Chapter 20 Rigid Body: Translation and Rotational Motion Kinematics for Fixed Axis Rotation
|
|
- Nelson Morton
- 7 years ago
- Views:
Transcription
1 Chapter 20 Rgd Body: Translaton and Rotatonal Moton Knematcs for Fxed Axs Rotaton 201 Introducton Constraned Moton: Translaton and Rotaton Rollng wthout slppng 5 Example 201 Bcycle Wheel Rollng Wthout Slppng 6 Example 202 Cylnder Rollng Wthout Slppng Down an Inclned Plane 9 Example 203 Fallng Yo-Yo 10 Example 204 Unwndng Drum Angular Momentum for a System of Partcles Undergong Translatonal and Rotatonal 12 Example 205 Earth s Moton Around the Sun Knetc Energy of a System of Partcles Rotatonal Knetc Energy for a Rgd Body Undergong Fxed Axs Rotaton 18 Appendx 20A Chasles s Theorem: Rotaton and Translaton of a Rgd Body 19
2 Chapter 20 Rgd Body: Translaton and Rotatonal Moton Knematcs for Fxed Axs Rotaton Hence I feel no shame n assertng that ths whole regon engrdled by the moon, and the center of the earth, traverse ths grand crcle amd the rest of the planets n an annual revoluton around the sun ear the sun s the center of the unverse Moreover, snce the sun remans statonary, whatever appears as a moton of the sun s really due rather to the moton of the earth Introducton Coperncus The general moton of a rgd body of mass m conssts of a translaton of the center of mass wth velocty V cm and a rotaton about the center of mass wth all elements of the rgd body rotatng wth the same angular velocty ω cm We prove ths result n Appendx A Fgure 201 shows the center of mass of a thrown rgd rod follows a parabolc trajectory whle the rod rotates about the center of mass Fgure 201 The center of mass of a thrown rgd rod follows a parabolc trajectory whle the rod rotates about the center of mass 202 Constraned Moton: Translaton and Rotaton We shall encounter many examples of a rollng object whose moton s constraned For example we wll study the moton of an object rollng along a level or nclned surface and the moton of a yo-yo unwndng and wndng along a strng We wll examne the constrant condtons between the translatonal quanttes that descrbe the moton of the center of mass, dsplacement, velocty and acceleraton, and the rotatonal quanttes that descrbe the moton about the center of mass, angular dsplacement, angular velocty and angular acceleraton We begn wth a dscusson about the rotaton and translaton of a rollng wheel 1 colaus Coperncus, De revolutonbus orbum coelestum (On the Revolutons of the Celestal Spheres), Book 1 Chapter
3 Fgure 202 Rollng Wheel Consder a wheel of radus R s rollng n a straght lne (Fgure 202) The center of mass of the wheel s movng n a straght lne at a constant velocty V cm Let s analyze the moton of a pont P on the rm of the wheel Let v P denote the velocty of a pont P on the rm of the wheel wth respect to reference frame O at rest wth respect to the ground (Fgure 203a) Let v P denote the velocty of the pont P on the rm wth respect to the center of mass reference frame O cm movng wth velocty V cm wth respect to at O (Fgure 203b) (You should revew the defnton of the center of mass reference frame n Chapter 1521) We can use the law of addton of veloctes (Eq1524) to relate these three veloctes, v P = v P + V cm (2021) Let s choose Cartesan coordnates for the translaton moton and polar coordnates for the moton about the center of mass as shown n Fgure 203 (a) (b) Fgure 203 (a) reference frame fxed to ground, (b) center of mass reference frame The center of mass velocty n the reference frame fxed to the ground s gven by V cm = V cm î (2022) 20-2
4 where V cm s the speed of the center of mass The poston of the center of mass n the reference frame fxed to the ground s gven by R cm (t) = (X cm, 0 + V cm t)î, (2023) where X cm, 0 s the ntal x -component of the center of mass at t = 0 The angular velocty of the wheel n the center of mass reference frame s gven by ω cm = ω cm ˆk (2024) where ω cm s the angular speed The pont P on the rm s undergong unform crcular moton wth the velocty n the center of mass reference frame gven by v P = Rω cm ˆθ (2025) If we want to use the law of addton of veloctes then we should express v P = Rω cm ˆθ n Cartesan coordnates Assume that at t = 0, θ(t = 0) = 0 e the pont P s at the top of the wheel at t = 0 Then the unt vectors n polar coordnates satsfy (Fgure 204) ˆr = snθî cosθ ĵ (2026) ˆθ = cosθî + snθ ĵ Therefore the velocty of the pont P on the rm n the center of mass reference frame s gven by v P = Rω cm ˆθ = Rω cm (cosθî snθ ĵ) (2027) Fgure 204 Unt vectors ow substtute Eqs (2022) and (2027) nto Eq (2021) for the velocty of a pont P on the rm n the reference frame fxed to the ground v P = Rω cm (cosθî + snθĵ) +V î cm = (V cm + Rω cm cosθ)î + Rω (2028) snθĵ cm 20-3
5 The pont P s n contact wth the ground when θ = π At that nstant the velocty of a pont P on the rm n the reference frame fxed to the ground s v P (θ = π ) = (V cm Rω cm )î (2029) What velocty does the observer at rest on the ground measure for the pont on the rm when that pont s n contact wth the ground? In order to understand the relatonshp between V cm and ω cm, we consder the dsplacement of the center of mass for a small tme nterval Δt (Fgure 205) Fgure 205 Dsplacement of center of mass n ground reference frame From Eq (2023) the x -component of the dsplacement of the center of mass s ΔX cm =V cm Δt (20210) The pont P on the rm n the center of mass reference frame s undergong crcular moton (Fgure 206) Fgure 206: Small dsplacement of pont on rm n center of mass reference frame 20-4
6 In the center of mass reference frame, the magntude of the tangental dsplacement s gven by the arc length subtended by the angular dsplacement Δθ = ω cm Δt, Δs = RΔθ = Rω cm Δt (20211) Case 1: f the x -component of the dsplacement of the center of mass s equal to the arc length subtended by Δθ, then the wheel s rollng wthout slppng or skddng, rollng wthout slppng for short, along the surface wth ΔX cm = Δs (20212) Substtute Eq (20210) and Eq (20211) nto Eq (20212) and dvde through by Δt Then the rollng wthout slppng condton becomes V cm = Rω cm, (rollng wthout slppng) (20213) Case 2: f the x -component of the dsplacement of the center of mass s greater than the arc length subtended by Δθ, then the wheel s skddng along the surface wth ΔX cm > Δs (20214) Substtute Eqs (20210) and (20211) nto Eq (20214) and dvde through by Δt, then V cm > Rω cm, (skddng) (20215) Case 3: f the x -component of the dsplacement of the center of mass s less than the arc length subtended by Δθ, then the wheel s slppng along the surface wth Argung as above the slppng condton becomes 2021 Rollng wthout slppng ΔX cm < Δs (20216) V cm < Rω cm, (slppng) (20217) When a wheel s rollng wthout slppng, the velocty of a pont P on the rm s zero when t s n contact wth the ground In Eq (2029) set θ = π, v P (θ = π ) = (V cm Rω cm )î = (Rω cm Rω cm )î = 0 (20218) Ths makes sense because the velocty of the pont P on the rm n the center of mass reference frame when t s n contact wth the ground ponts n the opposte drecton as the translatonal moton of the center of mass of the wheel The two veloctes have the 20-5
