Motion of System of Particles and Rigid Body | ISC Class 11 Physics Notes

This note covers rigid bodies, centre of mass, centre of gravity, linear momentum, angular motion, torque, angular momentum, equilibrium, the principle of moments, ladder problems, moment of inertia, radius of gyration, axes theorems, rotational dynamics and conservation of angular momentum.

What is a rigid body, and what kinds of motion can it have?

Definition: An ideal rigid body has an unchanging shape: the distance between every pair of its particles remains constant, even when forces act on it.

A particle is an ideal point mass with no size. An extended body has finite size and can be treated as a system of particles. For extended bodies, the particle model is often inadequate because it cannot describe changes in orientation.

No real body is truly rigid, since forces can deform it. In many situations these deformations are negligible, allowing wheels, beams and other objects to be treated as rigid bodies. The approximation concerns their shape, not whether they are moving.

In numerical values, m, cm, kg, g and s mean metre, centimetre, kilogram, gram and second. N denotes newton, the SI unit of force. Unit symbols and physical variables are distinguished by their context.

How do translation and rotation differ?

The axis of rotation is the line about which the body rotates. Particles on a fixed axis remain stationary. Other particles can describe circles of different radii. A spinning top can have a moving axis, but the rotational equations developed here concern a fixed axis unless otherwise specified.

How is the centre of mass of a particle system calculated?

The centre of mass, abbreviated CM, is the point whose position is the mass-weighted mean of the positions of the particles. A position coordinate gives a particle's signed location relative to a chosen origin. All coordinates in a calculation must use the same origin.

For two particles, let m₁ and m₂ be their masses and x₁ and x₂ their coordinates along the x-axis. Let M be their total mass and x_cm the centre-of-mass coordinate.

M = m₁ + m₂; x_cm = (m₁x₁ + m₂x₂)/M.

Equal masses place the centre of mass halfway between the particles. For unequal masses, it lies nearer the heavier particle. Masses weight the coordinates; simply averaging the coordinates is valid here only when the masses are equal.

What changes for many particles?

For N particles, the index i identifies a particle, mᵢ is its mass, and xᵢ, yᵢ and zᵢ are its three coordinates. The symbol Σ means addition over all N particles. The centre-of-mass coordinates are x_cm, y_cm and z_cm.

M = Σmᵢ; x_cm = Σmᵢxᵢ/M; y_cm = Σmᵢyᵢ/M; z_cm = Σmᵢzᵢ/M.

A vector has magnitude and direction; an arrow over a symbol identifies it. If r⃗ᵢ is a particle's position vector and R⃗_cm is the centre-of-mass position vector, then R⃗_cm = Σmᵢr⃗ᵢ/M. The sum in this expression is a vector sum.

Worked example 1. Three particles of masses 100 g, 150 g and 200 g occupy the vertices of an equilateral triangle of side 0.5 m. Their coordinates in metres are respectively (0, 0), (0.5, 0) and (0.25, 0.25√3). Find the centre of mass.

Formula: x_cm = Σmᵢxᵢ/M; y_cm = Σmᵢyᵢ/M. Substitute: M = 450 g; x_cm = [100(0) + 150(0.5) + 200(0.25)]/450; y_cm = [200(0.25√3)]/450.

Answer: x_cm = 5/18 m and y_cm = 1/(3√3) m. The common mass unit cancels in each ratio. The unequal masses mean that this point is not the geometric centre of the triangle.

Where is the centre of mass of a uniform rigid body?

A homogeneous body has uniformly distributed mass. Its symmetry can locate the centre of mass without adding the contributions of individual atoms. For a uniform rod, ring, disc or sphere, the centre of mass coincides with the geometric centre.

Why is a uniform rod's centre of mass at its midpoint?

Take the midpoint as origin and the rod along the x-axis. Each small mass element at a positive coordinate x has an equal partner at −x. Their contributions to the mass-weighted position cancel. Pairing all such elements puts the centre of mass at the origin.

