Chaper Roaional dynamics. Angular acceleraion Learning objecives: Wha do we mean by angular acceleraion? How can we calculae he angular acceleraion of a roaing objec when i speeds up or slows down? How can we calculae he number of urns a roaing objec makes in a cerain ime when i acceleraes uniformly? Angular speed and angular acceleraion There are many examples of roaing objecs around us. Examples include he Earh spinning on is axis, he wheels of a moving car, a flywheel in moion and he roaion of he armaure coil of an elecric moor. For any rigid body roaing abou a fixed axis: π Is angular speed, ω = T where T is is period of roaion. The uni of angular speed is he radian per second (i.e. rad s ). Is angular acceleraion,, is he change of is angular velociy per second. The uni of angular acceleraion is he radian per second per second (i.e. rad s ). Noes Angular speed is someimes expressed in revoluions per minue or rpm. To conver o rad s, π muliply by as here are radians per revoluion and 60 seconds per minue. 60 Angular velociy is angular speed in a cerain direcion of roaion. Figure A flywheel in moion
Link See A Physics A Topic. for more abou angular speed. When a flywheel roaes a consan angular speed, is angular acceleraion is zero because is π angular speed does no change. If ω is is angular speed, is period of roaion T =. If he flywheel has a radius r, a poin on he flywheel rim on a circular pah moves a a speed v = ωr. When a flywheel speeds up, every poin of he flywheel moves a increasing speed. If he flywheel speeds up from iniial angular speed o angular speed in ime, hen: is angular acceleraion he speed of any poin on is rim increases from speed u = r o speed v = r in ime, where r is he radius of he flywheel. The speeding-up process acceleraes such a poin along a circular pah of radius r. Therefore he acceleraion, a, of he poin along is pah may be calculaed from: a v u r r r r In general, for any roaing body, for any poin (small elemen ) in he body a perpendicular disance r from he axis of roaion, is speed v = ωr is acceleraion a = r Every par of a roaing objec experiences he same angular acceleraion. However, he acceleraion, a, a a poin in he body is along is circular pah (i.e. is linear acceleraion) and is proporional o r (because a = r). Noe Every poin of a roaing objec experiences a cenripeal acceleraion equal o ω r which acs direcly owards he cenre of roaion of ha poin. Worked example A flywheel is speeded up from 5.0 o.0 revoluions per minue in 00 s. The radius of he flywheel is 0.080 m. Calculae: a he angular acceleraion of he flywheel b he acceleraion of a poin on he rim along is circular pah. Soluion π a Iniial angular speed, 5.0 0. 5 rad s 60 π angular speed afer 00 s,.0. 5 rad s 60 angular acceleraion,.5 0.5 3 00 6.30 rad s
b Acceleraion angenial o he rim, 3 4 a r 6.30 0.080 5.00 0 4 m s Equaions for consan angular acceleraion These may be derived in much he same way as he equaions for sraigh line moion wih consan acceleraion From he definiion of angular acceleraion above, we obain by rearrangemen, = 0 + where 0 = iniial angular speed, and = angular speed a ime. The average value of he angular speed is obained by averaging he iniial and final values,. giving an average value of 0 The angle which he objec urns hrough, he angular displacemen, is equal o he average angular speed he ime aken which gives 0 3 The wo equaions above can be combined o eliminae o give 0 4 Alernaively, he firs wo equaions may be combined o eliminae, giving = 0 + Noe The four equaions above are direcly comparable wih he four linear dynamics equaions. The ask of using he angular equaions is made much easier by ranslaing beween linear and angular erms, as follows: Linear equaions Angular equaions displacemen s angular displacemen speed or velociy u v angular speed 0 acceleraion a angular acceleraion For example v = u + a = 0 + s = u + a = 0 + v = u + as = 0 + Link See AS Physics A Topic 8.3 for he linear dynamics equaions wih consan acceleraion.