7 same magntude so the vector sum s zero The observer at rest on the ground sees the contact pont on the rm at rest relatve to the ground Thus any frctonal force actng between the tre and the ground on the wheel s statc frcton because the two surfaces are nstantaneously at rest wth respect to each other Recall that the drecton of the statc frctonal force depends on the other forces actng on the wheel Example 201 Bcycle Wheel Rollng Wthout Slppng Consder a bcycle wheel of radus R that s rollng n a straght lne wthout slppng The velocty of the center of mass n a reference frame fxed to the ground s gven by velocty V cm A bead s fxed to a spoke a dstance b from the center of the wheel (Fgure 207) (a) Fnd the poston, velocty, and acceleraton of the bead as a functon of tme n the center of mass reference frame (b) Fnd the poston, velocty, and acceleraton of the bead as a functon of tme as seen n a reference frame fxed to the ground Fgure 207 Example 201 Fgure 208 Coordnate system for bead n center of mass reference frame Soluton: a) Choose the center of mass reference frame wth an orgn at the center of the wheel, and movng wth the wheel Choose polar coordnates (Fgure 208) The z - component of the angular velocty ω cm = dθ / dt > 0 Then the bead s movng unformly n a crcle of radus r = b wth the poston, velocty, and acceleraton gven by rb = b ˆr, v b = bω cm ˆθ, ab 2 = bω cm ˆr (20219) Because the wheel s rollng wthout slppng, the velocty of a pont on the rm of the wheel has speed v P = Rω cm Ths s equal to the speed of the center of mass of the wheel V cm, thus V cm = Rω cm (20220) ote that at t = 0, the angle θ = θ 0 = 0 So the angle grows n tme as 20-6
8 θ(t) = ω cm t = (V cm / R)t (20221) The velocty and acceleraton of the bead wth respect to the center of the wheel are then vb = bv cm R ˆθ, ab = bv 2 cm ˆr (20222) R 2 b) Defne a second reference frame fxed to the ground wth choce of orgn, Cartesan coordnates and unt vectors as shown n Fgure 209 Fgure 209 Coordnates of bead n reference frame fxed to ground Then the poston vector of the center of mass n the reference frame fxed to the ground s gven by R cm (t) = X cm î + R ĵ = V t î + R ĵ (20223) cm The relatve velocty of the two frames s the dervatve V cm = d R cm dt = dx cm dt î = V cm î (20224) Because the center of the wheel s movng at a unform speed the relatve acceleraton of the two frames s zero, A cm = d V cm = 0 (20225) dt Defne the poston, velocty, and acceleraton n ths frame (wth respect to the ground) by r b (t) = x b (t) î + y (t) ĵ, vb (t) = v b b,x (t) î + v (t) ĵ, b,y a(t) = ab,x (t) î + a (t) ĵ (20226) b,y Then the poston vectors are related by 20-7
9 r b (t) = R cm (t) + r b (t) (20227) In order to add these vectors we need to decompose the poston vector n the center of mass reference frame nto Cartesan components, rb (t) = b ˆr(t) = bsnθ(t) î + bcosθ(t) ĵ (20228) Then usng the relaton θ(t) = (V cm / R)t, Eq (20228) becomes r b (t) = R cm (t) + r b (t) = (V cm t î + R ĵ) + (bsnθ(t) î + bcosθ(t) ĵ) (20229) = V cm t + bsn((v cm / R)t) ( ) î + ( R + bcos((v / R)t) cm ) ĵ Thus the poston components of the bead wth respect to the reference frame fxed to the ground are gven by x b (t) = V cm t + bsn((v cm / R)t) (20230) y b (t) = R + bcos((v cm / R)t) (20231) A plot of the y -component vs the x -component of the poston of the bead n the reference frame fxed to the ground s shown n Fgure 2010 below usng the values V cm = 5 m s -1, R = 025 m, and b = 0125 m Ths path s called a cyclod We can dfferentate the poston vector n the reference frame fxed to the ground to fnd the velocty of the bead v b (t) = d r b dt (t) = d dt (V t + bsn((v / R)t)) î + d cm cm dt (R + bcos((v / R)t) )ĵ, (20232) cm v b (t) = (V cm + (b / R)V cos((v cm / R)t)) î ((b / R)V sn((v / R)t) )ĵ (20233) cm cm Fgure 2010 Plot of the y -component vs the x -component of the poston of the bead 20-8
10 Alternatvely, we can decompose the velocty of the bead n the center of mass reference frame nto Cartesan coordnates vb (t) = (b / R)V cm (cos((v cm / R)t) î sn((v / R)t) ĵ) (20234) cm The law of addton of veloctes s then v b (t) = V cm + v b (t), (20235) v b (t) = V cm î + (b / R)V cm (cos((v cm / R)t) î sn((v / R)t) ĵ), (20236) cm v b (t) = (V cm + (b / R)V cm cos((v cm / R)t)) î (b / R)sn((V / R)t) ĵ, (20237) cm n agreement wth our prevous result The acceleraton s the same n ether frame so a b (t) = a b = (b / R 2 )V cm2 ˆr = (b / R 2 )V 2 cm (sn((v cm / R)t) î + cos((v / R)t) ĵ) (20238) cm Example 202 Cylnder Rollng Wthout Slppng Down an Inclned Plane A unform cylnder of outer radus R and mass M wth moment of nerta about the center of mass I cm = (1/ 2)M R 2 starts from rest and rolls wthout slppng down an nclne tlted at an angle β from the horzontal The center of mass of the cylnder has dropped a vertcal dstance h when t reaches the bottom of the nclne Let g denote the gravtatonal constant What s the relaton between the component of the acceleraton of the center of mass n the drecton down the nclned plane and the component of the angular acceleraton nto the page of Fgure 2011? Fgure 2011 Example 202 Soluton: We begn by choosng a coordnate system for the translatonal and rotatonal moton as shown n Fgure
11 Fgure 2012 Coordnate system for rollng cylnder For a tme nterval Δt, the dsplacement of the center of mass s gven by Δ R cm (t) = ΔX cm î The arc length due to the angular dsplacement of a pont on the rm durng the tme nterval Δt s gven by Δs = RΔθ The rollng wthout slppng condton s ΔX cm = RΔθ If we dvde both sdes by Δt and take the lmt as Δt 0 then the rollng wthout slppng condton show that the x -component of the center of mass velocty s equal to the magntude of the tangental component of the velocty of a pont on the rm ΔX V cm = lm cm Δt 0 Δt = lm R Δθ Δt 0 Δt = Rω cm Smlarly f we dfferentate both sdes of the above equaton, we fnd a relaton between the x -component of the center of mass acceleraton s equal to the magntude of the tangental component of the acceleraton of a pont on the rm Example 203 Fallng Yo-Yo A cm = dv cm dt = R dω cm dt = Rα cm A Yo-Yo of mass m has an axle of radus b and a spool of radus R (Fgure 2013a) Its moment of nerta about the center of mass can be taken to be I = (1/ 2)mR 2 (the thckness of the strng can be neglected) The Yo-Yo s released from rest What s the relaton between the angular acceleraton about the center of mass and the lnear acceleraton of the center of mass? Soluton: Choose coordnates as shown n Fgure 2013b 20-10
12 Fgure 2013a Example 203 Fgure 2013b Coordnate system for Yo-Yo Consder a pont on the rm of the axle at a dstance r = b from the center of mass As the yo-yo falls, the arc length Δs = bδθ subtended by the rotaton of ths pont s equal to length of strng that has unraveled, an amount Δl In a tme nterval Δt, bδθ = Δl Therefore bδθ / Δt = Δl / Δt Takng lmts, notng that, V cm, y = dl / dt, we have that bω cm = V cm, y Dfferentatng a second tme yelds bα cm = A cm, y Example 204 Unwndng Drum Drum A of mass m and radus R s suspended from a drum B also of mass m and radus R, whch s free to rotate about ts axs The suspenson s n the form of a massless metal tape wound around the outsde of each drum, and free to unwnd (Fgure 2014) Gravty acts wth acceleraton g downwards Both drums are ntally at rest Fnd the ntal acceleraton of drum A, assumng that t moves straght down Fgure 2014 Example 204 Soluton: The key to solvng ths problem s to determne the relaton between the three knematc quanttes α A, α B, and a A, the angular acceleratons of the two drums and the 20-11
