DP Physics HL A.4 Rigid Body Mechanics

DP Physics HL A.4 Rigid Body Mechanics - Full notes

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DP Physics HL | A.4 Rigid Body Mechanics

DP PHYSICS · THEME A: SPACE, TIME AND MOTION

A.4

Rigid Body Mechanics

Full notes for HL · IB-style

Contents

1	Torque and equilibrium
2	Rotational kinematics
3	Rotational dynamics
4	Angular momentum
5	Rotational energy and rolling
6	Exam technique and mixed problems
7	Further practice
8	Investigating rotation
9	Common misconceptions and links
10	Exam-style question with markscheme
11	Check your understanding
12	Summary and topic map

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and do every Quick check before moving to the next section.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
calculate torque and analyse couples
apply conditions for translational and rotational equilibrium, including ladders and stability
use the equations of rotational kinematics and interpret rotational graphs
calculate moments of inertia and apply τ = Iα
use conservation of angular momentum and angular impulse
calculate rotational kinetic energy, including rolling without slipping

A.4 is an Additional Higher Level (AHL) subtopic. It is studied by HL students only.

Guiding questions

How can the understanding of linear motion be applied to rotational motion? How is the understanding of the forces and torques acting on a body used to predict its motion? How are the laws of conservation applied to rotating systems?

How this topic is examined

A.4 is an HL-only subtopic tested in Papers 1 and 2. Expect equilibrium (moments) problems, rotational kinematics, τ = Iα, angular momentum conservation and rolling energy problems.

Moments of inertia for shapes are given in questions when needed; learn the analogy table thoroughly.

Command term	What the examiner wants
State	a brief answer with no explanation
Outline	a brief account or summary
Describe	a detailed account without reasons
Explain	a detailed account including causes, reasons or mechanisms
Calculate	a numerical answer with working and correct units
Determine	obtain the only possible answer, often from a graph or data
Show that	obtain the given result, showing every step (give one more significant figure)
Estimate	an approximate value using reasonable assumptions
Sketch	a graph or diagram showing the general shape and key features
Deduce / Suggest	reach a conclusion from given information; propose a reasonable explanation

How to use these notes

Each section teaches one part of the syllabus in full. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section
Key point	the one idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
In the real world	where the physics shows up outside the classroom
Try this	something quick you can do yourself to test the idea
Worked example	full solutions with IB-style marks (M1 method, A1 answer, R1 reasoning)
Quick check	short questions to test yourself at the end of each section

1 Torque and equilibrium

The turning effect of forces.

Think about it

Why is it easier to open a door by pushing at the handle than near the hinges?

A rigid body is an extended object that keeps its shape. Unlike a point particle it can rotate as well as move. This section deals with the turning effect of forces and with bodies that are perfectly balanced.

1.1 Rigid bodies

A rigid body is an object whose shape and size do not change under the forces acting on it. Its motion can be described as the translation of its centre of mass plus rotation about an axis. All the points on a rotating rigid body share the same angular displacement, angular velocity and angular acceleration.

Quick check

1. Give two examples of rigid-body rotation.

Answers: 1. a spinning wheel; a door swinging on its hinges

1.2 Torque

The torque (moment) of a force about an axis is τ = Fr sin θ, where r is the distance from the axis to the point where the force acts and θ is the angle between F and r. Equivalently, τ = F × perpendicular distance from the axis to the line of action. Torque is measured in N m and is greatest when the force is perpendicular to r.

Figure 1. A force F acting at distance r from a pivot, at angle θ to the lever.

WORKED EXAMPLE 1 Spanner (standard)

A force of 40 N is applied at the end of a 0.25 m spanner at 60° to the handle. Calculate the torque.

Solution

τ = 40 × 0.25 × sin 60°M1

= 8.7 N mA1

WORKED EXAMPLE 2 Door (standard)

A door handle is 0.80 m from the hinges. Find the minimum force to produce a torque of 12 N m.

Solution

Minimum force when perpendicular: F = 12/0.80M1

= 15 NA1

Quick check

1. How could the torque be increased without increasing the force?

Answers: 1. use a longer lever or push perpendicular to it

1.3 Several torques on one body

When several forces act, find each torque with its sign (clockwise or anticlockwise) and add them. The resultant torque determines the angular acceleration; if it is zero the body is in rotational equilibrium.

WORKED EXAMPLE 3 Wheel with three forces (exam level)

A wheel of radius 0.50 m has tangential forces of 30 N (anticlockwise) and 20 N (clockwise) at its rim, and 40 N (clockwise) at 0.20 m from the axis. Find the resultant torque.

Solution

Anticlockwise: 30 × 0.50 = 15 N m; clockwise: 20 × 0.50 + 40 × 0.20 = 18 N mM1

Resultant = 3.0 N m clockwiseA1

1.4 Couples

A couple is a pair of equal, opposite, parallel forces whose lines of action do not coincide. A couple produces a torque Fd (d = perpendicular separation) but no resultant force, so it causes rotation without translation. The torque of a couple is the same about any axis.

