( ) rad ( 2.0 s) = 168 rad
|
|
- Audrey Chapman
- 5 years ago
- Views:
Transcription
1 .) α ω o 0 and ω 8.00 ω αt + ω o o t ω ω o α HO 9 Solution b.) ω ω o + αδθ o Δθ ω ω o α o Δθ 7. ev.3 ev π.) ω o.50, α 0.300, Δθ 3.50 ev π 7π ev ω ω o + αδθ o ω ω o + αδθ π ( ) ) θ( t) a + bt + ct 3 ω dθ dt d dt ( a + bt + ct 3 ) bt + 3ct and α dω d dt dt ( bt + 3ct ) b + 6ct 4.) α.50, Δθ 80.0, t 5.00 Δθ 80.0 αt Δθ αt + ω o t o ω o t ( ).50 ( ) ( ) ) D 0.00, ω o 0, ω 40, t 8.00 ω αt + ω o o α ω ω o t ( Δθ ω + ω o) t 6.) t 3.00, Δθ 6, ω ( ) 560 ( Δθ ω + ω o) t o ω o Δθ t ω ( 6 ) b.) ω αt + ω o o α ω ω o t
2 HO 9 Solution ( 7.) Δθ ω + ω o) t o () ubtituting () into () ω o Δθ t ω and () Δθ αt + ω o t Δθ αt + Δθ ω t t αt + Δθ ωt and Δθ ωt αt 8.) ω o 4.0, α 60, t 0 to t.0 then low down to ω 0 a Δθ 43 duing the peiod of acceleation Δθ α t + ω o t 60.0 ( ) (.0 ) 68 Δθ total Δθ + Δθ b.) afte.0 the angula velocity i ω α t + ω o 60.0 o ω o 44.0 fo the peiod of deceleation c.) ( Δθ ω + ω o) t o t Δθ ω + ω o ω α t + ω o o α ω ω o t ( ) ( 43 ) and t total t + t ) Δθ ev when blade low down fo ω to 0 ω ω o + αδθ o 0 ω + αδθ and α ω Δθ ω ev ( ) ω ω o + αδθ o 0 ω ( ) + αδθ and Δθ ( ω ) α 4ω 4 ev ω ( ev ) π 0.) Δθ 65 ev 30π, v 8.0 ev, v 4.0, D 0.80 o 0.40 ω v and ω v ω ω + αδθ o α ω ω Δθ π ( ) 4.50 b.) ω αt + ω o t ω ω α
3 HO 0 Solution.) 0.045, 0.005, v.00 The haft and the dik have the ae angula velocity ω v v ω ( ) ) t 3.00, 0.00, v 40.0, a 0.0 α a b.) ( ) v( 3 ) ω ω αt + ω o o ω o ω αt ( ) 350 c.) ( Δθ ω + ω o) t ( ) 85 d.) a 9.8 a v ω o ω a 7.0 ω αt + ω o o t ω ω o α ) D o 0.45, ω o 3.00 ev π ev 6.00π, α.50 ev π ev 3.00π b.) ω αt + ω o 3.00π.00 ( Δθ ω + ω o) t ( ) π π ( ) 3.57 Δθ 3.57 ev 3.75 ev π
4 HO 0 Solution 3.) (continued) c.) v ω 0.45 ( ) d.) a ω 8.3 ( 0.45 ) and a t α 0.45 ( ) 3.00π 4.00 a a t + a ) L 3 l L L L 3 I L i i 3 + L 3 l L L 3 3L 6 L 6 L 6 + l L 9 + 4L 9 + l I L 9 + 4L 9 + L L L 9 + L 4L L 36 + L L L 5.) L L 0.00 kg, L fo all fou phee L I i i 4 L L ( 0.00 kg) ( ) kg L b.) fo all fou phee L I 4 L i i L ( 0.00 kg) ( ) 0.03 kg 6.) 0.4 kg, L.0, L 3.0 L L fo all fou phee L I i i 4 L L ( 0.4 kg) ( 3.0 ).6 kg 7.) L 0.300, poke 0.30 kg, i.60 kg teat the i ha a hollow cylinde I MR and poke a od otated about thei end I 3 ML I I i + 8I poke i poke L I (.60 kg) ( ) ( 0.30 kg) ( ) 0. kg
5 8.) HO 0 Solution olid phee M 5.0 kg and R 0.0 about it cental axi I 5 MR ( 5.0 kg )( ) kg b.) hoop M 0.0 kg and R.5 about it cental axi pependicula to it diaete I MR ( 0.0 kg) (.5 ) 6.5 kg c.) hollow phee M 5.0 kg and R 3.00 about it cental axi I 3 MR ( 5.0 kg )( ) 90 kg d.) olid cylinde M 50 kg and R.5 about it cental axi pependicula to it diaete I MR ( 50 kg )(.5 ) 69 kg
6 HO Solution.) 4.00 kg y y 3.00 ω kg x y -.00 I i i I ( 4.00 kg) ( 3.00 ) + (.00 kg) (.00 ) + ( 3.00 kg) ( 4.00 ) 9 kg kg y K Iω ω ( 9 kg ) J b.) v ω 3.00 ( ) v ω.00 ( ).00 v 3 3 ω 4.00 ( ) K i v i v + v + v 3 3 K 4.00 kg ( ) kg ( ) kg ( ) J.) M x 0 x L - x I i i Mx + ( L x) it i at a iniu when di dx 0 ( ( ) ) 0 di dx d dx Mx + L x Mx + ( L x) ( ) Mx L + x 0 Mx L + x 0 x( M + ) L 0 x L M + the cente-of-a i x c i x i i M ( 0 ) + L M + L which i whee I i a iniu M +
7 .) (continued) b.) I M L + L L M + M + ( ) + L L I M L M + I M L M + ( ) + L L M L M + M + + L ( M + ) M L M + M + + L M + ( ) + L ( ) + L ( M + ) L ( M + ) 3 L + ( M + ) ( M + ) ( M + ) I M L ( M + ) + L ( ) M + M + M L + 3 L ( M + ) M + + ( ) ( M + ) I M L ( M + ) + M L + M L + 3 L M L + 3 L + ( M + ) ( M + ) L M L 3 L I M L + M L + M L + 3 L M L 3 L + 3 L ( M + ) ML ( M + ) ML ( ) ( M + ) I M L + M L M + HO Solution M + 3 L ( M + ) ( M + ) 3.) olid cylinde a M and iu R oll down an incline height h and v 0 uing Conevation of Enegy and the botto of the incline a a efeence y h and y 0 K +U g +W othe K +U g and 0 +U g + 0 K + 0 o U g K Mgy Mv + Iω Mv + MR v R Mgy Mv + 4 MR v R Mv + 4 Mv 3 4 Mv v 4 3 gy gh 3 4.) olid phee a M and iu R oll down an incline height h and v 0 uing Conevation of Enegy and the botto of the incline a a efeence y h and y 0 U g K o Mgy Mv + Iω Mv + 5 MR v R Mgy Mv + 0 MR v R v 0 7 gy 0gh 7 Mv + 0 Mv 7 0 Mv
