Physcs 231 Topc 8: Rotatonal Moton Alex Brown October 21-26 2015 MSU Physcs 231 Fall 2015 1
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Key Concepts: Rotatonal Moton Rotatonal Kneatcs Equatons of Moton for Rotaton Tangental Moton Knetc Energy and Rotatonal Inerta Moent of Inerta Rotatonal dynacs Rollng bodes Vector quanttes Mechancal Equlbru Angular Moentu Covers chapter 8 n Rex & Wolfson MSU Physcs 231 Fall 2015 4
Revew: Radans & Radus Crcuference= 2r s = r r s n radans! 360 o =2 rad = 6.28 rad (rad) = 2/360 o (deg) MSU Physcs 231 Fall 2015 5
Revew: Angular speed t f f l t0 t t t Average angular velocty Instantaneous Angular velocty Angular velocty : rad/s MSU Physcs 231 Fall 2015 6
Revew: Angular Lnear Veloctes s r s r s t r t v r lnear (tangental) velocty MSU Physcs 231 Fall 2015 7
Angular acceleraton Defnton: The change n angular velocty per te unt t f f l t0 t t t Average angular acceleraton Instantaneous angular acceleraton Unt: rad/s 2 MSU Physcs 231 Fall 2015 8
Angular Lnear Acceleratons v r v r v r t t velocty change n velocty Change n velocty per te unt a a r lnear or tangental acceleraton MSU Physcs 231 Fall 2015 9
Equatons of oton Lnear oton x(t) = x o + v o t+ ½at 2 v(t) = v o + at Angular oton (t) = o + o t+ ½t 2 (t) = o + t Angular oton = Rotatonal Moton! MSU Physcs 231 Fall 2015 10
exaple =0 A person s rotatng a wheel. The handle s ntally at rest at = 0 o. For 5s the wheel gets a constant angular acceleraton of 1 rad/s 2. After that The angular velocty s constant. Through what angle wll the wheel have rotated after 10s. Frst 5s: (5)= 0 + 0 t + ½t 2 (5) = 0 = 0 + (0)(5) + ½1(5) 2 = 12.5 + t = 0 + (1)(5) = 5 rad/s Next 5s: (5) = 12.5 + (5)(5) + 0 = 37.5 t 37.5 rad = 2150 o = 6.0 rev MSU Physcs 231 Fall 2015 11
A Rollng Con D d 0 = 18 rad/s = -1.8 rad/s d = 0.02 a) How long does the con roll before stoppng? b) What s the average angular velocty? c) How far (D) does the con roll before cong to rest? a) (t) = 0 + t 0 = 18-1.8t t=10 s b) = ( 0 + (10))/2 = (18 + 0)/2 = 18/2 = 9 rad/s = area under vs t curve = (10)(18)/2 = 90 rad c) D = s = r = (90) (0.02/2) = 0.9 MSU Physcs 231 Fall 2015 12
O d Top vew Torque It s uch easer to swng the door f the force F s appled as far away as possble (d) fro the rotaton axs (O). F Torque: The capablty of a force to rotate an object about an axs. Torque = F d (N) Torque s postve f the oton s counterclockwse (angle ncreases) Torque s negatve f the oton s clockwse (angle decreases) MSU Physcs 231 Fall 2015 13
Multple forces causng torque. F1 = 100 N Top vew 0.3 0.6 F 2 = 50 N Two persons try to go through a rotatng door at the sae te, one on the l.h.s. of the rotator and one the r.h.s. of the rotator. If the forces are appled as shown n the drawng, what wll happen? 1 = -F 1 d 1 = -(100)0.3 = -30 N 2 = F 2 d 2 = (50)0.6 = 30 N 0 N Nothng wll happen! The 2 torques are balanced. MSU Physcs 231 Fall 2015 14
