Class 11 · Physics · Part 1

Chapter 6: Systems of Particles and Rotational Motion

Revision notes, practice numericals, a question bank, assertion and reason questions and answers to the NCERT questions. Tap a group to open it; opening one closes the others.

6.1–6.2Rigid bodies and the centre of massTranslation, rotation, X = Σ m x / M

Rigid body motion

An ideal rigid body keeps the distances between all its particles fixed. Real solids are only approximately rigid, but the idea works well for wheels, tops and beams.
Pure translation: at any instant every particle has the same velocity. Rotation about a fixed axis: every particle moves in a circle whose centre is on the axis (a ceiling fan, a potter's wheel).
Motion can be a mixture of the two. The axis itself may move (a spinning top, an oscillating table fan), or the body may be fixed at a single point.

Rigid body motion: translation (same velocity everywhere) and rotation about a fixed axis (circles about the axis).

The centre of mass

For particles of masses m₁, m₂, … mₙ the centre of mass has coordinates X = Σ mᵢ xᵢ / M, Y = Σ mᵢ yᵢ / M, Z = Σ mᵢ zᵢ / M, where M = Σ mᵢ. In vector form R = Σ mᵢ rᵢ / M.
For a continuous body the sums become integrals, R = (1/M) ∫ r dm. For a uniform body with a centre of symmetry (sphere, cylinder, ring, cube, rod, disc) the centre of mass is at the geometric centre.
The centre of mass need not lie inside the body (a ring, a horseshoe). The result does not depend on the axes chosen, only its coordinates do.

Centre of mass of two particles: the mass-weighted average position (here X = x₁ + 0.75 (x₂ − x₁)).

Examples

Example 6.1: masses 100 g, 150 g and 200 g at the vertices of an equilateral triangle of side 0.5 m (the 100 g mass at the origin, the 150 g mass along the x-axis): X = (150 × 0.5 + 200 × 0.25)/450 = 5/18 m and Y = 200 × 0.25 √3 / 450 = 1/(3√3) m.
Example 6.2: a uniform triangular lamina has its centre of mass at the centroid (where the medians meet).
Example 6.3: an L-shaped lamina made of three unit squares has its centre of mass at (5/6, 5/6) m, found by treating each square as a point mass at its own centre.

6.3–6.4Motion of the centre of mass and linear momentumM A = F_ext, P = M V

Motion of the centre of mass

Differentiating M R = Σ mᵢ rᵢ twice: M A = Σ Fᵢ = F_ext. The internal forces occur in equal and opposite pairs (Newton's third law) and cancel in the sum.
So the centre of mass of a system moves like a single particle of mass M acted on by the total external force. Nothing about the internal forces or the internal motion is needed.
A projectile that bursts in mid-air: the fragments fly apart but their centre of mass keeps moving along the original parabola, because the only external force (gravity) is unchanged.

Linear momentum of a system

P = Σ pᵢ = M V, where V is the velocity of the centre of mass. Also dP/dt = F_ext.
If F_ext = 0, then P is constant: the total momentum is conserved and the centre of mass moves with constant velocity.
Examples: a nucleus decaying at rest (the fragments' momenta are equal and opposite); the two stars of a binary system, which orbit their common centre of mass; a man walking on a boat floating on still water (the boat moves the other way so that the centre of mass stays at rest).
In the centre of mass frame the total momentum is zero, which often simplifies collision and decay problems.

6.5The vector producta × b = a b sin θ, right-hand rule

Definition and properties

a × b = a b sin θ n̂, where θ is the angle between the vectors and n̂ is a unit vector perpendicular to both, given by the right-hand rule (turn the fingers from a to b; the thumb gives the direction). The magnitude equals the area of the parallelogram on a and b.
It is not commutative: a × b = − b × a. It is distributive: a × (b + c) = a × b + a × c. Also a × a = 0 (a null vector), and a × b = 0 if a ∥ b.
Unit vectors: î × ĵ = k̂, ĵ × k̂ = î, k̂ × î = ĵ, and î × î = 0.
In components a × b = (a_y b_z − a_z b_y) î + (a_z b_x − a_x b_z) ĵ + (a_x b_y − a_y b_x) k̂, the determinant of the 3 × 3 array with rows (î, ĵ, k̂), (a_x, a_y, a_z), (b_x, b_y, b_z).
Example 6.4: for a = 3 î − 4 ĵ + 5 k̂ and b = −2 î + ĵ − 3 k̂: a · b = −25 and a × b = 7 î − ĵ − 5 k̂.

The vector product a × b: perpendicular to a and b (right-hand rule); its magnitude is the area of the parallelogram.

6.6–6.7Angular velocity, torque and angular momentumv = ω × r, τ = r × F, l = r × p, dL/dt = τ

Angular velocity and acceleration

For rotation about a fixed axis the angular velocity ω = dθ/dt is a vector along the axis (right-hand rule). The linear velocity of a particle at position r is v = ω × r, of magnitude v = ω r⊥, where r⊥ is its perpendicular distance from the axis.
The angular acceleration is α = dω/dt. For a fixed axis it is the same for all particles.