13 lnear acceleraton of drum A Choose the postve y -axs pontng downward wth the orgn at the center of drum B After a tme nterval Δt, the center of drum A has undergone a dsplacement Δy An amount of tape Δl A = RΔθ A has unraveled from drum A, and an amount of tape Δl B = RΔθ B has unraveled from drum B Therefore the dsplacement of the center of drum A s equal to the total amount of tape that has unwound from the two drums, Δy = Δl A + Δl B = RΔθ A + RΔθ B Dvdng through by Δt and takng the lmt as Δt 0 yelds dy dt = R dθ A + R dθ B dt dt Dfferentatng a second tme yelds the desred relaton between the angular acceleratons of the two drums and the lnear acceleraton of drum A, d 2 y dt = R d 2 θ A + R d 2 θ B 2 dt 2 dt 2 a A, y = Rα A + Rα B 203 Angular Momentum for a System of Partcles Undergong Translatonal and Rotatonal We shall now show that the angular momentum of a body about a pont S can be decomposed nto two vector parts, the angular momentum of the center of mass (treated as a pont partcle) about the pont S, and the angular momentum of the rotatonal moton about the center of mass Consder a system of partcles located at the ponts labeled = 1,2,, The angular momentum about the pont S s the sum total L S = L S, = ( r S, m v ), (2031) where r S, s the vector from the pont S to the th partcle (Fgure 2015) satsfyng r S, = r S, cm + r cm,, (2032) v S, = V cm + v cm,, (2033) where v S, cm = V cm We can now substtute both Eqs (2032) and (2033) nto Eq (2031) yeldng total L S = ( r S, cm + r cm, ) m ( V cm + v cm, ) (2034) 20-12
14 Fgure 2015 Vector Trangle When we expand the expresson n Equaton (2034), we have four terms, L total S = ( r S,cm m vcm, ) + ( r S,cm m Vcm ) + ( r cm, m vcm, ) + ( r cm, m Vcm ) (2035) The vector r S, cm s a constant vector that depends only on the locaton of the center of mass and not on the locaton of the th partcle Therefore n the frst term n the above equaton, r S, cm can be taken outsde the summaton Smlarly, n the second term the velocty of the center of mass V cm s the same for each term n the summaton, and may be taken outsde the summaton, L total S = r S,cm m vcm, + r S,cm m V cm + ( r cm, m vcm, ) + m rcm, V cm (2036) The frst and thrd terms n Eq (2036) are both zero due to the fact that m rcm, = 0 m vcm, = 0 We frst show that m rcm, s zero We begn by usng Eq (2032), (2037) 20-13
15 (m rcm, ) = (m ( r r S,cm )) = m r (m ) r S,cm = m r m total rs,cm (2038) Substtute the defnton of the center of mass (Eq 1053) nto Eq (2038) yeldng The vanshng of 1 (m rcm, ) = m r m total m total m r = 0 (2039) m vcm, = 0 follows drectly from the defnton of the center of mass frame, that the momentum n the center of mass s zero Equvalently the dervatve of Eq (2039) s zero We could also smply calculate and fnd that m vcm, = m ( v V cm ) = m v V cm m = m total Vcm V cm m total = 0 (20310) We can now smplfy Eq (2036) for the angular momentum about the pont S usng the fact that, m T = m, and p sys = m Vcm T (n reference frame O ): total L S = r S, cm p sys + ( r cm, m vcm, ) (20311) Consder the frst term n Equaton (20311), r S,cm p sys ; the vector r S,cm s the vector from the pont S to the center of mass If we treat the system as a pont-lke partcle of mass m T located at the center of mass, then the momentum of ths pont-lke partcle s p sys = m Vcm T Thus the frst term s the angular momentum about the pont S of ths pont-lke partcle, whch s called the orbtal angular momentum about S, for the system of partcles L S orbtal = r S,cm p sys (20312) 20-14
16 Consder the second term n Equaton (20311), ( r cm, m vcm, ) ; the quantty nsde the summaton s the angular momentum of the th partcle wth respect to the orgn n the center of mass reference frame O cm (recall the orgn n the center of mass reference frame s the center of mass of the system), L cm, = r cm, m vcm, (20313) Hence the total angular momentum of the system wth respect to the center of mass n the center of mass reference frame s gven by L spn cm = L cm, = ( r cm, m vcm, ) (20314) a vector quantty we call the spn angular momentum Thus we see that the total angular momentum about the pont S s the sum of these two terms, total L S = L orbtal S + L spn cm (20315) Ths decomposton of angular momentum nto a pece assocated wth the translatonal moton of the center of mass and a second pece assocated wth the rotatonal moton about the center of mass n the center of mass reference frame s the key conceptual foundaton for what follows Example 205 Earth s Moton Around the Sun The earth, of mass 24 m e = kg and (mean) radus 6 R e = m, moves n a nearly crcular orbt of radus r s,e = m around the sun wth a perod T = days, and spns about ts axs n a perod T spn = 23 hr 56 mn, the axs orbt nclned to the normal to the plane of ts orbt around the sun by 235 (n Fgure 2016, the relatve sze of the earth and sun, and the radus and shape of the orbt are not representatve of the actual quanttes) Fgure 2016 Example
17 If we approxmate the earth as a unform sphere, then the moment of nerta of the earth about ts center of mass s 2 2 Icm = me Re (20316) 5 If we choose the pont S to be at the center of the sun, and assume the orbt s crcular, then the orbtal angular momentum s L S orbtal = r S,cm p sys = r s,e ˆr m e v cm ˆθ = rs,e m e v cm ˆk (20317) The velocty of the center of mass of the earth about the sun s related to the orbtal angular velocty by v = r ω, (20318) where the orbtal angular speed s cm s,e orbt ω orbt = 2π T orbt = 2π (36525 d)( s d 1 ) = rad s 1 The orbtal angular momentum about S s then orbtal L S = m e r 2 s,e ω orbt ˆk = ( kg)( m) 2 ( rad s 1 ) ˆk = ( kg m 2 s 1 ) ˆk (20319) (20320) The spn angular momentum s gven by L spn cm = I cm ω spn = 2 5 m R 2 ω e e spn ˆn, (20321) where ˆn s a unt normal pontng along the axs of rotaton of the earth and ω 2π 2π = = = rad s s spn 4 Tspn 5 1 (20322) The spn angular momentum s then L spn cm = 2 5 ( kg)( m) 2 ( rad s 1 ) ˆn = ( kg m 2 s 1 ) ˆn (20323) 20-16
18 The rato of the magntudes of the orbtal angular momentum about S to the spn angular momentum s greater than a mllon, L S m r ω r T = = =, (20324) L m R R T orbtal 2 2 e s,e orbt 5 s,e spn spn 2 2 cm (2 / 5) e e ωspn 2 e orbt as ths rato s proportonal to the square of the rato of the dstance to the sun to the radus of the earth The angular momentum about S s then total L ˆ ˆ S = me rs,e ωorbt k + mere ωspn n (20325) 5 The orbt and spn perods are known to far more precson than the average values used for the earth s orbt radus and mean radus Two dfferent values have been used for one day; n convertng the orbt perod from days to seconds, the value for the solar day, T = 86, 400s was used In convertng the earth s spn angular frequency, the sdereal solar day, Tsdereal = Tspn = 86,160s was used The two perods, the solar day from noon to noon and the sdereal day from the dfference between the tmes that a fxed star s at the same place n the sky, do dffer n the thrd sgnfcant fgure 204 Knetc Energy of a System of Partcles Consder a system of partcles The th partcle has mass m and velocty v wth respect to a reference frame O The knetc energy of the system of partcles s gven by 1 K = 2 m v 2 = 1 2 m v v = 1 m 2 ( v cm, + V cm ) ( v cm, + V cm ) where Equaton 1526 has been used to express v n terms of v cm, and the last dot product n Equaton (2041), V cm (2041) Expandng K = 1 2 m ( v cm, v, rel + V cm V cm + 2 v cm, V cm ) = 1 m 2 ( v cm, v, rel ) + 1 m 2 ( V cm V cm ) + = 1 2 m 2 v cm, + 1 m 2 V 2 cm + mv cm, V cm m vcm, V cm (2042) 20-17
