Chapter 20 Rgd Body: Translaton and Rotatonal Moton Knematcs for Fxed Axs Rotaton 20.1 Introducton... 1 20.2 Constraned Moton: Translaton and Rotaton... 1 20.2.1 Rollng wthout slppng... 5 Example 20.1 Bcycle Wheel Rollng Wthout Slppng... 6 Example 20.2 Cylnder Rollng Wthout Slppng Down an Inclned Plane... 9 Example 20.3 Fallng Yo-Yo... 10 Example 20.4 Unwndng Drum... 11 20.3 Angular Momentum for a System of Partcles Undergong Translatonal and Rotatonal... 12 Example 20.5 Earth s Moton Around the Sun... 15 20.4 Knetc Energy of a System of Partcles... 17 20.5 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
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. 1 Coperncus 20.1 Introducton The general moton of a rgd body of mass m conssts of a translaton of the center of mass wth velocty V and a rotaton about the center of mass wth all elements of the rgd body rotatng wth the same angular velocty ω. We prove ths result n Appendx A. Fgure 20.1 shows the center of mass of a thrown rgd rod follows a parabolc trajectory whle the rod rotates about the center of mass. Fgure 20.1 The center of mass of a thrown rgd rod follows a parabolc trajectory whle the rod rotates about the center of mass. 20.2 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 10. 20-1
Fgure 20.2 Rollng Wheel Consder a wheel of radus R s rollng n a straght lne (Fgure 20.2). The center of mass of the wheel s movng n a straght lne at a constant velocty V. 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 20.3a). Let v P denote the velocty of the pont P on the rm wth respect to the center of mass reference frame O movng wth velocty V wth respect to at O (Fgure 20.3b). (You should revew the defnton of the center of mass reference frame n Chapter 15.2.1.) We can use the law of addton of veloctes (Eq.15.2.4) to relate these three veloctes, v P = v P + V. (20.2.1) Let s choose Cartesan coordnates for the translaton moton and polar coordnates for the moton about the center of mass as shown n Fgure 20.3. (a) (b) Fgure 20.3 (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 = V î. (20.2.2) 20-2
where V 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 (t) = (X, 0 + V t)î, (20.2.3) where X, 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 = ω kˆ. (20.2.4) where ω 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 ˆ P = Rω θ. (20.2.5) If we want to use the law of addton of veloctes then we should express v P = Rω θˆ 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 20.4) rˆ = snθî cosθ ĵ. (20.2.6) θˆ = 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 θˆ = Rω (cosθî snθ ĵ). (20.2.7) Fgure 20.4 Unt vectors ow substtute Eqs. (20.2.2) and (20.2.7) nto Eq. (20.2.1) for the velocty of a pont P on the rm n the reference frame fxed to the ground v P = Rω (cosθî + snθĵ) +V î. (20.2.8) = (V + Rω cosθ)î + Rω snθˆ j 20-3
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 (θ = π ) = (V Rω )î. (20.2.9) P 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 and ω, we consder the dsplacement of the center of mass for a small tme nterval Δt (Fgure 20.5). Fgure 20.5 Dsplacement of center of mass n ground reference frame. From Eq. (20.2.3) the x -component of the dsplacement of the center of mass s ΔX =V Δt. (20.2.10) The pont P on the rm n the center of mass reference frame s undergong crcular moton (Fgure 20.6). Fgure 20.6: Small dsplacement of pont on rm n center of mass reference frame. 20-4
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 Δt, Δs = RΔθ = Rω Δt. (20.2.11) 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 = Δs. (20.2.12) Substtute Eq. (20.2.10) and Eq. (20.2.11) nto Eq. (20.2.12) and dvde through by Δt. Then the rollng wthout slppng condton becomes V = Rω, (rollng wthout slppng). (20.2.13) 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 > Δs. (20.2.14) Substtute Eqs. (20.2.10) and (20.2.11) nto Eq. (20.2.14) and dvde through by Δt, then V > Rω, (skddng). (20.2.15) 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 20.2.1 Rollng wthout slppng ΔX < Δs. (20.2.16) V < Rω, (slppng). (20.2.17) 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. (20.2.9) set θ = π, v P (θ = π ) = (V Rω )î = (Rω Rω )î = 0. (20.2.18) 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
