K²/R²); a = g sinθ/(1 + K²/R²); K²/R²: ring 1, disc ½, solid sphere ⅖, hollow sphere ⅔.

Common mistake / CET trap

The result of the race does not depend on M and R, only on the shape (K²/R²). And in pure rolling friction does not waste energy.

Shortcut

Memory clue: "ring slowest, sphere fastest (the smaller K²/R², the faster)".

Spot the difference

ObjectK² ÷ R²Acceleration (sin θ = 0.5)
Ring12.5 m/s²
Disc½3.33 m/s²
Solid sphere⅖3.57 m/s²

Practice questions

Easy
  1. 1Disc, h = 0.6 m: find v at the bottom.
  2. 2Ring, sin θ = 0.6: find a.
  3. 3What is K² ÷ R² for a hollow sphere?
Medium
  1. 4Solid sphere M = 5 kg, v = 2 m/s: find total KE.
  2. 5What percentage of a disc's total KE is translational?
  3. 6Solid sphere or ring — which reaches first?
Challenging
  1. 7Solid sphere, sin θ = 0.7: find a.
  2. 8Ring, h = 1.8 m: find v at the bottom.
Show answers (try them yourself first)
  1. 1v = √(2 × 10 × 0.6 ÷ 1.5) = 2.83 m/s
  2. 2a = 10 × 0.6 ÷ 2 = 3 m/s²
  3. 3K² ÷ R² = ⅔
  4. 4KE = ½ × 5 × 4 × 1.4 = 14 J
  5. 51 ÷ 1.5 = 66.7 %
  6. 6The solid sphere (smaller K²/R²).
  7. 7a = 10 × 0.7 ÷ 1.4 = 5 m/s²
  8. 8v = √(10 × 1.8) = 4.24 m/s

Quick revision of the whole lesson

ConceptKey formula / factUnit / dimensions
Angular velocityω = 2πn = 2π ÷ T; v = ω rrad/s; [T⁻¹]
Centripetal acceleration / forcea = v² ÷ r = ω² r; F = m v² ÷ rm/s²; N
Angular equationsω = ω₀ + α t; θ = ω₀ t + ½ α t²; ω² = ω₀² + 2 α θrad/s²; [T⁻²]
Flat roadv_max = √(μ r g)m/s; [L T⁻¹]
Well of deathv_min = √(r g ÷ μ)m/s; [L T⁻¹]
Banked roadtan θ = v² ÷ (r g); v_s = √(r g tan θ)θ : no unit
Conical pendulumT₀ cos θ = m g; T = 2π √(L cos θ ÷ g)N; s
Vertical circle (string)v_A = √(r g); v_B = √(5 r g); T_B − T_A = 6 m gm/s; N
Convex bridgeN = m g − m v² ÷ r; v_max = √(r g)N
Moment of inertiaring M R²; disc ½ M R²; solid sphere ⅖ M R²; hollow sphere ⅔ M R²kg m²; [M L²]
Axis theoremsI = I_C + M h²; I_z = I_x + I_ykg m²
Angular momentum / torqueL = I ω; τ = I α = dL/dt; I₁ω₁ = I₂ω₂kg m² s⁻¹; N m
Rolling motionKE = ½ M v² (1 + K² ÷ R²); a = g sin θ ÷ (1 + K² ÷ R²)J; m/s²

🎯 Most important points for the exam

  • Centripetal force is not a separate type of force; centrifugal force is a pseudo force.
  • Flat road = upper speed limit; well of death = lower speed limit.
  • tan θ = v² ÷ (r g) for both the banked road and the conical pendulum.
  • Minimum speeds at A, B, C and T_B − T_A = 6 m g; string versus rod.
  • I of common shapes, parallel and perpendicular axes theorems.
  • Conservation of angular momentum (τ = 0 ⇒ L constant).
  • Down a slope, the smaller K² ÷ R², the faster the body.
Study mantra

draw the figure → mark the forces → write the centripetal equation → apply the formula.