19 The last term n the thrd equaton n (2042) vanshes as we showed n Eq (2037) Then Equaton (2042) reduces to K = = 1 2 m 1 2 m 2 v cm, 2 v cm, m 2 V cm mtotal V cm 2 (2043) We nterpret the frst term as the knetc energy of the center of mass moton n reference frame O and the second term as the sum of the ndvdual knetc energes of the partcles of the system n the center of mass reference frame O cm At ths pont, t s mportant to note that no assumpton was made regardng the mass elements beng consttuents of a rgd body Equaton (2043) s vald for a rgd body, a gas, a frecracker (but K s certanly not the same before and after detonaton), and the sxteen pool balls after the break, or any collecton of objects for whch the center of mass can be determned 205 Rotatonal Knetc Energy for a Rgd Body Undergong Fxed Axs Rotaton The rotatonal knetc energy for the rgd body, usng v cm, = (r cm, ) ω cm ˆθ, smplfes to K rot = 1 2 I ω 2 (2051) cm cm Therefore the total knetc energy of a translatng and rotatng rgd body s K total = K trans + K rot = 1 2 mv cm 2 I ω 2 (2052) cm cm 20-18
20 Appendx 20A Chasles s Theorem: Rotaton and Translaton of a Rgd Body We now return to our descrpton of the translatng and rotatng rod that we frst consdered when we began our dscusson of rgd bodes We shall now show that the moton of any rgd body conssts of a translaton of the center of mass and rotaton about the center of mass We shall demonstrate ths for a rgd body by dvdng up the rgd body nto pont-lke consttuents Consder two pont-lke consttuents wth masses m 1 and m 2 Choose a coordnate system wth a choce of orgn such that body 1 has poston r 1 and body 2 has poston r 2 (Fgure 20A1) The relatve poston vector s gven by r 1,2 = r 1 r 2 (20A1) Fgure 20A1 Two-body coordnate system Recall we defned the center of mass vector, R cm, of the two-body system as R cm = m r m r2 2 (20A2) m 1 + m 2 In Fgure 20A2 we show the center of mass coordnate system Fgure 20A2 Poston coordnates wth respect to center of mass 20-19
21 The poston vector of the object 1 wth respect to the center of mass s gven by where r cm,1 = r 1 R cm = r 1 m r m r2 2 = m 1 + m 2 m1m 2 µ = m + m 1 2 m 2 m 1 + m 2 ( r 1 r 2 ) = µ m 1 r1,2, (20A3), (20A4) s the reduced mass In addton, the relatve poston vector between the two objects s ndependent of the choce of reference frame, r 12 = r 1 r 2 = ( r cm,1 + R cm ) ( r cm,2 + R cm ) = r cm,1 r cm,2 = r cm,1,2 Because the center of mass s at the orgn n the center of mass reference frame, (20A5) Therefore m rcm,1 1 + m rcm,2 2 = 0 m 1 + m 2 (20A6) m rcm,1 1 = m rcm,2 2 (20A7) m 1 rcm,1 = m 2 rcm,2 (20A8) The dsplacement of object 1 about the center of mass s gven by takng the dervatve of Eq (20A3), d r cm,1 = µ m 1 d r 1,2 (20A9) A smlar calculaton for the poston of object 2 wth respect to the center of mass yelds for the poston and dsplacement wth respect to the center of mass r cm,2 = r 2 R cm = µ m 2 r1,2, (20A10) d r cm,2 = µ m 2 d r 1,2 (20A11) Let = 1,2 An arbtrary dsplacement of the th object s gven respectvely by d r = d r cm, + d R cm, (20A12) 20-20
22 whch s the sum of a dsplacement about the center of mass d r cm, and a dsplacement of the center of mass d R cm The dsplacement of objects 1 and 2 are constraned by the condton that the dstance between the objects must reman constant snce the body s rgd In partcular, the dstance between objects 1 and 2 s gven by r 1,2 2 = ( r1 r 2 ) ( r 1 r 2 ) (20A13) Because ths dstance s constant we can dfferentate Eq (20A13), yeldng the rgd body condton that 0 = 2( r 1 r 2 ) (d r 1 d r 2 ) = 2 r 1,2 d r 1,2 (20A14) 20A1 Translaton of the Center of Mass The condton (Eq (20A14)) can be satsfed f the relatve dsplacement vector between the two objects s zero, dr 1,2 = dr 1 dr 2 = 0 (20A15) Ths mples, usng, Eq (20A9) and Eq (20A11), that the dsplacement wth respect to the center of mass s zero, dr cm,1 = dr cm,2 = 0 (20A16) Thus by Eq (20A12), the dsplacement of each object s equal to the dsplacement of the center of mass, dr = dr cm, (20A17) whch means that the body s undergong pure translaton 20A2 Rotaton about the Center of Mass ow suppose that d r 1,2 = d r 1 d r 2 0 The rgd body condton can be expressed n terms of the center of mass coordnates Usng Eq (20A9), the rgd body condton (Eq (20A14)) becomes 0 = 2 µ m 1 r1,2 d r cm,1 (20A18) Because the relatve poston vector between the two objects s ndependent of the choce of reference frame (Eq (20A5)), the rgd body condton Eq (20A14) n the center of mass reference frame s then gven by 0 = 2 r cm,1,2 d r cm,1 (20A19) 20-21
23 Ths condton s satsfed f the relatve dsplacement s perpendcular to the lne passng through the center of mass, r cm,1,2 dr cm,1 (20A20) By a smlar argument, r cm,1,2 d r cm,2 In order for these dsplacements to correspond to a rotaton about the center of mass, the dsplacements must have the same angular dsplacement Fgure 20A3 Infntesmal angular dsplacements n the center of mass reference frame In Fgure 20A3, the nfntesmal angular dsplacement of each object s gven by dθ 1 = d r cm,1 r cm,1, (20A21) dθ 2 = d r cm,2 r cm,2 (20A22) From Eq (20A9) and Eq (20A11), we can rewrte Eqs (20A21) and (20A22) as dθ 1 = µ m 1 d r 1,2 r cm,1, (20A23) dθ 2 = µ m 2 d r 1,2 r cm,2 (20A24) Recall that n the center of mass reference frame m 1 rcm,1 = m 2 rcm,2 (Eq (20A8)) and hence the angular dsplacements are equal, 20-22
24 dθ = dθ = dθ (20A25) 1 2 Therefore the dsplacement of the th object dr dffers from the dsplacement of the center of mass dr cm by a vector that corresponds to an nfntesmal rotaton n the center of mass reference frame d r = d R cm + d r cm, (20A26) We have shown that the dsplacement of a rgd body s the vector sum of the dsplacement of the center of mass (translaton of the center of mass) and an nfntesmal rotaton about the center of mass 20-23
Rotation Kinematics, Moment of Inertia, and Torque
Rotaton Knematcs, Moment of Inerta, and Torque Mathematcally, rotaton of a rgd body about a fxed axs s analogous to a lnear moton n one dmenson. Although the physcal quanttes nvolved n rotaton are qute
More informationGoals Rotational quantities as vectors. Math: Cross Product. Angular momentum
Physcs 106 Week 5 Torque and Angular Momentum as Vectors SJ 7thEd.: Chap 11.2 to 3 Rotatonal quanttes as vectors Cross product Torque expressed as a vector Angular momentum defned Angular momentum as a
More informationRotation and Conservation of Angular Momentum
Chapter 4. Rotaton and Conservaton of Angular Momentum Notes: Most of the materal n ths chapter s taken from Young and Freedman, Chaps. 9 and 0. 4. Angular Velocty and Acceleraton We have already brefly
More informationChapter 11 Torque and Angular Momentum
Chapter 11 Torque and Angular Momentum I. Torque II. Angular momentum - Defnton III. Newton s second law n angular form IV. Angular momentum - System of partcles - Rgd body - Conservaton I. Torque - Vector
More informationHomework: 49, 56, 67, 60, 64, 74 (p. 234-237)
Hoework: 49, 56, 67, 60, 64, 74 (p. 34-37) 49. bullet o ass 0g strkes a ballstc pendulu o ass kg. The center o ass o the pendulu rses a ertcal dstance o c. ssung that the bullet reans ebedded n the pendulu,
More informationwhere the coordinates are related to those in the old frame as follows.