Moment of a force: torque

τ = r × F, of magnitude τ = r F sin θ = r⊥ F = r F⊥. Its SI unit is N m and its dimensions are [M L² T⁻²] (the same as work, though torque is a vector and work is a scalar).
The torque is zero if F = 0, if r = 0, or if the line of action of F passes through the origin. Example 6.5: for r = î − ĵ + k̂ and F = 7 î + 3 ĵ − 5 k̂: τ = 2 î + 12 ĵ + 10 k̂ N m.

Torque τ = r × F: only the component of F perpendicular to r (F sin θ) turns the body.

Angular momentum

For a particle, l = r × p, of magnitude l = r p sin θ = m v r⊥. Differentiating, dl/dt = r × F = τ.
For a system, L = Σ lᵢ and dL/dt = τ_ext. The internal torques cancel because the internal forces are equal, opposite and (for central forces) along the line joining the particles.
Conservation: if τ_ext = 0, then L is constant. For a single particle, a constant velocity along a straight line gives a constant angular momentum (Example 6.6).
In components, l_x = y p_z − z p_y, l_y = z p_x − x p_z, l_z = x p_y − y p_x. For planar motion in the xy plane only l_z is non-zero.
A bicycle wheel spun on a rod held at its two ends and then tilted makes the rod and the person on a turntable rotate: angular momentum is a vector and is conserved as a vector.

6.8Equilibrium of a rigid bodyΣF = 0 and Στ = 0; couples; levers; the centre of gravity

Conditions of equilibrium

Translational equilibrium: Σ F = 0. Rotational equilibrium: Σ τ = 0. Both must hold for mechanical equilibrium (a rigid body can be in one without the other).
For forces in a single plane these are three scalar conditions: Σ Fₓ = 0, Σ F_y = 0 and Σ τ_z = 0. When Σ F = 0 the total torque is the same about every point, so the point may be chosen for convenience.
A body is in partial equilibrium if only one of the two conditions holds.

Couple

A couple is a pair of equal and opposite forces that do not act along one line. The net force is zero, so the centre of mass does not accelerate, but the torque is not: τ = F d, where d is the perpendicular distance between the lines of action. It is the same about every point (Example 6.7).
Examples: turning a key in a lock, a tap, a steering wheel.

A couple: two equal and opposite forces, not along one line. Net force 0, moment F d about any point.

Principle of moments and the centre of gravity

For a lever in equilibrium, F₁ d₁ = F₂ d₂: load arm × load = effort arm × effort. The ratio F₁/F₂ = d₂/d₁ is the mechanical advantage. A see-saw, a crowbar and a balance are examples.
The centre of gravity is the point about which the total gravitational torque is zero; it coincides with the centre of mass when g is uniform over the body. A uniform body can be balanced on a support under that point.
Example 6.8: a 70 cm bar of mass 4.00 kg rests on two knife-edges 10 cm from each end, with a 6.00 kg load at 30 cm. With g = 9.8 m s⁻², R₂ = (4 × 25 + 6 × 20) g/50 = 43 N and R₁ = 54.9 N (the book rounds to 55 N and 43 N).
Example 6.9: a 3 m, 20 kg ladder rests against a smooth wall with its foot 1 m from the wall. With g = 9.8 m s⁻²: N = W = 196 N, f = R = 34.6 N (from the moments about the foot) and the floor exerts a total force of about 199 N at about 80° to the horizontal.

Forces on a ladder against a smooth wall (the book's Example 6.9): the floor friction f balances the wall reaction R.

6.9Moment of inertiaI = Σ m r², K = ½ I ω², radius of gyration

Definition

For a body rotating about a fixed axis, each particle has speed v = ω r, so K = Σ ½ mᵢ vᵢ² = ½ (Σ mᵢ rᵢ²) ω². Define the moment of inertia I = Σ mᵢ rᵢ², where rᵢ is the perpendicular distance from the axis. Then K = ½ I ω².
I plays the role in rotation that mass plays in translation. Its SI unit is kg m² and its dimensions are [M L²]. It depends on the mass, its distribution and the axis, not on ω.
The radius of gyration k is defined by I = M k²: the distance at which the whole mass could be placed to give the same I. A flywheel has most of its mass at the rim (large I) so it resists changes in the angular speed of an engine.

Moments of inertia of common bodies (Table 6.1)

Thin ring, about its axis (perpendicular to its plane, through the centre): M R². Ring about a diameter: M R²/2.
Thin rod of length L, about a perpendicular axis through its midpoint: M L²/12.
Circular disc, about its axis: M R²/2. Disc about a diameter: M R²/4.
Hollow cylinder, about its axis: M R². Solid cylinder, about its axis: M R²/2. Solid sphere, about a diameter: 2 M R²/5.
Check: two masses M/2 at the ends of a light rod of length l, about its midpoint: I = 2 × (M/2)(l/2)² = M l²/4.

