1.1 KINEMATIC RELATIONSHIPS

Transcription

1 1.1 KINEMATIC RELATIONSHIPS Thoughout the Advanced Highe Physics couse calculus techniques will be used. These techniques ae vey poweful and knowledge of integation and diffeentiation will allow a deepe undestanding of the natue of physical phenomena. Kinematics is the study of the motion of points, making no efeence to what causes the motion. Displacement, velocity and acceleation ae addessed hee. Displacement The displacement, s, of a paticle is the length and diection fom the oigin to the paticle. The displacement of the paticle is a function of time: s = f(t) Conside a paticle moving along OX. At time t the paticle will be at point P. At time t +Δt paticle passes Q. Velocity aveage velocity Howeve the instantaneous velocity is diffeent, this is defined as: so v = ds dt Acceleation velocity changes by Δv in time Δt aveage acceleation Instantaneous acceleation: so a = dv dt 1

2 if then a = dv dt = d 2 s dt 2 Note: a change in velocity may esult fom a change in diection (e.g. unifom motion in a cicle - see late). 2

3 Mathematical Deivation of Equations of Motion fo Unifom Acceleation Two methods ae shown hee. One using the implementation of initial and final conditions (left column), the second using definite integals to the same effect (ight). Integate with espect to time: when t = 0 t = t ds dt = u so k = u ds dt = v [1] [1] integate again with espect to time: emembe that v = ds dt = u + at Substitute [1] apply initial conditions: when t = 0, s = 0 hence k = 0 [2] Equations 1 & 2 can now be combined: - - Squae both sides of [1]: [2] [3] is found by substituting [2], giving: 3

4 A useful fouth equation is EoM 4 This equation can be used to calculate displacement by using an aveage velocity when moving with a constant acceleation. Vaiable Acceleation If acceleation depends on time in a simple way, calculus can be used to solve the motion. This would look like a highe ode polynomial, fo example: Diffeentiating this expession twice will yield an acceleation which is still dependant on time! 4

5 Gaphs of Motion The slope o gadient of these gaphs povides useful infomation. Also the aea unde the gaph can have a physical significance. v = ds dt Displacement - time gaphs gadient = instantaneous velocity. Aea unde gaph - no meaning. a = dv dt Velocity - time gaphs gadient = instantaneous acceleation. Also s = v dt Aea unde v-t gaph gives the displacement. Calculations Involving Unifom Acceleations Examples of unifom acceleation ae: e c o on of pojec e ne he E h f ce, whee the acceleation a = g = 9.8 m s -2 vetically downwads ectilinea (i.e. staight line) motion e.g. vehicle acceleating along a oad. These have been coveed peviously; howeve a fulle mathematical teatment fo pojectiles is appopiate at this level. Conside the simple case of an object pojected with an initial velocity u at ight angles o he E h g on f e - (locally the field lines may be consideed paallel). a = g, time to tavel hoizontal distance s h is t apply s v = uvt a t2, uvt = 0 and a = g h h g g h h h h g and u h ae constants, s v and we have the equation of a paabola. The above poof and equations ae not equied fo examination puposes. 5

6 1.2 ANGULAR MOTION The Radian The adian is used when measuing a new quantity known as angula displacement, θ, measued in adians (ad). One adian epesents an ac with a length of one adius of that cicle. This is the displacement (in angle fom) aound the ac of a cicle, which has an equivalent angle in degees. Thee ae o π adians n h f c c e ( 8 ) Thee e o π n n f c c e ( 6 ) An angula displacement is theefoe linked to a linea displacement by 1 adius. θ Angula Velocity The angula velocity of a otating body is defined as the ate of change of angula displacement. θ whee is the angula velocity measued in adians pe second (ad s -1 ) Angula Acceleation whee he angula acceleation with units of ad s -2 We assume fo this couse that emains constant. Angula Motion Relationships Linea Quantity Relationship Angula Equivalent s θ θ u 0 v a t t 6

7 Angula Equations of Motion The deivation of the equations fo angula motion ae vey simila to those fo linea motion seen ealie. [1] θ [2] θ [3] You will note that these angula equations have exactly the same fom as the linea equations. Remembe that these equations only apply fo unifom (constant) angula acceleations. Unifom Motion in a Cicle Conside a paticle moving with unifom speed in a cicula path as shown hee. θ (Note: s is the ac swept out by the paticle and s = The otational speed v is constant, is also constant. T is the peiod of the motion and is the time taken to cove 2π adians. π T and π T since and s = π fo full otation Giving us:. 7