Center of gravty (center of ass) d pull 1 2 3 n F pull (sde vew) F gravty d gravty? = F pull d pull + F gravty d gravty We can assue that for the calculaton of torque due to gravty, all ass s concentrated n one pont: The center of gravty: the average poston of the ass d cg =( 1 d 1 + 2 d 2 + + n d n ) ( 1 + 2 + + n ) MSU Physcs 231 Fall 2015 15
Center of ass; ore general center of gravty = center f ass The center of gravty x CG x x 1,y 1 x 2,y 2 1 g 2 g x cg,y cg y CG y x 3,y 3 3 g Note 2: freely rotatng systes (I.e. not held fxed at soe pont) ust rotate around ther center of gravty. O MSU Physcs 231 Fall 2015 16
exaples: (ore n the book) Oxygen olecule relatve ass O s 16 H s 1 Where s the center of gravty? x Lcos 0.060 y Lsn 0.080 x CG x (16)(0) (1) x (1) x 1611 0.12 18 0.0067 y CG y (16)(0) (1) y (1)( y) 1611 0 MSU Physcs 231 Fall 2015 17
Clcker Quz Consder a hoop of ass 1 kg. The center of gravty s located at: a) The edge of the hoop b) The center of the hoop c) Depends on how the hoop s postoned relatve to the earth s surface. MSU Physcs 231 Fall 2015 18
Object n equlbru? Top vew d F p d CG Newton s 2nd law: F=a F p +(-F p )=a=0 No acceleraton of the center of ass up or down F p But the block starts to rotate! = F p d+ F p d= 2F p d (both clockwse) There s rotaton! MSU Physcs 231 Fall 2015 19
Torque: = F d Center of Gravty: x CG x y CG y Translatonal equlbru: F = a = 0 Rotatonal equlbru: =0 The center of gravty does not ove The object does not rotate Mechancal equlbru: F = a = 0 and =0 No oveent! MSU Physcs 231 Fall 2015 20
queston 20N 1 1 20N 20N COG Is ths freely rotatng object n echancal equlbru? a) Yes b) No Freely rotatng: can only rotate around ts COG: Rotatonal eq.: = -20x1 + 20x0 + 20x1 =0 Translatonal eq.: F= 20 20 +20 =+20 No translatonal equlbru: no echancal equlbru! MSU Physcs 231 Fall 2015 21
Torque: =Fd clcker quz 40N 20N 40N -2 0 1 2 A wooden bar s ntally balanced. Suddenly, 3 forces are appled, as shown n the fgure. Assung that the bar can only rotate, what wll happen (what s the su of torques)? a) the bar wll rean n balance b) the bar wll rotate counterclockwse c) the bar wll rotate clockwse MSU Physcs 231 Fall 2015 22
Soe thngs to keep n nd If an object s n equlbru, t does not atter where one chooses the axs of rotaton: the net torque ust always be zero. If t s gven that an object s at equlbru, you can choose the rotaton axs at a convenent locaton to solve the proble. The only part of the force that causes the rotaton s the coponent perpendcular to radus of the crcle. MSU Physcs 231 Fall 2015 23
ass: L Fallng Bar F g =g F p F p = F g cos = g cos perpendcular coponent In ths case for the torque = g cos (L/2) MSU Physcs 231 Fall 2015 24
T 1 T 2 0.75 0.25 Weght of board: w What s the tenson n each of the wres (n ters of w) gven that the board s n equlbru? Translatonal equlbru F = a = 0 T 1 +T 2 -w = 0 so T 1 =w-t 2 0 w Rotatonal equlbru = 0 T 1 0 (0.5)w + (0.75)T 2 =0 T 2 =0.5/(0.75 w) = 2/3w T 1 =1/3w T 2 =2/3w Note: I chose the rotaton pont at T 1, but could have chosen anywhere else. By choosng T 1 the torque due to T 1 Does not contrbute to the equaton for rotatonal equlbru MSU Physcs 231 Fall 2015 25