6.10–6.12Kinematics and dynamics of rotation about a fixed axisEquations of motion, τ = I α, L = I ω

Kinematics: the rotational equations

For a constant angular acceleration the equations of linear motion carry over: ω = ω₀ + α t, θ = θ₀ + ω₀ t + ½ α t², ω² = ω₀² + 2 α (θ − θ₀). Also v = ω r and the tangential acceleration is a = α r.
Example 6.11: a motor wheel speeds up from 1200 rpm to 3120 rpm in 16 s: α = (64π)/16 = 4π rad s⁻² and it makes 576 revolutions in that time.

Dynamics

Work by a torque: dW = τ dθ. Power: P = τ ω (compare P = F v).
Torque and angular acceleration: τ = I α (the rotational form of Newton's second law, for a fixed axis).
Table 6.2 (linear ↔ rotational): x ↔ θ, v ↔ ω, a ↔ α, m ↔ I, F ↔ τ, p ↔ L, F = m a ↔ τ = I α, K = ½ m v² ↔ K = ½ I ω², P = F v ↔ P = τ ω.
Example 6.12: a flywheel of mass 20 kg and radius 0.2 m with a steady pull of 25 N on a cord wound round it: τ = 5.0 N m, I = ½ M R² = 0.4 kg m², α = 12.5 rad s⁻². After 2 m of cord is unwound, the work done is 50 J and ω² = 250 rad² s⁻².

Angular momentum of a body rotating about a fixed axis

L_z = I ω (for a body symmetric about the axis; in general L is not parallel to ω). Then dL_z/dt = τ.
Conservation: if the external torque about the axis is zero, I ω = constant. When I decreases, ω increases.
Examples: a skater or a diver pulling the arms and legs in to spin faster; a person on a swivel chair or a turntable holding weights; an acrobat's somersault. The rotational kinetic energy K = L²/2I increases when I falls, because the person does internal work.

Try each problem first, then open the solution to check your working.

6.1–6.4Centre of mass and momentumMass-weighted averages and conservation of momentum · 6 problems

Q1Two masses on a line

A 2 kg mass is at x = 0 and a 6 kg mass is at x = 4 m. Find the position of the centre of mass.

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Given: m₁ = 2 kg at 0, m₂ = 6 kg at 4 m

Use: X = (m₁ x₁ + m₂ x₂)/(m₁ + m₂)

X = (2 × 0 + 6 × 4)/(2 + 6) = 24/8 = 3 m.

It lies 1 m from the heavier mass and 3 m from the lighter one.

Answer: 3 m from the 2 kg mass

Q2Three masses in a plane

Masses of 1 kg, 2 kg and 3 kg are at (0, 0), (3, 0) and (0, 4) (metres). Find the coordinates of the centre of mass.

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Given: m = 1, 2, 3 kg; positions as above

Use: X = Σ mᵢ xᵢ/M, Y = Σ mᵢ yᵢ/M

M = 6 kg.

X = (1 × 0 + 2 × 3 + 3 × 0)/6 = 1 m.

Y = (1 × 0 + 2 × 0 + 3 × 4)/6 = 2 m.

Answer: (1 m, 2 m)

Q3A carbon monoxide molecule

In a CO molecule the separation of the carbon (12 u) and oxygen (16 u) nuclei is 1.13 Å. How far is the centre of mass from the carbon nucleus?

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Given: m(C) = 12 u, m(O) = 16 u, separation 1.13 Å

Use: The centre of mass lies nearer the heavier atom

Take C at the origin: X = (12 × 0 + 16 × 1.13)/28 = 18.08/28 = 0.646 Å.

It lies 0.646 Å from C, i.e. nearer to the oxygen.

Answer: 0.646 Å from the carbon nucleus

Q4A man on a boat

A man of mass 60 kg walks 3 m along a boat of mass 120 kg that floats at rest on still water (neglect water resistance). How far does the boat move?

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Given: m = 60 kg, M = 120 kg, 3 m relative to the boat

Use: No external horizontal force: the centre of mass does not move

Let the boat move x backwards; the man then moves (3 − x) forward relative to the ground.

60 (3 − x) = 120 x, so 180 = 180 x and x = 1 m.

The boat moves 1 m in the direction opposite to the man's walk.

Answer: 1 m, opposite to the man

Q5A disc with a hole

A uniform disc of radius 0.6 m has a circular hole of radius 0.3 m cut out; the hole's centre is 0.3 m from the centre of the disc. How far, and on which side, is the centre of mass of the remainder?

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Given: R = 0.6 m, hole radius R/2, hole centre at R/2

Use: The hole has mass M/4; add it back as negative mass

Whole disc: mass M at 0. The hole: mass M/4 at +0.3 m. The rest has mass 3M/4 at x.

(3M/4) x + (M/4)(0.3) = 0, so x = −0.1 m (= −R/6).

The centre of mass is 0.1 m from the disc's centre on the side opposite to the hole.

Answer: 0.1 m, away from the hole

Q6A shell that explodes

A 5 kg shell moving at 10 m s⁻¹ explodes into a 2 kg piece and a 3 kg piece. The 2 kg piece continues in the same direction at 25 m s⁻¹. Find the velocity of the 3 kg piece, and the velocity of the centre of mass just after the explosion.