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 20.1 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. A bead s fxed to a spoke a dstance b from the center of the wheel (Fgure 20.7). (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 20.7 Example 20.1 Fgure 20.8 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 20.8). The z - component of the angular velocty ω = dθ / dt > 0. Then the bead s movng unformly n a crcle of radus r = b wth the poston, velocty, and acceleraton gven by 2 r b = b rˆ, v b = bω θˆ, a b = bω rˆ. (20.2.19) Because the wheel s rollng wthout slppng, the velocty of a pont on the rm of the wheel has speed v P = Rω. Ths s equal to the speed of the center of mass of the wheel V, thus V = Rω. (20.2.20) ote that at t = 0, the angle θ = θ 0 = 0. So the angle grows n tme as 20-6
θ(t) = ω t = (V / R)t. (20.2.21) The velocty and acceleraton of the bead wth respect to the center of the wheel are then bv bv 2 v = θˆ, a b b = rˆ. (20.2.22) R 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 20.9. Fgure 20.9 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 (t) = X î + R ĵ = V t î + R ˆ j. (20.2.23) The relatve velocty of the two frames s the dervatve d dx V = R = î = V î. (20.2.24) dt dt Because the center of the wheel s movng at a unform speed the relatve acceleraton of the two frames s zero, d V = A = 0. (20.2.25) dt Defne the poston, velocty, and acceleraton n ths frame (wth respect to the ground) by r (t) = x (t) î + y (t) ĵ, v (t) î + v (t) ĵ, a(t) = a (t) î + a (t) ĵ b b b b (t) = v b,x b,y b,x. (20.2.26) b,y Then the poston vectors are related by 20-7
r b (t) = R (t) + r b (t). (20.2.27) In order to add these vectors we need to decompose the poston vector n the center of mass reference frame nto Cartesan components, r b (t) = b rˆ(t) = bsnθ(t) î + bcosθ(t) ĵ. (20.2.28) Then usng the relaton θ(t) = (V / R)t, Eq. (20.2.28) becomes r b (t) = R (t) + r b (t) = (V t î + R ĵ) + (bsnθ(t) î + bcosθ(t) ĵ). (20.2.29) = V t + bsn((v / R)t)) î + (R + bcos((v / R)t)) 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 t + bsn((v / R)t) (20.2.30) y b (t) = R + bcos((v / R)t). (20.2.31) 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 20.10 below usng the values V = 5 m s -1, R = 0.25 m, and b = 0.125 m. We can dfferentate the poston vector n the reference frame fxed to the ground to fnd the velocty of the bead dr v b (t) = b dt (t) = d (V t + bsn((v / R)t)) î + d (R + bcos((v / R)t) )ĵ dt, (20.2.32) dt v (t) = (V + (b / R)V cos((v / R)t)) î ((b / R)V sn((v / R)t) )ˆj. (20.2.33) b Fgure 20.10 Plot of the y -component vs. the x -component of the poston of the bead 20-8
Alternatvely, we can decompose the velocty of the bead n the center of mass reference frame nto Cartesan coordnates v b (t) = (b / R)V (cos((v / R)t) î sn((v / R)t) ĵ). (20.2.34) The law of addton of veloctes s then v b (t) = V + v b (t), (20.2.35) v (t) = V î + (b / R)V (cos((v / R)t) î sn((v / R)t) ĵ), (20.2.36) b v (t) = (V + (b / R)V cos((v / R)t)) î (b / R)sn((V / R)t) ˆj, (20.2.37) b n agreement wth our prevous result. The acceleraton s the same n ether frame so 2 a (t) = = (b / R 2 )V rˆ = (b / R 2 )V 2 (sn((v / R)t) î + cos((v / R)t) ˆ b a b j). (20.2.38) When the bead s at the rm of the wheel, b = R, then the poston of the bead n the reference frame fxed to the ground s gven by r b (t) = (V t + Rsn((V / R)t)) î + R(1+ cos((v / R)t))) ĵ. (20.2.39) The path traced out by the bead n the reference frame fxed to the ground s called a cyclod. Example 20.2 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 = (1 / 2) MR 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 20.11? 20-9
