Unit 1: Rotational Dynamics and Oscillatory Motion [5 Hours]

Rotational dynamics studies how forces produce rotation and how mass distribution affects rotational motion. Oscillatory motion studies repeated motion about an equilibrium position, with the mass-spring system providing a standard example. This unit connects torque, moment of inertia, rotational kinetic energy, and angular momentum with the mathematical description, phase, and energy of spring oscillations.

Tested Topics

1. Oscillatory motion / mass-spring system (frequency, amplitude, velocity, acceleration, force constant) — 4x (2083, 2081, 2080, 2078)
2. Rotational kinetic energy and conservation of angular momentum — 2x (2081, 2080)
3. Moment of inertia — 2x (2083, 2081)

1.1 Moment of Inertia

Definition

The moment of inertia of a body about an axis is the measure of how strongly the body resists a change in its rotational motion about that axis.

A body is harder to rotate when more of its mass lies far from the axis. The amount of mass matters, but its distance from the axis matters even more.

Detailed Explanation

In straight-line motion, mass measures resistance to acceleration. In rotational motion, the corresponding quantity is moment of inertia. It depends on both the total mass and the way that mass is distributed around the axis of rotation.

For a single particle of mass m located at a perpendicular distance r from the axis:

I = mr²

For a system containing many particles:

I = Σm_ir_i²

For a continuous body, the mass is divided into very small elements dm:

I = ∫r² dm

Here:

• I = moment of inertia about the chosen axis
• m = mass of a particle
• r = perpendicular distance of the particle from the axis
• dm = very small element of mass

The SI unit of moment of inertia is kg m².

Why Mass Distribution Matters

Because distance appears as r², moving a small amount of mass farther from the axis can increase the moment of inertia considerably.

For example, compare two equal point masses:

• At distance r: I = mr²
• At distance 2r: I = m(2r)² = 4mr²

Doubling the distance makes the moment of inertia four times larger.

Moment of Inertia of Common Bodies

Radius of Gyration

The radius of gyration is the distance from the axis at which the entire mass of a body could be imagined to be concentrated without changing its moment of inertia.

If the radius of gyration is written as k_g:

I = Mk_g²

Therefore:

k_g = √(I/M)

Parallel-Axis Theorem

The parallel-axis theorem relates the moment of inertia about an axis through the center of mass to the moment of inertia about another parallel axis.

I = I_CM + Md²

Here I_CM is the moment of inertia about a parallel axis through the center of mass and d is the separation between the two axes.

Perpendicular-Axis Theorem

For a flat body lying in the xy-plane, if three mutually perpendicular axes meet at the same point:

I_z = I_x + I_y

This theorem applies to a plane lamina, meaning a thin flat body.

Example: Effect of Position on Moment of Inertia

Two particles, each of mass 2 kg, are placed 0.5 m from an axis. Their combined moment of inertia is:

I = Σmr²

I = 2(2)(0.5)²

I = 1.0 kg m²

If both particles are moved to 1.0 m from the axis:

I = 2(2)(1.0)² = 4.0 kg m²

The masses have not changed, but the moment of inertia becomes four times larger.

1.2 Torque

Definition

Torque is the turning effect produced by a force acting at a distance from an axis of rotation.

A force does not produce the same rotation everywhere. Its turning ability increases when it acts farther from the axis and when it acts more nearly perpendicular to the position vector.

Detailed Explanation

A force can change straight-line motion, but when the force acts on an object that can rotate, it can also produce angular acceleration. The quantity that measures this rotational effect is torque.

Torque is defined by the vector product:

τ = r × F

Its magnitude is:

τ = rF sin θ

Here:

• τ = torque
• r = distance from the axis to the point where the force is applied
• F = applied force
• θ = angle between r and F

The SI unit of torque is newton metre (N m).

Moment Arm or Lever Arm

The lever arm is the perpendicular distance from the axis of rotation to the line of action of the force.