A horizontal rod lies along the x-axis. Its midpoint is labelled O. Equal small mass elements, labelled dm, are shown at −x and x. Here dm means a very small element of mass.

See Fig. 6.8 in your NCERT textbook

The centre of mass need not lie within the material of the body. The centre of a uniform ring, for example, lies in its empty central region. What determines the position is the distribution of mass, rather than the presence of matter at that point.

Worked example 2. A uniform L-shaped lamina, meaning a thin flat plate, consists of three squares of side 1 m. Its total mass is 3 kg. The square centres have coordinates (0.5, 0.5), (1.5, 0.5) and (0.5, 1.5), in metres. Locate its centre of mass.

Formula: x_cm = Σmᵢxᵢ/M; y_cm = Σmᵢyᵢ/M. Each square has mass 1 kg. Substitute: x_cm = [1(0.5) + 1(1.5) + 1(0.5)]/3; y_cm = [1(0.5) + 1(0.5) + 1(1.5)]/3.

Answer: x_cm = y_cm = 5/6 m. Each square can be represented by its mass concentrated at its own centre when calculating the centre of mass of the complete plate.

How do external forces determine centre-of-mass motion?

Velocity is the time rate of change of position; acceleration is the time rate of change of velocity. Let t denote time, v⃗ᵢ a particle's velocity and a⃗ᵢ its acceleration. Write V⃗_cm and A⃗_cm for the velocity and acceleration of the centre of mass.

The notation d/dt means differentiation with respect to time. For the following treatment, each particle's mass and the total mass remain constant. This allows the masses to be taken outside the time derivatives.

Derivation: Equations for centre-of-mass motion

1. Start with the position relation M R⃗_cm = Σmᵢr⃗ᵢ, using one common coordinate system.
2. Differentiate once with respect to time: M V⃗_cm = Σmᵢv⃗ᵢ.
3. Differentiate again: M A⃗_cm = Σmᵢa⃗ᵢ. Newton's second law replaces each mass-times-acceleration term by the resultant force on that particle.
4. Internal forces, exerted by particles of the system on one another, cancel in equal and opposite pairs. External forces are exerted by bodies outside the system. Let F⃗_ext denote their resultant, or vector sum.

M A⃗_cm = F⃗_ext. The centre of mass moves as if the system's entire mass were concentrated there and all external forces acted there.

When is linear momentum conserved?

Linear momentum is mass multiplied by velocity. For a particle of mass m and velocity v⃗, p⃗ = mv⃗. For the whole system, P⃗ denotes total linear momentum.

P⃗ = Σmᵢv⃗ᵢ = M V⃗_cm; dP⃗/dt = F⃗_ext.

The law of conservation of linear momentum states that total momentum remains constant provided the total external force is zero. The centre of mass then moves with constant velocity, including the possibility of remaining at rest.

Internal motion can still be complicated. An exploding projectile's fragments separate, but their centre of mass follows the original parabolic path when gravity remains the only external force. Internal forces change individual motions without changing this centre-of-mass result.

The SI unit of linear momentum is kg m s⁻¹, where kg, m and s denote kilogram, metre and second. Its dimensions are [M L T⁻¹]. In dimensional expressions, M, L and T represent mass, length and time, rather than particular measured values.

How are angular displacement, velocity and acceleration related?

Angular displacement is the angle through which a body turns about its axis. Let θ be its angular position relative to a fixed reference direction. The SI unit of angular displacement is the radian, written rad. One complete revolution equals 2π rad, where π is the circle constant.

Angular velocity, ω, is the time rate of change of angular position. Angular acceleration, α, is the time rate of change of angular velocity.

ω = dθ/dt; α = dω/dt.

The SI unit of angular velocity is rad s⁻¹. The SI unit of angular acceleration is rad s⁻². Their dimensions are [T⁻¹] and [T⁻²], respectively; a plane angle is dimensionless.