Figure 2. A couple: two equal and opposite forces separated by distance d.

WORKED EXAMPLE 4 Steering wheel (standard)

A driver applies 20 N in opposite directions on either side of a steering wheel of diameter 0.40 m. Calculate the torque.

Solution

τ = Fd = 20 × 0.40M1

= 8.0 N mA1

1.5 Rotational equilibrium

A rigid body is in equilibrium when the resultant force is zero (translational equilibrium) and the resultant torque about any axis is zero (rotational equilibrium). Taking moments about the point where an unknown force acts removes that force from the equation.

Figure 3. A seesaw balanced by two children.

Figure 4. A uniform beam on two supports carrying a load.

Key point

In equilibrium, the resultant torque is zero about every point, so choose the point that eliminates the most unknowns.

WORKED EXAMPLE 5 Seesaw (standard)

A 30 kg child sits 1.5 m from the pivot. Where must a 45 kg child sit to balance?

Solution

30 × 1.5 = 45dM1

d = 1.0 m on the other sideA1

WORKED EXAMPLE 6 Beam (exam level)

For the beam shown, find the support forces.

Solution

Moments about the left support: Rᵣ × 4.0 = 500 × 1.0 + 200 × 2.0M1

Rᵣ = 225 N; Rₗ = 700 − 225 = 475 NA1 A1

1.6 Levers in the human body

Muscles, bones and joints act as levers. The biceps attaches very close to the elbow, so it must exert a much larger force than the load in the hand; the trade-off is that the hand moves faster and further.

Figure 5. The forearm as a lever: the biceps pulls up close to the elbow.

WORKED EXAMPLE 7 Biceps force (exam level)

Using the figure, find the force in the biceps needed to hold the forearm horizontal.

Solution

Moments about the elbow: F(0.04) = 15(0.15) + 50(0.33)M1

F = 18.75/0.04 = 469 NA1

Quick check

1. Why must the biceps force be so much larger than the load?

Answers: 1. its moment arm (0.04 m) is much smaller than the load's (0.33 m)

1.7 Cranes and counterweights

Tower cranes use a counterweight on the short arm so that the torques about the tower are nearly balanced, preventing the crane from toppling.

Figure 6. A tower crane with counterweight C.

WORKED EXAMPLE 8 Counterweight (exam level)

A crane lifts a 2000 kg load 10 m from the tower. Find the counterweight mass needed 6.0 m from the tower for balance.

Solution

M(6.0) = 2000(10)M1

M = 3.3 × 10³ kgA1

1.8 Finding the centre of mass

For a uniform symmetrical object, the centre of mass is at the geometric centre. For an irregular flat shape (lamina), hang it from two different points with a plumb line; the centre of mass lies where the two vertical lines cross.

Try this

Cut an irregular shape from card, hang it from a pin at two points and draw the plumb line each time. Check that the card balances on a pencil point at the crossing.

Quick check

1. Why does the plumb-line method work?

Answers: 1. a freely hanging body comes to rest with its centre of mass directly below the pivot

1.9 Stability of vehicles

A vehicle on a sideways slope tips over when the vertical line through its centre of mass passes outside the wheels. For track width w and centre of mass height h, the critical angle satisfies tan θ = (w/2)/h. Lowering h or widening w increases stability.

WORKED EXAMPLE 9 Tipping angle (exam level)

A van has track width 1.6 m and centre of mass 1.2 m above the road. Find the sideways slope at which it tips.

Solution

tan θ = 0.8/1.2 = 0.667M1

θ = 34°A1

Quick check

1. How does loading heavy goods on a van roof affect stability?

Answers: 1. raises the centre of mass, reducing the tipping angle

1.10 Ladders and walls

A ladder leaning against a smooth wall is a classic three-force (or four-force) problem: weight, normal reaction from the wall (horizontal), and the normal reaction and friction at the ground. Take moments about the foot of the ladder.

Figure 7. Forces on a uniform ladder resting against a smooth wall.

WORKED EXAMPLE 10 Ladder (exam level)

A uniform 6.0 m ladder of weight 300 N leans against a smooth wall at 60° to the ground. Find the reaction from the wall and the friction at the ground.

Solution

Moments about the foot: R(6.0 sin 60°) = 300(3.0 cos 60°)M1

R = 86.6 NA1

Horizontal equilibrium: friction = 86.6 N towards the wallA1

1.11 Centre of mass and stability

The weight of a body acts through its centre of mass. A body resting on a base topples if the vertical line through its centre of mass falls outside the base. A low centre of mass and a wide base increase stability. For a uniform block of width w and height h on a tilting surface, toppling begins when tan θ = w/h (if it does not slide first).