If the perpendicular distance is r_⊥:

τ = Fr_⊥

Since r_⊥ = r sin θ, this is equivalent to τ = rF sin θ.

Dependence of Torque on Angle

Direction of Torque

Torque is a vector. Its direction is determined by the right-hand rule. Curl the fingers of the right hand from r toward F. The thumb points in the direction of τ.

For planar problems, counterclockwise torque is commonly taken as positive and clockwise torque as negative.

Torque and Angular Acceleration

Torque builds directly on 1.1 Moment of Inertia. For a rigid body rotating about a fixed axis:

τ_net = Iα

Here α is angular acceleration.

This is the rotational counterpart of Newton's second law:

A greater net torque produces greater angular acceleration. A greater moment of inertia reduces angular acceleration for the same torque.

Example

A force of 20 N is applied perpendicular to a 0.30 m wrench.

τ = rF sin 90°

τ = 0.30 × 20 × 1

τ = 6 N m

If the same force acts at 30°:

τ = 0.30 × 20 × sin 30°

τ = 0.30 × 20 × 0.5

τ = 3 N m

The force has the same magnitude, but its turning effect is reduced because it is no longer perpendicular.

Torque and Angular Momentum

Torque also describes how angular momentum changes:

τ_ext = dL/dt

This relation becomes the basis of 1.4 Conservation of Angular Momentum. If the net external torque is zero, angular momentum does not change.

1.3 Rotational Kinetic Energy

Definition

Rotational kinetic energy is the energy possessed by a body because it is rotating.

Every small part of a rotating object is moving. The total kinetic energy of all those moving parts combines into one rotational-energy expression involving the body's moment of inertia and angular speed.

Detailed Explanation

Consider a rigid body rotating with angular speed ω. A particle of mass m_i at distance r_i from the axis has linear speed:

v_i = r_iω

Its kinetic energy is:

K_i = ½m_iv_i²

Substituting v_i = r_iω:

K_i = ½m_ir_i²ω²

The total kinetic energy is the sum over all particles:

K = Σ½m_ir_i²ω²

K = ½ω²Σm_ir_i²

From 1.1 Moment of Inertia:

I = Σm_ir_i²

Therefore:

K_rot = ½Iω²

This is the rotational counterpart of the translational expression K = ½mv².

Rolling Motion

A body that rolls without slipping has two types of motion at the same time:

• Translational motion: its center of mass moves with speed v_CM.
• Rotational motion: the body rotates about its center of mass with angular speed ω.

The total kinetic energy is:

K = ½Mv_CM² + ½I_CMω²

For rolling without slipping:

v_CM = Rω

Rolling Solid Sphere

For a solid sphere:

I = ⅖MR²

Since ω = v/R:

K = ½Mv² + ½(⅖MR²)(v²/R²)

K = ½Mv² + ⅕Mv²

K = ⁷⁄₁₀Mv²

Worked Example: Rolling Ball

A 0.3 kg solid ball of radius 0.1 m begins with speed 4 m/s. For a solid sphere:

I = ⅖mr²

For pure rolling:

ω = v/r = 4/0.1 = 40 rad/s

Translational kinetic energy:

K_trans = ½mv²

K_trans = ½(0.3)(4)² = 2.4 J

Moment of inertia:

I = ⅖(0.3)(0.1)² = 0.0012 kg m²

Rotational kinetic energy:

K_rot = ½Iω²

K_rot = ½(0.0012)(40)² = 0.96 J

Total initial kinetic energy:

K = 2.4 + 0.96 = 3.36 J

If the ball rolls upward without energy loss and eventually stops, its initial kinetic energy becomes gravitational potential energy:

mgh = 3.36

h = 3.36 / (0.3 × 9.8)

h ≈ 1.14 m

Data consistency note: the provided 2083 question states that the ball comes to rest at an elevation of 0.50 m and then asks how far up the slope it rolls. Under ideal rolling with the given initial speed, the calculated maximum vertical rise is approximately 1.14 m. A distance measured along the slope would also require the slope angle. The supplied wording therefore appears incomplete or inconsistent [verify].