Every particle of a rigid body has the same angular velocity during rotation about a fixed axis. Its linear speed v depends on its perpendicular distance r⊥ from that axis: v = ωr⊥. The linear velocity is tangential to the particle's circular path.

Which equations require constant angular acceleration?

Let θ₀ and ω₀ denote angular position and angular velocity at t = 0. For a fixed axis and constant α:

ω = ω₀ + αt; θ = θ₀ + ω₀t + ½αt²; ω² = ω₀² + 2α(θ − θ₀).

Note: Choose one rotational sense as positive and use it consistently. The constant-acceleration equations cannot be applied unchanged when angular acceleration varies with time.

Worked example 3. A motor wheel's angular speed increases uniformly from 1200 to 3120 revolutions per minute, abbreviated rpm, in 16 s. Find its angular acceleration and the number of revolutions completed. Use one revolution = 2π rad and one minute = 60 s.

Formula: α = (ω − ω₀)/t; Δθ = ω₀t + ½αt², where Δθ = θ − θ₀ is the angle turned. Substitute: ω₀ = 40π rad s⁻¹; ω = 104π rad s⁻¹; α = (104π − 40π)/16.

Answer: Over the 16 s interval, α = 4π rad s⁻²; Δθ = 40π(16) + ½(4π)(16²) = 1152π rad. Dividing by 2π rad per revolution gives 576 revolutions.

What is torque, and how is its direction determined?

Torque, or moment of a force, measures the turning effect of a force about a specified point. Let r⃗ be the position vector from the chosen origin to the point of application, F⃗ the force, and τ⃗ the torque.

τ⃗ = r⃗ × F⃗.

The symbol × between vectors denotes the vector product, or cross product. Its magnitude is the product of the two vector magnitudes and the sine of their included angle. Its direction is perpendicular to the plane containing those vectors.

If r and F are the magnitudes of r⃗ and F⃗, and φ is the angle between them, τ = rF sinφ = Fd. Here d = r sinφ is the moment arm, the perpendicular distance from the origin to the force's line of action.

How is the right-hand rule applied?

1. Place the position vector and force vector with their directions clearly marked.
2. Curl the fingers of the right hand from r⃗ towards F⃗ through the smaller angle.
3. The outstretched thumb gives the direction of τ⃗, perpendicular to their plane.
4. Reversing the order reverses the product: F⃗ × r⃗ = −r⃗ × F⃗.

The position vector r runs from O to P, where force F acts. The figure marks the angle between r and F and a dashed perpendicular from O to the force's line of action. The torque arrow is perpendicular to the shaded plane. Its angle label θ represents the angle called φ here.

See Fig. 6.18 in your NCERT textbook

The SI unit of torque is newton metre, N m. Its dimensions are [M L² T⁻²]. Although torque and energy have the same dimensions, torque is a vector and energy is a scalar, a quantity with magnitude alone.

A force produces zero torque about the origin if its line of action passes through that origin. This explains why pushing at a door's hinge line is ineffective for turning it, while a perpendicular push at its outer edge is effective.

What is angular momentum, and how does torque change it?

Angular momentum is the moment of linear momentum about a specified origin. For a particle with position vector r⃗ and momentum p⃗, denote its angular momentum by L⃗. Its definition and magnitude are:

L⃗ = r⃗ × p⃗; L = rp sinφ.

Here L and p are the magnitudes of angular momentum and linear momentum, and φ is the angle between r⃗ and p⃗. The right-hand rule determines the direction of L⃗. It is perpendicular to the plane containing r⃗ and p⃗.

The SI unit of angular momentum is kg m² s⁻¹. Its dimensions are [M L² T⁻¹]. The origin must be specified because the position vector, and hence angular momentum, is measured relative to it.

