Module Summary Sheets. Mechanics 3 (Version B: Reference to new book) Topic 4: Volumes of revolution and centres of mass by integration

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1 MEI Mathematics in Eucation an Inusty MEI Stuctue Mathematics Moule Summay Sheets (Vesion B: Refeence to new book) Topic : Cicula motion Topic : Elastic spings an stings Topic 3: Moelling oscillations Topic 4: Volumes of evolution an centes of mass by integation Topic 5: Dimensions an units Puchases have the licence to make multiple copies fo use within a single establishment Apil, 6 MEI, Oak House, 9 Epsom Cente, White Hose Business Pak, Towbige, Wiltshie. BA4 XG. Company No Englan an Wales Registee with the Chaity Commission, numbe 589 Tel: Fax:

2 Summay M3 Topic : Cicula Motion Chapte Pages -3 Example. Page 4 Execise A Q. 3, 8 Chapte Pages 7, 8 Example. Page 8 Chapte Pages 9-3 Example.4 Page 3 Execise B Q. Angula Spee an Velocities A foce is equie to keep a boy moving in a cicle as its velocity is always changing. If a paticle at P is moving in a cicle then the angle of OP with any fixe iection changes with time. The position is: x=, y= sin θ. Velocity: tangential component: θ v= = θ = ω t aial component : Acceleation : tangential component : θ = ω aial component: v θ = ω = the positive tangential iection Foces equie fo cicula motion θ. is witten as θ o ω t the positive aial iection Velocity Acceleation Newton's n Law ( F = ma) may be applie in the tangential an aial iections. Tangential component is m θ mv Raial component is m θ = mω = Vesion B: page Competence statements,, 3, 4, 5, 6, 7, 8 E.g. Ca A tavels oun a cicle of aius 3 m at 33 m s. Ca B tavels oun a cicle of aius 5 m at. a sec -. Which ca is moving the faste? If linea spees ae compae, Ca B: v = ω =..5 = 3 m s so ca A has geate linea spee. If angula spees ae compae, 33 Ca A: v = 33, = 3 ω = =. a s 3 So B has geate angula spee. E.g. The minute han of my watch is cm long. What is the spee of the tip of the han? =. m, ω = as pe hou = as s v = ms 3.5 m s ( s.f.) 36 E.g. What is the spee of the secon han, which is. cm long? 5. 3 v = m s. m s ( s.f.) E.g. A hoizontal, cicula ben on a ailway tack has a aius of 5 m an a tain of mass m kg negotiates it at 6 mph. What is the lateal foce, R N, between its wheels an the ails in tems of m? 6 mph 6.8 m s v 6.8 R= ma= m = =.44m 5 E.g. One en of a light inextensible sting of length.8 m is attache to a point V. The othe en is attache to the point O which is.4 m vetically below P. V A small smooth bea, B, of mass. kg is theae on θ.5 m the sting an moves in a.4 m hoizontal cicle, with cente O T N an aius.3 m. OB otates T N with constant angula O spee ω a s -..3 m B Fin the tension in the sting an ω..g N Res vet. T mg = 5 T =.g =.5 g( =.45...) 4 8 Res. Ho. T + Tsinθ = mω T = mω 5 8.5g ω = = ω = 8. ( s.f.)