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=24.4 o =20.0 o =24.4 o =45.6 o =24.4 o MSU Physcs 231 Fall 2015 27
Torque and angular acceleraton F Newton 2nd law: F=a r F= a Fr = r a Fr = r 2 Used a = r = F r = r 2 = I torque = I The angular acceleraton goes lnear wth the torque. I = r 2 = oent of nerta for one pont ass MSU Physcs 231 Fall 2015 28
More General for rotaton around a coon axs 2 r 1 r 2 r 3 3 1 = I Moent of nerta I: I = ( r 2 ) r s the dstance to the axs of rotaton I = r 2 = specal case when all ass s at the sae dstance r MSU Physcs 231 Fall 2015 29
Extended objects - hoop F M r I=( r 2 ) =( 1 + 2 + + n ) r 2 =M r 2 MSU Physcs 231 Fall 2015 30
A hoogeneous stck F Rotaton pont =( 1 r 12 + 2 r 22 + + n r n2 ) =( r 2 ) =I Moent of nerta I: I=( r 2 ) MSU Physcs 231 Fall 2015 31
Two nhoogeneous stcks r 5 5 F 18 Easy to rotate! =( r 2 ) 118r 2 5 5 F =( r 2 ) 310r 2 18 Dffcult to rotate MSU Physcs 231 Fall 2015 32
Deo: fghtng stcks MSU Physcs 231 Fall 2015 33
F A sple exaple A F B r r A and B have the sae total ass. If the sae torque s appled, whch one accelerates faster? Answer: A =I Moent of nerta I: I=( r 2 ) MSU Physcs 231 Fall 2015 34
0.2 kg The rotaton axs atters! 0.3 kg 0.5 I = ( r 2 ) =(0.2) 0.5 2 + (0.3) 0.5 2 + (0.2) 0.5 2 + (0.3) 0.5 2 = 0.5 kg 2 I = ( r 2 ) = (0.2) 0 + (0.3) 0.5 2 + (0.2) 0 + (0.3) 0.5 2 = 0.3 kg 2 MSU Physcs 231 Fall 2015 35
= I (copare to F= a ) Moent of nerta I: I = ( r 2 ) : angular acceleraton I depends on the shape of the object and the choce of rotaton axs!! MSU Physcs 231 Fall 2015 36
Soe coon cases R hoop/thn cylndrcal shell: I=MR 2 sold sphere I= 2 / 5 MR 2 R dsk/sold cylnder I= 1 / 2 MR 2 thn sphercal shell I= 2 / 3 MR 2 L Long thn rod wth rotaton axs through center: I= 1 / 12 ML 2 Long thn rod wth rotaton axs through end: I= 1 / 3 ML 2 L MSU Physcs 231 Fall 2015 37
ass: L Fallng Bars I bar =L 2 /3 F p = F g cos = g cos perpendcular coponent F g =g Copare the angular acceleraton for 2 bars of dfferent ass, but sae length. = I = L 2 /3 also = F p d = g(cos)l/2 So g(cos)l/2 = L 2 /3 and = 3g(cos)/(2L) Copare the angular acceleraton for 2 bars of sae ass, but dfferent length f L goes up, goes down! MSU Physcs 231 Fall 2015 38
r v Rotatonal knetc energy Consder a object rotatng wth constant velocty. Each pont oves wth velocty v. The total knetc energy s: 1 2 v 1 2 2 2 2 2 r 1 r 1 I 2 KE r = ½I 2 2 Conservaton of energy for rotatng object ust nclude: rotatonal and translatonal knetc energy and potental energy 2 2 [PE+KE t +KE r ] ntal = [PE+KE t +KE r ] fnal MSU Physcs 231 Fall 2015 39
Convenent way of wrtng KE for rotaton I = kr 2 object I k A Cylndrcal Mr 2 1 shell B Sold cylnder (1/2) r 2 0.5 C Thn sphercal (2/3) r 2 (2/3) shell D Sold sphere (2/5) r 2 (2/5) MSU Physcs 231 Fall 2015 40
rollng oton (no slppng) v rotaton r = v rotaton = v ths eans that =v/r MSU Physcs 231 Fall 2015 41
KE for rollng oton (no slppng) r = radus KE = KE (translatonal) + KE (rotatonal) = [½v 2 + ½I 2 ] = [½v 2 + ½ (kr 2 ) (v 2 /r 2 ) ] = [½v 2 +½kv 2 ] = (1+k)½v 2 used: I=kr 2 and =v/r KE = (1+k)½v 2 MSU Physcs 231 Fall 2015 42