Figure 8. A block is stable while the vertical line through its centre of mass stays inside its base.

WORKED EXAMPLE 11 Tilting block (exam level)

A uniform block 0.20 m wide and 0.50 m tall stands on a board that is slowly tilted. At what angle does it topple?

Solution

tan θ = 0.20/0.50 = 0.40M1

θ = 21.8°A1

In the real world

Double-decker buses are tilt-tested to about 28° before they are allowed on the road; their heavy engines and chassis keep the centre of mass low.

2 Rotational kinematics

Describing rotation.

2.1 Angular displacement, velocity and acceleration

Angular displacement θ is measured in radians. Angular velocity ω = Δθ/Δt (rad s⁻¹); angular acceleration α = Δω/Δt (rad s⁻²). A point at radius r has arc length s = rθ, speed v = rω and tangential acceleration a = rα.

WORKED EXAMPLE 12 rpm (standard)

A wheel turns at 1200 rpm. Find ω and the speed of a point 0.15 m from the axis.

Solution

ω = 2π × 1200/60 = 126 rad s⁻¹M1 A1

v = 126 × 0.15 = 18.8 m s⁻¹A1

Common mistake

Leaving ω in revolutions per minute. Always convert to rad s⁻¹ before using the equations.

2.2 Equations of uniform angular acceleration

For constant α, the rotational equivalents of the suvat equations apply, with θ, ωi, ωf, α and t.

WORKED EXAMPLE 13 Flywheel (exam level)

A flywheel accelerates uniformly from rest to 60 rad s⁻¹ in 12 s. Find α and the number of revolutions.

Solution

α = 60/12 = 5.0 rad s⁻²A1

θ = ½(60)(12) = 360 rad = 57 revolutionsM1 A1

WORKED EXAMPLE 14 Slowing wheel (exam level)

A wheel slows uniformly from 20 rad s⁻¹ to 8.0 rad s⁻¹ in 4.0 s. Find the angular deceleration and angle turned.

Solution

α = (8.0 − 20)/4.0 = −3.0 rad s⁻²A1

θ = ½(20 + 8.0)(4.0) = 56 radA1

2.3 Rotation of the Earth

Every point on Earth's surface rotates with the same angular velocity, ω = 2π/86 400 s = 7.27 × 10⁻⁵ rad s⁻¹, but the linear speed depends on the distance from the axis.

WORKED EXAMPLE 15 Speeds (standard)

Find the speed of a point on the equator (R = 6.37 × 10⁶ m) and at latitude 60°.

Solution

Equator: v = ωR = 463 m s⁻¹A1

60°: r = R cos 60°: v = 232 m s⁻¹A1

2.4 Angular displacement–time graphs

On a θ–t graph, a straight line means constant ω; a curve that gets steeper means angular acceleration.

Figure 9. Angular displacement of the same motor: a parabola while accelerating, then a straight line.

Quick check

1. What does the gradient of the straight part represent?

Answers: 1. the constant angular velocity, 30 rad s⁻¹

2.5 Graphs of rotational motion

As for linear motion: the gradient of a θ–t graph is ω; the gradient of an ω–t graph is α; the area under an ω–t graph is the angular displacement.

Figure 10. Angular velocity against time for a motor starting up.

WORKED EXAMPLE 16 Area (standard)

Find the angular displacement in the first 6.0 s.

Solution

½(3)(30) + 3(30)M1

= 135 radA1

3 Rotational dynamics

Torque causes angular acceleration.

Think about it

Why is it harder to start a heavy bicycle wheel spinning if its mass is at the rim rather than near the hub?

3.1 Moment of inertia

The moment of inertia I = Σmr² measures a body's resistance to angular acceleration (its rotational inertia). It depends on the total mass and on how far that mass is from the axis: moving mass further out increases I sharply.

Figure 11. Moments of inertia about a central axis for common shapes (given in questions when needed).

WORKED EXAMPLE 17 Point masses (standard)

Two 2.0 kg masses are fixed at the ends of a light 1.0 m rod rotating about its centre. Find I.

Solution

I = 2 × 2.0 × 0.50²M1

= 1.0 kg m²A1

WORKED EXAMPLE 18 Adding mass (exam level)

A 0.020 kg m² turntable has a 0.10 kg lump of clay placed 0.12 m from the axis. Find the new I.

Solution

I = 0.020 + 0.10 × 0.12²M1

= 0.0214 kg m²A1

3.2 Newton's second law for rotation

A resultant torque produces angular acceleration: τ = Iα. This is the rotational analogue of F = ma.

WORKED EXAMPLE 19 Disc (exam level)

A 4.0 kg solid disc of radius 0.30 m is acted on by a tangential force of 12 N at its rim. Find α.

