A.4 Rigid body mechanics · IB DP Physics

      
      IB DP Physics · Theme A · Space, time and motion
Rigid body mechanics.A.4 · Torque, rotational motion, moment of inertia and angular momentum
Higher level only: there is no standard level content in A.4

              Additional HL · 7 hours
              A.2 Forces and momentum ↗
            
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      The journey
Four parts.
Part 01
Torque and equilibrium\(\tau = Fr \sin \theta\), and when a body does not start to turn
Part 02
Rotational motionAngular displacement, velocity and acceleration
Part 03
Moment of inertia\(I = \Sigma m r^{2}\) and \(\tau = I \alpha\)
Part 04
Angular momentum and energy\(L = I \omega\), its conservation, and \(\frac{1}{2} I \omega^{2}\)
Every idea here has a twin in A.1 to A.3. Force becomes torque, mass becomes moment of inertia.
A.4 RIGID BODY MECHANICS · HL ONLY


      
      Part 01 · HL only
Torque and equilibrium.The turning effect of a force, and how to balance it
A.4 RIGID BODY MECHANICS · HL ONLY


      The turning effect of a force
Torque.\(\tau = F r \sin \theta\)
\(r\): from the axis to where the force acts
\(\theta\): the angle between the force and that line
Unit: N m
A sense: clockwise or counterclockwise
A SPANNER ON A NUT
A.4 RIGID BODY MECHANICS · HL ONLY
Three things matterHow hard you push, how far from the axis, and in what direction. A door shows all three. Push near the hinge and it hardly moves, however hard you push. Push at the handle, far from the hinge, and it swings easily. Push at the handle but straight towards the hinge, along the door, and nothing happens at all: \(\theta = 0\), so \(\sin \theta = 0\).
Two ways to read the equationGroup it as \((F \sin \theta) \times r\): only the part of the force at right angles to the handle turns the nut, and it acts at the full distance \(r\). Or group it as \(F \times (r \sin \theta)\): the whole force, times the perpendicular distance from the axis to the line along which the force acts. Both give the same number. Use whichever the diagram makes easier.
The unitN m is also what a joule is made of, but a torque is not an energy and is never written in joules.
The largest torque for a given force is when it acts at right angles to the handle: \(\sin 90^{\circ} = 1\) and \(\tau = Fr\).


      The story · Syracuse, third century BC
Background story · not assessed
A lever to move the world.
            Archimedes moving the Earth, from the Mechanics Magazine of 1824
          
People had used levers for thousands of years: the Egyptians moved obelisks of over a hundred tonnes with them
The earliest surviving writing on why they work is from the third century BC, and is credited to Archimedes
The boast attributed to him: give me a lever long enough and a place to rest it, and I shall move the world
Engraving: Mechanics Magazine, 1824, public domain.
A.4 RIGID BODY MECHANICS · HL ONLY


      No resultant torque
Rotational equilibrium.Resultant torque zero: clockwise torques = counterclockwise torques
About any axis you choose
For complete equilibrium, the resultant force is zero too
\(W_{1} d_{1} = W_{2} d_{2}\)
A BALANCED SEESAW
A.4 RIGID BODY MECHANICS · HL ONLY
Two conditions, two kinds of motionA body can move in two ways: its center of mass can accelerate, and it can start to spin. A resultant force causes the first. A resultant torque causes the second. A body is in equilibrium only if neither happens, so both must be zero. Two equal and opposite forces applied to opposite edges of a steering wheel add up to no force at all, yet the wheel turns: zero resultant force does not guarantee zero torque.
Choosing the axisIf a body is in equilibrium, the torques balance about every point, so you may take torques about whichever point is most convenient. Choose a point through which an unknown force acts: its distance from the axis is zero, so it drops out of the equation.
The pivot forceOn the seesaw the pivot pushes up with a force equal to the two weights together. It has no torque about the pivot, which is why it is missing from \(W_{1} d_{1} = W_{2} d_{2}\).
The weight of an extended body can be taken to act at one point, its center of mass. For a uniform plank that is the midpoint. You will not be asked to calculate where it is.