Derivation: Relation between torque and angular momentum

1. For a particle of constant mass, differentiate L⃗ = r⃗ × p⃗ about a fixed origin.
2. The product rule gives dL⃗/dt = (dr⃗/dt) × p⃗ + r⃗ × (dp⃗/dt).
3. Since dr⃗/dt = v⃗ and p⃗ = mv⃗, the first term is zero: the cross product of parallel vectors vanishes.
4. Newton's second law gives dp⃗/dt = F⃗. The remaining term is r⃗ × F⃗, which is torque.

dL⃗/dt = τ⃗. Torque is the time rate of change of angular momentum, just as force is the time rate of change of linear momentum.

How does the relation extend to a system?

Add the angular momenta of all particles vectorially to obtain the total angular momentum. Assume internal forces occur in equal and opposite pairs and act along the lines joining the particles. Their torques cancel, leaving dL⃗_total/dt = τ⃗_ext.

Here L⃗_total is the system's total angular momentum and τ⃗_ext its total external torque, both about the same fixed origin. Zero external torque therefore means constant total angular momentum, even when individual particles exchange angular momentum internally.

What conditions make a rigid body remain in equilibrium?

A rigid body is in mechanical equilibrium when its linear momentum and angular momentum do not change with time. It has neither linear nor angular acceleration. Two independent conditions must be satisfied:

ΣF⃗ = 0; Στ⃗ = 0.

The first gives translational equilibrium; the second gives rotational equilibrium. Equilibrium does not by itself require the body to be at rest. A body at rest in equilibrium is in static equilibrium.

For forces in one plane, resolve forces along two perpendicular directions and take moments about an axis perpendicular to that plane. Set both force sums and the signed moment sum to zero. Taking moments about a support eliminates the torque of its reaction force.

What is a couple?

A couple consists of equal and opposite parallel forces with different lines of action. Its resultant force is zero but its resultant torque is not. For force magnitude F and perpendicular separation d between the lines of action, its moment has magnitude Fd.

The moment of a couple is independent of the point about which moments are taken. Turning a bottle lid with the fingers provides an example. A couple also shows why zero resultant force alone does not guarantee complete mechanical equilibrium.

What is the principle of moments?

For rotational equilibrium, the sum of clockwise moments equals the sum of anticlockwise moments about the same point. In an ideal light lever, the pivot is the fulcrum; the load is the force being overcome, and the effort is the applied force.

Let F₁ be the load, F₂ the effort, and d₁ and d₂ their respective perpendicular moment arms. Then F₁d₁ = F₂d₂. The mechanical advantage is load divided by effort: F₁/F₂ = d₂/d₁. A longer effort arm allows a smaller effort to balance a larger load.

Worked example 4. A uniform horizontal bar is 70 cm long and has mass 4.00 kg. Supports are 10 cm from each end. A 6.00 kg load hangs 30 cm from the left end. Find the upward support reactions, using g = 9.8 m s⁻², where g is gravitational acceleration.

Formula: R₁ + R₂ = (4.00 + 6.00)g; 0.25R₁ − 0.25R₂ = 0.05(6.00g). Here R₁ and R₂ are the left and right reactions. The second equation takes moments about the bar's midpoint.

Substitute: R₁ + R₂ = 98.00 N; R₁ − R₂ = 11.76 N. Answer: R₁ = 54.88 N and R₂ = 43.12 N, approximately 55 N and 43 N. The load lies nearer the left support, consistent with its larger reaction.

How do centre of gravity and ladder equilibrium fit together?

The centre of gravity, abbreviated CG, is the point about which the total gravitational torque on a body is zero. The body's weight is the gravitational force on it. In uniform gravity, weight has magnitude Mg and can be treated as acting at the centre of gravity.

Centre of mass depends on mass distribution. Centre of gravity concerns gravitational forces. They coincide when the gravitational field does not vary from one part of the body to another. This condition is appropriate for a small body in a uniform gravitational field.

How can the centre of gravity be located?

Suspend a small irregular cardboard freely from one point and mark the vertical through the suspension point. Repeat from other points. The verticals intersect at the centre of gravity. In uniform gravity the same construction locates its centre of mass.

For a uniform ladder, its weight acts at its midpoint. A normal reaction is a contact force perpendicular to a surface; friction is a tangential contact force opposing relative sliding or its tendency. A smooth wall supplies no frictional force.