Chapter 2 - Cartesan Vectors and Tensors: Ther Algebra Defnton of a vector Examples of vectors Scalar multplcaton Addton of vectors coplanar vectors Unt vectors A bass of non-coplanar vectors Scalar product
More informationNMT EE 589 & UNM ME 482/582 ROBOT ENGINEERING. Dr. Stephen Bruder NMT EE 589 & UNM ME 482/582
NMT EE 589 & UNM ME 482/582 ROBOT ENGINEERING Dr. Stephen Bruder NMT EE 589 & UNM ME 482/582 7. Root Dynamcs 7.2 Intro to Root Dynamcs We now look at the forces requred to cause moton of the root.e. dynamcs!!
More informationFaraday's Law of Induction
Introducton Faraday's Law o Inducton In ths lab, you wll study Faraday's Law o nducton usng a wand wth col whch swngs through a magnetc eld. You wll also examne converson o mechanc energy nto electrc energy
More informationReview C: Work and Kinetic Energy
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department o Physcs 8.2 Revew C: Work and Knetc Energy C. Energy... 2 C.. The Concept o Energy... 2 C..2 Knetc Energy... 3 C.2 Work and Power... 4 C.2. Work Done by
More information21 Vectors: The Cross Product & Torque
21 Vectors: The Cross Product & Torque Do not use our left hand when applng ether the rght-hand rule for the cross product of two vectors dscussed n ths chapter or the rght-hand rule for somethng curl
More informationCHAPTER 8 Potential Energy and Conservation of Energy
CHAPTER 8 Potental Energy and Conservaton o Energy One orm o energy can be converted nto another orm o energy. Conservatve and non-conservatve orces Physcs 1 Knetc energy: Potental energy: Energy assocated
More information1 What is a conservation law?
MATHEMATICS 7302 (Analytcal Dynamcs) YEAR 2015 2016, TERM 2 HANDOUT #6: MOMENTUM, ANGULAR MOMENTUM, AND ENERGY; CONSERVATION LAWS In ths handout we wll develop the concepts of momentum, angular momentum,
More informationQ3.8: A person trying to throw a ball as far as possible will run forward during the throw. Explain why this increases the distance of the throw.
Problem Set 3 Due: 09/3/, Tuesda Chapter 3: Vectors and Moton n Two Dmensons Questons: 7, 8,, 4, 0 Eercses & Problems:, 7, 8, 33, 37, 44, 46, 65, 73 Q3.7: An athlete performn the lon jump tres to acheve
More informationChapter 9. Linear Momentum and Collisions
Chapter 9 Lnear Momentum and Collsons CHAPTER OUTLINE 9.1 Lnear Momentum and Its Conservaton 9.2 Impulse and Momentum 9.3 Collsons n One Dmenson 9.4 Two-Dmensonal Collsons 9.5 The Center of Mass 9.6 Moton
More informationPolitecnico di Torino. Porto Institutional Repository
Poltecnco d orno Porto Insttutonal Repostory [Proceedng] rbt dynamcs and knematcs wth full quaternons rgnal Ctaton: Andres D; Canuto E. (5). rbt dynamcs and knematcs wth full quaternons. In: 16th IFAC
More information8.5 UNITARY AND HERMITIAN MATRICES. The conjugate transpose of a complex matrix A, denoted by A*, is given by
6 CHAPTER 8 COMPLEX VECTOR SPACES 5. Fnd the kernel of the lnear transformaton gven n Exercse 5. In Exercses 55 and 56, fnd the mage of v, for the ndcated composton, where and are gven by the followng
More informationMean Molecular Weight
Mean Molecular Weght The thermodynamc relatons between P, ρ, and T, as well as the calculaton of stellar opacty requres knowledge of the system s mean molecular weght defned as the mass per unt mole of
More informationInner core mantle gravitational locking and the super-rotation of the inner core
Geophys. J. Int. (2010) 181, 806 817 do: 10.1111/j.1365-246X.2010.04563.x Inner core mantle gravtatonal lockng and the super-rotaton of the nner core Matheu Dumberry 1 and Jon Mound 2 1 Department of Physcs,
More informationSupport Vector Machines
Support Vector Machnes Max Wellng Department of Computer Scence Unversty of Toronto 10 Kng s College Road Toronto, M5S 3G5 Canada wellng@cs.toronto.edu Abstract Ths s a note to explan support vector machnes.
More informationLecture Topics. 6. Sensors and instrumentation 7. Actuators and power transmission devices. (System and Signal Processing) DR.1 11.12.
Lecture Tocs 1. Introducton 2. Basc knematcs 3. Pose measurement and Measurement of Robot Accuracy 4. Trajectory lannng and control 5. Forces, moments and Euler s laws 5. Fundamentals n electroncs and
More informationbenefit is 2, paid if the policyholder dies within the year, and probability of death within the year is ).
REVIEW OF RISK MANAGEMENT CONCEPTS LOSS DISTRIBUTIONS AND INSURANCE Loss and nsurance: When someone s subject to the rsk of ncurrng a fnancal loss, the loss s generally modeled usng a random varable or
More informationRecurrence. 1 Definitions and main statements
Recurrence 1 Defntons and man statements Let X n, n = 0, 1, 2,... be a MC wth the state space S = (1, 2,...), transton probabltes p j = P {X n+1 = j X n = }, and the transton matrx P = (p j ),j S def.
More informationLagrangian Dynamics: Virtual Work and Generalized Forces
Admssble Varatons/Vrtual Dsplacements 1 2.003J/1.053J Dynamcs and Control I, Sprng 2007 Paula Echeverr, Professor Thomas Peacock 4/4/2007 Lecture 14 Lagrangan Dynamcs: Vrtual Work and Generalzed Forces
More informationLecture 3: Force of Interest, Real Interest Rate, Annuity
Lecture 3: Force of Interest, Real Interest Rate, Annuty Goals: Study contnuous compoundng and force of nterest Dscuss real nterest rate Learn annuty-mmedate, and ts present value Study annuty-due, and
More informationA machine vision approach for detecting and inspecting circular parts
A machne vson approach for detectng and nspectng crcular parts Du-Mng Tsa Machne Vson Lab. Department of Industral Engneerng and Management Yuan-Ze Unversty, Chung-L, Tawan, R.O.C. E-mal: edmtsa@saturn.yzu.edu.tw
More informationv a 1 b 1 i, a 2 b 2 i,..., a n b n i.