      Worked example
A plank on two supports.
              Torques about the left-hand support
              \(F_{2} \times 3.0 = 200 \times 1.0 + 60 \times 1.5\)
              \(F_{2} = \dfrac{290}{3.0} = \class{res}{97 \,\mathrm{N}}\)
              Resultant force zero
              \(F_{1} + F_{2} = 200 + 60\)
              \(F_{1} = 260 - 97 = \class{res}{163 \,\mathrm{N}}\)
            
UNIFORM PLANK, WEIGHT 60 N
A.4 RIGID BODY MECHANICS · HL ONLY
Why that axisTwo forces are unknown. Taking torques about the left-hand support removes \(F_{1}\) from the equation, because it acts at the axis itself. One equation, one unknown. The second unknown then comes from the forces.
Where the plank’s weight actsThe plank is uniform, so its center of mass is at its midpoint, 1.5 m from each end. Its 60 N acts there.
Does it make sense?The load is nearer the left-hand support, so that support should carry more of it: 163 N against 97 N. As a check, take torques about the right-hand support instead: \(F_{1} \times 3.0 = 200 \times 2.0 + 60 \times 1.5 = 490\), so \(F_{1} = 163\) N again.
Leaving out the weight of the plank itself. If a question gives a mass or weight for the beam, it must appear in the torque equation.


      Quick check
Torque and equilibrium.01A 20 N force acts at right angles, 0.30 m from an axis. Torque?
6.0 N m
02The same force acts along the line to the axis. Torque?
Zero
03State the two conditions for equilibrium.
Zero resultant force; zero resultant torque
04400 N sits 1.5 m from a pivot. Where must 600 N sit to balance it?
1.0 m on the other side
05Why take torques about a point where an unknown force acts?
Its torque is zero there
06Where does the weight of a uniform beam act?
At its midpoint, the center of mass
PRESS → TO REVEAL EACH ANSWER


      
      Part 02 · HL only
Rotational motion.A.1 again, with angles in place of distances
A.4 RIGID BODY MECHANICS · HL ONLY


      Three angular quantities
Describing rotation.Angular displacement \(\Delta \theta\), in rad
Angular velocity \(\omega = \dfrac{\Delta \theta}{\Delta t}\), in rad s⁻¹
Angular acceleration \(\alpha = \dfrac{\Delta \omega}{\Delta t}\), in rad s⁻²
A point at radius \(r\): \(v = \omega r\) and \(a = \alpha r\)
A TURNING WHEEL
A.4 RIGID BODY MECHANICS · HL ONLY
Why anglesDifferent points on a turning wheel move at different speeds: the rim travels much faster than a point near the hub. But they all turn through the same angle in the same time. Describing the motion by angle gives one number for the whole body, which is what makes a rigid body simple to deal with.
Why radiansAn angle in radians is arc length divided by radius, \(\theta = \frac{s}{r}\). So the distance a point moves is \(s = r \theta\), its speed is \(v = r \omega\) and its acceleration along its path is \(a = r \alpha\), with no awkward constants. These links work only in radians. One revolution is \(2 \pi\) rad.
Two accelerations\(a = \alpha r\) is the acceleration along the circle, present only while the wheel is speeding up or slowing down. Every point also has the centripetal acceleration \(\omega^{2} r\) of A.2, towards the axis, whenever the wheel is turning.
To convert revolutions per minute to rad s⁻¹, multiply by \(\frac{2 \pi}{60}\). 300 rev min⁻¹ is 31.4 rad s⁻¹.


      Constant angular acceleration
The equations of rotational motion.
          Linear, from A.1RotationalMissing

          \(s = \dfrac{u + v}{2} t\)\(\Delta \theta = \dfrac{\omega_\mathrm{f} + \omega_\mathrm{i}}{2} t\)\(\alpha\)
\(v = u + at\)\(\omega_\mathrm{f} = \omega_\mathrm{i} + \alpha t\)\(\Delta \theta\)
\(s = ut + \dfrac{1}{2} a t^{2}\)\(\Delta \theta = \omega_\mathrm{i} t + \dfrac{1}{2} \alpha t^{2}\)\(\omega_\mathrm{f}\)
\(v^{2} - u^{2} = 2as\)\(\omega_\mathrm{f}^{2} - \omega_\mathrm{i}^{2} = 2 \alpha \Delta \theta\)\(t\)

        A.4 RIGID BODY MECHANICS · HL ONLY
Nothing new to learnReplace \(s\) with \(\Delta \theta\), \(u\) and \(v\) with \(\omega_\mathrm{i}\) and \(\omega_\mathrm{f}\), and \(a\) with \(\alpha\), and the four equations of A.1 become the four on the right. They come from the same reasoning, the definitions of velocity and acceleration with a constant acceleration, so they have the same form.
The same methodList the five quantities, \(\Delta \theta\), \(\omega_\mathrm{i}\), \(\omega_\mathrm{f}\), \(\alpha\) and \(t\). Three will be known and one wanted. The last column shows which equation leaves out the fifth.
The same conditionThe angular acceleration must be constant. If the torque on the body changes during the motion, these equations do not apply.
The data booklet writes the last one as \(\omega_\mathrm{f}^{2} = \omega_\mathrm{i}^{2} + 2 \alpha \Delta \theta\). It is the same equation.