Solution

I = ½(4.0)(0.30)² = 0.18 kg m²; τ = 12 × 0.30 = 3.6 N mM1

α = 3.6/0.18 = 20 rad s⁻²A1

WORKED EXAMPLE 20 Braking torque (exam level)

A disc (I = 0.050 kg m²) spinning at 40 rad s⁻¹ is stopped by a constant frictional torque in 8.0 s. Find the torque.

Solution

α = −40/8.0 = −5.0 rad s⁻²M1

τ = 0.050 × 5.0 = 0.25 N mA1

3.3 Falling mass and pulley

When a mass hangs from a string wrapped around a pulley that has a moment of inertia, apply F = ma to the mass, τ = Iα to the pulley, and link them with a = rα.

Figure 12. A mass hanging from a string wound around a pulley with moment of inertia I.

WORKED EXAMPLE 21 Pulley (exam level)

A 0.50 kg mass hangs from a string wound on a pulley (I = 0.020 kg m², r = 0.10 m). Find the acceleration of the mass and the tension.

Solution

mg − T = ma; Tr = I(a/r)M1

a = mg/(m + I/r²) = 4.905/2.5 = 1.96 m s⁻²M1 A1

T = I a/r² = 3.9 NA1

Common mistake

Taking T = mg. The mass accelerates, so the tension is less than its weight.

3.4 Composite bodies

Moments of inertia about the same axis add. A wheel can be modelled as a ring (rim) plus rods (spokes); a dumbbell as two point masses plus a rod.

WORKED EXAMPLE 22 Dumbbell (exam level)

Two 1.5 kg spheres (treated as points) are fixed to the ends of a 0.60 m uniform rod of mass 0.40 kg. Find I about the rod's centre.

Solution

Spheres: 2 × 1.5 × 0.30² = 0.27 kg m²M1

Rod: 0.40 × 0.60²/12 = 0.012; total 0.28 kg m²A1

3.5 Investigating τ = Iα

Apply different known torques to a wheel (using hanging masses on a string of known radius) and measure α. A graph of τ against α is a straight line with gradient I; a small intercept on the τ-axis reveals friction at the axle.

Figure 13. Torque against angular acceleration for a wheel. Gradient = I; intercept = frictional torque.

WORKED EXAMPLE 23 Reading the graph (exam level)

Use the graph to find I and the frictional torque.

Solution

Gradient = 0.050/10 = 5.0 × 10⁻³ kg m²M1 A1

Intercept = 0.010 N m frictional torqueA1

3.6 Linear and rotational analogues

Every linear quantity has a rotational partner. Learning this table lets you transfer methods from linear mechanics directly.

Linear	Rotational
displacement s	angular displacement θ
velocity v	angular velocity ω
acceleration a	angular acceleration α
mass m	moment of inertia I
force F	torque τ
F = ma	τ = Iα
momentum p = mv	angular momentum L = Iω
impulse FΔt	angular impulse τΔt
½mv²	½Iω²
P = Fv	P = τω

4 Angular momentum

Spinning things keep spinning.

4.1 Angular momentum and its conservation

Angular momentum L = Iω (kg m² s⁻¹). It is conserved when no resultant external torque acts. If the moment of inertia decreases, the angular velocity increases so that Iω stays constant.

Figure 14. A spinning skater pulls in her arms: I decreases and ω increases.

WORKED EXAMPLE 24 Skater (exam level)

A skater spins at 2.0 rev s⁻¹ with I = 3.6 kg m², then pulls in her arms so I = 1.2 kg m². Find the new rotation rate and comment on her kinetic energy.

Solution

I₁ω₁ = I₂ω₂: ω₂ = 3 × 2.0 = 6.0 rev s⁻¹M1 A1

Ek = L²/2I triples: she does work pulling her arms inR1

In the real world

Divers and gymnasts tuck their bodies to reduce their moment of inertia and spin faster, then extend to slow down before entering the water or landing.

4.2 Angular impulse

A torque acting for a time changes angular momentum: τΔt = ΔL. On a torque–time graph, the area is the angular impulse.

WORKED EXAMPLE 25 Starting a wheel (exam level)

A torque of 2.0 N m acts for 3.0 s on a wheel (I = 0.15 kg m²) initially at rest. Find its final angular velocity.

Solution

ΔL = 2.0 × 3.0 = 6.0 kg m² s⁻¹M1

ω = 6.0/0.15 = 40 rad s⁻¹A1

4.3 Angular momentum–time and torque–time graphs

The gradient of an angular momentum–time graph is the resultant torque. The area under a torque–time graph is the angular impulse, equal to the change in angular momentum.

Figure 15. Torque applied to a turntable; the shaded area is the angular impulse.

WORKED EXAMPLE 26 From the graph (exam level)

The turntable (I = 0.15 kg m²) starts at rest. Find its final angular velocity.