8 Angula acceleation and linea tangential acceleation Angula acceleation is given by Linea tangential acceleation is given by a t = dv dt when the otational speed v is changing. S nce then at any instant dv dt = giving a t = whee the diection of a t is at a tangent to the cicula path of adius. Radial (o Cental) Acceleation Conside a paticle undegoing cicula motion. The paticle tavels fom A to B in time Δ n w h speed v, thus u = v and Δ ( u) which gives Δv = v u. Now, Δ c θ aveage acceleation, Δ Δ Δ n θ n θ θ As θ tends to 0, aav tends to the instantaneous acceleation at point Q: θ n θ but, since when θ is small and is measued in adians sin θ = θ. since v = The diection of this acceleation is always towads the cente of the cicle. Note: This is not a unifom acceleation. Radial acceleation continuously changes diection. Its magnitude changes if the speed of otation changes. This motion is typical of many cicula types of motions (o simila) e.g. planetay o on, e ec on ob ng n c e n e ec on njec e gh ng e o n fo magnetic field. They ae all situations whee thee is a cental foce acting on the paticle. Thus any object pefoming cicula motion at unifom speed must have a constant cente-seeking o cental foce esponsible fo the natue of such motion. 8

9 Cental Foce Does a otating body eally have an inwad acceleation (and hence an inwad foce)? Agument: Most people have expeienced the sensation of being in a ca o a bus which n ng cone h gh pee The fee ng of be ng hown o he o e of he c e ey ong, e pec y f yo e ong he e Wh h ppen hee is that the fiction between youself and the seat is insufficient to povide the cental foce needed to deviate you fom the staight line path you wee following befoe the tun. In fact, instead of being thown outwads, you ae, in eality, continuing in a staight line while the ca moves inwads. Eventually you ae moved fom the staight line path by the inwad (cental) foce povided by the doo. Magnitude of the Foce and Thus cental foce, Examples since 1. A Ca on a Flat Tack If he c goe oo f, he c be w y ngen The foce of fiction is not enough to supply an adequate cental foce. 2. A Ca on a Banked Tack Fo tacks of simila suface popeties, a ca will be able to go faste on a banked tack befoe going off at a tangent because thee is a component of the nomal eaction as well as a component of fiction, F, supplying the cental foce. The cental foce is R sinθ + F cos θ which educes to R sinθ when the fiction is zeo. The analysis on the ight hand side is fo the fiction F equal to zeo. R he no e c on foce of he c on he c In the vetical diection thee is no acceleation: R cosθ = mg... 1 In the adial diection thee is a cental acceleation: R sinθ = mv Divide Eq. 2 by Eq. 1: n θ g (The eq on pp e o c e of b n ng nc ng aicaft tuning in hoizontal cicles). Remembe that we ae assuming that thee is no component povided by fiction. 9

10 1.3 ROTATIONAL DYNAMICS Moment of a foce The moment of a foce is the tuning effect it can poduce. Examples of moments ae: ng ong h n e cew e o e e off he of p n n, using a claw hamme to emove a nail fom a block of wood o leveing off a cap fom a bottle. Pushing a doo neae the handle than the hinge. The magnitude of the moment of a foce (o the tuning effect) = F is the foce applied d is the pependicula distance fom the diection of the foce to the tuning point The maximum tuning effect is also achieved when these ae at ight angles. Afte that we would need to conside that it depends on: Toque (i.e. sin 90 = 1) Fo cases whee a foce is applied and this causes otation about an axis, the moment of the foce is known as toque. Conside a foce F applied tangentially to the im of a disc which can otate about an axis O though its cente. The disc has adius. The toque T associated with this foce F is defined to be the foce multiplied by the adius.] T Toque has unit newton mete (N m) If the foce is not applied at a tangent to then T n θ is used Toque is a vecto quantity. The diection of the toque vecto is at ight angles to the plane containing both and F and lies along the axis of otation. (In the example shown in the diagam toque, T, points out of the page). A tangential foce acting on the im of an object will cause the object to otate; e.g. applying a push o a pull foce to a doo to open and close, poviding it ceates a non- 10