      Worked example
Speeding up and slowing down.
            From rest, \(\alpha = 2.5 \,\mathrm{rad} \,\mathrm{s}^{-2}\) for 8.0 s
            \(\omega_\mathrm{f} = \omega_\mathrm{i} + \alpha t = 0 + 2.5 \times 8.0\)
            \(\omega_\mathrm{f} = \class{res}{20 \,\mathrm{rad} \,\mathrm{s}^{-1}}\)
            \(\Delta \theta = \omega_\mathrm{i} t + \dfrac{1}{2} \alpha t^{2} = \dfrac{1}{2} \times 2.5 \times 8.0^{2}\)
            \(\Delta \theta = \class{res}{80 \,\mathrm{rad}}\)
          

            A disk at 30 rad s⁻¹ stops in 50 rad
            \(\omega_\mathrm{f}^{2} - \omega_\mathrm{i}^{2} = 2 \alpha \Delta \theta\)
            \(\alpha = \dfrac{\omega_\mathrm{f}^{2} - \omega_\mathrm{i}^{2}}{2 \Delta \theta} = \dfrac{0 - 30^{2}}{2 \times 50}\)
            \(\alpha = \class{res}{-9.0 \,\mathrm{rad} \,\mathrm{s}^{-2}}\)
          
A.4 RIGID BODY MECHANICS · HL ONLY
In revolutions80 rad is \(\frac{80}{2 \pi} = 12.7\) revolutions. Work in radians throughout and convert only at the end, if the question asks.
The minus signThe disk’s angular acceleration is negative because it is opposite to the angular velocity: the disk is slowing. As in A.1, choose one sense of rotation as positive at the start and keep to it.
A point on the rimIf the flywheel has a radius of 0.25 m, a point on its rim ends at \(v = \omega r = 20 \times 0.25 = 5.0\) m s⁻¹, and has traveled \(s = r \Delta \theta = 0.25 \times 80 = 20\) m round the circle.
Time for the disk to stop: \(t = \frac{\omega_\mathrm{f} - \omega_\mathrm{i}}{\alpha} = \frac{0 - 30}{-9.0} = 3.3\) s.


      Gradient and area
Graphs of rotational motion.Gradient of an \(\omega\)–\(t\) graph: the angular acceleration
Area under it: the angular displacement
Gradient of a \(\theta\)–\(t\) graph: the angular velocity
CONSTANT ANGULAR ACCELERATION
A.4 RIGID BODY MECHANICS · HL ONLY
The first equation, seenThe shaded shape is a trapezium of width \(t\) whose parallel sides are \(\omega_\mathrm{i}\) and \(\omega_\mathrm{f}\). Its area is their average times the width: \(\Delta \theta = \frac{\omega_\mathrm{f} + \omega_\mathrm{i}}{2} t\). This is exactly how the first equation of motion was found in A.1.
When the acceleration is not constant the graph is a curve, and the equations of the last slide fail, but the graph still works: the gradient of the tangent at any moment is the angular acceleration then, and the area under the curve is still the angle turned.
Part 4 adds one more: the area under a graph of torque against time is the change in angular momentum.


      Quick check
Rotational motion.01How many radians in one revolution?
\(2 \pi\)
02A wheel turns at 12 rad s⁻¹. Speed of a point 0.50 m from the axis?
6.0 m s⁻¹
03From 4.0 to 10 rad s⁻¹ in 3.0 s. Angular acceleration?
2.0 rad s⁻²
04And the angle turned in that time?
\(\frac{4.0 + 10}{2} \times 3.0 = 21 \,\mathrm{rad}\)
05What does the area under an \(\omega\)–\(t\) graph give?
Angular displacement
06What must be true for the four equations to apply?
Constant angular acceleration
PRESS → TO REVEAL EACH ANSWER


      
      Part 03 · HL only
Moment of inertia.How hard a body is to spin up, and Newton’s second law for rotation
A.4 RIGID BODY MECHANICS · HL ONLY