Conservation of Mechanical Energy in Rotational Motion

Mechanical energy remains constant when only conservative forces do work and dissipative effects are absent.

For a rotating or rolling system:

Initial kinetic energy + Initial potential energy = Final kinetic energy + Final potential energy

Conservation is applicable when conditions such as the following hold:

• the pivot is frictionless or frictional energy loss is negligible;
• rolling occurs without slipping when the rolling relation v = Rω is used;
• air resistance is negligible;
• the rope or connecting element is ideal when a wheel-and-mass system is analyzed;
• no external non-conservative agent adds or removes mechanical energy.

Worked Example: Wheel and Falling Mass

A wheel has radius R = 0.4 m and moment of inertia I = 1.2 kg m². A 2 kg mass descends 1.5 m while unwinding a rope from the wheel. The pivot is frictionless.

The loss in gravitational potential energy becomes both translational kinetic energy of the falling mass and rotational kinetic energy of the wheel:

mgh = ½mv² + ½Iω²

Because the rope does not slip:

v = Rω

Therefore:

mgh = ½mR²ω² + ½Iω²

mgh = ½(I + mR²)ω²

Using m = 2 kg, h = 1.5 m, R = 0.4 m, I = 1.2 kg m², and g = 9.8 m/s²:

2 × 9.8 × 1.5 = ½[1.2 + 2(0.4)²]ω²

29.4 = ½(1.52)ω²

ω² = 58.8/1.52

ω ≈ 6.22 rad/s

Rotational velocity of the wheel ≈ 6.22 rad/s.

1.4 Conservation of Angular Momentum

Definition

The law of conservation of angular momentum states that the total angular momentum of a system remains constant when the net external torque acting on the system is zero.

If nothing outside a rotating system provides a net twisting effect, the system cannot change its total angular momentum. It may change how its mass is distributed and how fast it rotates, but the product that represents angular momentum remains constant.

Detailed Explanation

For a particle, angular momentum is defined as:

L = r × p

where p = mv is linear momentum.

For a rigid body rotating about a fixed principal axis:

L = Iω

Here:

• L = angular momentum
• I = moment of inertia
• ω = angular velocity

The SI unit of angular momentum is kg m²/s.

Connection Between Torque and Angular Momentum

From 1.2 Torque:

τ_ext = dL/dt

If the net external torque is zero:

τ_ext = 0

Therefore:

dL/dt = 0

Hence:

L = constant

For a rotating rigid system:

I₁ω₁ = I₂ω₂

This relation shows the inverse connection between moment of inertia and angular speed when angular momentum is conserved:

I decreases → ω increases

I increases → ω decreases

Conditions for Conservation of Angular Momentum

Angular momentum of a chosen system is conserved when:

• the net external torque about the selected axis is zero;
• internal forces may act, but their torques cancel within the complete system;
• the system boundary includes all interacting parts whose angular momenta are being considered.

Zero external force is not required. What matters specifically is zero net external torque.

Worked Example: Ice Skater

The skater's body has moment of inertia:

I_body = 4 kg m²

Each arm has mass 5 kg, and initially its center of mass is 0.4 m from the body axis.

Treating each arm's mass as concentrated at its center of mass:

I_arms = 2mr²

I_arms = 2(5)(0.4)²

I_arms = 1.6 kg m²

Initial total moment of inertia:

I₁ = 4 + 1.6 = 5.6 kg m²

Initial rotation rate:

ω₁ = 0.5 rev/s

When the arms are pulled to the axis, their contribution to moment of inertia is taken as zero in the stated model:

I₂ = 4 kg m²

Using angular momentum conservation:

I₁ω₁ = I₂ω₂

5.6(0.5) = 4ω₂

ω₂ = 2.8/4

ω₂ = 0.70 rev/s

The rotation rate increases from 0.50 rev/s to 0.70 rev/s.