SECTION 8.4 COMPLEX VECTOR SPACES AND INNER PRODUCTS 455 8.4 COMPLEX VECTOR SPACES AND INNER PRODUCTS All the vector spaces we have studed thus far n the text are real vector spaces snce the scalars are
More informationDamage detection in composite laminates using coin-tap method
Damage detecton n composte lamnates usng con-tap method S.J. Km Korea Aerospace Research Insttute, 45 Eoeun-Dong, Youseong-Gu, 35-333 Daejeon, Republc of Korea yaeln@kar.re.kr 45 The con-tap test has the
More informationBERNSTEIN POLYNOMIALS
On-Lne Geometrc Modelng Notes BERNSTEIN POLYNOMIALS Kenneth I. Joy Vsualzaton and Graphcs Research Group Department of Computer Scence Unversty of Calforna, Davs Overvew Polynomals are ncredbly useful
More informationJet Engine. Figure 1 Jet engine
Jet Engne Prof. Dr. Mustafa Cavcar Anadolu Unversty, School of Cvl Avaton Esksehr, urkey GROSS HRUS INAKE MOMENUM DRAG NE HRUS Fgure 1 Jet engne he thrust for a turboet engne can be derved from Newton
More informationSPEE Recommended Evaluation Practice #6 Definition of Decline Curve Parameters Background:
SPEE Recommended Evaluaton Practce #6 efnton of eclne Curve Parameters Background: The producton hstores of ol and gas wells can be analyzed to estmate reserves and future ol and gas producton rates and
More informationInertial Field Energy
Adv. Studes Theor. Phys., Vol. 3, 009, no. 3, 131-140 Inertal Feld Energy C. Johan Masrelez 309 W Lk Sammamsh Pkwy NE Redmond, WA 9805, USA jmasrelez@estfound.org Abstract The phenomenon of Inerta may
More informationLinear Circuits Analysis. Superposition, Thevenin /Norton Equivalent circuits
Lnear Crcuts Analyss. Superposton, Theenn /Norton Equalent crcuts So far we hae explored tmendependent (resste) elements that are also lnear. A tmendependent elements s one for whch we can plot an / cure.
More informationThe OC Curve of Attribute Acceptance Plans
The OC Curve of Attrbute Acceptance Plans The Operatng Characterstc (OC) curve descrbes the probablty of acceptng a lot as a functon of the lot s qualty. Fgure 1 shows a typcal OC Curve. 10 8 6 4 1 3 4
More informationAnswer: A). There is a flatter IS curve in the high MPC economy. Original LM LM after increase in M. IS curve for low MPC economy
4.02 Quz Solutons Fall 2004 Multple-Choce Questons (30/00 ponts) Please, crcle the correct answer for each of the followng 0 multple-choce questons. For each queston, only one of the answers s correct.
More informationConversion between the vector and raster data structures using Fuzzy Geographical Entities
Converson between the vector and raster data structures usng Fuzzy Geographcal Enttes Cdála Fonte Department of Mathematcs Faculty of Scences and Technology Unversty of Combra, Apartado 38, 3 454 Combra,
More informationHALL EFFECT SENSORS AND COMMUTATION
OEM770 5 Hall Effect ensors H P T E R 5 Hall Effect ensors The OEM770 works wth three-phase brushless motors equpped wth Hall effect sensors or equvalent feedback sgnals. In ths chapter we wll explan how
More informationChapter 2. Lagrange s and Hamilton s Equations
Chapter 2 Lagrange s and Hamlton s Equatons In ths chapter, we consder two reformulatons of Newtonan mechancs, the Lagrangan and the Hamltonan formalsm. The frst s naturally assocated wth confguraton space,
More informationPSYCHOLOGICAL RESEARCH (PYC 304-C) Lecture 12
14 The Ch-squared dstrbuton PSYCHOLOGICAL RESEARCH (PYC 304-C) Lecture 1 If a normal varable X, havng mean µ and varance σ, s standardsed, the new varable Z has a mean 0 and varance 1. When ths standardsed
More informationCalculation of Sampling Weights
Perre Foy Statstcs Canada 4 Calculaton of Samplng Weghts 4.1 OVERVIEW The basc sample desgn used n TIMSS Populatons 1 and 2 was a two-stage stratfed cluster desgn. 1 The frst stage conssted of a sample
More informationLaws of Electromagnetism
There are four laws of electromagnetsm: Laws of Electromagnetsm The law of Bot-Savart Ampere's law Force law Faraday's law magnetc feld generated by currents n wres the effect of a current on a loop of
More informationSeries Solutions of ODEs 2 the Frobenius method. The basic idea of the Frobenius method is to look for solutions of the form 3
Royal Holloway Unversty of London Department of Physs Seres Solutons of ODEs the Frobenus method Introduton to the Methodology The smple seres expanson method works for dfferental equatons whose solutons
More informationIntroduction to Statistical Physics (2SP)
Introducton to Statstcal Physcs (2SP) Rchard Sear March 5, 20 Contents What s the entropy (aka the uncertanty)? 2. One macroscopc state s the result of many many mcroscopc states.......... 2.2 States wth
More informationSection 5.4 Annuities, Present Value, and Amortization
Secton 5.4 Annutes, Present Value, and Amortzaton Present Value In Secton 5.2, we saw that the present value of A dollars at nterest rate per perod for n perods s the amount that must be deposted today
More informationIDENTIFICATION AND CORRECTION OF A COMMON ERROR IN GENERAL ANNUITY CALCULATIONS
IDENTIFICATION AND CORRECTION OF A COMMON ERROR IN GENERAL ANNUITY CALCULATIONS Chrs Deeley* Last revsed: September 22, 200 * Chrs Deeley s a Senor Lecturer n the School of Accountng, Charles Sturt Unversty,
More informationKinetic Energy-Based Temperature Computation in Non-Equilibrium Molecular. Dynamics Simulation. China. Avenue, Kowloon, Hong Kong, China
Knetc Energy-Based Temperature omputaton n on-equlbrum Molecular Dynamcs Smulaton Bn Lu, * Ran Xu, and Xaoqao He AML, Department of Engneerng Mechancs, Tsnghua Unversty, Bejng 00084, hna Department of
More informationKinematic Analysis of Cam Profiles Used in Compound Bows. A Thesis. Presented to. The Graduate Faculty of the University of Missouri
Knematc Analyss of Cam Profles Used n Compound Bows A Thess Presented to The Graduate Faculty of the Unversty of Mssour In Partal Fulfllment of the Requrements for the Degree Master of Scence Andrew Joseph
More informationPoint cloud to point cloud rigid transformations. Minimizing Rigid Registration Errors
Pont cloud to pont cloud rgd transformatons Russell Taylor 600.445 1 600.445 Fall 000-014 Copyrght R. H. Taylor Mnmzng Rgd Regstraton Errors Typcally, gven a set of ponts {a } n one coordnate system and
More informationWhat is Candidate Sampling
What s Canddate Samplng Say we have a multclass or mult label problem where each tranng example ( x, T ) conssts of a context x a small (mult)set of target classes T out of a large unverse L of possble
More informationTHE DISTRIBUTION OF LOAN PORTFOLIO VALUE * Oldrich Alfons Vasicek
HE DISRIBUION OF LOAN PORFOLIO VALUE * Oldrch Alfons Vascek he amount of captal necessary to support a portfolo of debt securtes depends on the probablty dstrbuton of the portfolo loss. Consder a portfolo
More informationRing structure of splines on triangulations
www.oeaw.ac.at Rng structure of splnes on trangulatons N. Vllamzar RICAM-Report 2014-48 www.rcam.oeaw.ac.at RING STRUCTURE OF SPLINES ON TRIANGULATIONS NELLY VILLAMIZAR Introducton For a trangulated regon
More information+ + + - - This circuit than can be reduced to a planar circuit
MeshCurrent Method The meshcurrent s analog of the nodeoltage method. We sole for a new set of arables, mesh currents, that automatcally satsfy KCLs. As such, meshcurrent method reduces crcut soluton to
More informationPhysics 110 Spring 2006 2-D Motion Problems: Projectile Motion Their Solutions
Physcs 110 Sprn 006 -D Moton Problems: Projectle Moton Ther Solutons 1. A place-kcker must kck a football from a pont 36 m (about 40 yards) from the oal, and half the crowd hopes the ball wll clear the
More information1 Example 1: Axis-aligned rectangles
COS 511: Theoretcal Machne Learnng Lecturer: Rob Schapre Lecture # 6 Scrbe: Aaron Schld February 21, 2013 Last class, we dscussed an analogue for Occam s Razor for nfnte hypothess spaces that, n conjuncton
More informationThe Mathematical Derivation of Least Squares
Pscholog 885 Prof. Federco The Mathematcal Dervaton of Least Squares Back when the powers that e forced ou to learn matr algera and calculus, I et ou all asked ourself the age-old queston: When the hell
More informationSolution: Let i = 10% and d = 5%. By definition, the respective forces of interest on funds A and B are. i 1 + it. S A (t) = d (1 dt) 2 1. = d 1 dt.