      The rotational twin of mass
Moment of inertia.\(I = \Sigma m r^{2}\)
\(r\): the distance of each mass from the axis
Unit: \(\mathrm{kg} \,\mathrm{m}^{2}\)
Depends on how the mass is distributed, and on which axis
THE SAME MASS, TWO ARRANGEMENTS
A.4 RIGID BODY MECHANICS · HL ONLY
Why distance counts twiceA mass far from the axis has to move faster for the same angular velocity (\(v = \omega r\)), so it is harder to get going. And the force that accelerates it acts at that same distance from the axis, so the torque needed grows by the same factor again. The two effects together give \(r^{2}\). Doubling the distance of a mass from the axis makes it four times as hard to spin up.
A worked caseTwo 2.0 kg masses on a light rod, each 0.50 m from the axis: \(I = 2.0 \times 0.50^{2} + 2.0 \times 0.50^{2} = 1.0\) kg m². Slide them in to 0.25 m and \(I = 2 \times 2.0 \times 0.25^{2} = 0.25\) kg m²: a quarter as much, with the same mass.
One body, many moments of inertiaMass is a single number for a body. Moment of inertia is not: it depends on the axis. A long rod is easy to spin about its own length and much harder to spin about an axis across its middle.
Tightrope walkers carry a long pole. It gives them a large moment of inertia about the rope, so that they tip slowly and have time to correct.


      Given when needed
Some moments of inertia.
            Body of mass M, about its axis of symmetryI

            Hoop or thin ring, radius \(R\)\(M R^{2}\)
Solid disk or cylinder, radius \(R\)\(\tfrac{1}{2} M R^{2}\)
Solid sphere, radius \(R\)\(\tfrac{2}{5} M R^{2}\)
Thin rod, length \(L\), about its center\(\tfrac{1}{12} M L^{2}\)

          A hoop has all its mass at radius \(R\): the largest \(I\) for its size
A disk or sphere has much of its mass nearer the axis
The formula for a shape is provided in the question. Do not learn them.
A.4 RIGID BODY MECHANICS · HL ONLY
Where the fractions come fromEach is \(\Sigma m r^{2}\) added up over every particle of the body. For a hoop every particle is at the same distance \(R\), so the sum is simply \(M R^{2}\). For a disk the particles are at all distances from 0 to \(R\), and the average of \(r^{2}\) comes to half of \(R^{2}\). In a sphere even more of the mass lies close to the axis, and the fraction falls to two fifths.
Using themA solid disk of mass 4.0 kg and radius 0.30 m has \(I = \frac{1}{2} \times 4.0 \times 0.30^{2} = 0.18\) kg m². A hoop of the same mass and radius has twice as much, 0.36 kg m².
Adding bodiesMoments of inertia about the same axis add. A point mass \(m\) placed at radius \(r\) on a disk increases the total by \(m r^{2}\).
The smaller the fraction, the more easily the body rolls down a slope. That is the last worked example of this deck.


      Torque and angular acceleration
Newton’s second law for rotation.\(\tau = I \alpha\)
\(\tau\): the resultant torque
The rotational form of \(F = ma\)
A larger \(I\): a smaller \(\alpha\) for the same torque

            A disk, \(I = 0.18 \,\mathrm{kg} \,\mathrm{m}^{2}\), radius 0.30 m
            A 6.0 N force acts along its rim
            \(\tau = Fr = 6.0 \times 0.30 = 1.8 \,\mathrm{N} \,\mathrm{m}\)
            \(\alpha = \dfrac{\tau}{I} = \dfrac{1.8}{0.18}\)
            \(\alpha = \class{res}{10 \,\mathrm{rad} \,\mathrm{s}^{-2}}\)
          
A.4 RIGID BODY MECHANICS · HL ONLY
From \(F = ma\) to \(\tau = I \alpha\)Take one small mass \(m\) of the body, at radius \(r\). A force \(F\) along its path gives it an acceleration \(a = \alpha r\), so \(F = m \alpha r\). Multiply both sides by \(r\): the left is the torque \(Fr\) and the right is \(m r^{2} \alpha\). Add up over every particle and \(\tau = (\Sigma m r^{2}) \alpha = I \alpha\).
Resultant torqueAs with forces, it is the total that counts. If a driving torque of 5.0 N m acts on a wheel against a frictional torque of 2.0 N m, the resultant is 3.0 N m, and that is what goes into the equation.
Constant torque, constant \(\alpha\)The disk in the example gains 10 rad s⁻¹ every second for as long as the force acts, and the equations of Part 2 apply.
Putting the force into the equation in place of the torque. Find \(\tau = Fr \sin \theta\) first.