8 5.) HO Solution olid cylinde about an axi paallel to it cente of a and paing though it edge d R fo a olid cylinde I c MR and uing paallel-axi theoe I p I c + Md I p MR + MR 3 MR b.) hollow phee about an axi tangent to it uface d R fo a hollow phee I c 3 MR and uing paallel-axi theoe I p I c + Md I p 3 MR + MR 5 3 MR 6.) ey-go-ound M 640 kg and R 8.0 ω o 0 and ω ev 8 π ev π 4 uing the otational equivalent of the Wok-Enegy Theoe W ΔK Iω Iω o Iω MR ω 4 MR ω W ( 640 kg) ( ) π 4 7,000 J 7.) hollow phee R 0.00 and M.0 kg olling down an incline L 0.0 and θ 30 v 0 and uing botto of incline a a efeence y Linθ and y 0 uing Conevation of Enegy U K o Mgy g Mv + Iω Mv + 3 MR v R Mgy Mv + 3 MR v R Mv + 3 Mv 5 6 Mv v ( 0.0 )in30 5 gy 6gLinθ ω v 7.67 R L 8.) hollow cylinde R 0.40 and M.0 kg olling up an incline 30 θ 30 and v 4.00 uing botto of the incline a a efeence y 0 and y Linθ uing Conevation of Enegy K U o g Mv + Iω Mgy o Mv + ( MR ) v R Mv + ( MR ) v Mgy R Mv Mgy and y Linθ v g o L v ginθ in30
9 HO Solution 9.) fo a hollow phee I c 3 MR and fo a olid phee I c 5 MR d uing paallel-axi theoe I p I c + Md 3 MR 5 MR + Md 3 R 5 R d 0 5 R 6 5 R d d 4 5 R R 5 0.) β R β y R the ditance y that the cente of a fall i y R Rcoβ R( coβ) the oent of inetia of the hoop about it cente of a i I c MR uing paallel-axi theoe I p I c + Md, the oent of inetia about an axi paing though it edge i uing Conevation of Enegy I p MR + MR MR the potential enegy coe fo the fact that the cente of a fall a ditance y and uing the equilibiu poition a a efeence y y and y 0 thi i conveted into otational kinetic enegy fo the hoop which i otating about it i U g K o Mgy I pω and ω Mgy I p ω MgR ( coβ ) MR g( coβ) R
10 HO Solution.) L 4.00, 5.0 N O φ 90.0 φ 90 v τ v v inφ ( 4.00 ) ( 5.0 N)in90 60 N ( ccw) O 0.0 φ φ 0 v τ v v inφ ( 4.00 ) ( 5.0 N)in0 5 N ( ccw) O 30.0 φ φ 30 v τ v v inφ ( 4.00 ) ( 5.0 N)in30 30 N ( ccw) 60.0 O.00 φ φ 60 v τ v v inφ.00 ( ) 5.0 N ( )in -60 ( ) 6 N cw ( ) O φ φ 80 v τ v v inφ ( 4.00 ) ( 5.0 N)in( 80 ) 0.).0 N 30.0 O N φ φ 90 τ inφ ( 5.00 ) ( 8.00 N)in( 90 ) 40.0 N ( cw) φ φ 50 τ inφ (.00 ) (.00 N)in50.0 N ( ccw) τ net τ +τ 40.0 N +.0 N 8.0 N ( cw) 3.) 8.60 N, 4.30 N, φ φ 90 τ inφ ( ) 8.60 N ( )in 90 ( ).838 N cw ( ) φ φ 90 τ inφ ( ) 4.30 N ( )in90.49 N ccw ( ) τ net τ +τ.838 N +.49 N.49 N ( cw) 4.) I 3.50 kg ω o 0, ω 600 ev π in 0π in ev 60 ω αt + ω o o α ω ω o t 0π
11 HO Solution 4.) (continued) uing Newton nd Law fo otation τ net τ Iα τ Iα 3.50 kg ( ) N b.) K Iω ( 3.50 kg ) 0π 690 J 5.) 0.300, 50 N, I 4.00 kg φ τ inφ ( ) ( 50.0 N)in90 5 N ( ccw) uing Newton nd Law fo otation τ net τ Iα α τ I 5 N kg 6.) two dik D 0.075, M kg joined by cylindical hub D 0.00, M kg fo the dik R D and fo the hub R D I I dik + I hub M R + M R I ( kg) ( ) + ( kg) ( ) x 0-5 kg uing the lowet point a a efeence y h.0 and y 0 alo v ω 0 uing Conevation of Enegy K +U g +W othe K +U g o 0 +U g + 0 K + 0 o U g K gy Iω + v whee M + M ( kg) kg 0.05 kg ince the cod i attached to the hub ω v and gy R I v + R v I v R + v v I R + v gy I R + ( 0.05 kg) 9.8 (.00 ) 0.84 ( x 0-5 kg ) kg ( ) b.) the otational kinetic enegy i K ot Iω I v 0.84 R x ( 0-5 kg ) J the tanlational kinetic enegy i K tan v 0.05 kg ( ) J the faction of which i otational i K ot J o 96.4 % otational K ot + K tan J J
12 L 7.) CM HO Solution otation eult fo the toque due to the weight of the od Mg τ g inφ L Mgin ( 90 ) LMg ( cw) fo a od otating about it end I 3 ML and uing Newton nd Law fo otation τ net τ Iα LMg α τ I 3 g L o α 3 g L (cw) 3 ML b.) the tip i a ditance L fo the axi of otation o a α L 3 g L 3 g 8.) T T.0 kg,.5 kg and the pulley i hollow cylinde M.0 kg, R 0.5 fo a hollow cylinde I MR Looking at foce in the diection of otion and applying Newton nd Law and a Rα T T T I T g a a τ Iα () T a T a τ g τ Iα () g T a RT RT MR a R cobining () and () and (3) g a + a + Ma ( + + M )a a g + + M (.5 kg) kg +.5 kg +.0 kg 3.7 (3) T T Ma fo () α a R T a.0 kg.8 ( ) N fo () T g a ( g a).5 kg ( ) N
13 HO Solution 8.) (again) could ue enegy conevation to find the peed of block (and ) afte falling a ditance h and elating it to the acceleation a uing the kineatic elationhip v v o + aδx. uing the lowet point a a efeence and tating fo et y h, y 0, and v 0 K +U g +W othe K +U g o 0 +U g + 0 K + 0 o U g K the final kinetic enegy include the otational kinetic enegy of the pulley and the tanlational kinetic enegy of the block only block contibute to change in gavitational enegy gy ( + )v + Iω ( + )v + I v R gy ( + )v + MR v ( R + )v + Mv gy ( + + M )v and v gy + + M gh + + M () v v o + aδx o v v + aδy o v 0 + ah ah () cobining () and () ah gh + + M o a g + + M (the ae eult a peviouly obtained) α can be obtained fo a and the tenion obtained by conideing the foce on the block and applying Newton nd Law to each block
