hmwk-4-solutions-Fall-2006

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

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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
Background image of page 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
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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).
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 03/22/2008 for the course CH 339K taught by Professor Appling during the Fall '08 term at University of Texas at Austin.

Page1 / 14

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

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online