Summary quesions A flywheel acceleraes uniformly from res o rad s in 60 seconds. Calculae: a is angular acceleraion b i he angle i urned hrough in his ime ii he number of urns i made. A child s spinning op, spinning a a frequency of Hz, deceleraed uniformly o res in 50 s. Calculae: a is iniial angular speed b is angular deceleraion c he number of urns i made when i deceleraed. 3 A spin drier ub acceleraed uniformly from res o an angular speed of 00 revoluions per second in 50 seconds. Calculae: a he angle which he ub urned hrough in his ime b he number of urns i made as i acceleraed. 4 A vehicle wih wheels of diameer 0.45 m deceleraed uniformly from a speed of 4 m s o a sandsill in a disance of 60 m. Calculae: a he angular speed of each wheel when he vehicle was moving a 4 m s b i he ime aken by he vehicle o decelerae o a sandsill from a speed of 4 m s ii he number of urns each wheel made during he ime he vehicle slowed down. c he angular speed of he vehicle one second before i sopped.
. Momen of ineria Learning objecives: Wha is orque? Wha do we mean by momen of ineria? How does he angular acceleraion of a roaing objec depend on is momen of ineria? Torque To make a flywheel roae, a urning force mus be applied o i. The urning effec depends no jus on he force bu also on where i is applied. The orque of a urning force is he momen of he force abou he axis. Therefore, orque is defined as follows. Torque = force perpendicular disance from he axis o he line of acion of he force The uni of orque is he newon mere (N m). Figure Applying a orque Link See AS Physics A Topic 7.3 for he momen of a force. The ineria of an objec is is resisance o change of is moion. If a large orque is required o sar a flywheel urning, he flywheel mus have considerable ineria. In oher words, is resisance o change of is moion is high. Every objec has he propery of ineria because every objec has mass. However, he ineria of a roaing body depends on he disribuion of is mass as well as he amoun of mass. Consider a fla rigid body which can be roaed abou an axis perpendicular o is plane, as shown in Figure. Suppose i is iniially a res and a orque is applied o i o make i roae. Assuming here is no fricion on is bearing, he applied orque will increase is angular speed. When he orque is removed, he angular speed sops increasing so i urns a consan frequency once he orque is removed.
Figure A rigid body considered as a nework of poin masses The body in Figure may be hough of as a nework of poin masses, m, m, m 3, ec., a disances r, r, r 3, ec. from he axis. Each poin urns on a circular pah abou he axis. A angular speed, he speed of each poin is given by v = r so: he speed of m is r he speed of m is r, ec. When he body speeds up, every poin in i acceleraes. If he angular acceleraion of he body is, hen he acceleraion of each poin mass is given by a = r So: he acceleraion of m is r he acceleraion of m is r, ec. Using F = ma, he force needed o accelerae each poin mass is herefore given by: F = m r for m F = m r for m, ec. The momen needed for each poin mass o be given angular acceleraion is given by force F disance, r, since momen = force perpendicular disance from he poin mass o he axis. Thus: he momen for m = (m r )r he momen for m = (m r )r ec. The oal momen (orque T) needed o give he body angular acceleraion = he sum of he individual momens needed for all he poin masses. Hence orque T = (m r ) + (m r ) + (m 3 r 3 ) +.+ (m N r N ) = [(m r ) + (m r ) + (m 3 r 3 ) +.+ (m N r N )] = I where I = [(m r ) + (m r ) + (m 3 r 3 ) +.+ (m N r N )] is he momen of ineria of he body abou he axis of roaion. The summaion (m r ) + (m r ) + (m 3 r 3 ) +.+ (m N r N ) is wrien in shor form as mr (pronounced sigma m r squared )
The momen of ineria I of a body abou a given axis is defined as mr for all he poins in he body, where m i is he mass of each poin and r i is is perpendicular disance from he axis. The uni of I is kg m. In general, T is he resulan orque. For example, if a orque T is applied o a flywheel which is also aced on by a fricional orque T, he resulan orque is T T. Therefore when a body undergoes angular acceleraion, he resulan orque T acing on i is given by: T = I The derivaion of I = mr is no required in his specificaion. Worked example A flywheel of momen of ineria 0.45 kg m is acceleraed uniformly from res o an angular speed of 6.7 rad s in 4.8 s. Calculae he resulan orque acing on he flywheel during his ime. Soluion Angular acceleraion 6.7 0.4 rad s 4.8 Resulan orque T = I = 0.45.4 = 0.63 N m Momen of ineria and angular acceleraion When a resulan orque is applied o a body, he angular acceleraion,, of he body is given by = I T. Thus he angular acceleraion depends no jus on he orque T bu also on he momen of ineria I of he body abou he given axis which is deermined by he disribuion of mass abou he axis. Two bodies of equal mass disribued in differen ways will have differen values of I. For example, compare a hoop and a disc of he same mass, as in Figure 3.