14 HO 3 Solution.) M R R M fo unifo cylinde I MR M no lipping o acceleation a of the block and cylinde ae the ae and the angula acceleation of the pulley i elated to the acceleation by the elationhip a Rα and the linea velocity of the block and the cylinde ae the ae ue enegy conevation to find the peed of block M afte falling a ditance h and elating it to the acceleation a uing the kineatic elationhip v v o + aδy. uing the lowet point a a efeence and tating fo et y h, y 0, and v 0 K +U g +W othe K +U g o 0 +U g + 0 K + 0 o U g K the final kinetic enegy include the otational kinetic enegy of the pulley and the cylinde and the tanlational kinetic enegy of the block and cylinde only block M contibute to change in gavitational enegy Mgy ( M + M )v + I pulleyω + I cylindeω ( M )v + MR ω + M R Mgy Mv + 4 MR ω + 4 M ( 4R ) ω Mv + 4 MR v + MR v R R Mgy Mv + 4 MR v R + MR v R ( ) Mv + 4 MR v R + MR v 4R Mgy Mv + 4 Mv + 4 Mv Mv Mv 4 ( ) ω 3 Mgy Mv o v gy 3 gh 3 () v v o + aδy o v v + aδy o v 0 + ah ah () cobining () and () ah gh 3 o a g 3.) d D 0.500, 60 N. v o ω o 0, Δd 3.00, and t 4.00 φ D Δd 0.50 and Δθ Δθ αt + ω o t αt o α Δθ t ( ) ( ) b.) c.) d.) ( Δθ ω + ω o) t ω t o ω Δθ t ( ) W net ΔK K K o K and W net W v v d dcoθ o K dcoθ 60.0 N ( ) 3.00 ( )co0 80 J τ Iα o I τ α inφ α ( 0.50 )( 60.0 N)in kg
15 HO 3 Solution 3.) l 0.50, M kg, contant angula velocity ω Δθ π π Δt fo a lende od otating about one end I 3 Ml L Iω 3 Ml ω ( kg) ( ) π x 0-5 kg v.0 / 4.) P 3.00 kg, v.0, O φ v φ 43. v L v p v vinφ ( 3.00 kg) ( 8.00 ).0 in kg 5.) M a 8.0 kg, L a.8, R a 0.5, I body 0.40 kg, ω 0.60 ev when a ae extended I I body + I a I body + M L a a 0.40 kg + ( 8.0 kg )(.8 ).56 kg when a ae bought in I I body + I a I body + M a R a uing Conevation of Angula Moentu ( L L ) 0.40 kg + ( 8.0 kg) ( 0.5 ) 0.90 kg I ω I ω o ω I.56 kg ω 0.60 ev.7 ev I 0.90 kg 6.) Δθ ev, Δt 6.00, I g 00 kg, an 80.0 kg, 0, and.00 with an at cente I I g + I an I body + an 00 kg + ( 80.0 kg) ( 0) 00 kg afte an ove I I g + I an I body + an 00 kg + ( 80.0 kg) (.00) 50 kg ω Δθ Δt ev ev ) unifo dik ω 7.0 ev, M, R and unifo tick M, L R fo the dik I I dik MR and uing Conevation of Angula Moentu ( L L ) I ω I ω o ω I 00 kg ω 0.67 ev 0.3 ev I 50 kg fo the dik and tick I I dik + I tick MR + ML MR + M ( R) 5 6 MR uing Conevation of Angula Moentu ( L L ) I ω I ω o ω I I ω MR 5 6 MR 7.0 ev 4. ev
16 HO 3 Solution 8.) I I, I I, ω ω uing Conevation of Angula Moentu ( L L ) I ω I ω o ω I I ω I I ω ω 9.) ey-go-ound D 4., I g 760 kg, ω 0.80 and fou en 65 kg, 0, D. I I g 760 kg and I I g + I en I g kg + 4( 65 kg) (. ) 907 kg uing Conevation of Angula Moentu ( L L ) I ω I ω o ω I 760 kg ω I 907 kg 0.) b kg, v b 360, olid dik M 0.0 kg, R 0.300, ω 0 v b R the bullet becoe ebedded along a line 0.5 to the ight of the cente of the dik befoe tiking the dik the angula oentu of the bullet with epect to the axi of otation of the dik i L L b b v b b inφ b v b afte tiking the dik angula oentu of the dik and bullet i I I b + I dik b R b + MR uing Conevation of Angula Moentu ( L L ) o L I ω o ω L I b v b b R b + MR kg ω kg 0.5 ( ) ( ) ( ) kg ( ) 0.30 ( ) the tie to ake one coplete evolution i the peiod T and T π ω π
17 HO 3 Solution.) (again) the otation of the cylinde i due to the fictional foce between the cylinde and the uface that it i in contact with and the tenion in the cod doe not contibute Looking at foce in the diection of otion and applying Newton nd Law and a Rα f R T M T T I T g a a τ Iα () T f Ma g T Ma τ τ Iα τ Iα () Mg T Ma RT RT MR a R f R M ( R) α (3) T T Ma f R M R R ( ) a (4) f Ma cobining () and () and (3) and (4) Mg Ma + Ma + Ma + Ma 3Ma and a g 3 alo f Ma M g Mg 3 6 and ince f µ N µmg it follow that µ f Mg Mg 6 Mg
Dynamics of Rotational Motion
Dynamics of Rotational Motion Toque: the otational analogue of foce Toque = foce x moment am τ = l moment am = pependicula distance though which the foce acts a.k.a. leve am l l l l τ = l = sin φ = tan
More informationPhysics 1114: Unit 5 Hand-out Homework (Answers)
Physics 1114: Unit 5 Hand-out Homewok (Answes) Poblem set 1 1. The flywheel on an expeimental bus is otating at 420 RPM (evolutions pe minute). To find (a) the angula velocity in ad/s (adians/second),
More informationFormula Formula symbols Units. s = F A. e = x L. E = s ε. k = F δ. G = t γ. e = at. maximum load original cross sectional area. s M E = = N/m.