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Given: M = 5 kg, V = 10 m s⁻¹, m₁ = 2 kg, v₁ = 25 m s⁻¹

Use: Internal forces only: total momentum is conserved

5 × 10 = 2 × 25 + 3 v₂, so v₂ = 0: the 3 kg piece stops (drops vertically).

The centre of mass still moves at (2 × 25 + 3 × 0)/5 = 10 m s⁻¹, unchanged by the explosion.

Answer: 0; the centre of mass keeps 10 m s⁻¹

6.5–6.7Vector product, torque and angular momentumCross products and r × F · 6 problems

Q7Magnitude of a vector product

Two vectors of magnitudes 6 and 5 make an angle of 30° with each other. Find |a × b| and a · b.

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Given: a = 6, b = 5, θ = 30°

Use: |a × b| = a b sin θ, a · b = a b cos θ

|a × b| = 6 × 5 × 0.5 = 15.

a · b = 6 × 5 × (√3/2) = 25.98 ≈ 26.

Answer: 15; 26

Q8Vector product in components

Find a × b for a = î + 2ĵ + 3k̂ and b = 3î − ĵ + 2k̂, and show that it is perpendicular to a.

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Given: a = (1, 2, 3), b = (3, −1, 2)

Use: a × b = (a_y b_z − a_z b_y) î + (a_z b_x − a_x b_z) ĵ + (a_x b_y − a_y b_x) k̂

î: 2 × 2 − 3 × (−1) = 7. ĵ: 3 × 3 − 1 × 2 = 7. k̂: 1 × (−1) − 2 × 3 = −7.

So a × b = 7î + 7ĵ − 7k̂.

(a × b) · a = 7 + 14 − 21 = 0, so it is perpendicular to a (and similarly to b).

Answer: 7î + 7ĵ − 7k̂

Q9Torque from r and F

A force F = 3ĵ N acts at the point r = (2î + ĵ) m. Find the torque about the origin.

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Given: r = (2, 1, 0) m, F = (0, 3, 0) N

Use: τ = r × F

Only the k̂ component survives: τ_z = x F_y − y F_x = 2 × 3 − 1 × 0 = 6 N m.

τ = 6 k̂ N m, i.e. 6 N m out of the xy plane.

Answer: 6 N m along +z

Q10A spanner

A force of 40 N is applied to a spanner of length 0.25 m at 30° to the handle. Find the torque on the nut.

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Given: F = 40 N, r = 0.25 m, θ = 30°

Use: τ = r F sin θ

τ = 0.25 × 40 × sin 30° = 0.25 × 40 × 0.5 = 5 N m.

(At 90° the torque would be 10 N m, the maximum.)

Answer: 5 N m

Q11Angular momentum of a particle

(a) A 0.5 kg particle moves in a circle of radius 2 m at 4 m s⁻¹. Find its angular momentum about the centre. (b) A 1.5 kg particle moves at 2 m s⁻¹ along a straight line whose perpendicular distance from the origin is 3 m. Find its angular momentum about the origin.

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Given: (a) m = 0.5 kg, v = 4 m s⁻¹, r = 2 m; (b) m = 1.5 kg, v = 2 m s⁻¹, r⊥ = 3 m

Use: L = m v r⊥

(a) L = 0.5 × 4 × 2 = 4 kg m² s⁻¹.

(b) L = 1.5 × 2 × 3 = 9 kg m² s⁻¹.

In (b) it stays the same at every point of the path, because r⊥ does not change.

Answer: (a) 4; (b) 9 kg m² s⁻¹

Q12Torque and angular momentum

A constant torque of 6 N m acts for 4 s on a body that is initially at rest. Find the angular momentum it acquires.

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Given: τ = 6 N m, t = 4 s, L₀ = 0

Use: dL/dt = τ

L = τ t = 6 × 4 = 24 kg m² s⁻¹.

(If I = 3 kg m², then ω = L/I = 8 rad s⁻¹.)

Answer: 24 kg m² s⁻¹

6.8EquilibriumMoments, couples and ladders · 5 problems

Q13A see-saw

A 30 kg child sits 2 m from the pivot of a see-saw. A 40 kg child sits on the other side. How far from the pivot must the 40 kg child sit for balance?

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Given: m₁ = 30 kg, d₁ = 2 m, m₂ = 40 kg

Use: Principle of moments: m₁ d₁ = m₂ d₂

30 × 2 = 40 × d₂, so d₂ = 60/40 = 1.5 m.

Answer: 1.5 m

Q14Supports of a beam

A uniform beam 6 m long and of mass 30 kg rests on supports at its two ends. A 60 kg load is placed 2 m from the left end (g = 10 m s⁻²). Find the forces on the two supports.

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Given: L = 6 m, M = 30 kg, m = 60 kg at 2 m from the left

Use: Take moments about the left end; then ΣF = 0

Weights: beam 300 N at 3 m; load 600 N at 2 m.

R(right) × 6 = 300 × 3 + 600 × 2 = 2100, so R(right) = 350 N.

R(left) = 900 − 350 = 550 N.

