= + t ] can be used to calculate the angular

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1 WEEK-7 Reciaion PHYS 3 Mar 8, 8 Ch. 8 FOC:, 4, 6,, 3 &5. () Using Equaion 8. (θ = Arc lengh / Raius) o calculae he angle (in raians) ha each objec subens a your eye shows ha θ Moon = 9. 3 ra, θ Pea = 7. 3 ra, an θ Dime = 5 3 ra. Since θ Pea is less han θ Moon, he pea oes no compleely cover your view of he moon. However, since θ Dime is greaer han θ Moon, he ime oes compleely cover your view of he moon. 4. (a) An angular acceleraion of zero means ha he angular velociy has he same value a all imes, as in saemens A or B. However, saemen C is also consisen wih a zero angular acceleraion, because if he angular isplacemen oes no change as ime passes, hen he angular velociy remains consan a a value of ra/s. 6. (b) Since values are given for he iniial angular velociy ω, he final angular velociy ω, an he ime, Equaion 8.6 [ θ ( ω ω) isplacemen θ. = + ] can be use o calculae he angular. (b) Accoring o Equaion 8.9 (v T = rω), he angenial spee is proporional o he raius r when he angular spee ω is consan, as i is for he merry-go-roun. Thus, he angular. m spee of he secon chil is vt = (. m/s).4 m. 3. () Since he angular spee ω is consan, he angular acceleraion α is zero, accoring o Equaion 8.4. Since α = ra/s, he angenial acceleraion a T is zero, accoring o Equaion 8.. The cenripeal acceleraion a c, however, is no zero, since i is proporional o he square of he angular spee, accoring o Equaion 8., an he angular spee is no zero. 5. (a) The number N of revoluions is he isance s ravele ivie by he circumference πr of a wheel: N = s/(πr).

2 Problems: 9. SSM REASONING Equaion 8.4 = ( ) α ω ω / inicaes ha he average angular acceleraion is equal o he change in he angular velociy ivie by he elapse ime. Since he wheel sars from res, is iniial angular velociy is ω = ra/s. Is final angular velociy is given as ω =.4 ra/s. Since he average angular acceleraion is given as α =.3 ra/s, Equaion 8.4 can be solve o eermine he elapse ime. Solving Equaion 8.4 for he elapse ime gives ω ω.4 ra/s ra/s.3 ra/s = = = α 8. s. REASONING We know ha he final angular spee is ω = 4 ra / s. We also ha he iniial angular spee is ω = 4 ra / s an ha he ime uring which he change in angular spee occurs is = 5. s. Wih hese hree values, we can use θ = ( ω+ ω ) (Equaion 8.6) o calculae he angular isplacemen θ. Then, we can use ω = ω + α (Equaion 8.4) o eermine he angular acceleraion α. a. Using Equaion 8.6, we fin ha 3 ( ) ( )( ) θ = ω + ω = 4 ra/s + 4 ra/s 5. s = 4.6 ra b. Solving Equaion 8.4 for α, we fin ha ω ω 4 ra/s 4 ra/s α = = =. ra / s 5. s 5. SSM REASONING a. The ime for he wheels o come o a hal epens on he iniial an final velociies, ω an ω, an he angular isplacemen θ : θ = ( ω + ω) (see Equaion 8.6). Solving for he ime yiels θ = ω + ω b. The angular acceleraion α is efine as he change in he angular velociy, ω ω, ivie by he ime :

3 ω ω α = (8.4) a. Since he wheel comes o a res, ω = ra/s. Convering 5.9 revoluions o raians ( rev = π ra), he ime for he wheel o come o res is b. The angular acceleraion is π ra ( rev ) θ rev = = =. s ω + ω +. ra/s + ra/s ω ω ra/s. ra/s α = = =. ra/s. s 34. REASONING The angular acceleraion α gives rise o a angenial acceleraion a T accoring o a T = rα (Equaion 8.). Moreover, i is given ha a T = g, where g is he magniue of he acceleraion ue o graviy. Le r be he raial isance of he poin from he axis of roaion. Then, accoring o Equaion 8., we have Thus, g 9.8 m /s r = = = α. ra /s.87 m 4. REASONING The linear spee v wih which he bucke moves own he well is he same as he linear spee of he rope holing he bucke. The rope, in urn, is wrappe aroun he barrel of he han crank an unwins wihou slipping. This ensures ha he rope s linear spee is he same as he angenial spee vt = rω (Equaion 8.9) of a poin on he surface of he barrel, where ω an r are he barrel s angular spee an raius, respecively. Therefore, we have v = rω. When applie o he linear spee v of he crank hanle an he raius r of he circle he hanle raverses, Equaion 8.9 yiels v = r ω. We can use he same symbol ω o represen he angular spee of he barrel an he angular spee of he han crank, since boh make he same number of revoluions in any given amoun of ime. Lasly, we noe ha he raii r of he crank barrel an r of he han crank s circular moion are half of he respecive iameers =. m an =.4 m shown in he ex rawing.

4 Solving he relaions v = rω an v = rω for he angular spee ω an he linear spee v of he bucke, we obain v v vr v ω = = = = r r r or v The linear spee of he bucke, herefore, is ( ) v = ( ) (.4 m) (. m/s). m v = =.3 m/s 53. SSM REASONING AND From Equaion.4, he linear acceleraion of he moorcycle is v v. m/s m/s a = = =.44 m/s 9. s Since he ire rolls wihou slipping, he linear acceleraion equals he angenial acceleraion of a poin on he ouer ege of he ire: a = a T. Solving Equaion 8.3 for α gives α = a T r =.44 m/s.8 m = 8.7 ra/s Ch. 9: Problems 5, &, 5. SSM REASONING To calculae he orques, we nee o eermine he lever arms for each of he forces. These lever arms are shown in he following rawings:.5 m Axis 3. T W W = (.5 m) sin 3. a. Using Equaion 9., we fin ha he magniue of he orque ue o he weigh W is W.5 m Axis ( )( ) 3. T = (.5 m) cos 3. Magniue of orque = W = N.5 m sin 3 = 3 5 N m

5 b. Using Equaion 9., we fin ha he magniue of he orque ue o he hrus T is T ( )( ) Magniue of orque = T = 6 3 N.5 m cos 3 = 3 N m. REASONING AND The ne orque abou he axis in ex rawing (a) is Στ = τ + τ = F b F a = Consiering ha F = 3F, we have b 3a =. The ne orque in rawing (b) is hen Στ = F (. m a) F b = or. m a 3b = Solving he firs equaion for b, subsiuing ino he secon equaion an rearranging, gives a =. m an b =.3 m. REASONING The rawing shows he forces acing on he person. I also shows he lever arms for a roaional axis perpenicular o he plane of he paper a he place where he person s oes ouch he floor. Since he person is in equilibrium, he sum of he forces mus be zero. Likewise, we know ha he sum of he orques mus be zero. Taking upwar o be he posiive irecion, we have F + F W = FEET HANDS Remembering ha counerclockwise orques are posiive an using he axis an he lever arms shown in he rawing, we fin Subsiuing his value ino he balance-of-forces equaion, we fin F = W F = 584 N 39 N = 9 N FEET HANDS The force on each han is half he value calculae above, or 96 N. Likewise, he force on each foo is half he value calculae above, or 96 N.