Solution

Area = ½(1)(2) + 2(2) + ½(1)(2) = 6.0 N m sM1

ω = 6.0/0.15 = 40 rad s⁻¹A1

4.4 Collisions with rotating bodies

When a small object lands on a rotating platform, angular momentum is conserved (no external torque), but kinetic energy usually decreases, as in a perfectly inelastic collision.

WORKED EXAMPLE 27 Merry-go-round (exam level)

A merry-go-round (I = 500 kg m²) rotates at 1.0 rad s⁻¹. A 50 kg child jumps on at 2.0 m from the axis. Find the new angular velocity and the kinetic energy lost.

Solution

Inew = 500 + 50 × 4.0 = 700 kg m²; ω = 500/700 = 0.71 rad s⁻¹M1 A1

Ek: 250 J → 179 J: 71 J lostA1

4.5 Spinning tops and gyroscopes

Angular momentum is a vector along the axis of rotation. A rapidly spinning object resists changes to the direction of its axis, which is why spinning tops stay upright, bicycles are easier to balance at speed, and gyroscopes are used for navigation and in phones.

Quick check

1. Why is a moving bicycle easier to balance than a stationary one?

Answers: 1. the spinning wheels have angular momentum that resists tilting

4.6 Angular momentum in astronomy

A collapsing star or gas cloud conserves angular momentum: as its radius (and I) falls enormously, its rotation speeds up. Neutron stars (pulsars) can rotate hundreds of times per second. Kepler's second law is also a consequence of conservation of angular momentum.

Quick check

1. Why do collapsing gas clouds spin faster?

Answers: 1. I decreases, so ω increases to conserve L

5 Rotational energy and rolling

Energy stored in spinning.

5.1 Rotational kinetic energy

A rotating body stores kinetic energy Ek = ½Iω² = L²/2I. Flywheels use this to store energy.

WORKED EXAMPLE 28 Flywheel (standard)

A flywheel of I = 0.80 kg m² spins at 300 rad s⁻¹. Calculate its kinetic energy.

Solution

½(0.80)(300)²M1

= 3.6 × 10⁴ JA1

In the real world

Formula 1 cars have used flywheel and battery systems to recover braking energy; some city buses use flywheels to store energy between stops.

5.2 Gears and torque transmission

Meshing gears have the same speed at their rims, so ω₁r₁ = ω₂r₂. A smaller driven gear turns faster but with less torque; ideally the power τω is the same on both shafts. Bicycle gears and car gearboxes use this trade-off between torque and angular velocity.

Figure 16. Two meshing gears: the smaller gear turns twice as fast.

WORKED EXAMPLE 29 Gear ratio (exam level)

The driver gear (40 teeth) turns at 10 rad s⁻¹ with torque 6.0 N m. Find the angular velocity and torque of the 20-tooth driven gear (no losses).

Solution

ω = 10 × 40/20 = 20 rad s⁻¹A1

Power 60 W conserved: τ = 60/20 = 3.0 N mM1 A1

5.3 Flywheel energy storage

Because Ek = ½Iω², doubling the angular velocity quadruples the stored energy. Flywheels store energy in grid systems, buses and spacecraft; their maximum speed is limited by the strength of the material.

Figure 17. Energy stored in a flywheel (I = 0.80 kg m²) against angular velocity.

WORKED EXAMPLE 30 Energy released (exam level)

The flywheel slows from 300 rad s⁻¹ to 200 rad s⁻¹ in 5.0 s. Find the average power delivered.

Solution

ΔE = ½(0.80)(300² − 200²) = 2.0 × 10⁴ JM1

P = 4.0 kWA1

5.4 Work and power in rotation

The work done by a torque is W = τθ and the power is P = τω, analogous to W = Fs and P = Fv.

WORKED EXAMPLE 31 Engine (standard)

An engine delivers a torque of 250 N m at 3000 rpm. Calculate the power.

Solution

ω = 2π × 50 = 314 rad s⁻¹M1

P = 250 × 314 = 78.5 kWA1

5.5 Rolling without slipping

A rolling object has both translational and rotational kinetic energy, linked by v = rω. Objects with larger I/mr² roll down slopes more slowly because more of their energy goes into rotation.

Figure 18. Speed after descending a slope for objects with different moments of inertia.

WORKED EXAMPLE 32 Solid cylinder (exam level)

A solid cylinder rolls from rest down a slope of vertical height 1.2 m. Find its speed at the bottom.

Solution

mgh = ½mv² + ½(½mr²)(v/r)² = ¾mv²M1

v = √(4gh/3) = 3.96 m s⁻¹A1

Quick check

1. Which reaches the bottom of a slope first: a hoop or a solid sphere?

Answers: 1. the solid sphere (smaller I/mr²)

5.6 Rolling down an incline: acceleration

For an object rolling without slipping down a slope of angle θ, friction at the contact point provides the torque. Combining F = ma and τ = Iα gives a = g sin θ/(1 + I/mr²). Friction does no work when there is no slipping, so mechanical energy is conserved.