A Lit of foulae fo ecanical engineeing pinciple Foula Foula ybol Unit Ste Stain applied foce co ectionalaea cange in lengt oiginal lengt F A e x L Young odulu of elaticity te tain Stiffne foce extenion
More informationLECTURE 14. m 1 m 2 b) Based on the second law of Newton Figure 1 similarly F21 m2 c) Based on the third law of Newton F 12
CTU 4 ] NWTON W O GVITY -The gavity law i foulated fo two point paticle with ae and at a ditance between the. Hee ae the fou tep that bing to univeal law of gavitation dicoveed by NWTON. a Baed on expeiental
More informationTorque, Angular Momentum and Rotational Kinetic Energy
Toque, Angula Moentu and Rotational Kinetic Enegy In ou peious exaples that inoled a wheel, like fo exaple a pulley we wee always caeful to specify that fo the puposes of the poble it would be teated as
More information1121 T Question 1
1121 T1 2008 Question 1 ( aks) You ae cycling, on a long staight path, at a constant speed of 6.0.s 1. Anothe cyclist passes you, tavelling on the sae path in the sae diection as you, at a constant speed
More information06 - ROTATIONAL MOTION Page 1 ( Answers at the end of all questions )
06 - ROTATIONAL MOTION Page ) A body A of mass M while falling vetically downwads unde gavity beaks into two pats, a body B of mass ( / ) M and a body C of mass ( / ) M. The cente of mass of bodies B and
More informationSection 25 Describing Rotational Motion
Section 25 Decibing Rotational Motion What do object do and wh do the do it? We have a ve thoough eplanation in tem of kinematic, foce, eneg and momentum. Thi include Newton thee law of motion and two
More informationExam 3: Equation Summary
MAACHUETT INTITUTE OF TECHNOLOGY Depatment of Physics Physics 8. TEAL Fall Tem 4 Momentum: p = mv, F t = p, Fext ave t= t f t = Exam 3: Equation ummay = Impulse: I F( t ) = p Toque: τ =,P dp F P τ =,P
More informationRotational Motion: Statics and Dynamics
Physics 07 Lectue 17 Goals: Lectue 17 Chapte 1 Define cente of mass Analyze olling motion Intoduce and analyze toque Undestand the equilibium dynamics of an extended object in esponse to foces Employ consevation
More informationEN40: Dynamics and Vibrations. Midterm Examination Tuesday March
EN4: Dynaics and Vibations Midte Exaination Tuesday Mach 8 16 School of Engineeing Bown Univesity NME: Geneal Instuctions No collaboation of any kind is peitted on this exaination. You ay bing double sided
More informationTest 2 phy a) How is the velocity of a particle defined? b) What is an inertial reference frame? c) Describe friction.
Tet phy 40 1. a) How i the velocity of a paticle defined? b) What i an inetial efeence fae? c) Decibe fiction. phyic dealt otly with falling bodie. d) Copae the acceleation of a paticle in efeence fae
More informationPhysics C Rotational Motion Name: ANSWER KEY_ AP Review Packet
Linea and angula analogs Linea Rotation x position x displacement v velocity a T tangential acceleation Vectos in otational motion Use the ight hand ule to detemine diection of the vecto! Don t foget centipetal
More informationHoizontal Cicula Motion 1. A paticle of mass m is tied to a light sting and otated with a speed v along a cicula path of adius. If T is tension in the sting and mg is gavitational foce on the paticle then,
More informationb) (5) What average force magnitude was applied by the students working together?
Geneal Physics I Exam 3 - Chs. 7,8,9 - Momentum, Rotation, Equilibium Nov. 3, 2010 Name Rec. Inst. Rec. Time Fo full cedit, make you wok clea to the gade. Show fomulas used, essential steps, and esults
More informationChapter 19 Webassign Help Problems
Chapte 9 Webaign Help Poblem 4 5 6 7 8 9 0 Poblem 4: The pictue fo thi poblem i a bit mileading. They eally jut give you the pictue fo Pat b. So let fix that. Hee i the pictue fo Pat (a): Pat (a) imply
More informationForce & Motion: Newton s Laws
oce & otion: Newton Law ( t Law) If no net foce act on a body then the body velocity cannot change. Zeo net foce implie zeo acceleation. The ma of an object detemine how difficult it i to change the object
More informationRotational Motion. Every quantity that we have studied with translational motion has a rotational counterpart
Rotational Motion & Angula Momentum Rotational Motion Evey quantity that we have studied with tanslational motion has a otational countepat TRANSLATIONAL ROTATIONAL Displacement x Angula Position Velocity
More informationTranslation and Rotation Kinematics
Tanslation and Rotation Kinematics Oveview: Rotation and Tanslation of Rigid Body Thown Rigid Rod Tanslational Motion: the gavitational extenal foce acts on cente-of-mass F ext = dp sy s dt dv total cm
More informationCan a watch-sized electromagnet deflect a bullet? (from James Bond movie)
Can a peon be blown away by a bullet? et' ay a bullet of a 0.06 k i ovin at a velocity of 300 /. And let' alo ay that it ebed itelf inide a peon. Could thi peon be thut back at hih peed (i.e. blown away)?
More informationFrom Newton to Einstein. Mid-Term Test, 12a.m. Thur. 13 th Nov Duration: 50 minutes. There are 20 marks in Section A and 30 in Section B.