      Worked example · two bodies joined
A falling mass turns a wheel.
            \(m = 0.50 \,\mathrm{kg}\) · \(I = 0.020 \,\mathrm{kg} \,\mathrm{m}^{2}\) · \(r = 0.10 \,\mathrm{m}\)
            Mass: \(mg - T = ma\)
            Wheel: \(Tr = I \alpha\)  with  \(a = \alpha r\)
            \(a = \dfrac{mg}{m + \dfrac{I}{r^{2}}} = \dfrac{0.50 \times 9.8}{0.50 + 2.0}\)
            \(a = \class{res}{2.0 \,\mathrm{m} \,\mathrm{s}^{-2}}\)
            \(T = m(g - a) = \class{res}{3.9 \,\mathrm{N}}\)
          
A STRING WOUND ROUND THE RIM
A.4 RIGID BODY MECHANICS · HL ONLY
One law for each bodyThe hanging mass moves in a straight line, so it gets \(F = ma\): its weight down, the tension up. The wheel turns, so it gets \(\tau = I \alpha\): the only torque on it is the tension, acting at the rim. The string does not slip, so the acceleration of the mass and the angular acceleration of the wheel are tied together by \(a = \alpha r\).
The algebraFrom the wheel, \(T = \frac{I \alpha}{r} = \frac{I a}{r^{2}}\). Put that into the first equation: \(mg - \frac{I a}{r^{2}} = ma\), so \(a \left(m + \frac{I}{r^{2}}\right) = mg\).
Reading the result\(\frac{I}{r^{2}}\) behaves like an extra mass that has to be accelerated: here 2.0 kg, four times the hanging mass. So the mass falls at only a fifth of \(g\). The tension is less than the weight, as it must be for the mass to accelerate downwards.
Timing a mass as it falls a known distance gives \(a\), and from it the moment of inertia of the wheel: a standard experiment.


      Quick check
Moment of inertia.01A 3.0 kg point mass is 2.0 m from an axis. Moment of inertia?
12 kg m²
02It is moved to 1.0 m. Moment of inertia now?
3.0 kg m²: a quarter
03Which has the larger \(I\): a hoop or a disk of the same mass and radius?
The hoop
04A torque of 8.0 N m acts on \(I\) = 2.0 kg m². Angular acceleration?
4.0 rad s⁻²
05What is the unit of moment of inertia?
kg m²
06What links \(a\) and \(\alpha\) for a string round a wheel of radius \(r\)?
\(a = \alpha r\)
PRESS → TO REVEAL EACH ANSWER


      
      Part 04 · HL only
Angular momentum and energy.Two more conservation laws, for things that spin
A.4 RIGID BODY MECHANICS · HL ONLY


      The rotational twin of momentum
Angular momentum.\(L = I \omega\)
Unit: \(\mathrm{kg} \,\mathrm{m}^{2} \,\mathrm{s}^{-1}\)
Conserved unless a resultant torque acts from outside
If \(I\) falls, \(\omega\) must rise
A SKATER, SEEN FROM ABOVE
A.4 RIGID BODY MECHANICS · HL ONLY
Why the skater speeds upOn smooth ice almost no torque acts on the skater from outside, so \(I \omega\) cannot change. Pulling the arms in brings mass closer to the axis and lowers \(I\). The only way for the product to stay the same is for \(\omega\) to go up. Nothing pushes the skater round faster: the conservation law demands it.
The same law everywhereA diver tucks to somersault quickly and opens out to slow the spin before entering the water. A cloud of gas collapsing to form a star spins faster and faster as it shrinks. A neutron star, the collapsed core of a star that once turned every few weeks, can spin hundreds of times a second.
The conditionNo resultant external torque. Forces inside the system, such as the skater’s muscles pulling the arms in, cannot change its total angular momentum, just as internal forces cannot change the total linear momentum in A.2.
The guide expects problems where the moment of inertia of an extended body changes, and problems where two bodies couple together.


