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hmwk-4-solutions-Fall-2006

# hmwk-4-solutions-Fall-2006 - Homework 4 Solutions Problem 1...

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Homework 4. Solutions Problem 1. d) Arad ianisabout : 180 o πrad =57 o Problem 2. b) One revolution per minute is: rad sec = 1 rev min · 1 min 60 sec · 2 1 rev =0 . 105 rad s Problem 3. e) The angular speed is given by ω = 100 rev 10 sec · 2 1 rev =62 . 8 rad s Problem 4. d) The angle of rotation is given by θ = ω o t + 1 2 αt 2 Solving for α we get α = 2 t 2 ( θ ω o t )=6 . 66 rad s 2 Problem 5. d) The angular velocity is ω o rad/s ω = ω o αt The magnitude of the average angular acceleration is α = ω ω o t = 24 π 6 =4 π rad s 2 1

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Problem 6. b) The angular velocity is ω o =36 rad/s ω =24 rad/s ω = ω o αt Assuming the angular acceleration is constant, we fnd α = ω ω o t = 12 6 = 2 . 0 rad s 2 Problem 7. c) This problem involves non-constant acceleration. ThereFore, we integrate this to fnd the angular velocity, which upon integrating again we can fnd the angle through which the object rotates. α ( t )=(6 rad s 4 ) t 2 ω = ω o + Z t 0 αdt = ω o +2 t 3 θ = θ o + Z t 0 ωdt =0+ 2 t 4 4 =[(1 / 2) t 4 ] rad Problem 8. a) IF the angular velocity vector points out oF the page and the angular acceleration vector points into the page, the body is decelerating. Problem 9. b) Let v 1 = ωr 1 and let v 2 = 2 .Then v 1 v 2 = 1 ω r 1 2 = 2 r 1 r 1 =2 Problem 10. b) The moment oF inertia For a collection oF point masses is given by I = N X i =1 m i r i 2 2
I = m (0) 2 + m ( a ) 2 + m ( a ) 2 ++ m (2 a ) 2 I = ma 2 + ma 2 +4 ma 2 =6 ma 2 = 6(2)(1) 2 =12 kg · m 2 Problem 11. e) The density of the material is the same in both objects. Since the volumes are diFerent the masses will be diFerent. I 1 = 1 2 M 1 R 1 2 and 1 2 M 2 R 2 2 I 1 = 1 2 ρV 1 R 1 2 and 1 2 ρV 2 R 2 2 M 1 = ρπR 2 L and M 2 = ρπ (2 R ) 2 (2 L ) So I 2 I 1 = 1 2 M 2 R 2 2 1 2 M 1 R 1 2 1 2 ρπ (2 R ) 2 (2 L ) R 2 2 1 2 ρπR 2 LR 1 2 =32 Problem 12. e) R The moment of inertia of a solid uniform sphere is I = 2 5 MR 2 Using the parallel axis theorem we ±nd: I new = I + 2 = 2 5 2 + 2 = 7 5 2 Problem 13. d) F 2 r 2 r 1 F 1 The torque on an object is given by ± τ = ± r × ± F Let τ 1 = r 1 F 1 sinθ and τ 2 = r 2 F 2 sinθ . Then, τ 1 = (4)(5 sin 30 o )and τ 2 =( 2)( 5 sin 30 o ) 3

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Therefore, τ = τ 1 + τ 2 = (4)(5) 2 + (2)(5) 2 =10+5=15 N · m Problem 14. b) We know that τ = and ω = ω o + αt so τ = t So, ω = τt I so ω is proportional to 1 I Then, the moment of inertia of each object is: disk: I = 1 2 MR 2 ω disk 2 2 hoop: I = 2 ω hoop 1 2 sphere : I = 2 5 2 ω sphere 2 5 2 Therefore, the hoop will have the least angular velocity and the sphere will have the most angular velocity after a given time. Problem 15. d) The magnitude of the torque is τ = rF = So, α = I =0 . 4 rad/s 2 Calculate the time for which it takes the disk to rotate through an angle of π radians: θ = 1 2 αt 2 t =3 . 96 s Then, ω = αt = 4 10 (3 . 96) 1 . 6 rad/s Problem 16. c) Calculate the angle for which the saw rotates in 1 min (1 min = 60s).
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hmwk-4-solutions-Fall-2006 - Homework 4 Solutions Problem 1...

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