Object	I/mr²	a (θ = 20°) / m s⁻²
sliding block (no friction)	0	3.36
solid sphere	0.40	2.40
solid cylinder	0.50	2.24
hollow cylinder (hoop)	1.0	1.68

Figure 19. A cylinder rolling down a slope: friction f at the contact point provides the torque.

WORKED EXAMPLE 33 Cylinder (exam level)

Show that a solid cylinder rolling down a 20° slope has acceleration 2.24 m s⁻².

Solution

a = g sin 20°/(1 + 0.5)M1

= 3.36/1.5 = 2.24 m s⁻²AG

5.7 Linear and rotational energy compared

For a rolling object, the fraction of kinetic energy that is rotational is (I/mr²)/(1 + I/mr²). For a hoop half the kinetic energy is rotational; for a solid sphere only 2/7.

WORKED EXAMPLE 34 Energy split (exam level)

A 2.0 kg solid sphere rolls at 3.0 m s⁻¹. Find its translational, rotational and total kinetic energy.

Solution

Translational: ½(2.0)(3.0)² = 9.0 JA1

Rotational: ½(⅖mr²)(v/r)² = ⅕mv² = 3.6 J; total 12.6 JA1

5.8 Energy in the falling-mass pulley

Energy methods give the speed of a falling mass on a pulley directly: mgh = ½mv² + ½Iω², with ω = v/r.

WORKED EXAMPLE 35 Pulley energy (exam level)

A 0.40 kg mass falls 1.0 m from rest from a string wound on a cylinder (I = 0.010 kg m², r = 0.10 m). Find its speed.

Solution

3.92 = ½(0.40)v² + ½(0.010)(v/0.10)² = 0.70v²M1

v = 2.37 m s⁻¹A1

6 Exam technique and mixed problems

Bringing linear and rotational ideas together.

6.1 Choosing a method

For rotation, the same choices apply as in linear mechanics: torque and τ = Iα (with rotational suvat) when time is involved; energy (including ½Iω²) for speeds after a height change; angular momentum when parts move closer or further from the axis or objects land on rotating bodies.

Key point

Always state which quantity is conserved and why (no external torque, no friction).

6.2 Multi-step problem: wheel and string

Tension, torque and energy can all be tested in one question.

WORKED EXAMPLE 36 Yo-yo (exam level)

A yo-yo of mass 0.10 kg and I = 2.0 × 10⁻⁵ kg m² unwinds from a string around an axle of radius 0.010 m. Find (a) its linear acceleration; (b) its speed after falling 0.80 m.

Solution

(a) mg − T = ma; Tr = Ia/r: a = g/(1 + I/mr²) = 9.81/3.0 = 3.3 m s⁻²M1 A1

(b) v = √(2 × 3.27 × 0.80) = 2.3 m s⁻¹A1

6.3 Multi-step problem: rotating platform

Angular momentum and energy combine in collisions with rotating bodies.

WORKED EXAMPLE 37 Platform (exam level)

A student (point mass 60 kg) stands at the edge of a platform of radius 2.0 m and I = 400 kg m², rotating at 1.5 rad s⁻¹. She walks to the centre. Find the new angular velocity and the work she does.

Solution

I₁ = 400 + 60 × 4.0 = 640; I₂ = 400: ω₂ = 640 × 1.5/400 = 2.4 rad s⁻¹M1 A1

W = ½(400)(2.4)² − ½(640)(1.5)² = 432 JA1

6.4 Common exam errors

Examiners repeatedly see the same slips in rotation questions.

•Leaving ω in rpm or rev s⁻¹ instead of rad s⁻¹.
•Using the length of a lever instead of the perpendicular distance to the line of action.
•Assuming kinetic energy is conserved whenever angular momentum is.
•Writing T = mg for a mass accelerating on a pulley.
•Forgetting the rotational term ½Iω² for rolling objects.

7 Further practice

Graded problems across the topic.

7.1 Equilibrium problems

WORKED EXAMPLE 38 Shelf bracket (standard)

A uniform 1.2 m shelf of weight 40 N is hinged at the wall and supported by a cord attached to its outer end at 30° to the shelf. Find the tension in the cord.

Solution

Moments about the hinge: T sin 30° × 1.2 = 40 × 0.6M1

T = 40 NA1

WORKED EXAMPLE 39 Plank on two trestles (exam level)

A 5.0 m plank of weight 250 N rests on trestles 0.5 m from each end. A 700 N painter stands 1.0 m from the left end. Find the reaction at each trestle.