3 Summay M3 Topic : Cicula Motion Chapte Pages 5-9 Example.6 Page 9 Execise B Q. 9 Chapte Pages 6-8 Example.7 Page 8 Execise C Q. 3 Chapte Pages 3-35 Banke Tacks - fo motion in hoizontal cicle When thee is pefect banking thee will be no fiction o lateal foce ue to ails, etc., so the only hoizontal foce acting will be the esolve pat of the nomal eaction, R. - i.e. F = Given that F =, mv R sin θ = whee R = mg v g tanθ = If the angle of the banking is geate o less than that fo pefect banking, then the boy will stay on the cicula tack only if thee is fiction o lateal foce. If thee is fiction (i.e. a oa athe than ails) then it may be enough to hol the boy on that cicula motion o it may move to a cicula motion with a geate o lesse aius. Cicula motion with constant angula acceleation The equations fo motion with constant angula acceleation α ae ω= ω + αt ω = ω + αθ θ = + αt θ = ( ω + ω) t whee ω is the initial angula spee. Motion in a vetical cicle If a paticle is constaine to move in a vetical cicle (i.e. at the en of a igi o o on a wie) Using consevation of mechanical enegy: KE + GPE = const; If we take GPE = at the level of O E.g. A ca of mass kg is moving hoizontally at constant spee oun a banke tack which is a cicle of aius 5 m. The bank is angle at. Fin the foce on the wheels if the spee is (i). m s -, (ii) 3 m s -. (The iagam is opposite) No vetical acceleation R = mg+ Fsinθ mv Hoizontally: = Rsinθ + F mv ( mg + F sinθ) sinθ = + F mg sinθ + F ( sin θ + cos θ) = mg sinθ + F mv = F = mgsinθ (i) v=., = 5, θ = F = m(.35.35) = (ii) v = 3, = 5, θ = F = m( ) = 5 N N.B. You o not nee to know the positive iection of F befoe esolving. If F ha been taken positive up the plane you woul simply have obtaine the answe 5 N. E.g. A paticle is swung in a vetical cicle at the en of a light o of aius with initial spee u at the bottom of the cicle. Fin the conition fo the paticle to complete the cicle. The paticle completes the cicle if v > when θ = 8 i.e. = ( θ) Given v = u g cos then if v> an =, u > g( ) = 4 g so u > 4g Example.8 Page 3 Execise D enegy at A: mu mg enegy at P: mv mg Since enegy is conseve: mu mg = mv mg v = u g ( ) Vesion B: page 3 Competence statements,, 3, 4, 5, 6, 7, 8 O Diagam hee E.g. A small ing of mass m is theae onto a smooth vetical cicula wie of aius.5 m. The ing is pojecte fom the lowest point of the wie with spee. m s. How fa above the bottom of the wie oes the ing each? The ing stops when v = an the angle is θ, Enegy at A = mu mg.5 Enegy when ing stops = mg.5 Refee to the iagam opposite an conseving enegy: mu mg.5 = mg.5. m = + =.5 θ =.5mg So the height eache is.5 -.5cos =.5 m

4 Summay M3 Topic : Cicula Motion 3 Chapte Pages Chapte Page 37 Chapte Page 38 Chapte Page 39 Example. Page 4 Execise D Q. 5, 7 When vetical cicula motion beaks own (i.e. when the paticle may leave the cicula path) - a paticle on a light, inextensible sting; no ai esistance NL towas the cente gives T + mg = mω The paticle will emain moving in the cicle if T ω g Fig. If this is so when = (i.e at the top, with the maximum value fo cos θ) then ω g o v g whee v is the least spee an ω is the least angula velocity. -motion on the insie of a vetical cicle; no fiction NL towas the cente gives R+ mg = mω The paticle will emain moving in the cicle if R ω We equie g If this is so when = Fig. ( i.e. at the top, with the maxiumum value fo cos θ) then ω g o v g whee v is the least spee an ω is the least angula velocity. -motion on the outsie of a cicle; no fiction NL towas the cente gives mg R = mω The paticle will emain on the suface of the cicle if R ω We equie cos θ g Fig. 3 v o whee v is the spee when the angle is θ. g Vesion B: page 4 Competence statements,, 3, 4, 5, 6, 7, 8 E.g. A paticle is attache to the en of a light inextensible sting of length an swung in a vetical cicle with no ai esistance. Fin the least spee at the bottom of the cicle fo the paticle to complete cicula motion. Using the notation of Fig. an the wok opposite; If the spee at the bottom is u an that at the top is v then the consevation of enegy gives mu = mv + mg so v = u 4g Using the esult that v g eive opposite we equie u 4g g u 5g E.g. A paticle is pojecte up the insie of a smooth sphee of aius. Fin the least angula spee at the bottom of the cicle fo the paticle to complete cicula motion. Using the notation of Fig. an the wok opposite; If the angula spee at the bottom is Ω an that at the top is ω then the consevation of enegy gives 4g m Ω = m ω + mg so ω =Ω Using the esult that ω g eive opposite 4g 5g we equie Ω g Ω Note that the solution is vey simila to that above. All that has been one is to eplace u by Ω an v with ω. E.g. A paticle of mass.5 kg is gently isplace fom est fom the top of a soli smooth sphee of aius.5 m. Fin whee it leaves the suface. Refe to Fig 3. NL towas the cente gives mg R =.5.5 ω If θ is the angle whee the paticle leaves the suface then R= an at this point.5g =.5.5 ω so ω = g Loss of GPE is mgh = mg( cos θ) Gain in KE is mv = m( ω) Conseving enegy gives m ω = mg ( cos θ) g( cos θ) so ω = = 4g ( cos θ ) But ω = gcos θ so 4g( ) = gcos θ 6 = 4 = 3 θ = 48. (3 s.f. )