Fom Newton to Einstein Mid-Tem Test, a.m. Thu. 3 th Nov. 008 Duation: 50 minutes. Thee ae 0 maks in Section A and 30 in Section B. Use g = 0 ms in numeical calculations. You ma use the following epessions
More informationRotational Kinetic Energy
Add Impotant Rotational Kinetic Enegy Page: 353 NGSS Standad: N/A Rotational Kinetic Enegy MA Cuiculum Famewok (006):.1,.,.3 AP Phyic 1 Leaning Objective: N/A, but olling poblem have appeaed on peviou
More informationExam 3: Equation Summary
MAACHUETT INTITUTE OF TECHNOLOGY Depatment of Physics Physics 8. TEAL Fall Tem 4 Momentum: p = mv, F t = p, Fext ave t= t f t = Exam 3: Equation ummay = Impulse: I F( t ) = p Toque: τ =,P dp F P τ =,P
More informationHW 7 Help. 60 s t. (4.0 rev/s)(1 min) 240 rev 1 min Solving for the distance traveled, we ll need to convert to radians:
HW 7 Help 30. ORGANIZE AND PLAN We ae given the angula velocity and the time, and we ae asked to ind the distance that is coveed. We can ist solve o the angula displacement using Equation 8.3: t. The distance
More informationPhysics 111 Lecture 5 Circular Motion
Physics 111 Lectue 5 Cicula Motion D. Ali ÖVGÜN EMU Physics Depatment www.aovgun.com Multiple Objects q A block of mass m1 on a ough, hoizontal suface is connected to a ball of mass m by a lightweight
More informationCircular Motion. Mr. Velazquez AP/Honors Physics
Cicula Motion M. Velazquez AP/Honos Physics Objects in Cicula Motion Accoding to Newton s Laws, if no foce acts on an object, it will move with constant speed in a constant diection. Theefoe, if an object
More informationLecture 13. Rotational motion Moment of inertia
Lectue 13 Rotational motion Moment of inetia EXAM 2 Tuesday Mach 6, 2018 8:15 PM 9:45 PM Today s Topics: Rotational Motion and Angula Displacement Angula Velocity and Acceleation Rotational Kinematics
More informationSolutions Practice Test PHYS 211 Exam 2
Solution Pactice Tet PHYS 11 Exam 1A We can plit thi poblem up into two pat, each one dealing with a epaate axi. Fo both the x- and y- axe, we have two foce (one given, one unknown) and we get the following
More informationHomework Set 4 Physics 319 Classical Mechanics. m m k. x, x, x, x T U x x x x l 2. x x x x. x x x x
Poble 78 a) The agangian i Hoewok Set 4 Phyic 319 Claical Mechanic k b) In te of the cente of a cooinate an x x1 x x1 x xc x x x x x1 xc x xc x x x x x1 xc x xc x, x, x, x T U x x x x l 1 1 1 1 1 1 1 1
More informationPhysics 231 Lecture 21
Physics 3 Lectue Main points o today s lectue: Angula momentum: L Newton s law o univesal gavitation: GMm F PE GMm Keple s laws and the elation between the obital peiod and obital adius. T π GM 4 3 Rolling
More informationQuiz 6--Work, Gravitation, Circular Motion, Torque. (60 pts available, 50 points possible)
Name: Class: Date: ID: A Quiz 6--Wok, Gavitation, Cicula Motion, Toque. (60 pts available, 50 points possible) Multiple Choice, 2 point each Identify the choice that best completes the statement o answes
More informationSeat: PHYS 1500 (Fall 2006) Exam #2, V1. After : p y = m 1 v 1y + m 2 v 2y = 20 kg m/s + 2 kg v 2y. v 2x = 1 m/s v 2y = 9 m/s (V 1)
Seat: PHYS 1500 (Fall 006) Exa #, V1 Nae: 5 pt 1. Two object are oving horizontally with no external force on the. The 1 kg object ove to the right with a peed of 1 /. The kg object ove to the left with
More informationTP A.4 Post-impact cue ball trajectory for any cut angle, speed, and spin
technical poof TP A.4 Pot-impact cue ball tajectoy fo any cut anle, peed, and pin uppotin: The Illutated Pinciple of Pool and Billiad http://billiad.colotate.edu by Daid G. Alciatoe, PhD, PE ("D. Dae")
More informationPhysics Sp Exam #4 Name:
Phyic 160-0 Sp. 017 Ea #4 Nae: 1) A coputer hard dik tart ro ret. It peed up with contant angular acceleration until it ha an angular peed o 700 rp. I it coplete 150 revolution while peeding up, what i
More informationt α z t sin60 0, where you should be able to deduce that the angle between! r and! F 1
PART III Problem Problem1 A computer dik tart rotating from ret at contant angular acceleration. If it take 0.750 to complete it econd revolution: a) How long doe it take to complete the firt complete
More informationPrinciples of Physics I
Pinciples of Physics I J. M. Veal, Ph. D. vesion 8.05.24 Contents Linea Motion 3. Two scala equations........................ 3.2 Anothe scala equation...................... 3.3 Constant acceleation.......................
More informationSection 26 The Laws of Rotational Motion
Physics 24A Class Notes Section 26 The Laws of otational Motion What do objects do and why do they do it? They otate and we have established the quantities needed to descibe this motion. We now need to
More informationLINEAR MOMENTUM Physical quantities that we have been using to characterize the motion of a particle
LINEAR MOMENTUM Physical quantities that we have been using to chaacteize the otion of a paticle v Mass Velocity v Kinetic enegy v F Mechanical enegy + U Linea oentu of a paticle (1) is a vecto! Siple
More informationPhysics 201 Lecture 18
Phsics 0 ectue 8 ectue 8 Goals: Define and anale toque ntoduce the coss poduct Relate otational dnamics to toque Discuss wok and wok eneg theoem with espect to otational motion Specif olling motion (cente
More informationPhysics 1A (a) Fall 2010: FINAL Version A 1. Comments:
Physics A (a) Fall 00: FINAL Vesion A Name o Initials: Couse 3-digit Code Comments: Closed book. No wok needs to be shown fo multiple-choice questions.. A helicopte is taveling at 60 m/s at a constant
More informationChapter 7-8 Rotational Motion
Chapte 7-8 Rotational Motion What is a Rigid Body? Rotational Kinematics Angula Velocity ω and Acceleation α Unifom Rotational Motion: Kinematics Unifom Cicula Motion: Kinematics and Dynamics The Toque,
More informationPHYS 1114, Lecture 21, March 6 Contents:
PHYS 1114, Lectue 21, Mach 6 Contents: 1 This class is o cially cancelled, being eplaced by the common exam Tuesday, Mach 7, 5:30 PM. A eview and Q&A session is scheduled instead duing class time. 2 Exam
More information- 5 - TEST 1R. This is the repeat version of TEST 1, which was held during Session.