Answer: Left 550 N; right 350 N

Q15Moment of a couple

(a) Two forces of 15 N each, equal and opposite, act along parallel lines 0.4 m apart. Find the moment of the couple. (b) What force is needed in each of the forces of a couple to give a moment of 12 N m with the lines 0.3 m apart?

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Given: (a) F = 15 N, d = 0.4 m; (b) τ = 12 N m, d = 0.3 m

Use: τ = F d

(a) τ = 15 × 0.4 = 6 N m.

(b) F = 12/0.3 = 40 N.

Answer: (a) 6 N m; (b) 40 N

Q16A ladder against a smooth wall

A 5 m ladder of mass 20 kg rests with its top on a smooth vertical wall and its foot on rough ground 3 m from the wall (so its top is 4 m high). Find the wall reaction, the friction at the foot and the least μ for the ladder not to slip (g = 10 m s⁻²).

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Given: L = 5 m, M = 20 kg, base 3 m, height 4 m

Use: Moments about the foot, with W at the middle (1.5 m from the wall horizontally)

Vertical: N = W = 200 N.

Moments about the foot: R × 4 = 200 × 1.5, so R = 75 N.

Horizontal: friction f = R = 75 N.

μ ≥ f/N = 75/200 = 0.375.

Answer: R = f = 75 N; μ ≥ 0.375

Q17A balanced metre stick

A uniform metre stick of mass 200 g is pivoted at the 40 cm mark. What mass hung at the 10 cm mark balances it?

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Given: M = 200 g at 50 cm; pivot at 40 cm; the mass m at 10 cm

Use: Moments about the pivot

Stick: 200 g × (50 − 40) cm = 2000 g cm (clockwise).

Hanging mass: m × (40 − 10) = 30 m (anticlockwise).

30 m = 2000, so m = 66.7 g.

Answer: 66.7 g

6.9Moment of inertiaI = Σ m r² and Table 6.1 · 5 problems

Q18Four masses at the corners of a square

Four 0.5 kg masses are at the corners of a square of side 0.4 m. Find the moment of inertia about an axis through the centre and perpendicular to the square (neglect the frame).

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Given: m = 0.5 kg each, side 0.4 m

Use: I = Σ mᵢ rᵢ²

The distance from the centre to a corner: r² = (0.4/√2)² = 0.08 m².

I = 4 × 0.5 × 0.08 = 0.16 kg m².

Answer: 0.16 kg m²

Q19Three bodies of the same mass and radius

A ring, a disc and a solid sphere each have M = 3 kg and R = 0.2 m. Find I for the ring and the disc about their axes, and for the sphere about a diameter.

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Given: M = 3 kg, R = 0.2 m

Use: Table 6.1

Ring: I = M R² = 3 × 0.04 = 0.12 kg m².

Disc: I = ½ M R² = 0.06 kg m².

Sphere: I = (2/5) M R² = 0.4 × 3 × 0.04 = 0.048 kg m².

Answer: 0.12, 0.06 and 0.048 kg m²

Q20A rod about its midpoint

A thin uniform rod of length 1.2 m and mass 3 kg rotates about an axis through its midpoint and perpendicular to its length. Find I.

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Given: L = 1.2 m, M = 3 kg

Use: I = M L²/12

I = 3 × 1.44/12 = 0.36 kg m².

Answer: 0.36 kg m²

Q21Radius of gyration of a disc

Find the radius of gyration of a uniform disc of radius 0.5 m about its axis.

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Given: R = 0.5 m

Use: I = M k² and I = ½ M R²

M k² = ½ M R², so k = R/√2 = 0.5/1.414 = 0.354 m.

Answer: 0.354 m

Q22Rotational kinetic energy of a flywheel

A flywheel of moment of inertia 2 kg m² rotates at 300 rpm. Find its kinetic energy.

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Given: I = 2 kg m², 300 rpm

Use: K = ½ I ω²

ω = 2π × 300/60 = 10π = 31.4 rad s⁻¹.

K = ½ × 2 × (10π)² = 100 π² = 987 J.

Answer: ≈ 987 J

6.10–6.12Rotational kinematics and dynamicsω, α, τ = I α, power and conservation of angular momentum · 6 problems

Q23Uniform angular acceleration

A wheel starts from rest with α = 3 rad s⁻². Find its angular speed and the angle turned after 8 s, and the number of revolutions.

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Given: ω₀ = 0, α = 3 rad s⁻², t = 8 s

Use: ω = ω₀ + α t, θ = ω₀ t + ½ α t²

ω = 3 × 8 = 24 rad s⁻¹.

θ = ½ × 3 × 64 = 96 rad.

Revolutions = 96/2π = 15.3.

Answer: 24 rad s⁻¹; 96 rad; 15.3 rev

Q24A fan slowing down

A ceiling fan running at 600 rpm is switched off and stops uniformly in 20 s. Find the angular deceleration and the number of revolutions it makes.

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Given: ω₀ = 600 rpm = 20π rad s⁻¹, ω = 0, t = 20 s

Use: ω = ω₀ − α t, ω² = ω₀² − 2 α θ

α = 20π/20 = π rad s⁻² (about 3.14 rad s⁻²).