Figure 3 Disribuion of mass Abou he axis shown: he momen of ineria of he hoop is simply MR where M is is mass and R is is radius. This is because all he mass of he hoop is a he same disance (= R) from he axis. The momen of ineria, I, abou he axis shown, mr, is herefore jus MR for he hoop. he momen of ineria of he disc abou he same axis is less han MR because he mass of he disc is disribued beween he cenre and he rim. Deailed heory shows ha he value of I for he disc abou he axis shown is MR. In general: he furher he mass is disribued from he axis, he greaer is he momen of ineria abou ha axis. The momen of ineria abou a given axis of an objec wih a simple geomerical shape can be calculaed using an appropriae mahemaical formula for ha shape and axis. In general erms, such formulae include geomerical facors as well as mass. For example: he hoop as shown in Figure 3 has a momen of ineria given by MR he disc shown in Figure 3 has a momen of ineria given by MR a uniform beam of lengh L and mass M has a momen of ineria abou an axis perpendicular o is lengh given by: ML if he axis is hrough is cenre ML if he axis is hrough is end. 3 The formulae above and for oher bodies wih simple shapes are derived using he mahemaical echnique of inegraion o obain mr for all poins in he body. The emphasis in physics and engineering is on he use raher han he derivaion of he formulae. By using he appropriae formula, he momen of ineria of a simple shape can be calculaed from is dimensions and mass.
Then, when subjeced o orque, he effec on he moion can be deermined. Apar from a circular hoop, you are no expeced o recall he formula for he momen of ineria of any objec and quesions will supply, if necessary, any such formula. Worked example A solid circular drum of mass 4.0 kg and radius 0.5 m is roaing a an angular speed of rad s abou an axis as shown in Figure 4 when a braking orque is applied o i which brings i o res in 5.8 s. Figure 4 Calculae: a is angular deceleraion when he braking orque is applied b he momen of ineria of he drum abou he axis shown c he resulan orque ha causes i o decelerae. Momen of ineria of drum abou he axis shown = MR Soluion a Angular acceleraion 0 3.8 rad s 5.8 b I = MR = 0.5 4.0 0.5 = 0.045 kg m c Resulan orque T = I = 0.045 3.8 = 0.7 N m
Summary quesions The roaing par of an elecric fan has a momen of ineria of 0.68 kg m. The roaing par is acceleraed uniformly from res o an angular speed of 3.7 rad s in 9. s. Calculae he resulan orque acing on he fan during his ime. A solid circular disc of mass 7.4 kg and radius 0.090 m is mouned on an axis as in Figure. A force of 7.0 N is applied angenially o he disc a is rim as shown in Figure o accelerae he disc from res. a Show ha: i he momen of ineria of he disc abou his axis is 0.030 kg m ii he orque applied o he disc is 0.63 N m. b The force of 7.0 N is applied for 5.0 s. Calculae: i he angular acceleraion of he disc a he end of his ime ii he number of urns made by he disc in his ime. Momen of ineria of disc abou he axis shown = MR 3 Figure 5 shows cross-secions of wo discs X and Y which have he same mass and radius. Sae and explain which disc has he greaer momen of ineria abou he axis shown. Figure 5 4 A flywheel is acceleraed by a consan orque for 8 s from res. During his ime i makes 36 urns. I hen slows down o a sandsill 9 s afer he orque is removed, making 87 urns during his ime. a i Show ha in he ime i acceleraed, is angular acceleraion was.40 rad s. ii Show ha in he ime i slowed down, is angular deceleraion was 0.3 rad s. b The orque applied o i when i acceleraed was 6 N m. i Show ha he fricional orque ha slowed i down was. N m. ii Calculae he momen of ineria of he flywheel.