11 zeo esulting toque. The distance fom the axis of otation is an impotant measuement when calculating toque. It is instuctive to measue the elative foces equied to open a doo by pulling with a sping balance fistly at the handle and then pulling in the middle of the doo. Anothe example would be a toque wench which is used to o e he whee n on c o ce n gh ne pec f e by he manufactue. As with linea motion, an unbalanced toque will esult in an angula acceleation, wheeas balanced toques will esult in constant angula velocity. In the above diagam if thee ae no othe foces then the foce F will cause the object to begin otating. Inetia The magnitude of the linea acceleation poduced by a given unbalanced foce will depend on the mass of the object, also known as its inetia. The wod inetia can be loosely descibed as esistance to change in motion of an object. Objects with a lage mass ae difficult to stat moving and once moving ae difficult to stop. Moment of Inetia The moment of inetia, I, of an object can be descibed as its esistance to change in its angula motion. The moment of inetia fo otational motion is analogous to the mass, m, fo linea motion. The moment of inetia of an object depends on the mass and the distibution of the mass about the axis of otation. Fo a mass, m, at a distance,, fom the axis of otation the moment of inetia of this mass is given by: I unit of Moment of Inetia, kg m 2 Example A vey light od has two 0.8 kg masses each at a distance of 50 cm fom the axis of otation. The moment of inetia of each mass is m 2 = 0.8 x = 0.2 kg m 2 giving a total moment of inetia I = 0.4 kg m 2. Notice that we assume that all the mass is at the 50 cm distance. The small moment of inetia of the light od has been ignoed. Anothe example is a hoop, with vey light spokes connecting the hoop to an axis of otation though the cente of the hoop and pependicula to the plane of the hoop, e.g. a bicycle wheel. Almost all the mass of the hoop is at a distance R, whee R is the adius of the hoop. Hence, I = M R 2 whee M is the total mass of the hoop. 11

12 Fo objects whee all the mass can be consideed to be at the same distance fom the axis of otation this equation I = m 2 can be used diectly. Howeve most objects do not have all thei mass at a single distance fom the axis of otation and we must conside the distibution of the mass. 12

13 Moment of inetia and mass distibution Conside a small paticle of the disc as shown. This paticle of mass m is at a distance fom the axis of otation 0. The contibution of this mass to the moment of inetia of the whole object (in this case a disc) is given by the mass m multiplied by 2. To obtain the moment of inetia of the disc we need to conside all the paticles of the disc, each at thei diffeent distances. Any object can be consideed to be made of n paticles each of mass m. Each paticle is at a paticula adius fom the axis of otation. The moment of inetia of the object is detemined by the summation of all these n paticles e.g.. Calculus methods ae used to detemine the moments of inetia of extended objects. In this couse, moments of inetia of extended objects, about specific axes, will be given. Some examples include: It can be shown that the moment of inetia of a unifom od of length L and total mass M though its cente is M, but the moment of inetia of the same od though its end is M, i.e. fou times lage. This is because it is hade to make the od otate about an axis at the end than an axis though its middle because thee ae now moe paticles at a geate distance fom the axis of otation. Toque and Moment of Inetia An unbalanced toque will poduce an angula acceleation. As discussed above, the moment of inetia of an object is the opposition to a change in its angula motion. Thus the angula acceleation,, poduced by a given toque, T, will depend on the moment of inetia, I, of that object. T I 13

14 Angula Momentum The angula momentum L of a paticle about an axis is defined as the moment of momentum. p c e of o e -1 about the point O. The linea momentum p = m v. The moment of p = m v ( is pependicula to v). 2, since v =. Thus the angula momentum of this paticle, L = m v = m Fo a igid object about a fixed axis the angula momentum L will be the summation of all the individual angula momenta. Thus the angula momentum L of an object is given by Σ (m 2 ) This can be witten as Σ (m 2 ), since all the individual pats of the object will have the same angula velocity,. Also, we have I = (m 2 ). Thus the angula momentum of a igid body is: I the units of L ae kg m 2 s -1. Notice that the angula momentum of a igid object about a fixed axis depends on the moment of inetia. Angula momentum is a vecto quantity. The diection of this vecto is at ight angles to the plane containing v (since p = m v and mass is scala) and and lies along the axis of otation. Fo inteest only, in the above example L is out of the page. (Consideation of the vecto natue of T and L will not be equied fo assessment puposes.) Consevation of angula momentum The total angula momentum befoe an impact will equal the total angula momentum afte impact poviding no extenal toques ae acting. You will meet a vaiety of poblems which involve use of the consevation of angula momentum duing collisions fo thei solution. Rotational Kinetic Enegy The otational kinetic enegy of a igid object is also dependant on the moment of inetia of that object.fo an object of moment of inetia, I,otating unifomly at ad s -1 the otational kinetic enegy is given by: 14

15 Enegy and wok done If a toque, T, is applied though an angula displacement, θ, hen wok done = T θ Doing wok poduces a tansfe of enegy wok done = ΔEk 15

16 Summay and Compaison of Linea and Angula Equations Quantity Linea Motion Angula Motion acceleation a ( ) velocity ( ) displacement ( θ) θ momentum p I kinetic enegy I New on Secon w T I I Laws Consevation of linea momentum b b b b Consevation of angula momentum I I b I I b Consevation of linea kinetic enegy - Consevation of angula kinetic enegy Tθ I - I Some Moments of Inetia (fo efeence) Thin disc about an axis though its cente I = 1 2 and pependicula to the disc. M R2 R = adius of disc Thin od about its cente I = 1 12 ML2 L = length of od Thin hoop about its cente I = M R 2 R = adius of hoop Sphee about its cente I = 2 5 M R2 R = adius of sphee Whee M is the total mass of the object in each case. 16