θ = ω₀²/(2α) = 400π²/(2π) = 200π rad.

Revolutions = 200π/2π = 100.

Answer: π rad s⁻² deceleration; 100 revolutions

Q25Torque on a disc

A uniform disc of mass 10 kg and radius 0.5 m is acted on by a torque of 5 N m about its axis. Find the angular acceleration.

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Given: M = 10 kg, R = 0.5 m, τ = 5 N m

Use: τ = I α, I = ½ M R²

I = ½ × 10 × 0.25 = 1.25 kg m².

α = τ/I = 5/1.25 = 4 rad s⁻².

Answer: 4 rad s⁻²

Q26A cord on a solid cylinder

A cord wound round a solid cylinder of mass 8 kg and radius 0.25 m is pulled with a steady force of 20 N. Find the angular acceleration and the acceleration of the cord (no slipping).

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Given: M = 8 kg, R = 0.25 m, F = 20 N

Use: τ = F R = I α, a = α R

τ = 20 × 0.25 = 5 N m; I = ½ × 8 × 0.0625 = 0.25 kg m².

α = 5/0.25 = 20 rad s⁻².

a = α R = 20 × 0.25 = 5 m s⁻².

Answer: 20 rad s⁻²; 5 m s⁻²

Q27Power of a motor

A motor exerts a torque of 50 N m on a shaft turning at 1500 rpm. Find the power delivered.

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Given: τ = 50 N m, 1500 rpm

Use: P = τ ω

ω = 2π × 1500/60 = 50π = 157.1 rad s⁻¹.

P = 50 × 157.1 = 7854 W ≈ 7.85 kW.

Answer: ≈ 7.85 kW

Q28A skater pulls in her arms

A skater spins at 2 rev s⁻¹ with I = 3.0 kg m². She pulls in her arms so that I becomes 1.2 kg m². Find her new angular speed and the ratio of the new to the old kinetic energy.

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Given: I₁ = 3.0 kg m², ω₁ = 2 rev s⁻¹, I₂ = 1.2 kg m²

Use: No external torque: I₁ ω₁ = I₂ ω₂

ω₂ = 3.0 × 2/1.2 = 5 rev s⁻¹.

K₂/K₁ = (½ I₂ ω₂²)/(½ I₁ ω₁²) = (1.2 × 25)/(3.0 × 4) = 2.5.

The extra kinetic energy comes from the work done by her muscles.

Answer: 5 rev s⁻¹; K₂/K₁ = 2.5

Answers to every exercise question of the NCERT chapter. Try each one first, then tap Show answer.

6.1Exercises 6.1 to 6.17Centre of mass, torque, equilibrium and rotation · 17 questions

Q16.1 Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body?

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For a uniform body with a centre of symmetry, the centre of mass is at the geometric centre: (i) the centre of the sphere, (ii) the midpoint of the axis of the cylinder, (iii) the centre of the ring, and (iv) the point where the diagonals of the cube meet.

No. The centre of mass of a ring lies at its centre, where there is no material. The same is true of a hollow sphere or a bent bar.

Q26.2 In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 Å (1 Å = 10⁻¹⁰ m). Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.

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Let the mass of H be m, so that of Cl is 35.5 m. Take the Cl nucleus as the origin and the H nucleus at 1.27 Å.

X = (35.5 m × 0 + m × 1.27)/(36.5 m) = 1.27/36.5 = 0.0348 Å.

The centre of mass is about 0.035 Å from the chlorine nucleus, on the line joining the two nuclei: only about one thirty-seventh of the separation, so it lies very close to the heavier chlorine atom.

Q36.3 A child sits stationary at one end of a long trolley moving uniformly with a speed V on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system?

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The floor is smooth, so there is no external horizontal force on the system (trolley + child). The forces between the child and the trolley are internal.

By M A = F_ext = 0, the velocity of the centre of mass cannot change. The speed of the CM stays V, however the child runs about. (The trolley's own speed changes while the child runs, but the centre of mass is unaffected.)

Q46.4 Show that the area of the triangle contained between the vectors a and b is one half of the magnitude of a × b.

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Let the vectors a = OP and b = OQ be two sides of the triangle OPQ, with angle θ between them.

Drop a perpendicular from Q to OP: its length is the height h = b sin θ. The base is OP = a.

Area = ½ × base × height = ½ a b sin θ = ½ |a × b|, since |a × b| = a b sin θ. (The parallelogram on a and b has area |a × b|, and the triangle is half of it.)

Q56.5 Show that a · (b × c) is equal in magnitude to the volume of the parallelepiped formed on the three vectors a, b and c.

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Let the parallelepiped have edges a, b and c. The base is the parallelogram on b and c, whose area is |b × c|, and the vector b × c is perpendicular to the base.

The height is the component of a along the perpendicular to the base: h = a cos φ, where φ is the angle between a and b × c, i.e. h = a · n̂ with n̂ = (b × c)/|b × c|.

Volume = (area of base) × (height) = |b × c| (a · n̂) = |a · (b × c)|.

