Unit 4: Dynamics

Introduction

Dynamics is the branch of mechanics that deals with the motion of bodies under the action of forces. In this chapter we build on the concepts of kinematics and introduce the quantitative tools needed to predict how forces change motion. The discussion proceeds from linear momentum and impulse, through the conservation of momentum, to the direct application of Newton’s three laws, then to rotational effects (moment, torque, equilibrium), and finally to the empirical laws governing solid friction.

1. Linear Momentum and Impulse

Definition of Linear Momentum

Linear momentum (p) of a particle is defined as the product of its mass (m) and its velocity (v). Because velocity is a vector, momentum is also a vector quantity:

p = m v

Where:

m = mass (kg)
v = velocity vector (m s⁻¹)
p = linear momentum (kg m s⁻¹)

Impulse

When a force acts over a finite time interval, it produces an impulse that changes the momentum of the object. Impulse (J) is defined as the integral of force with respect to time; for a constant force it simplifies to:

J = F Δt

The impulse‑momentum theorem states that the impulse equals the change in momentum:

J = Δp = p_final – p_initial

Combining the two expressions gives the useful form:

F Δt = m Δv

Example 1: Impulse on a Cricket Ball

A cricket ball of mass 0.16 kg is moving at 15 m s⁻¹ toward a batsman. The batsman exerts an average force of 300 N for 0.02 s in the opposite direction. Calculate the final speed of the ball.

Initial momentum: p_i = m v_i = 0.16 × 15 = 2.4 kg m s⁻¹ (toward batsman).
Impulse exerted by bat (opposite direction): J = F Δt = 300 × 0.02 = 6 N s. Since it opposes motion, impulse is negative relative to initial direction: J = –6 kg m s⁻¹.
Final momentum: p_f = p_i + J = 2.4 – 6 = –3.6 kg m s⁻¹ (negative indicates reversal).
Final speed: v_f = p_f / m = –3.6 / 0.16 = –22.5 m s⁻¹. Magnitude = 22.5 m s⁻¹ away from the batsman.

2. Conservation of Linear Momentum

Principle

If the net external force acting on a system is zero, the total linear momentum of the system remains constant:

Σ p_initial = Σ p_final

For a two‑particle system this becomes:

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

Where u₁, u₂ are initial velocities and v₁, v₂ final velocities.

Applications

Collisions – elastic (kinetic energy also conserved) and inelastic (momentum conserved only).
Explosions – a single body at rest splits into fragments; total momentum stays zero.
Rocket propulsion – exhaust gases carry momentum opposite to the rocket’s motion, giving the rocket forward thrust.

Example 2: Inelastic Collision

A 0.5 kg trolley moving at 2 m s⁻¹ collides and sticks to a stationary 1.0 kg trolley. Find the common velocity after impact.

Initial momentum: p_i = (0.5)(2) + (1.0)(0) = 1.0 kg m s⁻¹.
Final mass: M = 0.5 + 1.0 = 1.5 kg.
Using conservation: p_i = M v_f → v_f = p_i / M = 1.0 / 1.5 ≈ 0.667 m s⁻¹ in the original direction.

Example 3: Rocket Thrust (Simplified)

A rocket ejects exhaust gases at a rate of 20 kg s⁻¹ with a relative speed of 2500 m s⁻¹. Compute the thrust.

Thrust equals rate of momentum change of exhaust: F = ṁ v_rel = 20 × 2500 = 50 000 N.

3. Application of Newton's Laws

First Law – Law of Inertia

An object remains at rest or moves with constant velocity unless acted upon by a net external force.

Example: A book lying on a table stays at rest because the normal force balances its weight.

Second Law – Force and Acceleration

The net force acting on a body equals the rate of change of its momentum. For constant mass this reduces to the familiar form:

F = m a

Where F is net force (N), m mass (kg), a acceleration (m s⁻²).

Example: A 10 kg cart is pulled by a 40 N force on a frictionless surface gives acceleration a = F/m = 40 a 40 N force on a frictionless surface. Its acceleration is a = F/m = 40/10 = 4 m s⁻².



Third Law – Action–Reaction
For every action there is an equal and opposite reaction: F_AB = –F_BA. The forces act on different bodies.
Example: When you push a wall, the wall pushes back on your hand with equal magnitude.

Applications of Newton's Laws

Tension in a string – Consider a mass m hanging from a light string. The tension T equals weight mg when the system is static.
Friction – Opposes relative motion; discussed in detail later.
Banked roads – The normal force provides a component toward the centre of the turn, allowing a vehicle to negotiate a curve without relying on friction.
Atwood machine – Two masses m₁ and m₂ connected over a pulley. Acceleration a = (m₂ – m₁)g / (m₁ + m₂) (assuming massless, frictionless pulley).