17 Objects Rolling down an Inclined Plane When an object such as a sphee o cylinde is allowed to un down a slope, the E p at the top, (m g h), will be conveted to both linea ( ) and angula ( I ) kinetic enegy. An equation fo the enegy of the motion (assume no slipping) is given below. g h I The above fomula can be used in an expeimental detemination of the moment of inetia of a cicula object. Example A solid cylinde is allowed to oll fom est down a shallow slope of length 2.0 m. The height of the slope is 0.02 m, the time taken to oll down the slope is 7.8 s. The mass of the cylinde is 10 kg and its adius is 0.10 m. Using this infomation about the motion of the cylinde and the equation above, calculate the moment of inetia of the cylinde. Solution ( ) ( ) E p E ne c E o on g h I 9 8 ( 5 ) I (5 ) ( 96 ) g 17

18 The Flywheel Example The flywheel shown below compises a solid cylinde mounted though its cente and is fee to otate in the vetical plane. Flywheel: mass = 25 kg adius = 0.30 m. Mass of hanging weight = 2.5 kg The hanging weight is eleased. This esults in an angula acceleation of the flywheel. Assume that the effects of fiction ae negligible. (a) Calculate the angula acceleation of the flywheel. (b) Calculate the angula velocity of the flywheel just as the weight eaches gound level. Solution (a) We need to know the moment of inetia of the flywheel: I M R I 5 ( ) 5 g Now conside the foces involved. The weight of the hanging mass (mg) is esponsible fo the acceleation of the hanging mass as it descends (given by ma) and the tangential foce (F T ) applied to the flywheel leading to its tangential acceleation. g T a is linea a of mass (and tangential a of flywheel) g - T g - T ng cce e on of f ywhee ( nce ) Now Toque, And so Finally (b) To calculate the angula velocity we will need to know θ, the angula displacement fo a length of ope 2.0 m long being unwound. giving: θ 6 67 o = 0 2 = o θ =? 2 = x 5.44 x

19 = 5.44 ad s -2 2 = θ = 6.67 ad = 8.52 ad s -1 Fictional Toque Example The fiction acting at the axle of a bicycle wheel can be investigated as follows. The wheel, of mass 1.2 kg and adius 0.50 m, is mounted so that it is fee to otate in the vetical plane. A diving toque is applied and when the wheel is otating at 5.0 evs pe second the diving toque is emoved. The wheel then takes 2.0 minutes to stop. (a) Assuming that all the spokes of the wheel ae vey light and the adius of the wheel is 0.50 m, calculate the moment of inetia of the wheel. (b) Calculate the fictional toque which causes the wheel to come to est. (c) The effective adius of the axle is 1.5 cm. Calculate the foce of fiction acting at the axle. (d) Calculate the kinetic enegy lost by the wheel. Whee has this enegy gone? Solution (a) In this case I fo wheel = MR 2 I ( 5 ) I g (b) To find fictional toque we need the angula acceleation since T I t = 120 s -1 0 = 5.p.s. = 31.4 ad s = ad s -2 T I (- 6 ) = N m This is a fictional foce and so a negative value is sensible! (c) Toque and foce elated by: T ( = 1.5 cm = m) T N again negative value indicates foce opposing motion. (d) All initial kinetic enegy has been lost and so: When the wheel stops Ek(ot) = 0. This 148 J will have changed to heat in axle due to the wok done by the foce of fiction. the 19

20 Consevation of Angula Momentum Example A tuntable, which is otating on fictionless beaings, otates at an angula speed of 15 evolutions pe minute. A mass of 60 g is dopped fom est just above the disc at a distance of 0.12 m fom the axis of otation though its cente. As a esult of this impact, it is obseved that the ate of otation of the disc is educed to 10 evolutions pe minute. (a) (b) Use this infomation and the pinciple of consevation of angula momentum to calculate the moment of inetia of the disc. Show by calculation whethe this is an elastic o inelastic collision. Solution (a) moment of inetia of disc = I moment of inetia of 60 g mass = m 2 ( e p c e =0.12 m) = 0.06 x (0.12) 2 Imass = 8.64 x 10-4 kg m 2 initial angula velocity = o = 15 ev min -1 = o = 1.57 ad s -1 final angula velocity = = 10 ev min -1 = 1.05 ad s -1 total angula momentum befoe impact = total angula momentum afte impact I (I I ) I 57 (I ) 5 5 I 9 7 I g (b) Ek befoe impact = I 74 - ( 57) Ek afte impact = (I 6 - ( 5) Ek diffeence = 7.1 x 10-4 J Thus the collision is inelastic. The enegy diffeence will be changed to heat. 20