Ladder AB joins foot A on the floor to B on the wall. The wall-floor corner is C and midpoint D carries downward weight W. The wall force F₁ points away from the wall. At A, normal reaction N is upward and friction F points towards the wall; F₂ is their resultant.

See Fig. 6.27 in your NCERT textbook

Worked example 5. A uniform ladder of length 3 m and mass 20 kg rests in equilibrium against a frictionless vertical wall. Its foot is 1 m from the wall on a rough horizontal floor. Find the wall reaction and the floor-force components. Take g = 9.8 m s⁻².

Formula: N = W; F = F₁; F₁h = W(0.5 m). Here W is weight, N the floor's normal reaction, F its frictional force, F₁ the wall reaction and h the height reached. Taking moments about the foot eliminates both floor-force moments.

Substitute: h = √(3² − 1²) = 2√2 m; W = 20(9.8) = 196.0 N; F₁ = 196.0/(4√2). Answer: N = 196.0 N; F₁ = F = 34.6 N. The resultant floor reaction is √(N² + F²) = 199.0 N, approximately 80° above the horizontal towards the wall.

What do moment of inertia and radius of gyration measure?

Moment of inertia, I, measures a body's rotational inertia about a specified axis. For particles of masses mᵢ at perpendicular distances rᵢ from that axis, I = Σmᵢrᵢ². Each mass contributes according to the square of its distance from the axis.

Moment of inertia depends on the mass, shape, size and mass distribution of the body, as well as the position and orientation of the rotation axis. Unlike mass, it is not a single fixed property independent of the chosen axis.

The SI unit of moment of inertia is kg m². Its dimensions are [M L²]. For a given rigid body and axis, I does not depend on angular speed. A larger value means greater resistance to changing the rotational motion under a given torque.

Derivation: Kinetic energy of a rotating rigid body

1. Let K be the body's rotational kinetic energy, the energy associated with its rotation. Each particle has speed vᵢ = rᵢω, where ω is common to all particles.
2. The particle's kinetic energy is ½mᵢvᵢ² = ½mᵢrᵢ²ω².
3. Add the particle energies: K = ½Σmᵢrᵢ²ω².
4. Take the common factor ω² outside the sum and identify Σmᵢrᵢ² as I.

K = ½Iω². This parallels the translational kinetic energy ½Mv² of a body of mass M moving with speed v. The SI unit of kinetic energy is the joule, J.

What is the meaning of radius of gyration?

The radius of gyration, k, is the distance from the axis at which a single particle of mass equal to the body's total mass would have the same moment of inertia as the body.

I = Mk²; k = √(I/M).

The SI unit of radius of gyration is metre. Its dimension is [L]. It describes the distribution of mass relative to an axis and need not equal the body's geometrical radius. A flywheel's large moment of inertia helps it resist sudden increases or decreases in rotational speed.

Which standard moments of inertia and axes theorems are useful?

A moment-of-inertia formula is incomplete unless the axis is specified. In the following table, M is total mass, R is radius and L is rod length. Here L denotes length only; elsewhere in the note it denotes the magnitude of angular momentum.

The hollow-cylinder expression MR² treats its wall as thin, with its mass at radius R. A thin hollow sphere, meaning a uniform spherical shell of negligible thickness, has I = 2MR²/3 about a diameter. These shell assumptions matter when selecting a formula.

What does the parallel-axis theorem state?

The parallel-axis theorem states that I = I_cm + Md². Here I_cm is the moment of inertia about an axis through the centre of mass, I is the moment about a parallel axis, and d is the perpendicular distance between the axes.

For a uniform thin rod, move the perpendicular axis from its midpoint to one end. The separation is L/2, so I_end = ML²/12 + M(L/2)² = ML²/3. Here I_end is the moment of inertia about the end axis.

Draw a straight rod and mark its midpoint. Draw parallel axes perpendicular to the rod through the midpoint and one end. Label the rod length L and the separation of the axes L/2.