Example 4: Atwood Machine
Let m₁ = 3 kg, m₂ = 5 kg, g = 9.8 m s⁻².
Acceleration: a = (5 – 3) × 9.8 / (3 + 5) = 2 × 9.8 / 8 = 19.6 / 8 = 2.45 m s⁻² (direction: heavier mass descends).
Tension in the string: Using T = m₁(g + a) (for the lighter mass ascending): T = 3 × (9.8 + 2.45) = 3 × 12.25 = 36.75 N.

4. Moment, Torque and Equilibrium
Moment of a Force (Torque)
The tendency of a force to rotate a body about a point or axis is measured by the moment (torque). For a force F applied at a position vector r relative to the pivot, the torque vector is:
τ = r × F
Magnitude (when r and F are perpendicular): τ = r F sinθ. Units: newton‑metre (N m).

Equilibrium Conditions
A rigid body is in mechanical equilibrium when both the net force and the net torque acting on it are zero:
Σ F = 0 (translational equilibrium)
Σ τ = 0 (rotational equilibrium)

Types of Equilibrium

Translational equilibrium – No linear acceleration; the body’s centre of mass moves with constant velocity (or is at rest).
Rotational equilibrium – No angular acceleration; the body’s orientation does not change.


Centre of Mass
The centre of mass (CM) is the point where the total mass of a system can be considered to be concentrated for the purpose of analysing translational motion. For a system of particles:
R_cm = ( Σ m_i r_i ) / ( Σ m_i )
For a uniform rigid body, the CM coincides with its geometric centre.

Applications

Beam balance – The torques produced by weights on either side of the fulcrum must balance: W₁ d₁ = W₂ d₂.
See‑saw – A classic illustration of torque balance; children of different weights sit at distances inversely proportional to their masses.
Crane – The load torque must be countered by the counterweight torque to keep the crane in equilibrium.


Example 5: See‑saw Balance
Child A (mass 30 kg) sits 2.0 m from the pivot. Child B (mass 20 kg) sits on the opposite side. Find the distance d_B required for balance.
Set torques equal (taking clockwise as positive): m_A g d_A = m_B g d_B → d_B = (m_A d_A) / m_B = (30 × 2.0) / 20 = 60 / 20 = 3.0 m.
Thus Child B must sit 3.0 m from the pivot.

5. Solid Friction
Laws of Solid Friction (Empirical)

Friction opposes the relative motion (or tendency of motion) between two surfaces in contact.
The magnitude of the frictional force is directly proportional to the normal reaction: f = μ R, where μ is the coefficient of friction.
Friction is independent of the apparent area of contact (for dry, clean surfaces).
Friction is approximately independent of the sliding velocity (for moderate speeds).


Types of Friction

Static friction (f_s) – acts when surfaces are at rest relative to each other; varies from zero up to a maximum value f_s,max = μ_s R.
Kinetic (sliding) friction (f_k) – acts when surfaces slide; approximately constant: f_k = μ_k R.
Rolling friction (f_r) – resists rolling motion; usually much smaller than sliding friction.


Verification Methods

Inclined plane method – Place a block on a plane of adjustable angle θ. The block just begins to slide when tan θ = μ_s. Measuring the critical angle gives the static coefficient.
Block on plank method – Pull a block across a horizontal plank with a spring balance; the force required to maintain constant speed equals kinetic friction, giving μ_k = F / (mg).


Coefficients – Typical Values


Surface Pair μ_s (static) μ_k (kinetic)


Rubber on dry concrete 1.0 0.8
Steel on steel (lubricated) 0.15 0.1
Wood on wood 0.25–0.5 0.2–0.4
Ice on ice 0.1 0.03



Methods to Reduce Friction

Lubrication – Introduces a fluid layer that converts solid‑solid contact to fluid‑film shear, drastically lowering μ.
Ball bearings / roller bearings – Replace sliding contact with rolling contact, exploiting the much lower rolling friction.
Streamlined shapes – Reduce fluid drag (relevant when motion involves fluids) and can also minimize contact area in certain designs.
Surface polishing – Decreases microscopic interlocking, lowering both static and kinetic coefficients.


Example 6: Calculating Friction on an Incline
A 5 kg block rests on a rough incline of angle 30°. The coefficient of static friction is μ_s = 0.4. Determine whether the block will slide.

Weight component parallel to plane: W_∥ = mg sinθ = 5 × 9.8 × sin30° = 5 × 9.8 × 0.5 = 24.5 N

Surface Pair	μ_s (static)	μ_k (kinetic)
Rubber on dry concrete	1.0	0.8
Steel on steel (lubricated)	0.15	0.1
Wood on wood	0.25–0.5	0.2–0.4
Ice on ice	0.1	0.03