Rigid bodies and centre of mass
A spanner tossed in the air spins in a complicated way, yet one special point follows a smooth parabola. Link to earlier classes : until now every object was a point; now we treat objects of real size. Simple definition : A rigid body keeps the distances between all its particles fixed. The centre of mass (CM) is the point where the whole mass can be taken to be concentrated for translational motion.
Rule 1: Position of the centre of mass — R = Σ mᵢrᵢ ÷ M (M = total mass). For two particles on a line : x_cm = (m₁x₁ + m₂x₂) ÷ (m₁ + m₂). It lies nearer the heavier particle. For a uniform symmetric body it is at the geometric centre.
- Easy : Masses 1 kg at x = 0 and 3 kg at x = 4 m ⇒ x_cm = (0 + 12) ÷ 4 = 3 m.
- Medium : Three particles of 2 kg each at (0, 0), (6, 0), (0, 6) ⇒ CM = (2, 2).
- Exam : A uniform rod of length L and a uniform disc of radius r each have the CM at the centre; for a uniform triangular plate it lies at the intersection of the medians, one third of the height from the base.
Rule 2: Removed part — A body with a hole can be treated as the full body plus a negative mass at the hole.
- Easy : A uniform disc of mass 10 kg has a 2 kg piece cut out centred at x = 0.4 m. Remaining mass 8 kg ⇒ x_cm = (10 × 0 − 2 × 0.4) ÷ 8 = −0.1 m (shifted away from the hole).
- Medium : If a disc of mass 9m is the full disc and the removed part has mass m at a distance d from the centre, the remaining CM is at distance d ÷ 8 on the opposite side (m × d ÷ 8m).
- Exam : For a system of particles the position of the CM depends only on the masses and positions, not on the choice of axes.
R = Σ mr ÷ M; CM is a weighted average of positions.
Taking the plain average of the positions and forgetting the masses.
Memory clue : "CM — the balancing point." It moves closer to the heavy side.
Spot the difference
| Body | Centre of mass |
|---|---|
| Uniform rod | middle point |
| Uniform disc or sphere | centre |
| Triangular plate | intersection of medians |
Motion of the centre of mass and linear momentum
When fireworks burst in the sky, the pieces fly everywhere but the centre of mass keeps following the original parabola. Link to earlier classes : Newton's laws now work for a whole system of particles.
The CM moves as if all the mass were concentrated there and all the external forces acted there; internal forces do not affect it.
Rule 1: Equations — V = P ÷ M (P = total momentum), M a_cm = F_ext. If F_ext = 0, the velocity of the CM is constant and the total momentum is conserved.
- Easy : A 2 kg and a 3 kg particle moving at 5 m s⁻¹ and 2 m s⁻¹ in the same direction have P = 10 + 6 = 16 kg m s⁻¹, V = 16 ÷ 5 = 3.2 m s⁻¹.
- Medium : A shell bursts in the air; the CM of the fragments continues in the same parabola.
- Exam : Two ice skaters on frictionless ice pull on a rope; they meet at the CM, which never moves since the external force is zero.
Rule 2: Kinetic energy — K = K′ + ½MV² (K′ = kinetic energy about the CM; ½MV² = kinetic energy of the CM).
- Easy : A rolling ball has both kinds : translational (½MV²) and rotational about its CM.
- Medium : If the CM is at rest, K = K′.
- Exam : Internal explosions change K′ but not the CM motion.
V = P ÷ M and M a_cm = F_ext; internal forces cancel out.
Thinking an internal explosion can move the CM.
Memory clue : "Only outsiders move the CM."
Spot the difference
| Force | Effect on CM |
|---|---|
| External | accelerates the CM |
| Internal | no effect on the CM |
Vector product and angular velocity
A screw turned clockwise goes in; turned anticlockwise it comes out — the direction of rotation is linked to a direction along the axis. Link to earlier classes : the dot product gave a scalar; now we meet a product that gives a vector.
The vector (cross) product A × B is a vector perpendicular to both A and B, found by the right-hand rule. Angular velocity is a vector along the axis of rotation.