5 Summay M3 Topic : Elastic stings an spings Chapte Pages 5-56 Example. Page 55 Execise A Q., 4 Example.3 Page 58 Execise B Q. 7 Chapte Pages 66-7 Example.5 Page 68 Execise C Chapte Pages Example.6 Page 74 Execise D Stings an spings We sometimes moel a sting to be inextensible. This is not always the case: stings an spings which stetch ae sai to be elastic. The length when thee is no foce attache to it is calle the natual length. When stetche the incease in length is calle the extension. A sping may be compesse. A sting cannot be compesse. Hooke's Law The tension in an elastic sping is popotional to the extension. If a sping is compesse the thust is popotional to the ecease in length of the sping. If the natual length is l an the extension is x, esulting fom a foce T, then T = kx. whee k is the stiffness. O we wite T = x l whee is calle the moulus of elasticity. Wok an Enegy The total wok one in stetching a sting by x fom its natual length l is given by Wok one = kx = x. l This is known as the elastic potential enegy. (E.P.E.) The Pinciple of Consevation of Enegy The pinciple of consevation of enegy may be use when the enegy changes only involve (some of) KE, GPE an EPE. Vetical motion When a paticle attache to a sping is elease fom a point othe than the equilibium position then the pinciples of consevation of enegy may be applie. Consevation of enegy gives: Incease in KE + incease in GPE + incease in EPE = Cae must be taken when an elastic sting is aise above its equilibium position as it will go slack athe than compess. Vesion B: page 5 Competence statements h,, 3, 4, 5 E.g. A piece of light sting is 5 cm long. Expeiments have inicate that = 8 N. A mass of 5 kg is hung fom the sting. Fin the extension. 5g.5 T = 5 g = x x= =.36 l 8 Extension is 3.6 cm (3 s.f.). E.g. A light sping of natual length 5 cm is attache to a ceiling. When a mass of 4 g is hung in equilibium, the extension is 5 cm. Fin (i) the tension in the sping, (ii) the stiffness of the sping, (iii) the moulus of elasticity of the sping. Take g = m s. (i) T = mg =.4g =.4 N (ii) T = kx.4 =.5k k =.8 N m x 5 T= x = T = = (iii).4.4n x x 5 E.g. Fin the EPE stoe in the example above when the system is in equilibium kx = = =. J E.g. Suppose, in the example above, the mass is pulle own a futhe cm an elease fom est. Does the sting go slack? Total extension is 5 cm. EPE stoe is.8.5 =.9 J GPE gaine by mass in being aise 5 cm is.4.5 =.6 J As EPE >GPE the mass is still moving upwas when the sting goes slack. Its spee is given by.4 v =.9.6 so v =.5; spee is. m s (3 s.f.) A light buffe of natual length 5 cm has a moulus of elasticity kn. (i) What foce will compess it to half its length? (ii) How much enegy is stoe in the buffe when it is compesse to cm..5 (i) T = x= = 5kN l.5 5 (ii) EPE = (.5.) = 9 J.5 E.g. A light sping of stiffness 5 N m can poject a 5 g stone 5 m vetically. How much compession in the sping is equie? mgh kx = mgh x = k = =.33 m 5