- 5 - TEST 1R This is the epeat vesion of TEST 1, which was held duing Session. This epeat test should be attempted by those students who missed Test 1, o who wish to impove thei mak in Test 1. IF YOU
More informationPhysics 110. Exam #1. September 30, 2016
Phyic 110 Exa #1 Septebe 30, 016 Nae Pleae ead and follow thee intuction caefully: Read all poble caefully befoe attepting to olve the. You wok ut be legible, and the oganization clea. You ut how all wok,
More informationChapter 8. Accelerated Circular Motion
Chapte 8 Acceleated Cicula Motion 8.1 Rotational Motion and Angula Displacement A new unit, adians, is eally useful fo angles. Radian measue θ(adians) = s = θ s (ac length) (adius) (s in same units as
More informationConditions for equilibrium (both translational and rotational): 0 and 0
Leon : Equilibriu, Newton econd law, Rolling, Angular Moentu (Section 8.3- Lat tie we began dicuing rotational dynaic. We howed that the rotational inertia depend on the hape o the object and the location
More informationPhysics 111 Lecture 10. SJ 8th Ed.: Chap Torque, Energy, Rolling. Copyright R. Janow Spring basics, energy methods, 2nd law problems)
hysics Lectue 0 Toque, Enegy, Rolling SJ 8th Ed.: Chap 0.6 0.9 Recap and Oveview Toque Newton s Second Law fo Rotation Enegy Consideations in Rotational Motion Rolling Enegy Methods Second Law Applications
More informationSPH4U Magnetism Test Name: Solutions
SPH4U Magneti et Nae: Solution QUESION 1 [4 Mak] hi and the following two quetion petain to the diaga below howing two cuent-caying wie. wo cuent ae flowing in the ae diection (out of the page) a hown.
More informationTRAVELING WAVES. Chapter Simple Wave Motion. Waves in which the disturbance is parallel to the direction of propagation are called the
Chapte 15 RAVELING WAVES 15.1 Simple Wave Motion Wave in which the ditubance i pependicula to the diection of popagation ae called the tanvee wave. Wave in which the ditubance i paallel to the diection
More informationPHYSICS 2210 Fall Exam 4 Review 12/02/2015
PHYSICS 10 Fall 015 Exa 4 Review 1/0/015 (yf09-049) A thin, light wire is wrapped around the ri of a unifor disk of radius R=0.80, as shown. The disk rotates without friction about a stationary horizontal
More informationPHY 211: General Physics I 1 CH 10 Worksheet: Rotation
PHY : General Phyic CH 0 Workheet: Rotation Rotational Variable ) Write out the expreion for the average angular (ω avg ), in ter of the angular diplaceent (θ) and elaped tie ( t). ) Write out the expreion
More informationLecture 23: Central Force Motion
Lectue 3: Cental Foce Motion Many of the foces we encounte in natue act between two paticles along the line connecting the Gavity, electicity, and the stong nuclea foce ae exaples These types of foces
More information1131 T Question 1
1131 T1 2008 Question 1 ( aks) You ae cycling, on a long staight path, at a constant speed of 6.0.s 1. Anothe cyclist passes you, taelling on the sae path in the sae diection as you, at a constant speed
More informationMAGNETIC FIELD INTRODUCTION
MAGNETIC FIELD INTRODUCTION It was found when a magnet suspended fom its cente, it tends to line itself up in a noth-south diection (the compass needle). The noth end is called the Noth Pole (N-pole),
More informationMon , (.12) Rotational + Translational RE 11.b Tues.
Mon..-.3, (.) Rotational + Tanlational RE.b Tue. EP0 Mon..4-.6, (.3) Angula Momentum & Toque RE.c Tue. Wed..7 -.9, (.) Toque EP RE.d ab Fi. Rotation Coue Eval.0 Quantization, Quiz RE.e Mon. Review fo Final
More informationKEPLER S LAWS AND PLANETARY ORBITS
KEPE S AWS AND PANETAY OBITS 1. Selected popeties of pola coodinates and ellipses Pola coodinates: I take a some what extended view of pola coodinates in that I allow fo a z diection (cylindical coodinates
More informationMagnetic Dipoles Challenge Problem Solutions
Magnetic Dipoles Challenge Poblem Solutions Poblem 1: Cicle the coect answe. Conside a tiangula loop of wie with sides a and b. The loop caies a cuent I in the diection shown, and is placed in a unifom
More informationMultiple choice questions [100 points] As shown in the figure, a mass M is hanging by three massless strings from the ceiling of a room.
Multiple choice questions [00 points] Answe all of the following questions. Read each question caefully. Fill the coect ule on you scanton sheet. Each coect answe is woth 4 points. Each question has exactly
More informationCh. 4: FOC 9, 13, 16, 18. Problems 20, 24, 38, 48, 77, 83 & 115;
WEEK-3 Recitation PHYS 3 eb 4, 09 Ch. 4: OC 9, 3,, 8. Pobles 0, 4, 38, 48, 77, 83 & 5; Ch. 4: OC Questions 9, 3,, 8. 9. (e) Newton s law of gavitation gives the answe diectl. ccoding to this law the weight
More informationω = θ θ o = θ θ = s r v = rω
Unifom Cicula Motion Unifom cicula motion is the motion of an object taveling at a constant(unifom) speed in a cicula path. Fist we must define the angula displacement and angula velocity The angula displacement
More informationb) (5) What is the magnitude of the force on the 6.0-kg block due to the contact with the 12.0-kg block?
