Physics 207: Lecture 25. Announcements

Transcription

1 Physcs 07: Lecture Announcements Exm 3 revew sesson: TODAY from 8:00 9:30 pm; here Exm 3 on Frdy, y 6th Brng photo-id, scentfc clcultor, pencl, erser Topcs from Chpter 8, 9, nd 10 otton ecp Tody s Agend ny body dynmcs exmples ollng down n nclne Bowlng bll: sldng to rollng Atwood s chne wth mssve pulley Torque due to grvty 1 Torque ecll the defnton of torque: τ = rf θ = r F sn φ = r sn φ F τ = r p F r p = dstnce of closest pproch r r ore generlly τ = F r F r φ F φ F θ φ r r p Pge 1

2 Work & Power The work done by torque τ ctng through dsplcement θ s gven by: W = τθ The power provded by constnt torque s therefore gven by: P dw d = = τ θ = τω dt dt 3 ecp of otton About fxed rotton xs, you cn lwys wrte τ = Iα where τ s the torque, I s the moment of nert, nd α s the ngulr ccelerton. For dscrete pont prtcles, I = Σm r The prllel xs theorem lets you clculte the moment of nert bout n xs prllel to n xs through the C f you know I C : D I PAALLEL = I C + D x C I PAALLEL If the object s ccelertng, we cn stll use τ = Iα provded tht we re consderng rottons bout n xs through the C. I C L 4 Pge

3 Work & netc Energy: ecll the Work/netc Energy Theorem: = W NET Ths s true n generl, nd hence pples to rottonl moton s well s lner moton. So for n object tht rottes bout fxed xs: ( f ω ) WNET 1 = I ω = For objects tht re trnsltng nd rottng 1 1 = I( ω ω ) + ( V, V, ) = W f cm f cm NET Fllng weght & pulley A mss m s hung by strng tht s wrpped round pulley of rdus ttched to hevy flywheel. The moment of nert of the pulley + flywheel s I. The strng does not slp on the pulley. Strtng t rest, how long does t tke for the mss to fll dstnce L. α I T m mg L 6 Pge 3

4 Fllng weght & pulley... For the hngng mss use F = m mg - T = m For the pulley + flywheel use τ = Iα α I τ = T = Iα elze tht = α T = I T Now solve for usng the bove equtons. m mg m = g m + I L 7 Fllng weght & pulley... Usng 1-D knemtcs (Lecture ) we cn solve for the tme requred for the weght to fll dstnce L: L = 1 t t L = α T I where m = g m + I m mg L 8 Pge 4

5 ollng An object wth mss, rdus, nd moment of nert I rolls wthout slppng down plne nclned t n ngle θ wth respect to horzontl. Wht s ts ccelerton? Consder C moton nd rotton bout the C seprtely when solvng ths problem I θ 9 ollng... Sttc frcton f cuses rollng. It s n unknown, so we must solve for t. Frst consder the free body dgrm of the object nd use F NET = A C : In the x drecton g sn θ -f= A Now consder rotton bout the C nd use τ = Iα relzng tht τ = f nd A = α y f x f A = I f A = I θ g 10 Pge

6 ollng... We hve two equtons: gsnθ - f = ma f = I A We cn combne these to elmnte f: sn θ A = g + I I For sphere: A A = g sn θ + = gsn θ 7 θ 11 Sldng to ollng A bowlng bll of mss nd rdus s thrown wth ntl velocty v 0. It s ntlly not rottng. After sldng wth knetc frcton long the lne for dstnce D t fnlly rolls wthout slppng nd hs new velocty v f. The coeffcent of knetc frcton between the bll nd the lne s µ. Wht s the fnl velocty, v f, of the bll? v f = ω ω v 0 f = µg D 1 Pge 6

7 Sldng to ollng... Whle sldng, the force of frcton wll ccelerte the bll n the -x drecton: F = -µg = so = -µg The speed of the bll s therefore v = v 0 - µgt () Frcton lso provdes torque bout the C of the bll. Usng τ = Iα nd rememberng tht I = / for sold sphere bout n xs through ts C: τ = µ g = α µg α = µ g ω = ω0 + αt = t (b) x v f = ω ω v 0 f = µg D 13 We hve two equtons: Sldng to ollng... µg v = v0 µ gt () ω = t (b) Usng (b) we cn solve for t s functon of ω: t = ω µ g Pluggng ths nto () nd usng v f = ω (the condton for rollng wthout slppng): vf = v 7 0 Doesn t depend on µ,, g!! x v f = ω ω v 0 f = µg D 14 Pge 7

8 Lecture, Act 1 ottons A bowlng bll (unform sold sphere) rolls long the floor wthout slppng. Wht s the rto of ts rottonl knetc energy to ts trnsltonl knetc energy? 1 1 () (b) (c) ecll tht I = for sold sphere bout n xs through ts C: 1 Lecture, Act 1 Soluton The totl knetc energy s prtly due to rotton nd prtly due to trnslton (C moton). = 1 I ω + 1 V rottonl trnsltonl 16 Pge 8

9 Lecture, Act 1 Soluton = 1 I ω + 1 V Snce t rolls wthout slppng: ω= V rottonl Trnsltonl OT TANS 1 Iω = 1 V V = V = 17 Lecture, Act ottons A bll nd box hve the sme mss nd re movng wth the sme velocty cross horzontl floor. The bll rolls wthout slppng nd the box sldes wthout frcton. They encounter n upwrd slope n the floor. Whch one mkes t frther up the hll before stoppng? () bll (b) box (c) sme 18 Pge 9

10 Lecture, Act Soluton The bll nd box wll stop when ther ntl knetc energes hve been converted to grvttonl potentl energy (mgh). The ntl knetc energy of the box s = 1 mv = 1 mv + 1 The ntl knetc energy of the bll s I ω bgger v ω v 19 Lecture, Act Soluton Snce the bll hs more ntl knetc energy, t wll go hgher! 0 Pge 10

11 Atwoods chne wth ssve Pulley: A pr of msses re hung over mssve dsk-shped pulley s shown. Fnd the ccelerton of the blocks. For the hngng msses use F = m -m 1 g + T 1 = -m 1 -m g + T = m α y x For the pulley use τ = Iα = I T 1 T = I 1 T 1 - T = (Snce I = 1 for dsk) m 1 m 1 g m m g 1 Atwoods chne wth ssve Pulley... We hve three equtons nd three unknowns (T 1, T, ). Solve for. y -m 1 g + T 1 = -m 1 (1) -m g + T = m () α x T 1 -T = 1 (3) T 1 T m m = 1 m + m + g 1 m 1 m 1 g m m g Pge 11

12 Torque due to Grvty As we now know τ = Iα where τ = r X F Tke the rotton xs to be long the z drecton (s usul) nd recll tht τ So: τ = τ Z, = r X, F Y, -F X, r Y, = x (-m g) - 0 = g m x = gx τ = gx cm cm (lso obtn v closest pproch method) z-xs r 4 x y m 4 F 4 r 1 F 1 y m 1 x Where: = m m3 r 3 r m F 3 F 3 Torque due to Grvty... But ths s the sme expresson we would get f we were to fnd the C... y τ = NET gx cm = m C 4 Pge 1

13 Torque due to Grvty......nd ssume tht ll of the mss ws locted there! So for the purpose of fgurng out the torque due to grvty, you cn tret n object s though ll of ts mss were locted t the center of mss. y τ = NET gx cm = m r cm x cm g Exmple problem: Yo-yo Problem done on the bord 6 Pge 13