      Worked example
A skater pulls in their arms.
            \(I\) falls from 4.0 to 1.6 kg m² · \(\omega_{1} = 3.0 \,\mathrm{rad} \,\mathrm{s}^{-1}\)
            \(I_{1} \omega_{1} = I_{2} \omega_{2}\)
            \(\omega_{2} = \dfrac{I_{1} \omega_{1}}{I_{2}} = \dfrac{4.0 \times 3.0}{1.6}\)
            \(\omega_{2} = \class{res}{7.5 \,\mathrm{rad} \,\mathrm{s}^{-1}}\)
          

            Rotational kinetic energy, \(E_\mathrm{k} = \tfrac{1}{2} I \omega^{2}\)
            Before: \(\tfrac{1}{2} \times 4.0 \times 3.0^{2} = 18 \,\mathrm{J}\)
            After: \(\tfrac{1}{2} \times 1.6 \times 7.5^{2} = \class{res}{45 \,\mathrm{J}}\)
            Work done by the skater: 27 J
          
A.4 RIGID BODY MECHANICS · HL ONLY
Momentum conserved, energy notThe angular momentum is 12 kg m² s⁻¹ before and after. The kinetic energy has more than doubled. That is not a contradiction: no outside torque acted, so angular momentum had to be conserved, but the skater’s muscles did work. Each arm is moving in a circle and must be pulled inwards against its tendency to carry straight on. That work becomes rotational kinetic energy.
Using \(E_\mathrm{k} = \frac{L^{2}}{2I}\)With \(L\) fixed at 12, the energy is \(\frac{12^{2}}{2I}\): 18 J when \(I = 4.0\) and 45 J when \(I = 1.6\). This form shows at once that, at constant angular momentum, a smaller moment of inertia means more kinetic energy.
Assuming that kinetic energy is conserved because angular momentum is. Check each separately.


      The story · Paris, 1852
Background story · not assessed
A wheel that shows the Earth turning.
            Foucault’s gyroscope, with its microscope
          
A fast-spinning wheel in freely turning rings: no torque can reach it
So its angular momentum, and the direction of its axis, stay fixed
In 1852 Léon Foucault watched one through a microscope and saw the laboratory, and the Earth, turn slowly round it
He could watch for 8 to 10 minutes before friction slowed the wheel
He named it from the Greek for “to see the rotation”.
Photograph: Stéphane Magnenat, public domain.
A.4 RIGID BODY MECHANICS · HL ONLY


      Worked example · two bodies join
One disk dropped onto another.
            Spinning: \(I_{1} = 0.20 \,\mathrm{kg} \,\mathrm{m}^{2}\) at \(12 \,\mathrm{rad} \,\mathrm{s}^{-1}\)
            Dropped on: \(I_{2} = 0.10 \,\mathrm{kg} \,\mathrm{m}^{2}\), at rest
            \(I_{1} \omega_{1} = (I_{1} + I_{2}) \omega\)
            \(\omega = \dfrac{I_{1} \omega_{1}}{I_{1} + I_{2}} = \dfrac{0.20 \times 12}{0.30}\)
            \(\omega = \class{res}{8.0 \,\mathrm{rad} \,\mathrm{s}^{-1}}\)
          
ON THE SAME AXLEPRESS → TO DROP
A.4 RIGID BODY MECHANICS · HL ONLY
A collision, in rotationThis is the joined-trolleys problem of A.2 turned through a right angle. Two bodies end up moving together, the total momentum before equals the total after, and the shared speed comes from dividing by the total inertia. Friction between the disks is an internal force of the system and cannot change its total angular momentum.
Energy is lostBefore: \(\frac{1}{2} \times 0.20 \times 12^{2} = 14.4\) J. After: \(\frac{1}{2} \times 0.30 \times 8.0^{2} = 9.6\) J. A third of the kinetic energy has become thermal energy as the disks rubbed against each other until they turned at the same rate. It is an inelastic collision.
If the second disk were already spinning the opposite way, its angular momentum would go in with a minus sign.


      Changing angular momentum
Angular impulse.\(\Delta L = \tau \Delta t = \Delta (I \omega)\)
A resultant torque acting for a time changes the angular momentum
\(\tau \Delta t\): the angular impulse
The area under a torque–time graph

            A torque of 3.0 N m acts for 4.0 s
            on a wheel of \(I = 2.0 \,\mathrm{kg} \,\mathrm{m}^{2}\)
            \(\Delta L = \tau \Delta t = 3.0 \times 4.0\)
            \(\Delta L = \class{res}{12 \,\mathrm{kg} \,\mathrm{m}^{2} \,\mathrm{s}^{-1}}\)
            \(\Delta \omega = \dfrac{\Delta L}{I} = \dfrac{12}{2.0} = \class{res}{6.0 \,\mathrm{rad} \,\mathrm{s}^{-1}}\)
          