6 Summay M3 Topic 3: Moelling oscillations Chapte 3 Pages 8-87 Example 3. Page 87 Execise 3A Q. 3 Chapte 3 Pages Example 3. Page 89 Execise 3B Oscillating motion Suppose a paticle oscillates about a cental position, O, on the x-axis. The paticle moves between two points, x = a an x = +a. The istance a is calle the amplitue of the motion. The motion epeats itself in a cyclic fashion. The numbe of cycles pe secon is calle the fequency, enote by v. The motion epeats itself afte time T. The time inteval, T, is calle the peio: it is the time fo a complete cycle of the motion. The fequency is the ecipocal of the peio. So v = T The S.I. unit fo fequency is the Hetz. Hetz is one cycle pe secon. Simple Hamonic Motion The acceleation is popotional to the magnitue of the isplacement fom the cente point of the motion an is iecte towas this cente point. If x is the isplacement the SHM equation is x = ω x which gives v =ω ( a x ) In aition: x is iecte towas O, so the foce causing the action must also be iecte towas O. x (an theefoe the foce) = when x =. x has a maximum value at the extemes x=± a. The maximum velocity is when x = an is ± aω. The peio, T =. ω SHM as a function of time. Any motion of the fom x = asin is a solution of the SHM equation since x = asin v = x = aωcos x = aω sin ω t = ω x an so is a solution of the SHM equation The solution above is a paticula solution of the moe geneal fom x = C sin( ω t + ε) (See next page fo a list of expessions fo SHM.) Also, given v = x = aωcos ω t, v = a ω cos v = a ( sin ω t) = a x So v = a x E.g. The height, h m, of wate in a habou may be moelle ove a shot time by simple hamonic motion, with peio hous. At high wate the epth of wate is 6 m geate than the epth at low wate. ω = = as pe hou. T 6 Amplitue of oscillation = 3 m. Maximum spee = aω = metes pe hou. =.6 cm pe minute (3.s.f.). 6 Assuming h = when t =, h = 3sin t 6 Checking gives h = cos t, h = sin t h = h = ω h as equie. 6 E.g. An oscillating weight on the en of a light sping completes 48 cycles pe minute. The amplitue is estimate as 5 cm. If the oscillation begins at the equilibium point, wite an equation of motion an fin the geatest spee. Let x be the isplacement fom the equilibium point v = 48 cpm = Hz T = ω = = = ( = 5.65) T x =.5sin t =.5sin.6 t 6 Max v = aω =.5 m s (3.s.f.) x = ω x = x Equation of motion: 5.7 (4 E.g. If t = at the mile of the motion then when t =, x =. Using x= Csin( + ε) we have = asin ε ε = giving x= asin If t = at one en of the motion then when t =, x = ±a. Again, using x = Csin( + ε ) s.f.) Taking x = a when t = a = asin ε ε = giving x = asin + = acos Taking x = a when t = x = acos Vesion B: page 6 Competence statements o,, 3, 4, 5, 6, 7, 8, 9,