Exaple. 1 Consder a ball (sphere) and a block gong down the sae 1-hgh slope. The ball rolls and both objects do not feel frcton. If both have ass 1kg, what are ther veloctes at the botto (I.e. whch one arrves frst?). The radus of the ball s 0.4. Block: [½v 2 +gy] ntal = [½v 2 + gy] fnal v=2gh 0 + gh = ½v 2 + 0 so v = 4.43 /s MSU Physcs 231 Fall 2015 43
Exaple. 1 Consder a ball and a block gong down the sae 1-hgh slope. The ball rolls and both objects do not feel frcton. If both have ass 1kg, what are ther veloctes at the botto (I.e. whch one arrves frst?). The radus of the ball s 0.4. Ball: [½(1+k)v 2 +gy] ntal = [½(1+k)v 2 + gy] fnal 0 + gh = ½(1+k)v 2 + 0 v=[2gh/(1+k)] wth k = 2/5 so v = 3.74 /s slower than the block snce part of the energy goes nto rotatonal KE MSU Physcs 231 Fall 2015 44
Rollng quz 1 whch gets to the botto frst A) Sold sphere B) Sold cylnder C) Hoop (thn cylndrcal shell) MSU Physcs 231 Fall 2015 45
Rollng quz 2 whch gets to the botto frst A) Kart B) Sold sphere C) Sold cylnder D) Hoop (thn cylndrcal shell) MSU Physcs 231 Fall 2015 46
Rollng quz 3 whch gets to the botto frst A) Can of energy drnk B) Sold sphere C) Sold cylnder D) Hoop (thn cylndrcal shell) MSU Physcs 231 Fall 2015 47
Whch one goes fastest (h=1) object k v Fastest (rank) Block sldng 0 4.43 1 Cylndrcal shell 1 3.13 5 Sold cylnder 1/2 3.61 3 Thn sphercal shell 2/3 3.42 4 Sold sphere 2/5 3.74 2 The larger the oent of nerta, the slower t rolls! The avalable KE s spread over rotatonal and translatonal Or: besdes ovng the object, you spend energy rotatng t MSU Physcs 231 Fall 2015 48
MSU Physcs 231 Fall 2015 49 Angular oentu 0 L then 0 f 0 0 0 t L t L L I L t I I t I I Conservaton of angular oentu: If the net torque equals zero, the angular oentu L does not change I = I f f
Sun: radus: R s = (7) 10 5 k Perod of rotaton: 25 days Neutron star Supernova exploson Neutron star: radus: R n = 10 k What s frequency of rotaton? I sphere =(2/5)MR 2 s = sun n = neutron star (assue no ass s lost so M s =M n ) s = 2/(25 days x 24 x 3600) = (2.9) 10-6 rad/s Conservaton of angular oentu: I s s = I n n n = (R s /R n ) 2 s = 1.4E+04 rad/s f n = ( n /2 ) = 2200 Hz! MSU Physcs 231 Fall 2015 50
The spnnng lecturer (sple odel) A lecturer (60 kg) s rotatng on a platfor wth =2 rad/s (1 rev/s). He s holdng two 1 kg asses 0.8 away fro hs body. He then puts the asses close to hs body (R=0.4 ). How fast he wll rotate? Is angular oentu conversed? Is knetc energy conserved? 0.4 0.8 MSU Physcs 231 Fall 2015 51
The spnnng lecturer (sple odel) A lecturer (60 kg) s rotatng on a platfor wth =2 rad/s (1 rev/s). He s holdng two 1 kg asses 0.8 away fro hs body. He then puts the asses close to hs body (R=0.4 ). I =2 M w R 2 w (two weghts) = 1.28 kg 2 0.8 0.4 I f =2M w (R w /2) 2 = 0.32 kg 2 (two weghts) = 6.28 rad/s I = I f f f = (I /I f ) = 25.1 rad/s KE = (1/2)I ( ) 2 = 25 KE f = (1/2)I ( f ) 2 = 100 MSU Physcs 231 Fall 2015 52
Conservaton laws: In a closed syste: Conservaton of energy E Conservaton of lnear oentu p Conservaton of angular oentu L MSU Physcs 231 Fall 2015 53