Chapter 9 Revew problems 9.1 Interest rate measurement Example 9.1. Fund A accumulates at a smple nterest rate of 10%. Fund B accumulates at a smple dscount rate of 5%. Fnd the pont n tme at whch the forces
More informationBrigid Mullany, Ph.D University of North Carolina, Charlotte
Evaluaton And Comparson Of The Dfferent Standards Used To Defne The Postonal Accuracy And Repeatablty Of Numercally Controlled Machnng Center Axes Brgd Mullany, Ph.D Unversty of North Carolna, Charlotte
More informationLeast Squares Fitting of Data
Least Squares Fttng of Data Davd Eberly Geoetrc Tools, LLC http://www.geoetrctools.co/ Copyrght c 1998-2016. All Rghts Reserved. Created: July 15, 1999 Last Modfed: January 5, 2015 Contents 1 Lnear Fttng
More informationn + d + q = 24 and.05n +.1d +.25q = 2 { n + d + q = 24 (3) n + 2d + 5q = 40 (2)
MATH 16T Exam 1 : Part I (In-Class) Solutons 1. (0 pts) A pggy bank contans 4 cons, all of whch are nckels (5 ), dmes (10 ) or quarters (5 ). The pggy bank also contans a con of each denomnaton. The total
More informationRisk-based Fatigue Estimate of Deep Water Risers -- Course Project for EM388F: Fracture Mechanics, Spring 2008
Rsk-based Fatgue Estmate of Deep Water Rsers -- Course Project for EM388F: Fracture Mechancs, Sprng 2008 Chen Sh Department of Cvl, Archtectural, and Envronmental Engneerng The Unversty of Texas at Austn
More informationMulti-Robot Tracking of a Moving Object Using Directional Sensors
Mult-Robot Trackng of a Movng Object Usng Drectonal Sensors Xaomng Hu, Karl H. Johansson, Manuel Mazo Jr., Alberto Speranzon Dept. of Sgnals, Sensors & Systems Royal Insttute of Technology, SE- 44 Stockholm,
More informationExtending Probabilistic Dynamic Epistemic Logic
Extendng Probablstc Dynamc Epstemc Logc Joshua Sack May 29, 2008 Probablty Space Defnton A probablty space s a tuple (S, A, µ), where 1 S s a set called the sample space. 2 A P(S) s a σ-algebra: a set
More information4 Cosmological Perturbation Theory
4 Cosmologcal Perturbaton Theory So far, we have treated the unverse as perfectly homogeneous. To understand the formaton and evoluton of large-scale structures, we have to ntroduce nhomogenetes. As long
More informationDEFINING %COMPLETE IN MICROSOFT PROJECT
CelersSystems DEFINING %COMPLETE IN MICROSOFT PROJECT PREPARED BY James E Aksel, PMP, PMI-SP, MVP For Addtonal Informaton about Earned Value Management Systems and reportng, please contact: CelersSystems,
More informationHedging Interest-Rate Risk with Duration
FIXED-INCOME SECURITIES Chapter 5 Hedgng Interest-Rate Rsk wth Duraton Outlne Prcng and Hedgng Prcng certan cash-flows Interest rate rsk Hedgng prncples Duraton-Based Hedgng Technques Defnton of duraton
More information"Research Note" APPLICATION OF CHARGE SIMULATION METHOD TO ELECTRIC FIELD CALCULATION IN THE POWER CABLES *
Iranan Journal of Scence & Technology, Transacton B, Engneerng, ol. 30, No. B6, 789-794 rnted n The Islamc Republc of Iran, 006 Shraz Unversty "Research Note" ALICATION OF CHARGE SIMULATION METHOD TO ELECTRIC
More information1. Measuring association using correlation and regression
How to measure assocaton I: Correlaton. 1. Measurng assocaton usng correlaton and regresson We often would lke to know how one varable, such as a mother's weght, s related to another varable, such as a
More informationHÜCKEL MOLECULAR ORBITAL THEORY
1 HÜCKEL MOLECULAR ORBITAL THEORY In general, the vast maorty polyatomc molecules can be thought of as consstng of a collecton of two electron bonds between pars of atoms. So the qualtatve pcture of σ
More informationDescription of the Force Method Procedure. Indeterminate Analysis Force Method 1. Force Method con t. Force Method con t
Indeternate Analyss Force Method The force (flexblty) ethod expresses the relatonshps between dsplaceents and forces that exst n a structure. Prary objectve of the force ethod s to deterne the chosen set
More informationCHOLESTEROL REFERENCE METHOD LABORATORY NETWORK. Sample Stability Protocol
CHOLESTEROL REFERENCE METHOD LABORATORY NETWORK Sample Stablty Protocol Background The Cholesterol Reference Method Laboratory Network (CRMLN) developed certfcaton protocols for total cholesterol, HDL
More informationUPGRADE YOUR PHYSICS
Correctons March 7 UPGRADE YOUR PHYSICS NOTES FOR BRITISH SIXTH FORM STUDENTS WHO ARE PREPARING FOR THE INTERNATIONAL PHYSICS OLYMPIAD, OR WISH TO TAKE THEIR KNOWLEDGE OF PHYSICS BEYOND THE A-LEVEL SYLLABI.
More informationProject Networks With Mixed-Time Constraints
Project Networs Wth Mxed-Tme Constrants L Caccetta and B Wattananon Western Australan Centre of Excellence n Industral Optmsaton (WACEIO) Curtn Unversty of Technology GPO Box U1987 Perth Western Australa
More informationFace Verification Problem. Face Recognition Problem. Application: Access Control. Biometric Authentication. Face Verification (1:1 matching)
Face Recognton Problem Face Verfcaton Problem Face Verfcaton (1:1 matchng) Querymage face query Face Recognton (1:N matchng) database Applcaton: Access Control www.vsage.com www.vsoncs.com Bometrc Authentcaton
More informationCompaction of the diamond Ti 3 SiC 2 graded material by the high speed centrifugal compaction process
Archves of Materals Scence and Engneerng Volume 8 Issue November 7 Pages 677-68 Internatonal Scentfc Journal publshed monthly as the organ of the Commttee of Materals Scence of the Polsh Academy of Scences
More informationThe Greedy Method. Introduction. 0/1 Knapsack Problem
The Greedy Method Introducton We have completed data structures. We now are gong to look at algorthm desgn methods. Often we are lookng at optmzaton problems whose performance s exponental. For an optmzaton
More informationLecture 3: Annuity. Study annuities whose payments form a geometric progression or a arithmetic progression.