7 Summay M3 Topic 3: Moelling oscillations Chapte 3 Pages 99-5 Execise 3C Q., Chapte 3 Pages 7-9 Example 3.6 Page 7 Execise 3C Q. 8 Chapte 3 Pages 5-9 Example 3.7 Page 7 Execise 3D Q. 3 Chapte 3 Pages 9-4 Execise 3D Q. 3 Altenative foms of the solution of the SHM equation Motion in a staight line of the fom x = Asin ω t + Bcos o x = Csin( ω t +ε) o x = C cos( ω t +ε') satisfy the equation x = ω an the motion is SHM. Cicula motion an SHM If a paticle is moving at constant angula spee ω oun a cicle of aius a whose cente is O an the x-axis is a iamete, then the x cooinate of its position is given by The simple penulum Oscillating Spings If a paticle is suspene fom a pefectly elastic light sping of natual length l, an moulus of elasticity, then the equation of motion of oscillation is given by x = ml whee x is the isplacement fom the equilibium position. This is the stana equation fo SHM with ω = ; This gives a peio of ml x x = acos θ whee θ = an ω is the angula spee. x= acos an x = aωsin So x= aω cos = ω x Use x= asin when t = at the cente of motion. Use x= acos when t = at the exteme of the motion in the positive iection. In tangential iection: ml θ = mg sinθ gθ Fo small θ,sinθ θ giving θ l This equation of motion is the SHM equation. l The solution has peio T =. g θ R The solution of the equation may be g witten as θ = αsin t + ε whee mg l α is the maximum amplitue. Note that the equation fo motion is the angula isplacement of the penulum. x ml Consie the motion of a paticle in a staight line whee the isplacement, x, is given by x = sin t + 3cos t x = cos t sin t x = sin t cos t = sin t 3cos + t = x So the motion is SHM. Also x= sin t + 3cos t 4 4 x= sin t cos ε+ cos t sinε 4 4 whee cosε = an sin ε = 3 = 3 ancos ε = ε =.983 (3 s.f.) 3 x= 3sin t+.983 (3 s.f.) 4 E.g. A simple penulum is in the fom of a light o of length. m with a heavy paticle at the en. It is pulle to one sie though an angle of fom the vetical an then elease fom est. Fin (i) the peio, (ii) the time to each the lowest point, (iii) the angula spee at the lowest point. (Take g = 9.8 m s -.) g (i) ω = = ω= l Peio =. s (3 s.f.) ω (ii) Motion is moelle by θ = αcos whee α = a 8 At lowest point θ = cos= t= sec. ω (iii) Spee = aω = =.9975 a s. E.g. How long shoul be a simple penulum fo a secon tick. T g 4g secon peio so = l =.993 m 4 4 Vesion B: page 7 Competence statements M3 o,, 3, 4, 5, 6, 7, 8, 9,

8 Summay M3 Topic 4: Volumes of evolution an centes of mass by integation Chapte 4 Pages 3-38 Example 4. Page 35 Execise 4A Q. (v),4 Chapte 4 Pages 44-5 Example 4.5 Page 49 Execise 4B Q. 3, 6 Chapte 4 Pages Execise 4B Q. 8 Chapte 4 Page 6 Execise 4C Volume of evolution about the x-axis If the cuve y = f(x) between the oinates x = a an x = b is otate though 36 about the x-axis then the volume of the esulting soli is given by: b V = y x a Note that y must be eplace by f(x) befoe integating. Volume of evolution about the y-axis If the cuve x = f(y) between y = c an y = is otate though 36 about the y-axis then the volume of the esulting soli is given by: V = x y c Note that x must be eplace by f(y) befoe integating. Centes of mass (c.o.m) Fo a volume of evolution about the x-axis: b b x y x= xy x, y= a Fo a volume of evolution about the y-axis: Centes of mass of plane egions Fo the lamina which is the aea between the cuve y = f(x), the x-axis, x = a an x = b. Fo the lamina which is the aea between the cuve x = f(y), the y-axis, y = c an y =. x y x c A = whee A x y y = c xy y c Vesion B: page 8 Competence statements g,, 3, 4, 5 a y x y= yx y, x =. c b xy x b x a A = whee A y x y b = y a y a c E.g.. The line y = x + fom x = to x = is otate though 36 about the x-axis. Fin the volume of the soli fome. ( ) ( ) V = y x= x+ x= x + x+ x 3 x 8 6 = + x + x = + 4+ = E.g.. The cuve y = x fom y = to y = 3 is otate though 36 about the y-axis. Fin the volume of the soli fome. 3 3 ( ) V = x y= y+ y 3 y 9 = + y 3 6 = + + = Centes of mass of stana shapes Soli shape Volume Height of c.o.m. above plane base Soli hemisphee, aius Soli cone, height h, aius h h 3 4 Hollow shape Cuve Height of c.o.m. suface aea above base Hollow hemisphee, aius Hollow cone, height h, aius l h 3 Composite boies may be mae by combining simple shapes in some way. E.g. The c.o.m. of a fustum which is the lowe half of a soli cone, height h. The height of the fustum is h /. Let the mass of cone be 8m. Then mass of cone cut off is m an mass of fustum is 7m. O h / h / h h h 8m = 7mx + m h= 7x + h 7x = h x = h x