Rule 1: Cross product — |A × B| = AB sin θ. A × B = −(B × A) (not commutative). For unit vectors : î × ĵ = k̂, ĵ × k̂ = î, k̂ × î = ĵ, and A × A = 0. Magnitude equals the area of the parallelogram formed by A and B.
- Easy : A = 3, B = 4, θ = 30° ⇒ |A × B| = 12 × 0.5 = 6.
- Medium : A = 2î and B = 3ĵ ⇒ A × B = 6k̂.
- Exam : Area of the triangle formed by the vectors A = 3î and B = 4ĵ is ½|A × B| = ½ × 12 = 6.
Rule 2: Angular velocity — ω = dθ ÷ dt along the axis; the linear velocity of a particle is v = ω × r, so v = ωr for a circle of radius r. In rotation about a fixed axis every particle has the same ω.
- Easy : A wheel turns 5 rev in 2 s ⇒ ω = 5 × 2π ÷ 2 = 15.7 rad s⁻¹.
- Medium : A point 0.2 m from the axis has v = 15.7 × 0.2 = 3.14 m s⁻¹.
- Exam : The tip of a 10 cm fan blade at 600 rpm : ω = 600 × 2π ÷ 60 = 62.8 rad s⁻¹; v = 6.28 m s⁻¹.
|A × B| = AB sin θ; direction by the right-hand rule; v = ω × r.
Writing A × B = B × A.
Memory clue : "Curl the fingers from A to B; the thumb gives A × B."
Spot the difference
| Product | Result | Commutative? |
|---|---|---|
| A · B | scalar | yes |
| A × B | vector | no (A × B = −B × A) |
Torque and angular momentum
A door is easy to open by pushing at the handle and almost impossible near the hinge. Link to earlier classes : the turning effect of a force of class 8 now becomes a vector, the torque.
Torque is the turning effect of a force about a point. Angular momentum is the rotational analogue of linear momentum.
Rule 1: Torque — τ = r × F, magnitude τ = rF sin θ (= force × perpendicular distance from the axis). Unit : N m.
- Easy : F = 20 N at 0.3 m perpendicular to the arm ⇒ τ = 6 N m.
- Medium : The same force at 30° to the arm : τ = 20 × 0.3 × 0.5 = 3 N m.
- Exam : A force of 10 N applied at 0.5 m to a spanner at 90° gives 5 N m; the same torque with a spanner of length 1 m needs only 5 N.
Rule 2: Angular momentum — L = r × p = r p sin θ; for a particle in a circle L = mvr. For a system : dL ÷ dt = τ_ext. If τ_ext = 0, L is conserved.
- Easy : m = 2 kg, v = 3 m s⁻¹, r = 0.5 m on a circle ⇒ L = 3 kg m² s⁻¹.
- Medium : A planet moving in a central force has no torque, so its angular momentum is conserved; it moves faster when nearer the Sun.
- Exam : For a body of mass 0.5 kg moving in a straight line at 4 m s⁻¹ with perpendicular distance 2 m from a point, L = 0.5 × 4 × 2 = 4 kg m² s⁻¹.
τ = r × F; L = r × p; dL ÷ dt = τ_ext.
Using the full distance r instead of the perpendicular distance, and forgetting that torque depends on the point chosen.
Memory clue : "Torque = force × lever arm." A longer lever needs less force.
Spot the difference
| Linear | Rotational |
|---|---|
| force F | torque τ |
| momentum p | angular momentum L |
| F = dp ÷ dt | τ = dL ÷ dt |
Equilibrium of a rigid body, couple and centre of gravity
A see-saw balances when the heavier child sits closer to the middle. Link to earlier classes : the principle of moments of class 8 becomes a general law.
A rigid body is in equilibrium when the total external force is zero (translational) and the total external torque is zero (rotational). A couple is a pair of equal and opposite forces with different lines of action; it gives a torque but no net force. The centre of gravity is the point about which the total gravitational torque is zero.
Rule 1: Conditions — ΣF = 0 and Στ = 0. Principle of moments for a lever : F₁d₁ = F₂d₂.
- Easy : A 40 N weight at 1.5 m on one side balances a force F at 0.5 m on the other : F = 40 × 1.5 ÷ 0.5 = 120 N.