Lecture 3: Annuty Goals: Learn contnuous annuty and perpetuty. Study annutes whose payments form a geometrc progresson or a arthmetc progresson. Dscuss yeld rates. Introduce Amortzaton Suggested Textbook
More informationFINANCIAL MATHEMATICS. A Practical Guide for Actuaries. and other Business Professionals
FINANCIAL MATHEMATICS A Practcal Gude for Actuares and other Busness Professonals Second Edton CHRIS RUCKMAN, FSA, MAAA JOE FRANCIS, FSA, MAAA, CFA Study Notes Prepared by Kevn Shand, FSA, FCIA Assstant
More informationExperiment 5 Elastic and Inelastic Collisions
PHY191 Experment 5: Elastc and Inelastc Collsons 8/1/014 Page 1 Experment 5 Elastc and Inelastc Collsons Readng: Bauer&Westall: Chapter 7 (and 8, or center o mass deas) as needed 1. Goals 1. Study momentum
More informationSection 2 Introduction to Statistical Mechanics
Secton 2 Introducton to Statstcal Mechancs 2.1 Introducng entropy 2.1.1 Boltzmann s formula A very mportant thermodynamc concept s that of entropy S. Entropy s a functon of state, lke the nternal energy.
More informationModule 2 LOSSLESS IMAGE COMPRESSION SYSTEMS. Version 2 ECE IIT, Kharagpur
Module LOSSLESS IMAGE COMPRESSION SYSTEMS Lesson 3 Lossless Compresson: Huffman Codng Instructonal Objectves At the end of ths lesson, the students should be able to:. Defne and measure source entropy..
More informationGRAVITY DATA VALIDATION AND OUTLIER DETECTION USING L 1 -NORM
GRAVITY DATA VALIDATION AND OUTLIER DETECTION USING L 1 -NORM BARRIOT Jean-Perre, SARRAILH Mchel BGI/CNES 18.av.E.Beln 31401 TOULOUSE Cedex 4 (France) Emal: jean-perre.barrot@cnes.fr 1/Introducton The
More informationActuator forces in CFD: RANS and LES modeling in OpenFOAM
Home Search Collectons Journals About Contact us My IOPscence Actuator forces n CFD: RANS and LES modelng n OpenFOAM Ths content has been downloaded from IOPscence. Please scroll down to see the full text.
More informationWe are now ready to answer the question: What are the possible cardinalities for finite fields?
Chapter 3 Fnte felds We have seen, n the prevous chapters, some examples of fnte felds. For example, the resdue class rng Z/pZ (when p s a prme) forms a feld wth p elements whch may be dentfed wth the
More informationFigure 1. Inventory Level vs. Time - EOQ Problem
IEOR 54 Sprng, 009 rof Leahman otes on Eonom Lot Shedulng and Eonom Rotaton Cyles he Eonom Order Quantty (EOQ) Consder an nventory tem n solaton wth demand rate, holdng ost h per unt per unt tme, and replenshment
More informationFixed income risk attribution
5 Fxed ncome rsk attrbuton Chthra Krshnamurth RskMetrcs Group chthra.krshnamurth@rskmetrcs.com We compare the rsk of the actve portfolo wth that of the benchmark and segment the dfference between the two
More informationTexas Instruments 30X IIS Calculator
Texas Instruments 30X IIS Calculator Keystrokes for the TI-30X IIS are shown for a few topcs n whch keystrokes are unque. Start by readng the Quk Start secton. Then, before begnnng a specfc unt of the
More informationAn Integrated Semantically Correct 2.5D Object Oriented TIN. Andreas Koch
An Integrated Semantcally Correct 2.5D Object Orented TIN Andreas Koch Unverstät Hannover Insttut für Photogrammetre und GeoInformaton Contents Introducton Integraton of a DTM and 2D GIS data Semantcs
More informationReduced magnetohydrodynamic equations with coupled Alfvén and sound wave dynamics
PHYSICS OF PLASMAS 14, 10906 007 Reduced magnetohydrodynamc equatons wth coupled Alfvén and sound wave dynamcs R. E. Denton and B. Rogers Department of Physcs and Astronomy, Dartmouth College, Hanover,
More informationThe Full-Wave Rectifier
9/3/2005 The Full Wae ectfer.doc /0 The Full-Wae ectfer Consder the followng juncton dode crcut: s (t) Power Lne s (t) 2 Note that we are usng a transformer n ths crcut. The job of ths transformer s to
More informationHow To Understand The Results Of The German Meris Cloud And Water Vapour Product
Ttel: Project: Doc. No.: MERIS level 3 cloud and water vapour products MAPP MAPP-ATBD-ClWVL3 Issue: 1 Revson: 0 Date: 9.12.1998 Functon Name Organsaton Sgnature Date Author: Bennartz FUB Preusker FUB Schüller
More informationLoop Parallelization
- - Loop Parallelzaton C-52 Complaton steps: nested loops operatng on arrays, sequentell executon of teraton space DECLARE B[..,..+] FOR I :=.. FOR J :=.. I B[I,J] := B[I-,J]+B[I-,J-] ED FOR ED FOR analyze
More informationTHE METHOD OF LEAST SQUARES THE METHOD OF LEAST SQUARES
The goal: to measure (determne) an unknown quantty x (the value of a RV X) Realsaton: n results: y 1, y 2,..., y j,..., y n, (the measured values of Y 1, Y 2,..., Y j,..., Y n ) every result s encumbered
More informationCHAPTER 14 MORE ABOUT REGRESSION
CHAPTER 14 MORE ABOUT REGRESSION We learned n Chapter 5 that often a straght lne descrbes the pattern of a relatonshp between two quanttatve varables. For nstance, n Example 5.1 we explored the relatonshp
More informationCausal, Explanatory Forecasting. Analysis. Regression Analysis. Simple Linear Regression. Which is Independent? Forecasting
Causal, Explanatory Forecastng Assumes cause-and-effect relatonshp between system nputs and ts output Forecastng wth Regresson Analyss Rchard S. Barr Inputs System Cause + Effect Relatonshp The job of
More informationAn interactive system for structure-based ASCII art creation
An nteractve system for structure-based ASCII art creaton Katsunor Myake Henry Johan Tomoyuk Nshta The Unversty of Tokyo Nanyang Technologcal Unversty Abstract Non-Photorealstc Renderng (NPR), whose am
More informationsubstances (among other variables as well). ( ) Thus the change in volume of a mixture can be written as
Mxtures and Solutons Partal Molar Quanttes Partal molar volume he total volume of a mxture of substances s a functon of the amounts of both V V n,n substances (among other varables as well). hus the change
More informationCHAPTER 5 RELATIONSHIPS BETWEEN QUANTITATIVE VARIABLES
CHAPTER 5 RELATIONSHIPS BETWEEN QUANTITATIVE VARIABLES In ths chapter, we wll learn how to descrbe the relatonshp between two quanttatve varables. Remember (from Chapter 2) that the terms quanttatve varable
More informationAn Alternative Way to Measure Private Equity Performance
An Alternatve Way to Measure Prvate Equty Performance Peter Todd Parlux Investment Technology LLC Summary Internal Rate of Return (IRR) s probably the most common way to measure the performance of prvate
More information5.74 Introductory Quantum Mechanics II
MIT OpenCourseWare http://ocw.mt.edu 5.74 Introductory Quantum Mechancs II Sprng 9 For nformaton about ctng these materals or our Terms of Use, vst: http://ocw.mt.edu/terms. 4-1 4.1. INTERACTION OF LIGHT
More informationNordea G10 Alpha Carry Index
Nordea G10 Alpha Carry Index Index Rules v1.1 Verson as of 10/10/2013 1 (6) Page 1 Index Descrpton The G10 Alpha Carry Index, the Index, follows the development of a rule based strategy whch nvests and
More information7.5. Present Value of an Annuity. Investigate
7.5 Present Value of an Annuty Owen and Anna are approachng retrement and are puttng ther fnances n order. They have worked hard and nvested ther earnngs so that they now have a large amount of money on
More information