9 Summay M3 Topic 5: Dimensions an units Chapte 5 Pages 68-7 Example 5. Page 7 Chapte 5 Page 7 Execise 5A Chapte 5 Pages 7 Example 5.3 Page 73 Execise 5B Q. The thee funamental imensions Mass Length Time M L T Othe imensions [Aea] means the imensions of aea which is L. [Aea] = L [Volume] = L 3 [Spee] = LT [Acceleation] = LT - [Foce] = MLT In any equation o fomula with units thee must be imensional consistency. Dimensionless Quantities All numbes, incluing an e, angles an tigonometical atios have no imensions. Using imensions to check elationships When all quantities ae given in algebaic fom, imensional consistency can be use to check answes to poblems. The imensions of a quantity help to etemine which units ae appopiate. E.g. KE = mv [ KE] mv M( LT ) = = = ML T E.g. v = u + at [v] = [u + at] Dimensions of L.H.S. ae LT Dimensions of R.H.S. ae LT an LT.T = LT So the fomula has imensional consistency. E.g. Justify the imensional consistency of the fomula mv mu = mghsin θ whee the vaiables have the usual meaning. [ mgh θ ] L ML mv = mu = M. = T T L ML sin = M..L = T T E.g. Given that pessue = foce pe unit aea, fin the imensions of pessue. Foce Pessue = [ Pessue ] = Aea MLT = = ML T L [ Foce] [ Aea] Chapte 5 Pages Example 5.5 Page 75 Execise 5B Q. Fining the fom of a elationship It is sometimes possible to etemine the fom of a elationship just by looking at the imensions of the quantities. Thee ae thee funamental quantities - Mass, Length an Time.. Moelling assumptions ae mae in the usual way. 3. The metho can only be use when a quantity can be witten as a pouct of powes of othe quantities. Thee ae estictions on the numbe of quantities involve. Vesion B: page 9 Competence statements q,, 3, 4, 5, 6 E.g. the fequency, f, of the note emitte by an ogan pipe of length a when the ai pessue is p an the ai ensity is is believe to obey a law of the fom f = ka α p β γ whee k is a imensionless constant. Fin the values of α, β an γ. [ ] α β γ α β γ α β γ f = ka p f = a p = a p β ( ) ( ) α 3 γ T = L ML T ML Equating: Fo T: = β Fo L: = α β 3γ Fo M: = β + γ Solving gives β =, γ =, α = k p f = ka p =. a

AH Mechanics Checklist (Unit 2) AH Mechanics Checklist (Unit 2) Circular Motion

AH Mechanics Checklist (Unit 2) AH Mechanics Checklist (Unit 2) Circular Motion AH Mechanics Checklist (Unit ) AH Mechanics Checklist (Unit ) Cicula Motion No. kill Done 1 Know that cicula motion efes to motion in a cicle of constant adius Know that cicula motion is conveniently descibed