Fgure 20.11 Example 20.2 Soluton: We begn by choosng a coordnate system for the translatonal and rotatonal moton as shown n Fgure 20.12. Fgure 20.12 Coordnate system for rollng cylnder For a tme nterval Δt, the dsplacement of the center of mass s gven by ΔR (t) = ΔX 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 = 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 Δθ V = lm = lm R = Rω. Δt 0 Δt Δt 0 Δt 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 20.3 Fallng Yo-Yo dv dω A = = R = Rα. dt dt A Yo-Yo of mass m has an axle of radus b and a spool of radus R (Fgure 20.13a). 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? 20-10
Soluton: Choose coordnates as shown n Fgure 20.13b. Fgure 20.13a Example 20.3 Fgure 20.13b 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 = dl / dt, we have that, y bω = V. Dfferentatng a second tme yelds bα = A., y, y Example 20.4 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 20.14). 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 20.14 Example 20.4 20-11
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 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 dθ A dθ B = R + R. dt 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 d 2 θ A d 2 θ B = R + R dt 2 dt 2 dt 2 a A, y = Rα A + Rα B. 20.3 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 = L = ( r m v ), (20.3.1) S, L total S S, =1 =1 where r S, s the vector from the pont S to the th partcle (Fgure 20.15) satsfyng r = r + r, (20.3.2) S, S,, v = V + v, (20.3.3) S,, where v S, = V. We can now substtute both Eqs. (20.3.2) and (20.3.3) nto Eq. (20.3.1) yeldng 20-12
L total ( + ) m ( V r S, r, + v, ). (20.3.4) = S =1 Fgure 20.15 Vector Trangle When we expand the expresson n Equaton (20.3.4), we have four terms, = ( m ) ( r V ) S, + m S, L total r v S, =1 =1 + ( r m v, ) + ( r m V )., =1 =1, (20.3.5) The vector r S, s a constant vector that depends only on the locaton of the center of th mass and not on the locaton of the partcle. Therefore n the frst term n the above equaton, r S, can be taken outsde the summaton. Smlarly, n the second term the velocty of the center of mass V s the same for each term n the summaton, and may be taken outsde the summaton, total L = r r S, S m S, v + m V, =1 =1 + ( r, m v, ) + m r V., =1 =1 (20.3.6) The frst and thrd terms n Eq. (20.3.6) are both zero due to the fact that =1 m r, = 0 =1 m v, = 0. We frst show that m r, s zero. We begn by usng Eq. (20.3.2), =1 (20.3.7) 20-13
(m ) (m ( )) r =, r rs, =1 =1 = =. r total m r r r (m ) S, m m S, =1 =1 =1 (20.3.8) Substtute the defnton of the center of mass (Eq. 10.5.3) nto Eq. (20.3.8) yeldng 1 total (m r ) = m = 0. (20.3.9), m r m r total =1 =1 m =1 The vanshng of m v = 0 follows drectly from the defnton of the center of mass, =1 frame, that the momentum n the center of mass s zero. Equvalently the dervatve of Eq. (20.3.9) s zero. We could also smply calculate and fnd that m v, = ( v V ) m = m v V m (20.3.10) total total = m V V m = 0. We can now smplfy Eq. (20.3.6) for the angular momentum about the pont S usng the fact that, m T = m, and p = m T V (n reference frame O ): =1 sys = r sys p + ( r m v ). (20.3.11), L total S S,, =1 Consder the frst term n Equaton (20.3.11), r S, p sys ; the vector r S, 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 = m T V. Thus the frst term s the angular momentum about the pont S of ths sys pont-lke partcle, whch s called the orbtal angular momentum about S, orbtal = r S, p sys. L S (20.3.12) for the system of partcles. 20-14