.3 Roaional kineic energy Learning objecives: Wha does he kineic energy of a roaing objec depend on? How much work does a orque do when i makes a roaing objec urn? How can we measure he momen of ineria of a flywheel? Kineic energy To make a body which is iniially a res roae abou a fixed axis, i is necessary o apply a orque o he body. The orque does work on he body and, as long as he applied orque exceeds he fricional orque, he work done increases he kineic energy of he body and he faser he body roaes. The kineic energy, E K, of a body roaing a angular speed is given by: E K = I where I is is momen of ineria abou he axis of roaion. To prove his equaion, consider he body as a nework of poin masses m, m, m 3, m 4, ec. When he body roaes a angular speed, he speed of each poin mass is given by v = r (see Topic.). speed of m = r, where r is he disance of m from he axis speed of m = r, where r is he disance of m from he axis, ec, speed of m N = r N, where r N is he disance of m N from he axis. Since he kineic energy of each poin mass is given by mv, hen kineic energy of m = m v = m r, kineic energy of m = m v = m r, ec kineic energy of m N = m N v N = m N r N. Hence he oal kineic energy = m r + m r +.. + m N r N = [(m r ) + (m r ) +.+ (m N r N )] = I since I = [(m r ) + (m r ) +.+ (m N r N )] The equaion E K = I enables us o calculae how much energy a roaing objec sores due o is roaional moion. In addiion, i shows ha he kineic energy of a roaing objec is proporional o: is momen of ineria abou he axis of roaion
he square of is angular speed. Flywheels are used in many machines and engines so ha he moving pars coninue o move when he load on he machine increases and i has o do more work. For example, when a meal press is used o make a shaped objec from a shee of hin meal, he press is able o do he necessary work because he flywheel keeps i moving. Flywheels are also fied in some vehicles o sore kineic energy when he vehicle brakes are applied and i slows down. Insead of energy being ransferred as hea o he surroundings, some of he vehicle s kineic energy is ransferred o an on-board flywheel o be reurned o he vehicle when he acceleraor pedal is pressed. In 009 Formula One moor racing cars sared o be fied wih a Kineic Energy Recovery Sysem (KERS) based on he flywheel principle. Link See AS Physics A Topics 0. and 0.3 for kineic energy, work and power. Work done The work done W by a consan orque T when he body is urned hrough angle is given by W = T This can be seen by considering ha he orque is due o a force F acing a a perpendicular disance d from he axis of roaion. The force acs hrough a disance s = d when i urns he body hrough angle. Therefore, he work done by he force W = Fs = Fd = T (as T = Fd). Assuming here is no fricional orque, he applied orque T = I. Therefore, he work done W by he applied orque is given by W = T = (I). The dynamics equaion for angular moion = 0 + gives = 0, so W = I = I( 0 ) = I I 0 = he gain of kineic energy. Therefore, in he absence of fricion, he work done by he orque is equal o he gain of roaional kineic energy of he body. The number of urns for an angular displacemen in radians = Power For a body roaing a consan angular speed, he power, P, delivered by a orque T acing on he body is given by P = T We can see how his equaion arises by considering a consan orque T acing on a body for a ime.