When does the perpendicular-axis theorem apply?

The perpendicular-axis theorem applies to a plane lamina. Let x and y be mutually perpendicular axes in its plane, meeting at a point. Let z pass through that same point perpendicular to the plane. Then I_z = I_x + I_y, where the subscripts identify the axes.

For a uniform disc, take two perpendicular diameters through its centre. Symmetry gives I_x = I_y, while I_z = MR²/2. Thus each diametral moment is MR²/4. The theorem relates three intersecting axes and should not be applied to an arbitrary three-dimensional body.

How do rotational dynamics and angular momentum conservation work?

For rotation about a fixed axis, use the components of torque and angular momentum along that axis. In the scalar equations below, τ is the signed external torque about the axis and L is the angular momentum component along it.

L = Iω; τ = dL/dt. For constant I, differentiation gives τ = Iα. This is the rotational counterpart of Newton's second law. For the same torque, the angular acceleration is smaller when the moment of inertia is larger.

Let W denote work done, s the distance moved along an applied force, and P the rate of doing work, or power. In this section W denotes work, rather than weight. For fixed-axis rotation, dW = τ dθ; P = τω. For constant torque, W = τΔθ.

How do linear and rotational quantities compare?

Worked example 6. A flywheel of mass 20 kg and radius 20 cm rotates on a fixed axle with frictionless bearings. Treat it as a uniform solid disc. A light cord wound around its rim is pulled tangentially with a steady force of 25 N without slipping. Starting from rest, find its angular acceleration and the work and kinetic energy after 2 m of cord unwinds.

Formula: I = MR²/2; τ = FR; α = τ/I; W = Fs. Substitute: R = 0.20 m; I = 20(0.20)²/2 = 0.4 kg m²; τ = 25(0.20) = 5.0 N m.

Answer: α = 12.5 rad s⁻² and W = 25(2) = 50 J. The angle turned is Δθ = s/R = 10 rad. Hence ω² = 2αΔθ = 250 rad² s⁻², and K = ½(0.4)(250) = 50 J. With no frictional loss, the work becomes rotational kinetic energy.

When does changing shape change angular speed?

The law of conservation of angular momentum states that a system's total angular momentum remains constant provided its total external torque is zero. For the axial scalar relation this gives I₁ω₁ = I₂ω₂, with subscripts 1 and 2 denoting initial and final states.

For a person on a swivel chair, neglect friction in the rotating mechanism and keep the feet off the ground. Stretching the arms increases the moment of inertia and reduces angular speed. Bringing the arms closer to the body produces the opposite effect.

A diver, acrobat or skater uses the same principle. The condition is zero external torque; a change in moment of inertia alone is not sufficient to justify conservation. Also, angular momentum and angular velocity are not necessarily parallel vectors for arbitrary bodies and axes.

Glossary

• Rigid body — An ideal body whose interparticle distances remain constant despite forces acting on it.
• Centre of mass — The point whose position is the mass-weighted mean of the positions of all particles.
• Centre of gravity — The point about which the total gravitational torque on the body is zero.
• Angular velocity — The time rate of change of angular position, directed along the instantaneous rotation axis.
• Angular acceleration — The time rate of change of angular velocity of a rotating body.
• Torque — The vector product of position with force, expressing the force's turning effect about an origin.
• Moment arm — The perpendicular distance from the chosen origin to a force's line of action.
• Angular momentum — The vector product of a particle's position vector with its linear momentum about an origin.
• Couple — Two equal and opposite parallel forces acting along different lines, producing a nonzero resultant torque.
• Mechanical equilibrium — The condition in which both total linear momentum and total angular momentum remain unchanged.
• Moment of inertia — The sum of particle masses multiplied by squared perpendicular distances from a specified rotation axis.
• Radius of gyration — The distance at which the body's entire mass would reproduce its moment of inertia about the axis.