Angular Momentum versus Rotational Kinetic Energy

Conservation of angular momentum does not automatically mean rotational kinetic energy is constant.

With L constant:

ω = L/I

Substituting into K = ½Iω²:

K = L²/(2I)

Therefore, if a skater reduces I while keeping L constant, rotational kinetic energy increases. The additional energy comes from work done by the skater's muscles while pulling the arms inward.

1.5 Oscillation of a Spring: Frequency, Period, Amplitude, Phase Angle and Energy

Definition

Oscillatory motion is repeated to-and-fro motion of a body about an equilibrium position. A mass attached to an ideal spring performs Simple Harmonic Motion (SHM) when the restoring force is directly proportional to displacement and directed toward equilibrium.

A displaced spring tries to pull or push the attached mass back toward its central position. The mass passes through the center because of its inertia, stretches or compresses the spring on the opposite side, and the cycle repeats.

Detailed Explanation

This topic has several connected quantities, so the motion is developed from the force law first and then used to obtain displacement, frequency, velocity, acceleration, phase, and energy.

Oscillatory Motion and Simple Harmonic Motion

All Simple Harmonic Motion is oscillatory, but not every oscillation is Simple Harmonic Motion.

Mass-Spring System and Hooke's Law

Consider a mass m attached to a spring on a frictionless horizontal surface. Let x represent displacement from the equilibrium position.

For an ideal spring, Hooke's law gives:

F = −kx

Here:

• F = restoring force
• k = spring constant or force constant, measured in N/m
• x = displacement from equilibrium
• the negative sign means the force is opposite to displacement.

If the mass is displaced to the right, x is positive and the spring force acts left. If the mass is displaced to the left, x is negative and the restoring force acts right.

Differential Equation of a Mass-Spring System

Newton's second law gives:

F = ma

Since a = d²x/dt² and F = −kx:

m(d²x/dt²) = −kx

Rearranging:

m(d²x/dt²) + kx = 0

Dividing by m:

d²x/dt² + (k/m)x = 0

Define the angular frequency ω by:

ω² = k/m

Therefore:

d²x/dt² + ω²x = 0

This is the standard differential equation of Simple Harmonic Motion.

Solution of the Differential Equation

A general solution is:

x(t) = A cos(ωt + φ)

An equivalent sine form can also be used:

x(t) = A sin(ωt + φ')

where:

• A = amplitude
• ω = angular frequency
• t = time
• φ = phase angle or initial phase

The sine and cosine forms describe the same physical motion. The value of the phase constant changes according to the initial position and velocity.

Amplitude

Amplitude is the maximum magnitude of displacement of the oscillating body from its equilibrium position.

If the endpoints are x = +A and x = −A, the total distance between the two extreme positions is:

Distance between endpoints = 2A

Therefore:

A = half the distance between endpoints

Angular Frequency

For a spring-mass system:

ω² = k/m

Therefore:

ω = √(k/m)

Angular frequency is measured in radians per second (rad/s).

Frequency

Frequency is the number of complete oscillations performed per second.

It is represented by f and measured in hertz (Hz).

Angular frequency and ordinary frequency are related by:

ω = 2πf

Therefore:

f = ω/(2π)

For a mass-spring system:

f = (1/2π)√(k/m)

This expression shows:

• larger spring constant k → larger frequency;
• larger mass m → smaller frequency.

Period

Period is the time required for one complete oscillation.

It is represented by T and measured in seconds.

Frequency and period are reciprocals:

T = 1/f

Since ω = 2πf:

T = 2π/ω

For the mass-spring system:

T = 2π√(m/k)

Time Between Opposite Endpoints

Motion from one extreme position to the opposite extreme covers half an oscillation:

Time from +A to −A = T/2

Motion from an extreme position to equilibrium covers one quarter of an oscillation:

Time from ±A to 0 = T/4

Phase and Phase Angle

The phase describes the state of an oscillator within its cycle. For:

x = A cos(ωt + φ)

the total phase at time t is:

θ = ωt + φ

The constant φ is the phase angle or initial phase. It specifies the state of the oscillator at t = 0.