A.4 RIGID BODY MECHANICS · HL ONLY
The twin of impulseIn A.2, \(F \Delta t = \Delta p\): a force acting for a time changes the momentum. Here \(\tau \Delta t = \Delta L\). The reasoning is the same, and so is the usefulness: when a torque varies during the time it acts, you can still find the total change in angular momentum from the area under the torque–time graph.
Torque as a rateRearranged, \(\tau = \frac{\Delta L}{\Delta t}\): torque is the rate of change of angular momentum. If \(I\) is constant this is \(I \frac{\Delta \omega}{\Delta t} = I \alpha\), the second law of Part 3. Written with \(L\) it still holds when the moment of inertia changes.
And conservationPut \(\tau = 0\) and \(\Delta L = 0\): with no resultant torque the angular momentum cannot change. The conservation law is a special case of this equation.
The units agree: \(\mathrm{N} \,\mathrm{m} \,\mathrm{s} = \mathrm{kg} \,\mathrm{m}^{2} \,\mathrm{s}^{-1}\).


      Energy of a spinning body
Rotational kinetic energy and rolling.\(E_\mathrm{k} = \dfrac{1}{2} I \omega^{2} = \dfrac{L^{2}}{2I}\)
Rolling without slipping: \(v = \omega r\)
Total: \(\dfrac{1}{2} m v^{2} + \dfrac{1}{2} I \omega^{2}\)
A RACE DOWN A SLOPE
A.4 RIGID BODY MECHANICS · HL ONLY
Two kinds of kinetic energy at onceA rolling body is moving along and turning. Its center of mass carries \(\frac{1}{2} m v^{2}\), as if all the mass were there, and the turning about the center adds \(\frac{1}{2} I \omega^{2}\). If it does not slip, the point touching the ground is momentarily at rest, which fixes \(v = \omega r\).
Why the sphere winsBoth start with the same potential energy per kilogram. Both must share it between moving and turning. The hoop, with all its mass at the rim, has to put half into rotation. The solid sphere needs only two sevenths for rotation, leaving more for getting down the slope. Neither the mass nor the radius matters: every solid sphere beats every hoop.
Friction without lossFriction is what makes the body turn instead of slide. But the point of contact does not move while it is touching, so the friction does no work, and mechanical energy is conserved.
Combined rotation and translation is limited to rolling without slipping.


      Worked example
A sphere rolls down a slope.
            Solid sphere, \(I = \tfrac{2}{5} m r^{2}\) · height fallen \(h = 1.0 \,\mathrm{m}\)
            \(mgh = \dfrac{1}{2} m v^{2} + \dfrac{1}{2} I \omega^{2}\)  with  \(\omega = \dfrac{v}{r}\)
            \(mgh = \dfrac{1}{2} m v^{2} + \dfrac{1}{5} m v^{2} = \dfrac{7}{10} m v^{2}\)
            \(v = \sqrt{\dfrac{10 g h}{7}} = \sqrt{\dfrac{10 \times 9.8 \times 1.0}{7}}\)
            \(v = \class{res}{3.7 \,\mathrm{m} \,\mathrm{s}^{-1}}\)
          
A hoop, \(I = m r^{2}\): \(v = \sqrt{gh} = 3.1 \,\mathrm{m} \,\mathrm{s}^{-1}\)
Sliding with no friction: \(v = \sqrt{2gh} = 4.4 \,\mathrm{m} \,\mathrm{s}^{-1}\)
The mass and the radius have canceled.
A.4 RIGID BODY MECHANICS · HL ONLY
Where \(\frac{1}{5}\) comes from\(\frac{1}{2} I \omega^{2} = \frac{1}{2} \times \frac{2}{5} m r^{2} \times \frac{v^{2}}{r^{2}} = \frac{1}{5} m v^{2}\). The \(r^{2}\) in the moment of inertia cancels the \(r^{2}\) from \(\omega = \frac{v}{r}\). That is why the radius does not matter, and since \(m\) is in every term, the mass does not either.
The three speeds comparedA block sliding without friction keeps all its energy as motion along the slope and arrives fastest. A rolling sphere arrives with five sevenths of its energy in that form, and a hoop with half. The more of the mass that is far from the axis, the more of the energy the rotation takes.
The method is conservation of energy from A.3 with one extra term. Write the term, substitute \(\omega = \frac{v}{r}\), and solve.