Common errors and misconceptions

• Misconception: The centre of mass must contain matter. Correct: A uniform ring has its centre of mass at its geometric centre, in empty space.
• Misconception: Internal forces accelerate the centre of mass of an isolated system. Correct: Internal forces cancel in the total force; centre-of-mass acceleration depends on external forces.
• Misconception: Every point of a rotating rigid body has the same linear speed. Correct: Angular velocity is common, but linear speed depends on perpendicular distance from the axis.
• Misconception: Any nonzero force gives a nonzero moment about a point. Correct: Its moment vanishes when the force's line of action passes through that point.
• Misconception: Zero resultant force guarantees complete equilibrium. Correct: The total torque must also vanish; a couple has zero resultant force but a nonzero torque.
• Misconception: Moment of inertia is determined by mass alone. Correct: Mass distribution and the position and orientation of the axis also determine it.
• Misconception: The perpendicular-axis theorem applies to any solid object. Correct: It applies to a plane lamina with the three axes meeting at one point.
• Misconception: Constant angular momentum means constant angular speed. Correct: Angular speed can change when moment of inertia changes while their product remains constant.

Exam-style questions with model answers

Q1. Define a rigid body and state the velocity condition for pure translation. [2 marks]

1. A rigid body is an ideal body in which the distance between every pair of particles remains constant.
2. In pure translation, all particles of the body have the same velocity at any given instant.

Q2. Two particles have constant masses m₁ and m₂, coordinates x₁ and x₂, velocities v₁ and v₂, and accelerations a₁ and a₂ along one axis. Write the expressions for their centre-of-mass position, velocity and acceleration. Define M. [3 marks]

1. The total mass is M = m₁ + m₂. The centre-of-mass coordinate is x_cm = (m₁x₁ + m₂x₂)/M, measured from the same origin as both particle coordinates.
2. Differentiating with respect to time, with masses constant, gives the centre-of-mass velocity V_cm = (m₁v₁ + m₂v₂)/M.
3. Differentiating once more gives the centre-of-mass acceleration A_cm = (m₁a₁ + m₂a₂)/M, the mass-weighted mean of the particle accelerations.

Q3. Define torque as a vector, give its magnitude and SI unit, and state when a nonzero force has zero torque about an origin. [3 marks]

1. Torque is τ⃗ = r⃗ × F⃗, where r⃗ runs from the origin to the point where force F⃗ acts. Its direction follows the right-hand rule.
2. Its magnitude is τ = rF sinφ, where φ is the angle between r⃗ and F⃗. The SI unit is newton metre, N m.
3. A nonzero force has zero torque about the origin when its line of action passes through that origin, making the perpendicular moment arm zero.

Q4. Derive the kinetic energy of a rigid body rotating about a fixed axis. Let particle i have mass mᵢ and perpendicular distance rᵢ from the axis, and let the common angular velocity be ω. [5 marks]

1. Each particle moves in a circle centred on the axis. Its linear speed is vᵢ = rᵢω, so particles at different distances can have different speeds.
2. The kinetic energy of particle i is ½mᵢvᵢ². Substituting the speed gives ½mᵢrᵢ²ω² for that particle.
3. The body's total rotational kinetic energy K is the sum of its particle energies: K = ½Σmᵢrᵢ²ω², with Σ denoting summation over all particles.
4. Since the body is rigid, the same angular velocity applies to every particle. Therefore K = ½ω²Σmᵢrᵢ².
5. Define moment of inertia about the stated axis as I = Σmᵢrᵢ². The required expression is K = ½Iω².

Q5. A uniform ladder of length 3 m and mass 20 kg is at rest against a smooth vertical wall. Its foot is 1 m from the wall on a rough horizontal floor. Take g = 9.8 m s⁻². Calculate the wall reaction, floor normal reaction and frictional force. [5 marks]

1. The ladder's weight is W = mg = 20 × 9.8 = 196.0 N. Since it is uniform, this downward force acts at its midpoint.
2. The height reached is h = √(3² − 1²) = 2√2 m. The horizontal distance of the midpoint from the foot is 0.5 m.
3. Let N be the upward floor normal reaction. Vertical equilibrium gives N − W = 0, so N = 196.0 N.
4. Let F₁ be the horizontal wall reaction. Taking moments about the foot gives F₁(2√2) = 196.0(0.5), hence F₁ = 34.6 N.
5. Horizontal equilibrium requires floor friction F = F₁ = 34.6 N. It acts towards the wall, opposing the foot's tendency to slide away.