Solution

Moments about the left trestle (at 0.5 m): R(4.0) = 700(0.5) + 250(2.0)M1

Right = 212.5 N; left = 950 − 212.5 = 737.5 NA1 A1

Quick check

1. How far can the painter walk towards the right end before the plank tips?

Answers: 1. when the left reaction becomes zero: 700x = 250(2.0), x = 0.71 m beyond the right trestle, i.e. past the end (it does not tip)

7.2 Rotational kinematics problems

WORKED EXAMPLE 40 Hard drive (standard)

A disc spins up from rest to 7200 rpm in 3.0 s. Find its angular acceleration and the revolutions made.

Solution

ω = 754 rad s⁻¹; α = 251 rad s⁻²M1 A1

θ = ½(754)(3.0) = 1131 rad = 180 revolutionsA1

WORKED EXAMPLE 41 Washing machine (exam level)

A drum of radius 0.25 m spins at 1200 rpm. Find the centripetal acceleration of clothes at the rim in multiples of g.

Solution

ω = 126 rad s⁻¹; a = ω²r = 3950 m s⁻²M1

= 400gA1

7.3 Dynamics and angular momentum problems

WORKED EXAMPLE 42 Grindstone (exam level)

A grindstone (I = 0.40 kg m²) slows from 30 rad s⁻¹ to rest in 15 s when a tool is pressed on it. Find the frictional torque and the work done by friction.

Solution

α = 2.0 rad s⁻²; τ = 0.80 N mM1 A1

W = ½(0.40)(30)² = 180 JA1

WORKED EXAMPLE 43 Neutron star (exam level)

A star of radius 7.0 × 10⁸ m rotating once every 25 days collapses to a neutron star of radius 10 km. Assuming it remains a uniform sphere with the same mass, find the new period.

Solution

I ∝ R², so T ∝ R²M1

T = 25 × 86 400 × (10⁴/7.0 × 10⁸)² = 4.4 × 10⁻⁴ sA1

7.4 Energy problems

WORKED EXAMPLE 44 Rolling up a slope (exam level)

A solid sphere rolls without slipping at 4.0 m s⁻¹ onto an upward slope. Find the vertical height it reaches.

Solution

Total Ek = (7/10)mv² = mghM1

h = 0.7 × 16/9.81 = 1.14 mA1

WORKED EXAMPLE 45 Bicycle wheel (standard)

A bicycle wheel (I = 0.12 kg m², r = 0.35 m) rolls at 6.0 m s⁻¹. Find its rotational kinetic energy.

Solution

ω = 6.0/0.35 = 17.1 rad s⁻¹M1

E = ½(0.12)(17.1)² = 17.6 JA1

7.5 Mixed linear and rotational problems

Many HL questions link rotation to earlier topics: circular motion, energy and momentum.

WORKED EXAMPLE 46 Car wheel (exam level)

A car of mass 1200 kg accelerates from rest to 20 m s⁻¹. Each of its four wheels has I = 0.80 kg m² and radius 0.30 m. Find the total kinetic energy and the percentage that is rotational.

Solution

Translational: ½(1200)(20)² = 240 kJ; ω = 66.7 rad s⁻¹M1

Rotational: 4 × ½(0.80)(66.7)² = 7.1 kJA1

Total 247 kJ; rotational 2.9%A1

WORKED EXAMPLE 47 Spinning ice skater energy (exam level)

A skater with I = 4.0 kg m² spinning at 3.0 rad s⁻¹ pulls in her arms to I = 1.6 kg m². Find the work she does.

Solution

ω₂ = 4.0 × 3.0/1.6 = 7.5 rad s⁻¹M1

W = ½(1.6)(7.5)² − ½(4.0)(3.0)² = 45 − 18 = 27 JA1

8 Investigating rotation

Practical work.

8.1 Measuring a moment of inertia

Wrap a string around an axle, attach a falling mass, and time its fall through a known height. From the acceleration (s = ½at²) and the axle radius, find α and the tension, then I = τ/α. Alternatively use energy conservation.

WORKED EXAMPLE 48 From timing (exam level)

A 0.20 kg mass falls 0.80 m from rest in 2.0 s from an axle of radius 0.020 m. Find the moment of inertia of the wheel.

Solution

a = 2s/t² = 0.40 m s⁻²; T = m(g − a) = 1.88 NM1

α = a/r = 20 rad s⁻²; I = Tr/α = 1.9 × 10⁻³ kg m²A1

Try this

Roll a hollow and a solid cylinder of the same mass and radius down the same ramp and time them. Use the times to compare their moments of inertia.

8.2 Testing conservation of angular momentum

Drop a ring of known I onto a rotating turntable and measure ω before and after with a light gate or rotation sensor. Compare I₁ω₁ with (I₁ + I₂)ω₂.