At t = 0:

x(0) = A cos φ

The phase angle is normally expressed in radians.

Velocity in Simple Harmonic Motion

Starting from:

x = A cos(ωt + φ)

Differentiate with respect to time:

v = dx/dt

v = −Aω sin(ωt + φ)

The maximum speed occurs when sin(ωt + φ) = ±1:

v_max = Aω

Velocity can also be related directly to displacement. Using sin²θ + cos²θ = 1:

v² = ω²(A² − x²)

Therefore:

v = ±ω√(A² − x²)

The sign depends on the direction of motion.

Acceleration in Simple Harmonic Motion

Differentiate velocity with respect to time:

a = dv/dt

a = −Aω² cos(ωt + φ)

Because x = A cos(ωt + φ):

a = −ω²x

Since ω² = k/m:

a = −(k/m)x

The maximum magnitude of acceleration occurs at the endpoints:

a_max = Aω²

Restoring Force

Since F = ma:

F = m(−ω²x)

F = −mω²x

For a spring, mω² = k, so:

F = −kx

The force is zero at equilibrium and has maximum magnitude at the extreme positions.

Phase Relationships Among Displacement, Velocity and Acceleration

Motion at Important Positions

Energy of a Mass-Spring Oscillator

An ideal mass-spring oscillator continuously exchanges kinetic energy and elastic potential energy.

Elastic Potential Energy

For a spring displaced by x:

U = ½kx²

At an extreme position x = ±A:

U_max = ½kA²

Kinetic Energy

Kinetic energy is:

K = ½mv²

Using:

v² = ω²(A² − x²)

and mω² = k:

K = ½k(A² − x²)

Total Mechanical Energy

Total energy is:

E = K + U

E = ½k(A² − x²) + ½kx²

Therefore:

E = ½kA²

For an ideal frictionless oscillator, this total remains constant.

Worked Example: Differential Equation and Solution

For a block of mass m attached to a spring of constant k:

Restoring force:

F = −kx

Newton's second law:

m(d²x/dt²) = −kx

Therefore:

d²x/dt² + (k/m)x = 0

Taking:

ω = √(k/m)

the equation becomes:

d²x/dt² + ω²x = 0

Its general solution is:

x = A cos(ωt + φ)

where A and φ are fixed by the initial position and initial velocity.

Worked Example: Frequency, Amplitude and Spring Constant

An oscillating block has mass 250 g = 0.250 kg. It takes 0.2 s to move from one endpoint to the other. The endpoints are 50 cm apart.

Endpoint to opposite endpoint takes half a period:

T/2 = 0.2 s

T = 0.4 s

Frequency:

f = 1/T = 1/0.4

f = 2.5 Hz

The distance between endpoints equals 2A:

2A = 0.50 m

A = 0.25 m

For a mass-spring system:

ω = 2πf

ω = 2π(2.5) = 5π rad/s

Using:

k = mω²

k = 0.250(5π)²

k = 6.25π²

k ≈ 61.7 N/m

Worked Example: Force Constant, Frequency, Velocity and Acceleration

A spring stretches 10⁻⁴ m under a force of 20 N.

From Hooke's law:

k = F/x

k = 20/(10⁻⁴)

k = 2.0 × 10⁵ N/m

A 2 kg block is attached to the spring, displaced 0.2 m, and released. Therefore:

A = 0.20 m

Angular frequency:

ω = √(k/m)

ω = √[(2.0 × 10⁵)/2]

ω = √(10⁵)