If he body urns hrough an angle in his ime, he work done W by he orque is given by W = T Since he power P delivered by he orque is he rae of work done by he orque, hen P = W Hence P = W T = = T where he angular speed = If a roaing body is aced on by an applied orque T which is equally opposed by a fricional orque T F, he resulan orque is zero so is angular speed is consan. The power P delivered by he applied orque T = The work done per second by he fricional orque T F = work donew ime aken T F = TF T = = T In his siuaion, he rae of ransfer of energy due o he applied force is equal o he rae of ransfer of energy o he surroundings by he fricional force. So he roaing body does no gain any kineic energy. Worked example A flywheel is roaing a an angular speed of 0 rad s on a fixed axle which is mouned on fricionless bearings. The momen of ineria of he flywheel and he axle abou he axis of roaion is 0.068 kg m. a Calculae he roaional kineic energy of he flywheel when i roaes a 0 rad s. b When a braking orque of.4 N m is applied o is rim, he flywheel is brough o res. Calculae he number of urns he flywheel makes as i deceleraes o a sandsill. Soluion a E K = I = 0.5 0.068 0 = 490 J b The work done by he braking orque = loss of kineic energy E K Therefore T = E K hence.4 = 490 Hence = 490 350 = 350 rad = = 56 urns.4 π Experimen o measure he momen of ineria I of a flywheel The experimen described below is no par of he specificaion for he opion bu i reinforces and brings ogeher many of he ideas you have covered so far. In addiion, i provides an opporuniy o do some pracical work. An objec of known mass M hanging from a sring is used o accelerae he flywheel from res as shown in Figure.
Figure Measuring he momen of ineria of a flywheel The following measuremens need o be made: The disance, h, fallen by he objec from release o when he sring unwraps iself from he axle of he flywheel The diameer, d, of he axle The ime aken,, for he sring o unwrap The measuremens and he known mass M are used in he following calculaions: The number of urns made by he flywheel as he sring unwraps, N h πd The maximum angular speed of he flywheel, = he average angular speed = The speed of he objec a he insan he sring unwraps, v = The kineic energy gained by he flywheel, E KF = I, o be calculaed in erms of I The kineic energy gained by he objec of mass M, E KO = Mv d where v = The poenial energy los by he objec of mass M, E P = Mgh The momen of ineria of he flywheel can be calculaed from he equaion below, assuming fricion on he flywheel is negligible. he oal gain of kineic energy, E KF + E KO = he loss of poenial energy, E P I + Mv = Mgh d πn
Summary quesions A flywheel is roaing a an angular speed of 0 rad s on a fixed axle which is mouned on fricionless bearings. The momen of ineria of he flywheel and he axle abou he axis of roaion is 0.048 kg m. Calculae: a he roaional kineic energy of he flywheel when i roaes a 0 rad s b i he orque needed o accelerae he flywheel from res o an angular speed of 0 rad s in 5.0 s ii he angle which he flywheel urns hrough in his ime while i is being acceleraed. A 0.65 kg objec hanging from a sring is used o accelerae a flywheel on fricionless bearings from res, as shown in Figure. The objec falls hrough a verical disance of.9 m in 4.6 s which is he ime he sring akes o unwrap from he axle which has a diameer of 8.5 mm. Calculae: a he poenial energy los by he objec in descending.9 m b he kineic energy of he objec 4 s afer i was released from res c i he kineic energy gained by he flywheel ii he momen of ineria of he flywheel. 3 A ball released a he op of a slope rolls down he slope and coninues on a fla horizonal surface unil i sops. Discuss he energy changes of he ball from he momen i is released o when i sops. 4 A flywheel fied o a vehicle gains kineic energy when he vehicle slows down and sops. The kineic energy of he flywheel is used o make he vehicle sar moving again. a The flywheel is a uniform seel disc of diameer 0.3 m and hickness 0.08 m. Calculae: i he mass of he disc ii he momen of ineria of he flywheel. densiy of seel = 7800 kg m 3 ; momen of ineria of a uniform flywheel = MR b i Calculae he kineic energy of he flywheel when i is roaing a 3000 revoluions per minue. ii The kineic energy of he flywheel can be convered o kineic energy of moion of he vehicle in 30 s. Esimae he average power ransferred from he flywheel.