- Medium : Two children weighing 300 N and 200 N sit on a see-saw, the lighter at 3 m from the pivot. The heavier sits at 200 × 3 ÷ 300 = 2 m.
- Exam : A uniform 4 m plank weighing 100 N rests on supports 1 m from each end : each support carries 50 N.
Rule 2: Couple — Torque of a couple = F × d (d = perpendicular distance between the forces); it is the same about every point.
- Easy : Two 5 N forces 0.2 m apart ⇒ τ = 1 N m.
- Medium : Turning a tap or a steering wheel with both hands applies a couple.
- Exam : For a rigid body, the CM and the centre of gravity coincide if the gravitational field is uniform over the body.
equilibrium needs ΣF = 0 and Στ = 0; a couple gives torque without net force.
Checking only the forces, and ignoring the torque.
Memory clue : "Clockwise moments = anticlockwise moments."
Spot the difference
| Equilibrium kind | Condition |
|---|---|
| Translational | ΣF = 0 |
| Rotational | Στ = 0 |
Moment of inertia
An ice skater spins faster when she pulls her arms in; a flywheel resists changes in its spin. Link to earlier classes : mass measured resistance to a change in linear motion; the moment of inertia measures resistance to a change in rotation.
The moment of inertia (I) of a body about an axis is Σ mr², where r is each particle's perpendicular distance from the axis. Unit : kg m². Rotational kinetic energy is K = ½Iω².
Rule 1: Common values — Ring about its axis : I = MR². Solid disc about its axis : I = ½MR². Thin rod about its centre : I = ML² ÷ 12; about an end : ML² ÷ 3. Solid sphere about a diameter : I = 2MR² ÷ 5. Hollow sphere : 2MR² ÷ 3.
- Easy : A ring of mass 2 kg and radius 0.5 m ⇒ I = 2 × 0.25 = 0.5 kg m².
- Medium : A disc of mass 4 kg and radius 0.5 m ⇒ I = ½ × 4 × 0.25 = 0.5 kg m².
- Exam : A 1.2 kg rod 1 m long : about its centre I = 1.2 ÷ 12 = 0.1 kg m²; about an end I = 1.2 ÷ 3 = 0.4 kg m².
Rule 2: Two theorems — Parallel axes : I = I_cm + Md² (d = distance between the parallel axes). Perpendicular axes (flat bodies) : I_z = I_x + I_y. Radius of gyration k : I = Mk².
- Easy : A ring about a point on its rim : I = MR² + MR² = 2MR².
- Medium : A disc about a tangent in its plane : I = ¼MR² + MR² = 5MR² ÷ 4.
- Exam : For a flat ring about a diameter I_x = I_y = I_z ÷ 2 = MR² ÷ 2.
I = Σ mr²; the farther the mass from the axis, the larger the I.
Using the formula about the centre when the axis is elsewhere, without the parallel-axes term.
Memory clue : "Ring 1, Disc 1/2, Sphere 2/5." The mass farthest from the axis has the largest factor.
Spot the difference
| Body (axis through centre) | I |
|---|---|
| Ring | MR² |
| Solid disc | ½MR² |
| Solid sphere | 2MR² ÷ 5 |
| Thin rod | ML² ÷ 12 |
Kinematics of rotation about a fixed axis
A ceiling fan takes time to speed up after it is switched on, and keeps rotating after it is switched off. Link to earlier classes : the equations of linear motion have exact rotational twins.
For constant angular acceleration α the rotational equations are the same as the linear ones, with x → θ, v → ω, a → α.
Rule 1: Equations — ω = ω₀ + αt; θ = ω₀t + ½αt²; ω² = ω₀² + 2αθ. Link : v = ωr, a_t = αr.
- Easy : A wheel starting from rest with α = 2 rad s⁻² for 5 s : ω = 10 rad s⁻¹; θ = ½ × 2 × 25 = 25 rad.
- Medium : A fan at 300 rpm (31.4 rad s⁻¹) stops uniformly in 10 s ⇒ α = −3.14 rad s⁻²; number of turns = (31.4 × 10 ÷ 2) ÷ 2π = 25.