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L I MSU Physcs 231 Fall 2015 55
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queston A B d A d B Two people are on a see saw. Person A s lghter than person B. If person A s sttng at a dstance d A fro the center of the seesaw, s t possble to fnd a dstance d B so that the seesaw s n balance? A) Yes B) No If M A =0.5M B and d A =3, then to reach balance, d B =? A) 1 B) 1.5 C) 2 D) 2.5 E) 3 MSU Physcs 231 Fall 2015 58
The drecton of the rotaton axs L L The rotaton n the horzontal plane s reduced to zero: There ust have been a large net torque to accoplsh ths! (ths s why you can rde a bke safely; a wheel wants to keep turnng n the sae drecton.) The conservaton of angular oentu not only holds for the agntude of the angular oentu, but also for ts drecton. MSU Physcs 231 Fall 2015 59
Rotatng a bke wheel! L A person (p) on a platfor that can freely rotate s holdng a spnnng bke wheel (b) and then turns the wheel around. What wll happen? L Intal: angular oentu: L = I b b Closed syste, so L ust be conserved. Fnal: L f = - I b b + I p p = L p = 2( I b /I p ) b MSU Physcs 231 Fall 2015 60
The spnnng lecturer (ore detaled odel) A lecturer (60 kg) s rotatng on a platfor wth =2 rad/s (1 rev/s). He s holdng two 1 kg asses 0.8 away fro hs body. He then puts the asses close to hs body (R=0.0 ). Estate how fast he wll rotate ( ar =2.5 kg). 0.8 0.4 I = 0.5 M lec R 2 (body) +2(M w R w2 ) (two weghts) +2(0.33M ar 0.8 2 ) (two ars) = 1.2 + 1.3 + 1.0 = 3.5 kg 2 I f = 0.5 M lec R 2 = 1.2 kg 2 MSU Physcs 231 Fall 2015 61
The spnnng lecturer (ore detaled odel) A lecturer (60 kg) s rotatng on a platfor wth =2 rad/s (1 rev/s). He s holdng two 1 kg asses 0.8 away fro hs body. He then puts the asses close to hs body (R=0.0 ). Estate how fast he wll rotate ( ar =2.5 kg). 0.8 0.4 I = 0.5 M lec R 2 (body) +2(M w R w2 ) (two weghts) +2(0.33M ar 0.8 2 ) (two ars) = 1.2 + 1.3 + 1.0 = 3.5 kg 2 I f = 0.5 M lec R 2 = 1.2 kg 2 MSU Physcs 231 Fall 2015 62
The spnnng lecturer (ore detaled odel) A lecturer (60 kg) s rotatng on a platfor wth =2 rad/s (1 rev/s). He s holdng two 1 kg asses 0.8 away fro hs body. He then puts the asses close to hs body (R=0.0 ). Estate how fast he wll rotate ( ar =2.5 kg). 0.8 0.4 Conservaton of angular o. I I = I f f f = (I /I f ) f = (3.5/1.2) 2 = 18.3 rad/s (approx 3 rev/s) For detals see next slde KE = KE f? A=yes B=no MSU Physcs 231 Fall 2015 63
Detals of spnnng lecturer I ntal =I body +2I ass +2I ars I body = 0.5M lec R 2 lec (sold cylndrcal shell) M=60, R=0.2 I ass =M ass R 2 ass (pont-lke ass at radus R) M=1 R=0.8 I ars =1/3M ar R 2 ar (ar s lke a long rod wth rotaton axs through one of ts ends) M=2.5 kg R=0.8 I ntal =0.5M lec R lec2 +2(M ass R ass2 )+2(0.33M ar R ar2 ) =1.2+1.3+1.0= 3.5 kg 2 I fnal =0.5M lec R 2 =1.2 kg 2 (assued that asses and ars do not contrbute to oent of nerta anyore) Conservaton of angular o. I =I f f 3.5*2=1.2* f = was gven to be 2 rad/s f =18.3 rad/s (approx 3 rev/s) KE = (1/2)I ( ) 2 = 69 KE f = 201 MSU Physcs 231 Fall 2015 64