.4 Angular momenum Learning objecives: Wha is angular momenum and why is i imporan? Wha do we mean by conservaion of angular momenum? How do he equaions for angular momenum and linear momenum compare wih each oher? Spin a work Figure A spinning ice skaer An ice skaer spinning rapidly is a dramaic sigh. The skaer urns slowly a firs, hen quie suddenly goes ino a rapid spin. This sudden change is brough abou by he skaer pulling boh arms (and possibly a leg!) owards he axis of roaion. In his way he momen of ineria of he skaer abou he axis is reduced. As a resul, he skaer spins faser. To slow down, he skaer only needs o srech his or her arms and maybe a leg. In his way, he momen of ineria is increased. So he skaer slows down. To undersand such effecs, consider a roaing body wih no resulan orque on i. Provided is momen of ineria says he same, hen is angular speed does no change. This can be seen by rewriing he equaion T =I where is he angular acceleraion. If he resulan orque T is zero, hen he angular acceleraion is zero so he angular speed is consan. In more general erms, he equaion T = may be wrien as d Iω T d where d is he formal mahemaical way of wriing change per uni ime, d
The quaniy I is he angular momenum of he roaing body. So he above equaion ells us ha resulan orque T is he rae of change of angular momenum of he roaing objec. Angular momenum of a roaing objec = I where I is he momen of ineria of he body abou he axis of roaion and is is angular speed. The uni of angular momenum is kg m rad s or N m s as explained below in Noe. When he resulan orque is zero, hen consan. d Iω d 0 which means is angular momenum I is In he ice skaer example above, he momen of ineria of he ice skaer suddenly decreases when he or she pulls heir arms in. Since he angular momenum is consan, he sudden decrease in he momen of ineria causes he angular speed o increase. In specific erms, if he momen of ineria changes from I o I causing he angular speed o change from o such ha I = I Noes For a roaing objec whose momen of ineria I does no change: d Iω T d d I I because he angular acceleraion d d d If he objec undergoes uniform angular acceleraion from res o reach angular speed in I ime, he resulan orque acing on i is T. Hence is angular momenum I = T. This equaion shows ha he uni of angular momenum can also be given as he uni of orque he uni of ime i.e. N m s. Pulsars Pulsars are rapidly spinning sars. They were firs discovered by asronomers in 967. Regular pulses of radio energy were deeced from hese sars, which some asronomers called LGM sars. I seemed as if lile green men were rying o conac us! Tha hypohesis was soon abandoned when i was shown ha pulsars are in fac rapidly roaing neuron sars which emi radio energy in a beam a an angle o he axis. Each ime he beam sweeps round o poin owards Earh, radio energy is direced owards us, raher like a ligh beam from a lighhouse. Figure A pulsar
Neuron sars are he remnans of large sars. When a massive sar runs ou of fuel, a huge explosion akes place. The remnans of he explosion are pulled in ogeher by heir graviaional aracion, perhaps equivalen o a mass equal o he Sun shrinking o only 5 km or so in diameer. The momen of ineria is herefore made much smaller so he angular speed increases. The pulse frequency from a pulsar is of he order of o 0 Hz, so he rae of roaion is of ha order, much much greaer han he Sun s rae which is abou once every 5 days! Conservaion of angular momenum In boh of he above examples, he ice skaer and he pulsar, he angular momenum afer he change is equal o he angular momenum before he change because he resulan orque in each case is zero. In oher words, he angular momenum is conserved. Where a sysem is made up of more han one spinning body, hen when wo of he bodies inerac (e.g. collide), one migh lose angular momenum o he oher. If he resulan orque on he sysem is zero, he oal amoun of angular momenum mus say he same. Examples of he conservaion of angular momenum Capure of spinning saellie If a spinning saellie is aken on board a space repair laboraory, he whole laboraory is se spinning. The angular momenum of he saellie is ransferred o he laboraory when he saellie is aken on board and