      The twins, side by side
Linear and rotational.
          LinearRotational

          Position, velocity, acceleration\(s\), \(v\), \(a\)\(\theta\), \(\omega\), \(\alpha\)
InertiaMass \(m\)Moment of inertia \(I\)
Cause of accelerationForce: \(F = ma\)Torque: \(\tau = I \alpha\)
Momentum\(p = mv\)\(L = I \omega\)
Impulse\(F \Delta t = \Delta p\)\(\tau \Delta t = \Delta L\)
Kinetic energy\(\tfrac{1}{2} m v^{2} = \dfrac{p^{2}}{2m}\)\(\tfrac{1}{2} I \omega^{2} = \dfrac{L^{2}}{2I}\)

        A.4 RIGID BODY MECHANICS · HL ONLY


      Quick check
Angular momentum and energy.01\(I\) = 0.50 kg m², \(\omega\) = 6.0 rad s⁻¹. Angular momentum?
3.0 kg m² s⁻¹
02When is angular momentum conserved?
No resultant external torque
03A spinning star shrinks to half its moment of inertia. Its angular velocity?
It doubles
04And its rotational kinetic energy?
It doubles: \(\frac{L^{2}}{2I}\)
05Rotational kinetic energy for \(I\) = 0.50 kg m² at 6.0 rad s⁻¹?
\(\frac{1}{2} \times 0.50 \times 6.0^{2} = 9.0 \,\mathrm{J}\)
06Which reaches the bottom of a slope first: a rolling hoop or a rolling solid sphere?
The sphere
PRESS → TO REVEAL EACH ANSWER


      Summary
Rigid body mechanics on one slide.Torque
\(\tau = F r \sin \theta\)Equilibrium: resultant torque and resultant force both zero
Rotational motion
\(\omega_\mathrm{f} = \omega_\mathrm{i} + \alpha t\)The equations of A.1, with \(\Delta \theta\), \(\omega\) and \(\alpha\)
Moment of inertia
\(I = \Sigma m r^{2}\)Depends on where the mass is, and on the axis
Second law
\(\tau = I \alpha\)Angular impulse: \(\Delta L = \tau \Delta t\)
Angular momentum
\(L = I \omega\)Conserved when no resultant external torque acts
Energy
\(E_\mathrm{k} = \frac{1}{2} I \omega^{2} = \frac{L^{2}}{2I}\)Rolling: add \(\frac{1}{2} m v^{2}\), with \(v = \omega r\)
A.4 RIGID BODY MECHANICS · HL ONLY


      Summary
Key words.TorqueThe turning effect of a force about an axis
Rotational equilibriumZero resultant torque
Center of massWhere the mass can be taken to be for linear motion
Angular velocityRate of change of angle, in rad s⁻¹
Angular accelerationRate of change of angular velocity
Moment of inertiaResistance to angular acceleration: \(\Sigma m r^{2}\)
Angular momentum\(I \omega\)
Angular impulseTorque × time: the change in angular momentum
Rolling without slipping\(v = \omega r\)
A.4 RIGID BODY MECHANICS · HL ONLY


    
Linear, from A.1	Rotational	Missing
\(s = \dfrac{u + v}{2} t\)	\(\Delta \theta = \dfrac{\omega_\mathrm{f} + \omega_\mathrm{i}}{2} t\)	\(\alpha\)
\(v = u + at\)	\(\omega_\mathrm{f} = \omega_\mathrm{i} + \alpha t\)	\(\Delta \theta\)
\(s = ut + \dfrac{1}{2} a t^{2}\)	\(\Delta \theta = \omega_\mathrm{i} t + \dfrac{1}{2} \alpha t^{2}\)	\(\omega_\mathrm{f}\)
\(v^{2} - u^{2} = 2as\)	\(\omega_\mathrm{f}^{2} - \omega_\mathrm{i}^{2} = 2 \alpha \Delta \theta\)	\(t\)
Body of mass M, about its axis of symmetry	I
Hoop or thin ring, radius \(R\)	\(M R^{2}\)
Solid disk or cylinder, radius \(R\)	\(\tfrac{1}{2} M R^{2}\)
Solid sphere, radius \(R\)	\(\tfrac{2}{5} M R^{2}\)
Thin rod, length \(L\), about its center	\(\tfrac{1}{12} M L^{2}\)
	Linear	Rotational
Position, velocity, acceleration	\(s\), \(v\), \(a\)	\(\theta\), \(\omega\), \(\alpha\)
Inertia	Mass \(m\)	Moment of inertia \(I\)
Cause of acceleration	Force: \(F = ma\)	Torque: \(\tau = I \alpha\)
Momentum	\(p = mv\)	\(L = I \omega\)
Impulse	\(F \Delta t = \Delta p\)	\(\tau \Delta t = \Delta L\)
Kinetic energy	\(\tfrac{1}{2} m v^{2} = \dfrac{p^{2}}{2m}\)	\(\tfrac{1}{2} I \omega^{2} = \dfrac{L^{2}}{2I}\)