Consder the second term n Equaton (20.3.11), (r m v ) ; the quantty nsde,, =1 the summaton s the angular momentum of the th partcle wth respect to the orgn n the center of mass reference frame O (recall the orgn n the center of mass reference frame s the center of mass of the system), L r = m v. (20.3.13),,, 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 = L =, =1 =1 r, m v ( ). (20.3.14), 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, L total S = L orbtal S + L spn. (20.3.15) 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 20.5 Earth s Moton Around the Sun The earth, of mass m e = 5.97 10 24 kg and (mean) radus R e = 6.38 10 6 m, moves n a nearly crcular orbt of radus r s,e = 1.50 10 11 m around the sun wth a perod T orbt = 365.25 days, and spns about ts axs n a perod T spn = 23 hr 56 mn, the axs nclned to the normal to the plane of ts orbt around the sun by 23.5 (n Fgure 20.16, the relatve sze of the earth and sun, and the radus and shape of the orbt are not representatve of the actual quanttes). Fgure 20.16 Example 20.5 20-15
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 e e I = m R. (20.3.16) 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 orbtal L rˆ m v θ ˆ = r m v k ˆ S = r S, p sys = r s,e e s,e e. (20.3.17) The velocty of the center of mass of the earth about the sun s related to the orbtal angular velocty by v = r s,e ω orbt, (20.3.18) where the orbtal angular speed s 2π 2π ω orbt = = Torbt (365.25 d)(8.640 10 4 s d 1 ) (20.3.19) = 1.991 10 7 rad s 1. The orbtal angular momentum about S s then L orbtal S = m e r s,e kˆ 2 ω orbt = (5.97 10 24 kg)(1.50 10 11 m) 2 (1.991 10 7 rad s 1 ) kˆ (20.3.20) = (2.68 10 40 kg m 2 s 1 ) kˆ. The spn angular momentum s gven by L spn = I ω 2 2 = m R ω nˆ, (20.3.21) spn e e 5 spn where nˆ s a unt normal pontng along the axs of rotaton of the earth and 2π 2π 5 1 ω spn = = = 7.293 10 rad s. (20.3.22) T 8.616 10 4 s spn The spn angular momentum s then spn 2 L 1 = (5.97 10 24 kg)(6.38 10 6 m) 2 (7.293 10 5 rad s ) nˆ 5 (20.3.23) = (7.10 10 33 kg m 2 s 1 ) nˆ. 20-16
The rato of the magntudes of the orbtal angular momentum about S to the spn angular momentum s greater than a mllon, orbtal L m r 2 ω r 2 e s,e orbt 5 s,e T S = = spn = 3.77 10 6, (20.3.24) spn 2 2 L (2/ 5) m R ω 2 R T e e spn 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 2 ˆ 2 2 L S = me r s,e ω orbt k + me R e ω spn nˆ. (20.3.25) 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 solar = 86,400s was used. In convertng the earth s spn angular frequency, the sdereal day, T sdereal = T spn = 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. 20.4 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 K = 1 2 m v = 1 m v v 2 2 1 m ( v +V ) ( v 2 =,, + V ). (20.4.1) where Equaton 15.2.6 has been used to express the last dot product n Equaton (20.4.1), n terms of v and V v,. Expandng! V V!! K = 1 m (!! v v + + 2 v! V ),,, 2 1!! 1!!! m v v V V v V! 2 ( ) + m 2 m ( ) + (20.4.2) =,,, = 1 2 1 2 +, 2 m V + 2 m v! v! V m,. 20-17
The last term n the thrd equaton n (20.4.2) vanshes as we showed n Eq. (20.3.7). Then Equaton (20.4.2) reduces to K = 1 2 1 m v +, 2 2 2 m V 1 2 1 total = m V 2 v + m., 2 2 (20.4.3) We nterpret the frst term as the sum of the ndvdual knetc energes of the partcles of the system n the center of mass reference frame O and the second term as the knetc energy of the center of mass moton n reference frame O. At ths pont, t s mportant to note that no assumpton was made regardng the mass elements beng consttuents of a rgd body. Equaton (20.4.3) 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. 20.5 Rotatonal Knetc Energy for a Rgd Body Undergong Fxed Axs Rotaton The rotatonal knetc energy for the rgd body, usng v = (r ) ω θˆ, smplfes to,, 1 2 K = I ω. (20.5.1) rot 2 Therefore the total knetc energy of a translatng and rotatng rgd body s 1 2 1 2 = K + K = mv + I ω. (20.5.2) 2 2 K total trans rot 20-18