sopped. Unless rocke moors are used o preven i from urning, hen he whole laboraory would spin. Figure 3 The space shule Endeavour aking a communicaions saellie on board In his example: he oal angular momenum before he saellie is aken on board = I, where I is he momen of ineria of he saellie and is is iniial angular speed he oal angular momenum afer he saellie has been aken on board and sopped = (I + I ), where is he final angular speed and (I + I ) is he oal momen of ineria abou he axis of roaion of he saellie as I is he momen of ineria of he Space Shule. According o he conservaion of momenum, (I + I ) = I Rearranging his equaion gives I = I I
This rearranged equaion shows ha he angular momenum gained by he Space Shule is equal o he angular momenum los by he saellie. Objec dropped ono a urnable If a small objec is dropped ono a freely roaing urnable, as shown in Figure 4, so ha i sicks o he urnable, he objec gains angular momenum and he urnable loses angular momenum. Figure 4 Measuring he momen of ineria of a freely roaing urnable In his example: he oal angular momenum of he urnable before he objec is dropped ono i = I, where I is he momen of ineria of he urnable and is is iniial angular speed he oal angular momenum afer he objec has been dropped ono he urnable = (I + I ), where is he final angular speed and (I + I ) is he oal momen of ineria abou he axis of roaion of he urnable. If he mass of he objec is m and is perpendicular disance from he axis of roaion of he urnable is r, hen he momen of ineria I of he objec abou he axis of roaion = mr. According o he conservaion of momenum, (I + mr ) = I Rearranging his equaion gives mr Hence I can be found by measuring m, r, and. I Noe The angular momenum of a poin mass is defined as is momenum is disance from he axis of roaion. For a poin mass m roaing a angular speed a disance r from he axis, is momenum is mr (as is speed v = r) so is angular momenum is mr. For a nework of poin masses m, m,, m N which make up a rigid body, he oal angular momenum = (m r ) + (m r ) + + (m N r N ) Link See A Physics A Topic. for momenum. = [(m r ) + (m r ) +.+ (m N r N )] = I
Comparison of linear and roaional moion Applied physics When analysing a roaional dynamics siuaion, i is someimes useful o compare he siuaion wih an equivalen linear siuaion. For example, he linear equivalen of a orque T used o change he angular speed of flywheel is a force F used o change he speed of an objec moving along a sraigh line. If he change akes place in ime he change of momenum of he objec = F he change of angular momenum of he flywheel = T Table summarises he comparison beween linear and roaional moion. Table Comparison beween linear and roaional moion Linear moion displacemen s speed and velociy v acceleraion a mass m momenum mv force F F = ma F = d mv d Roaional moion angular displacemen angular speed angular acceleraion momen of ineria I angular momenum I orque T = Fd T = I T = I d d kineic energy mv kineic energy I work done = Fs power = Fv work done = T power = T
Summary quesions A vehicle wheel has a momen of ineria of 5.0 0 kg m and a radius of 0.30 m. a Calculae he angular momenum of he wheel when he vehicle is ravelling a a speed of 7 m s. b When he brakes are applied, he vehicle speed decreases from 30 m s o zero in 9.0 s. Calculae he resulan orque on he wheel during his ime. A meal disc X on he end of an axle roaes freely a 40 revoluions per minue. The momen of ineria of he disc and he axle is 0.044 kg m. a Calculae he angular momenum of he disc and he axle. Figure 5 b Afer a second disc Y ha is iniially saionary is engaged by X, boh discs roae a 60 revoluions per second. Calculae he momen of ineria of Y. c Show ha he oal loss of kineic energy is 4.6 J. 3 A pulsar is a collapsed sar ha roaes very rapidly. Explain why a slowly roaing sar ha collapses roaes much faser as a resul of he collapse. 4 A fricionless urnable is se roaing a a seady angular speed of 0 rad s. A small, 0. kg, mass is dropped ono he urnable from res jus above i, a a disance of 0.4 m from he cenre of he urnable. As a resul, he angular speed of he urnable decreases o 8 rad s. Calculae he momen of ineria of he urnable abou is axis of roaion.