Geneal Physics I Exam 2 - Chs. 4,5,6 - Foces, Cicula Motion, Enegy Oct. 13, 2010 Name Rec. Inst. Rec. Time Fo full cedit, make you wok clea to the gade. Show fomulas used, essential steps, and esults with
More information) 1.5"10 11 m. ( )( 1.99 "10 30 kg)
Exaple 1: a.) What i the foce of gaity between a gazelle with a a of 100 kg and a lion with a a that i 50 kg if the lion i lying in wait 100 ete fo the gazelle? b.) What would happen to the foce of gaity
More informationPhysics 1A (b) Fall 2010: FINAL Version A 1. Comments:
Physics A (b) Fall 00: FINAL Vesion A Name o Initials: Couse 3-digit Code Comments: Closed book. No wok needs to be shown fo multiple-choice questions.. An 80 kg man is one fouth of the way up a 0 m ladde
More informationLab 10: Newton s Second Law in Rotation
Lab 10: Newton s Second Law in Rotation We can descibe the motion of objects that otate (i.e. spin on an axis, like a popelle o a doo) using the same definitions, adapted fo otational motion, that we have
More informationExperiment 09: Angular momentum
Expeiment 09: Angula momentum Goals Investigate consevation of angula momentum and kinetic enegy in otational collisions. Measue and calculate moments of inetia. Measue and calculate non-consevative wok
More informationCircular Motion & Torque Test Review. The period is the amount of time it takes for an object to travel around a circular path once.
Honos Physics Fall, 2016 Cicula Motion & Toque Test Review Name: M. Leonad Instuctions: Complete the following woksheet. SHOW ALL OF YOUR WORK ON A SEPARATE SHEET OF PAPER. 1. Detemine whethe each statement
More informationCHAPTER 6: UNIFORM CIRCULAR MOTION AND GRAVITATION
College Physics Student s Manual Chapte 6 CHAPTER 6: UIORM CIRCULAR MOTIO AD GRAVITATIO 6. ROTATIO AGLE AD AGULAR VELOCITY. Sei- taile tucks hae an odoete on one hub of a taile wheel. The hub is weighted
More informationPhysics 4A Solutions to Chapter 10 Homework
Physics 4A Solutions to Chapter 0 Homework Chapter 0 Questions: 4, 6, 8 Exercises & Problems 6, 3, 6, 4, 45, 5, 5, 7, 8 Answers to Questions: Q 0-4 (a) positive (b) zero (c) negative (d) negative Q 0-6
More informationContent 5.1 Angular displacement and angular velocity 5.2 Centripetal acceleration 5.3 Centripetal force. 5. Circular motion.
5. Cicula otion By Liew Sau oh Content 5.1 Angula diplaceent and angula elocity 5. Centipetal acceleation 5.3 Centipetal foce Objectie a) expe angula diplaceent in adian b) define angula elocity and peiod
More informationNiraj Sir. circular motion;; SOLUTIONS TO CONCEPTS CHAPTER 7
SOLUIONS O CONCEPS CHAPE 7 cicula otion;;. Distance between Eath & Moon.85 0 5 k.85 0 8 7. days 4 600 (7.) sec.6 0 6 sec.4.85 0 v 6.6 0 8 05.4/sec v (05.4) a 0.007/sec.7 0 /sec 8.85 0. Diaete of eath 800k
More informationPHYS 1410, 11 Nov 2015, 12:30pm.
PHYS 40, Nov 205, 2:30pm. A B = AB cos φ x = x 0 + v x0 t + a 2 xt 2 a ad = v2 2 m(v2 2 v) 2 θ = θ 0 + ω 0 t + 2 αt2 L = p fs µ s n 0 + αt K = 2 Iω2 cm = m +m 2 2 +... m +m 2 +... p = m v and L = I ω ω
More informationChapter 13 Gravitation
Chapte 13 Gavitation In this chapte we will exploe the following topics: -Newton s law of gavitation, which descibes the attactive foce between two point masses and its application to extended objects
More informationr dt dt Momentum (specifically Linear Momentum) defined r r so r r note: momentum is a vector p x , p y = mv x = mv y , p z = mv z
Moentu, Ipulse and Collisions Moentu eeyday connotations? physical eaning the tue easue of otion (what changes in esponse to applied foces) d d ΣF ( ) dt dt Moentu (specifically Linea Moentu) defined p
More informationPhysics 201. Professor P. Q. Hung. 311B, Physics Building. Physics 201 p. 1/1
Physics 201 p. 1/1 Physics 201 Professor P. Q. Hung 311B, Physics Building Physics 201 p. 2/1 Rotational Kinematics and Energy Rotational Kinetic Energy, Moment of Inertia All elements inside the rigid
More informationPhysics 2A Chapter 10 - Moment of Inertia Fall 2018
Physics Chapte 0 - oment of netia Fall 08 The moment of inetia of a otating object is a measue of its otational inetia in the same way that the mass of an object is a measue of its inetia fo linea motion.
More informationPerhaps the greatest success of his theory of gravity was to successfully explain the motion of the heavens planets, moons, &tc.
AP Phyic Gavity Si Iaac Newton i cedited with the dicovey of gavity. Now, of coue we know that he didn t eally dicove the thing let face it, people knew about gavity fo a long a thee have been people.
More informationWhen a mass moves because of a force, we can define several types of problem.
Mechanics Lectue 4 3D Foces, gadient opeato, momentum 3D Foces When a mass moves because of a foce, we can define seveal types of poblem. ) When we know the foce F as a function of time t, F=F(t). ) When
More informationHO 25 Solutions. = s. = 296 kg s 2. = ( kg) s. = 2π m k and T = 2π ω. kg m = m kg. = 2π. = ω 2 A = 2πf
HO 5 Soution 1.) haronic ociator = 0.300 g with an idea pring T = 0.00 T = π T π π o = = ( 0.300 g) 0.00 = 96 g = 96 N.) haronic ociator = 0.00 g and idea pring = 140 N F = x = a = d x dt o the dipaceent
More informationChapter 10 Solutions
Chapter 0 Solutions 0. (a) α ω ω i t.0 rad/s 4.00 rad/s 3.00 s θ ω i t + αt (4.00 rad/s )(3.00 s) 8.0 rad 0. (a) ω ω π rad 365 days π rad 7.3 days *0.3 ω i 000 rad/s α 80.0 rad/s day 4 h day 4 h h 3600
More informationDepartment of Physics, Korea University Page 1 of 5
Name: Depatment: Student ID #: Notice ˆ + ( 1) points pe coect (incoect) answe. ˆ No penalty fo an unansweed question. ˆ Fill the blank ( ) with ( ) if the statement is coect (incoect). ˆ : coections to
More information( ) Physics 1401 Homework Solutions - Walker, Chapter 9
Phyic 40 Conceptual Quetion CQ No Fo exaple, ey likely thee will be oe peanent deoation o the ca In thi cae, oe o the kinetic enegy that the two ca had beoe the colliion goe into wok that each ca doe on
More informationRE 11.e Mon. Review for Final (1-11) HW11: Pr s 39, 57, 64, 74, 78 Sat. 9 a.m. Final Exam (Ch. 1-11)
Mon. Tue. We. ab i..4-.6, (.) ngula Momentum Pincile & Toque.7 -.9, (.) Motion With & Without Toque Rotation Coue Eval.0 Quantization, Quiz RE.c EP RE. RE.e Mon. Review fo inal (-) HW: P 9, 57, 64, 74,
More informationAP-C WEP. h. Students should be able to recognize and solve problems that call for application both of conservation of energy and Newton s Laws.