Quick check

1. Name two sources of error in this experiment.

Answers: 1. friction at the axle; ring not dropped centrally

9 Common misconceptions and links

Students often think…	Actually…
torque depends only on force	it depends on force, distance and angle
a heavier object always has a larger moment of inertia	the distribution of mass about the axis matters
angular momentum is conserved when kinetic energy is	L is conserved with no external torque; kinetic energy can change
a couple produces translation	a couple produces rotation only
rolling objects have only translational kinetic energy	they also have rotational kinetic energy

How A.4 connects to the rest of the course

Topic	Connection
A.1 Kinematics	rotational analogues of suvat
A.2 Forces and momentum	Newton's laws and momentum conservation
A.3 Work, energy and power	rotational energy and power
D.1 Gravitational fields	orbital angular momentum and Kepler's second law
E.5 Fusion and stars	spinning neutron stars

10 Exam-style question with markscheme

Try this under exam conditions (about 20 minutes) before reading the markscheme.

A solid cylinder of mass 2.0 kg and radius 0.10 m can rotate about its axis. A light string wound round it supports a 0.40 kg mass that is released from rest.

(a) Calculate the moment of inertia of the cylinder. [1]

(b) Show that the acceleration of the falling mass is about 2.8 m s⁻². [3]

(d) Calculate the rotational kinetic energy at this point. [2]

(e) Explain, with reference to energy, why the mass falls with an acceleration less than g. [2]

Markscheme

Part	Answer	Marks
(a)	I = ½(2.0)(0.10)² = 0.010 kg m²	A1
(b)	mg − T = ma; Tr = I a/r	M1
	a = mg/(m + I/r²) = 3.92/(0.40 + 1.0)	M1
	= 2.80 m s⁻²	A1
(c)	v = √(2 × 2.80 × 1.0) = 2.37 m s⁻¹	M1
	ω = v/r = 23.7 rad s⁻¹	A1
(d)	½(0.010)(23.7)²	M1
	2.8 J	A1
(e)	some gravitational potential energy goes into rotational kinetic energy	R1
	less kinetic energy for the mass, so smaller acceleration	R1

Examiner's comments. In (b), the most common error was writing T = mg. In (e), good answers related the energy split to the moment of inertia.

11 Check your understanding

Section A: multiple choice

1. The unit of moment of inertia is

A. kg m² B. kg m s⁻¹ C. N m D. kg m² s⁻¹

2. Torque is greatest when the force is applied

A. along the lever arm B. at the pivot C. perpendicular to the lever arm, far from the pivot D. parallel to the axis

3. The rotational analogue of mass is

A. torque B. moment of inertia C. angular momentum D. angular velocity

4. A skater pulls in her arms. Her angular momentum

A. increases B. decreases C. becomes zero D. stays the same

5. A solid sphere and a hoop of equal mass and radius roll down a slope. First to the bottom is

A. the hoop B. the sphere C. both together D. it depends on the mass

6. Angular impulse equals

A. change in angular momentum B. torque C. work done D. angular acceleration

7. For rolling without slipping

A. v = ω/r B. v = rω² C. v = rα D. v = rω

8. Power delivered by a torque τ at angular velocity ω is

A. τ/ω B. ½τω² C. τω D. Iω

Section B: structured

9. A uniform 6.0 m ladder of weight 300 N leans against a smooth wall at 60° to the ground. Find the reaction from the wall. [4]

10. A turntable of I = 0.020 kg m² spins at 3.3 rad s⁻¹. A 0.10 kg lump of clay drops on at 0.12 m. Find the new angular velocity. [3]

11. Explain why a hollow cylinder accelerates down a slope more slowly than a solid one. [3]

12. A wheel slows uniformly from 20 rad s⁻¹ to 8 rad s⁻¹ in 4.0 s. Find the angular deceleration and angle turned. [3]

Answers

1. A 2. C 3. B 4. D 5. B 6. A 7. D 8. C

9. Moments about base: R(6 sin 60°) = 300(3 cos 60°): R = 86.6 N.

10. Inew = 0.0214; ω = 0.066/0.0214 = 3.1 rad s⁻¹.

11. Larger I/mr²: more energy goes to rotation, so less translational kinetic energy at each height.

12. α = −3.0 rad s⁻²; θ = ½(20 + 8)(4) = 56 rad.

12 Summary and topic map

Idea	Remember
Torque	τ = Fr sin θ; couples rotate only
Equilibrium	ΣF = 0 and Στ = 0
Kinematics	rotational suvat, v = rω
Dynamics	τ = Iα; I = Σmr²
Angular momentum	L = Iω; conserved with no external torque
Energy	½Iω²; rolling adds ½mv²; P = τω

Before the exam, make sure you…

•take moments about the point with most unknowns
•convert rpm to rad s⁻¹
•link linear and angular quantities with r
•include rotational energy in rolling problems

Rigid Body Mechanics on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 20. Topic map for Rigid Body Mechanics.

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