- Exam : A point 0.4 m from the axis of a wheel with α = 2 rad s⁻² has a tangential acceleration 0.8 m s⁻².
replace x, v, a by θ, ω, α; v = ωr.
Mixing rpm with rad s⁻¹.
Memory clue : "θ ω α — twins of x v a."
Spot the difference
| Linear | Rotational |
|---|---|
| v = u + at | ω = ω₀ + αt |
| s = ut + ½at² | θ = ω₀t + ½αt² |
| v² = u² + 2as | ω² = ω₀² + 2αθ |
Dynamics of rotation about a fixed axis
A heavy flywheel needs a strong push to speed up; a rolling ball is easier to start than a rolling block of the same mass. Link to earlier classes : F = ma now has its rotational twin.
The torque on a body about a fixed axis produces angular acceleration in proportion to its moment of inertia.
Rule 1: Equations — τ = Iα; work W = τθ; power P = τω; kinetic energy K = ½Iω²; angular momentum L = Iω.
- Easy : τ = 4 N m, I = 2 kg m² ⇒ α = 2 rad s⁻².
- Medium : A flywheel (I = 0.5 kg m²) accelerates at 10 rad s⁻² under a torque τ = 0.5 × 10 = 5 N m; at ω = 20 rad s⁻¹ it has K = ½ × 0.5 × 400 = 100 J.
- Exam : A 2 kg uniform disc of radius 0.2 m rotates freely. A 1 N tangential force at the rim gives τ = 0.2 N m; I = ½ × 2 × 0.04 = 0.04 kg m²; α = 5 rad s⁻².
Rule 2: Rolling motion — For rolling without slipping v_cm = ωR and K = ½mv² (1 + k² ÷ R²). On an incline of angle θ : a = g sin θ ÷ (1 + I ÷ mR²).
- Easy : A solid sphere rolling on a 30° incline: I = 2mR² ÷ 5 ⇒ a = 9.8 × 0.5 ÷ 1.4 = 3.5 m s⁻².
- Medium : A solid cylinder (I = ½mR²): a = 4.9 ÷ 1.5 = 3.27 m s⁻²; a ring (I = mR²): a = 4.9 ÷ 2 = 2.45 m s⁻². The sphere reaches the bottom first, then the cylinder, then the ring.
- Exam : A rolling disc has K = ½mv² + ¼mv² = ¾mv²; ⅓ of K is rotational.
τ = Iα, L = Iω, K = ½Iω²; a rolling body has both kinds of kinetic energy.
Using a = g sin θ for a rolling body (it ignores the rotational part).
Memory clue : "Smaller I wins the race." Sphere beats cylinder beats ring.
Spot the difference
| Linear | Rotational |
|---|---|
| F = ma | τ = Iα |
| p = mv | L = Iω |
| K = ½mv² | K = ½Iω² |
Conservation of angular momentum
A skater draws her arms in and spins faster; a diver tucks in to somersault. Link to earlier classes : the conservation laws of the earlier chapters get one more member.
If the total external torque on a system is zero, its angular momentum stays constant : Iω = constant for a fixed axis.
Rule 1: The law — I₁ω₁ = I₂ω₂. When I decreases, ω increases and the rotational kinetic energy ½Iω² increases (the extra energy comes from the work done by the person).
- Easy : A skater of I = 3 kg m² at 2 rad s⁻¹ pulls her arms in to I = 1 kg m² ⇒ ω = 6 rad s⁻¹.
- Medium : Kinetic energy increases from ½ × 3 × 4 = 6 J to ½ × 1 × 36 = 18 J.
- Exam : A turntable (I = 2 kg m², ω = 4 rad s⁻¹) has a 1 kg lump of clay dropped at 0.5 m from the axis. I₂ = 2 + 0.25 = 2.25 ⇒ ω₂ = 2 × 4 ÷ 2.25 = 3.56 rad s⁻¹.
no external torque ⇒ Iω constant; smaller I ⇒ faster spin.
Saying that rotational kinetic energy is also conserved.
Memory clue : "Arms in, spin up."
Spot the difference
| Quantity | Conserved when |
|---|---|
| Linear momentum | external force is zero |
| Angular momentum | external torque is zero |