Q66.6 Find the components along the x, y, z axes of the angular momentum l of a particle, whose position vector is r with components x, y, z and momentum is p with components pₓ, p_y and p_z. Show that if the particle moves only in the x-y plane the angular momentum has only a z-component.

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l = r × p = (y p_z − z p_y) î + (z pₓ − x p_z) ĵ + (x p_y − y pₓ) k̂, so

l_x = y p_z − z p_y
l_y = z pₓ − x p_z
l_z = x p_y − y pₓ

If the particle moves only in the x-y plane, z = 0 and p_z = 0. Then l_x = 0 and l_y = 0, leaving only l_z = x p_y − y pₓ.

Q76.7 Two particles, each of mass m and speed v, travel in opposite directions along parallel lines separated by a distance d. Show that the vector angular momentum of the two-particle system is the same whatever the point about which the angular momentum is taken.

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Let the particles have momenta p and −p (equal magnitudes, opposite directions) and position vectors r₁ and r₂ from any chosen origin.

Total angular momentum L = r₁ × p + r₂ × (−p) = (r₁ − r₂) × p.

The vector r₁ − r₂ joins the two particles; it does not depend on the origin. So L does not depend on the origin. Its magnitude is |r₁ − r₂| p sin θ = p × (perpendicular distance between the lines) = m v d, and its direction is perpendicular to the plane of the two paths.

Q86.8 A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in Fig. 6.33. The angles made by the strings with the vertical are 36.9° and 53.1° respectively. The bar is 2 m long. Calculate the distance d of the centre of gravity of the bar from its left end.

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I take the book's figure as follows: the left string is at 36.9° to the vertical, the right string at 53.1°, and W acts at distance d from the left end (cos 36.9° = 0.8, sin 36.9° = 0.6; cos 53.1° = 0.6, sin 53.1° = 0.8).

Fig. 6.33 (Ex 6.8) as read here: left string 36.9° and right string 53.1° to the vertical; W acts at distance d from the left end.

Horizontal balance: T₁ sin 36.9° = T₂ sin 53.1°, so 0.6 T₁ = 0.8 T₂ and T₁ = (4/3) T₂.

Vertical balance: T₁ cos 36.9° + T₂ cos 53.1° = W, so 0.8 × (4/3) T₂ + 0.6 T₂ = W. This gives T₂ = 0.6 W and T₁ = 0.8 W.

Moments about the left end: T₂ cos 53.1° × 2 = W × d, so d = 0.6 × 0.6 × 2 = 0.72 m.

Q96.9 A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.

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Take g = 9.8 m s⁻². The weight W = 1800 × 9.8 = 17 640 N acts at the centre of gravity, 1.05 m behind the front axle (so 0.75 m in front of the back axle).

Let R(front) and R(back) be the total forces on the front and back wheel pairs. Then R(front) + R(back) = W. Moments about the front axle: R(back) × 1.8 = W × 1.05, so R(back) = 10 290 N and R(front) = 7350 N.

Each front wheel: 7350/2 = 3675 N. Each back wheel: 10 290/2 = 5145 N.

Q106.10 Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time?

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Hollow cylinder: I₁ = M R². Solid sphere: I₂ = (2/5) M R² (Table 6.1).

With the same torque τ, α = τ/I, so α₂/α₁ = I₁/I₂ = 5/2. After the same time ω = α t, so ω₂/ω₁ = 5/2.

The solid sphere acquires the greater angular speed (2.5 times that of the cylinder).

Q116.11 A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad s⁻¹. The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis?

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I = ½ M R² = ½ × 20 × (0.25)² = 0.625 kg m².

K = ½ I ω² = ½ × 0.625 × (100)² = 3125 J.

L = I ω = 0.625 × 100 = 62.5 kg m² s⁻¹ (J s).

Q126.12 (a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 2/5 times the initial value? Assume that the turntable rotates without friction. (b) Show that the child's new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?

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(a) No external torque acts, so I₁ ω₁ = I₂ ω₂. With I₂ = (2/5) I₁: ω₂ = (5/2) × 40 = 100 rev/min.

(b) K₂/K₁ = (½ I₂ ω₂²)/(½ I₁ ω₁²) = (2/5) × (5/2)² = 5/2 = 2.5. The kinetic energy increases to 2.5 times.

The increase comes from the work done by the child: his muscles do work in pulling his arms in against the (outward) tendency of the arms. This energy is supplied by the internal chemical energy of the child.

Q136.13 A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N? What is the linear acceleration of the rope? Assume that there is no slipping.

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I = M R² = 3 × (0.4)² = 0.48 kg m². The torque τ = F R = 30 × 0.4 = 12 N m.

α = τ/I = 12/0.48 = 25 rad s⁻².

With no slipping, the linear acceleration of the rope = α R = 25 × 0.4 = 10 m s⁻².

Q146.14 To maintain a rotor at a uniform angular speed of 200 rad s⁻¹, an engine needs to transmit a torque of 180 N m. What is the power required by the engine? (Note: uniform angular velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter frictional torque.) Assume that the engine is 100% efficient.

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P = τ ω = 180 × 200 = 36 000 W = 36 kW.

