( ) rad ( 2.0 s) = 168 rad

Size: px

Start display at page:

Download "( ) rad ( 2.0 s) = 168 rad"

Audrey Chapman
5 years ago
Views:

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