Q6. State the parallel-axis and perpendicular-axis theorems, including the condition on the body for the latter. [2 marks]

1. Parallel-axis theorem: I = I_cm + Md², where M is mass and d separates parallel axes, one through the centre of mass.
2. Perpendicular-axis theorem: I_z = I_x + I_y for a plane lamina, with mutually perpendicular x and y axes in its plane and z perpendicular through their intersection.

Q7. A motor wheel speeds up uniformly from 1200 rpm to 3120 rpm in 16 s. Using one revolution = 2π rad and one minute = 60 s, find its angular acceleration and the revolutions completed. [4 marks]

1. Convert the initial angular speed: ω₀ = 1200 × 2π/60 = 40π rad s⁻¹.
2. Convert the final angular speed: ω = 3120 × 2π/60 = 104π rad s⁻¹.
3. Uniform angular acceleration is α = (ω − ω₀)/t = (104π − 40π)/16 = 4π rad s⁻².
4. The angle turned is Δθ = ω₀t + ½αt² = 1152π rad. The number of revolutions is Δθ/(2π) = 576.

Q8. A person spins on a swivel chair with feet off the ground. Neglect friction in the rotating mechanism and all other external torques about the vertical axis. Explain why drawing the arms inward increases angular speed. [3 marks]

1. The total external torque about the vertical axis is zero under the stated conditions. Therefore the angular momentum about that axis remains constant.
2. Drawing the arms inward brings some mass closer to the rotation axis. This reduces the moment of inertia I of the person and chair.
3. Angular momentum is Iω, where ω is angular velocity. Since this product remains constant, the decrease in I produces an increase in angular speed.

Key takeaways

• The centre of mass is a mass-weighted position; symmetry places it at the geometric centre of a uniform rod.
• The centre of mass responds to total external force, while internal forces can change the motions of individual particles.
• A rotating rigid body's particles share angular velocity, but their linear speeds depend on their distances from the rotation axis.
• Torque depends on both force and perpendicular moment arm; angular momentum is defined relative to a chosen origin.
• Mechanical equilibrium requires both zero resultant external force and zero resultant external torque on the rigid body.
• Moment of inertia describes mass distribution about a specified axis and determines rotational kinetic energy together with angular speed.
• The parallel-axis theorem uses a centre-of-mass axis; the perpendicular-axis theorem applies to a plane lamina.
• Angular momentum remains constant when total external torque vanishes; angular speed can still change if moment of inertia changes.

Test yourself

Where is the centre of mass of two equal point masses?

It lies at the midpoint of the straight line joining the two masses.

Can a system's particles accelerate while its centre of mass has constant velocity?

Yes. Internal forces can accelerate individual particles while zero total external force keeps the centre-of-mass velocity constant.

When do centre of mass and centre of gravity coincide?

They coincide when the gravitational field is uniform throughout the body.

What happens to a force's torque when its line of action passes through the origin?

The torque about that origin is zero because the perpendicular moment arm vanishes.

Why is zero resultant force insufficient for rigid-body equilibrium?

A nonzero resultant torque can still change the body's angular momentum, as happens with a couple.

What must be specified alongside a moment-of-inertia formula?

The body's mass distribution and the position and orientation of the rotation axis must be specified.

What condition permits the standard constant-acceleration rotational equations?

They require rotation about a fixed axis with angular acceleration remaining constant during the interval.

Does angular momentum conservation require angular speed to remain constant?

No. The product of moment of inertia and angular velocity can remain constant while both factors change.