ω ≈ 316.23 rad/s

Frequency:

f = ω/(2π)

f = 316.23/(2π)

f ≈ 50.3 Hz

At x = 0.12 m:

v = ±ω√(A² − x²)

v = ±316.23√[(0.20)² − (0.12)²]

v = ±316.23√(0.0256)

v = ±316.23(0.16)

|v| ≈ 50.6 m/s

The block was released from the positive side and is passing x = +0.12 m for the first time while moving toward equilibrium, so velocity is negative:

v ≈ −50.6 m/s

Acceleration:

a = −ω²x

a = −(10⁵)(0.12)

a = −1.2 × 10⁴ m/s²

Dependence of Oscillation on Mass and Spring Constant

Since:

f = (1/2π)√(k/m)

and:

T = 2π√(m/k)

For ideal Simple Harmonic Motion, frequency and period do not depend on amplitude.

Extra: Topics Asked in Exams But Not in Syllabus

These topics have appeared in past papers. They are not listed in the syllabus but you should know them.

Rigid and Non-Rigid Body (2080 exam)

A rigid body is an ideal body in which the distance between any two particles remains constant even when forces act on the body. Its shape and size therefore remain unchanged during the motion being considered.

A non-rigid body is a body whose shape, size, or internal distances can change appreciably when forces act on it.

Perfectly rigid bodies do not exist in nature because every real material deforms to some extent. The rigid-body model is used when the deformation is small enough to be neglected.

Exam Questions

1. A ball of mass 0.3 kg and radius 0.1 m rolls along a ground with a speed of 4 m/s. It comes to a rest at an elevation of 0.50 m. How far up the slope does it roll? It is given that the moment of inertia of the ball is I = 2mr²/5. (2083)
2. Develop differential equation for the oscillation of a mass-spring system and write its solution. (2081)
3. Suppose the body of an ice skater has a moment of inertia 4 kgm² and her arms have a mass of 5 kg each with the center of mass at 0.4 m from her body. She starts to turn at 0.5 rev/s on the point of her skate with her arms outstretched. She then pulls her arms inward so that their center of mass is at the axis of her body. What will be her speed of rotation? (2081)
4. Distinguish rigid and non-rigid body. Derive an expression for rotational kinetic energy and discuss the conditions for conservation of energy. A wheel of radius 0.4 m and moment of inertia 1.2 kg-m², pivoted at the center, is free to rotate without friction. A rope is wound around it and a 2-kg weight is attached to the rope. When the weight has descended 1.5 m from its starting position, find the rotational velocity of the wheel. (2080)
5. An oscillating block of mass 250 g takes 0.2 sec to move between the endpoints of the motion, which are 50 cm apart. Find the frequency and amplitude of the motion. What is the force constant of the spring? (2080)
6. A given spring stretches 10⁻⁴ m when a force of 20 N pulls on it. A 2-kg block attached to it on a frictionless surface is pulled to the right 0.2 m and released. (a) What is frequency of oscillation of the block? (b) What are the velocity and acceleration when x = 0.12 m, on the block's first passing this point? (2078)

Whole Chapter Summary

Moment of Inertia

Moment of inertia is the rotational equivalent of mass. It measures resistance to angular acceleration and depends strongly on how far the body's mass lies from the chosen rotation axis, following I = Σmr².

Torque

Torque is the turning effect of force and is given by τ = rF sin θ. It produces angular acceleration according to τ_net = Iα and changes angular momentum according to τ_ext = dL/dt.

Rotational Kinetic Energy

A rotating body has kinetic energy K_rot = ½Iω². A rolling body simultaneously has translational and rotational energy, so both terms must be included in its total kinetic energy.

Conservation of Angular Momentum

When net external torque is zero, angular momentum remains constant. For a fixed-axis rotating system, I₁ω₁ = I₂ω₂, so decreasing moment of inertia increases angular speed.

Oscillation of a Spring

An ideal spring-mass system performs Simple Harmonic Motion because F = −kx. Its motion is described by x = A cos(ωt + φ), with frequency and period determined by mass and spring constant, while kinetic and potential energies continuously exchange at constant total energy.

Key Formulas and Points Sheet

Last-Minute Revision Sheet