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 20A.1). The relatve poston vector s gven by r = r 1,2 1 r. 2 (20.A.1) Fgure 20A.1 Two-body coordnate system. Recall we defned the center of mass vector, R, of the two-body system as R r r = m + m 1 1 2 2. m 1 + m 2 (20.A.2) In Fgure 20A.2 we show the center of mass coordnate system. Fgure 20A.2 Poston coordnates wth respect to center of mass 20-19
The poston vector of the object 1 wth respect to the center of mass s gven by where r m 1 r 1 + m 2 r 2 m 2 µ = r R = r 1 = (r r ) = r (20.A.3) m 1 + m 2 m 1 + m 2 m 1,1 1 1 2 1,2, m1m 2 µ =, (20. A. 4) m 1 + m2 s the reduced mass. In addton, the relatve poston vector between the two objects s ndependent of the choce of reference frame, r = r r = ( r + R ) ( r +R ) = r r = r (20.A.5). 12 1 2,1,2,1,2,1,2 Because the center of mass s at the orgn n the center of mass reference frame, Therefore m 1 r,1 + m 2 m 1 + m 2 r,2 = 0. (20.A.6) m 1 r,1 = m 2 r,2 (20.A.7) m 1 r,1 = m 2 r,2. (20.A.8) The dsplacement of object 1 about the center of mass s gven by takng the dervatve of Eq. (20.A.3), dr,1 = µ dr m 1,2. (20.A.9) 1 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 µ r R = r (20.A.10) =,,2 2 1,2 m2 µ d = dr 1,2. r,2 m 2 (20.A.11) Let = 1,2. An arbtrary dsplacement of the th object s gven respectvely by d r = d r + d R, (20.A.12), 20-20
whch s the sum of a dsplacement about the center of mass dr and a dsplacement of, the center of mass dr. 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 2 = ( r r ) ( r r ). (20.A.13) 1,2 1 2 1 2 Because ths dstance s constant we can dfferentate Eq. (20.A.13), yeldng the rgd body condton that 0 = 2(r r ) (dr dr ) = 2 r d (20.A.14) 20A.1. Translaton of the Center of Mass 1 2 1 2 1,2 The condton (Eq. (20.A.14)) can be satsfed f the relatve dsplacement vector between the two objects s zero, d r = dr d r = 0. (20.A.15) 1,2 1 2 Ths mples, usng, Eq. (20.A.9) and Eq. (20.A.11), that the dsplacement wth respect to the center of mass s zero, d r = d r = 0. (20.A.16),1 Thus by Eq. (20.A.12), the dsplacement of each object s equal to the dsplacement of the center of mass, d = dr, (20.A.17) whch means that the body s undergong pure translaton. 20A.2 Rotaton about the Center of Mass r ow suppose that d = dr dr 0. The rgd body condton can be expressed n r 1,2 1 2 terms of the center of mass coordnates. Usng Eq. (20.A.9), the rgd body condton (Eq. (20.A.14)) becomes 0 = 2 µ r m 1,2 dr,1. (20.A.18) 1 Because the relatve poston vector between the two objects s ndependent of the choce of reference frame (Eq. (20.A.5)), the rgd body condton Eq. (20.A.14) n the center of mass reference frame s then gven by,2 r 1,2 0 = 2r,1,2 dr,1. (20.A.19) 20-21
Ths condton s satsfed f the relatve dsplacement s perpendcular to the lne passng through the center of mass, r,1,2 dr. (20.A.20),1 By a smlar argument, r,1,2 dr. In order for these dsplacements to correspond to a,2 rotaton about the center of mass, the dsplacements must have the same angular dsplacement. Fgure 20A.3 Infntesmal angular dsplacements n the center of mass reference frame In Fgure 20A.3, the nfntesmal angular dsplacement of each object s gven by dr,1 dθ 1 = r,1 dr,2 dθ 2 = r,2, (20.A.21). (20.A.22) From Eq. (20.A.9) and Eq. (20.A.11), we can rewrte Eqs. (20.A.21) and (20.A.22) as dθ 1 = µ m 1 dθ 2 = µ m 2 dr 1,2, r,1 dr 1,2. r,2 (20.A.23) (20.A.24) Recall that n the center of mass reference frame m 1 r,1 = m 2 r,2 and hence the angular dsplacements are equal, (Eq. (20.A.8)) 20-22
dθ 1 = dθ 2 = dθ. (20.A.25) Therefore the dsplacement of the th object dr dffers from the dsplacement of the center of mass dr by a vector that corresponds to an nfntesmal rotaton n the center of mass reference frame d r = d R + d r (20.A.26)., 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
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