AP-C WEP 1. Wok a. Calculate the wok done by a specified constant foce on an object that undegoes a specified displacement. b. Relate the wok done by a foce to the aea unde a gaph of foce as a function
More informatione.g: If A = i 2 j + k then find A. A = Ax 2 + Ay 2 + Az 2 = ( 2) = 6
MOTION IN A PLANE 1. Scala Quantities Physical quantities that have only magnitude and no diection ae called scala quantities o scalas. e.g. Mass, time, speed etc. 2. Vecto Quantities Physical quantities
More informationMidterm Exam #2, Part A
Physics 151 Mach 17, 2006 Midtem Exam #2, Pat A Roste No.: Scoe: Exam time limit: 50 minutes. You may use calculatos and both sides of ONE sheet of notes, handwitten only. Closed book; no collaboation.
More informationPHYSICS 220. Lecture 08. Textbook Sections Lecture 8 Purdue University, Physics 220 1
PHYSICS 0 Lectue 08 Cicula Motion Textbook Sections 5.3 5.5 Lectue 8 Pudue Univesity, Physics 0 1 Oveview Last Lectue Cicula Motion θ angula position adians ω angula velocity adians/second α angula acceleation
More informationPhysics 6A. Practice Final (Fall 2009) solutions. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB
Phyic 6A Practice inal (all 009) olution or Capu Learning Aitance Service at UCSB . A locootive engine of a M i attached to 5 train car, each of a M. The engine produce a contant force that ove the train
More informationPhysics 4A Chapter 8: Dynamics II Motion in a Plane
Physics 4A Chapte 8: Dynamics II Motion in a Plane Conceptual Questions and Example Poblems fom Chapte 8 Conceptual Question 8.5 The figue below shows two balls of equal mass moving in vetical cicles.
More informationGravity. David Barwacz 7778 Thornapple Bayou SE, Grand Rapids, MI David Barwacz 12/03/2003
avity David Bawacz 7778 Thonapple Bayou, and Rapid, MI 495 David Bawacz /3/3 http://membe.titon.net/daveb Uing the concept dicued in the peceding pape ( http://membe.titon.net/daveb ), I will now deive
More informationCHAPTER 5: Circular Motion; Gravitation
CHAPER 5: Cicula Motion; Gavitation Solution Guide to WebAssign Pobles 5.1 [1] (a) Find the centipetal acceleation fo Eq. 5-1.. a R v ( 1.5 s) 1.10 1.4 s (b) he net hoizontal foce is causing the centipetal
More informationChapter 6. NEWTON S 2nd LAW AND UNIFORM CIRCULAR MOTION. string
Chapte 6 NEWTON S nd LAW AND UNIFORM CIRCULAR MOTION 103 PHYS 1 1 L:\103 Phy LECTURES SLIDES\103Phy_Slide_T1Y3839\CH6Flah 3 4 ting Quetion: A ball attached to the end of a ting i whiled in a hoizontal
More informationFARADAY'S LAW. dates : No. of lectures allocated. Actual No. of lectures 3 9/5/09-14 /5/09
FARADAY'S LAW No. of lectues allocated Actual No. of lectues dates : 3 9/5/09-14 /5/09 31.1 Faaday's Law of Induction In the pevious chapte we leaned that electic cuent poduces agnetic field. Afte this
More informationChapter 4. Newton s Laws of Motion
Chapte 4 Newton s Laws of Motion 4.1 Foces and Inteactions A foce is a push o a pull. It is that which causes an object to acceleate. The unit of foce in the metic system is the Newton. Foce is a vecto
More informationImpulse and Momentum
Impule and Momentum 1. A ca poee 20,000 unit of momentum. What would be the ca' new momentum if... A. it elocity wee doubled. B. it elocity wee tipled. C. it ma wee doubled (by adding moe paenge and a
More informationAH Mechanics Checklist (Unit 2) AH Mechanics Checklist (Unit 2) Circular Motion
AH Mechanics Checklist (Unit ) AH Mechanics Checklist (Unit ) Cicula Motion No. kill Done 1 Know that cicula motion efes to motion in a cicle of constant adius Know that cicula motion is conveniently descibed
More informationChapter 6. NEWTON S 2nd LAW AND UNIFORM CIRCULAR MOTION
Chapte 6 NEWTON S nd LAW AND UNIFORM CIRCULAR MOTION Phyic 1 1 3 4 ting Quetion: A ball attached to the end of a ting i whiled in a hoizontal plane. At the point indicated, the ting beak. Looking down
More informationChapter 10. Rotation of a Rigid Object about a Fixed Axis
Chapter 10 Rotation of a Rigid Object about a Fixed Axis Angular Position Axis of rotation is the center of the disc Choose a fixed reference line. Point P is at a fixed distance r from the origin. A small
More informationAngular velocity and angular acceleration CHAPTER 9 ROTATION. Angular velocity and angular acceleration. ! equations of rotational motion
Angular velocity and angular acceleration CHAPTER 9 ROTATION! r i ds i dθ θ i Angular velocity and angular acceleration! equations of rotational motion Torque and Moment of Inertia! Newton s nd Law for
More information(a) Calculate the apparent weight of the student in the first part of the journey while accelerating downwards at 2.35 m s 2.
Chapte answes Heineann Physics 1 4e Section.1 Woked exaple: Ty youself.1.1 CALCULATING APPARENT WEIGHT A 79.0 kg student ides a lift down fo the top floo of an office block to the gound. Duing the jouney
More informationPHY 171 Practice Test 3 Solutions Fall 2013
PHY 171 Practice et 3 Solution Fall 013 Q1: [4] In a rare eparatene, And a peculiar quietne, hing One and hing wo Lie at ret, relative to the ground And their wacky hairdo. If hing One freeze in Oxford,
More information