Q156.15 From a uniform disk of radius R, a circular hole of radius R/2 is cut out. The centre of the hole is at R/2 from the centre of the original disc. Locate the centre of gravity of the resulting flat body.

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Let the mass per unit area be σ. The whole disc has mass M = σ π R². The hole has area π (R/2)² = πR²/4, so its mass is M/4, and the remainder has mass 3M/4.

Take the centre of the original disc as the origin and the centre of the hole at x = +R/2. The centre of mass of the whole disc is at 0 and is the combination of the remainder (3M/4 at x) and the hole (M/4 at R/2):

0 = [(3M/4) x + (M/4)(R/2)]/M, so x = −R/6.

The centre of gravity of the resulting body is at R/6 from the centre of the original disc, on the side opposite to the hole.

Q166.16 A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick?

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The stick is uniform, so its weight M g acts at the 50.0 cm mark. The coins have a combined mass of 10 g at the 12.0 cm mark, and the balance (pivot) is at 45.0 cm.

Moments about the pivot: M × (50.0 − 45.0) = 10 × (45.0 − 12.0), so 5 M = 330 and M = 66 g.

Q176.17 Assume that the mean speed of an oxygen molecule is 500 m/s, its mass 5.30 × 10⁻²⁶ kg and its moment of inertia 1.94 × 10⁻⁴⁶ kg m². If the rotational kinetic energy of the molecule is two-thirds of its translational kinetic energy, find its angular velocity.

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Translational kinetic energy: K(trans) = ½ m v² = ½ × 5.30 × 10⁻²⁶ × (500)² = 6.63 × 10⁻²¹ J.

Rotational kinetic energy: K(rot) = (2/3) K(trans) = 4.42 × 10⁻²¹ J.

K(rot) = ½ I ω², so ω = √(2 K(rot)/I) = √(2 × 4.42 × 10⁻²¹/1.94 × 10⁻⁴⁶) = √(4.55 × 10²⁵) = 6.7 × 10¹² rad s⁻¹ (about 6.8 × 10¹² with the book's rounding).

Rigid body motion

The centre of mass

Examples

Motion of the centre of mass

Linear momentum of a system

Definition and properties

Angular velocity and acceleration

Moment of a force: torque

Angular momentum

Conditions of equilibrium

Couple

Principle of moments and the centre of gravity

Definition

Moments of inertia of common bodies (Table 6.1)

Kinematics: the rotational equations

Dynamics

Angular momentum of a body rotating about a fixed axis

Q1Two masses on a line

Q2Three masses in a plane

Q3A carbon monoxide molecule

Q4A man on a boat

Q5A disc with a hole

Q6A shell that explodes

Q7Magnitude of a vector product

Q8Vector product in components

Q9Torque from r and F

Q10A spanner

Q11Angular momentum of a particle

Q12Torque and angular momentum

Q13A see-saw

Q14Supports of a beam

Q15Moment of a couple

Q16A ladder against a smooth wall

Q17A balanced metre stick

Q18Four masses at the corners of a square

Q19Three bodies of the same mass and radius

Q20A rod about its midpoint

Q21Radius of gyration of a disc

Q22Rotational kinetic energy of a flywheel

Q23Uniform angular acceleration

Q24A fan slowing down

Q25Torque on a disc

Q26A cord on a solid cylinder

Q27Power of a motor

Q28A skater pulls in her arms

Q16.1 Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body?

Q36.3 A child sits stationary at one end of a long trolley moving uniformly with a speed V on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system?

Q46.4 Show that the area of the triangle contained between the vectors a and b is one half of the magnitude of a × b.

Q56.5 Show that a · (b × c) is equal in magnitude to the volume of the parallelepiped formed on the three vectors a, b and c.

Q66.6 Find the components along the x, y, z axes of the angular momentum l of a particle, whose position vector is r with components x, y, z and momentum is p with components pₓ, py and pz. Show that if the particle moves only in the x-y plane the angular momentum has only a z-component.

Q76.7 Two particles, each of mass m and speed v, travel in opposite directions along parallel lines separated by a distance d. Show that the vector angular momentum of the two-particle system is the same whatever the point about which the angular momentum is taken.

Q96.9 A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.

Q116.11 A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad s⁻¹. The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis?

Q136.13 A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N? What is the linear acceleration of the rope? Assume that there is no slipping.

Q156.15 From a uniform disk of radius R, a circular hole of radius R/2 is cut out. The centre of the hole is at R/2 from the centre of the original disc. Locate the centre of gravity of the resulting flat body.

Q166.16 A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick?

Q176.17 Assume that the mean speed of an oxygen molecule is 500 m/s, its mass 5.30 × 10⁻²⁶ kg and its moment of inertia 1.94 × 10⁻⁴⁶ kg m². If the rotational kinetic energy of the molecule is two-thirds of its translational kinetic energy, find its angular velocity.

Q66.6 Find the components along the x, y, z axes of the angular momentum l of a particle, whose position vector is r with components x, y, z and momentum is p with components pₓ, p_y and p_z. Show that if the particle moves only in the x-y plane the angular momentum has only a z-component.