HW09-solutions - lee(cyl327 – HW09 – gentle –(56245 1...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: lee (cyl327) – HW09 – gentle – (56245) 1 This print-out should have 33 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A ladybug sits at the outer edge of a merry- go-round, and a gentleman bug sits halfway between her and the axis of rotation. The merry-go-round makes a complete revolution once each second. What is the gentleman bug’s angular speed? 1. the same as the ladybug’s correct 2. twice the ladybug’s 3. impossible to determine 4. half the ladybug’s Explanation: Angular speed is the same for every point on the merry-go-round. 002 (part 1 of 5) 10.0 points A wheel starts from rest with constant angu- lar acceleration of 4 . 8 rad / s 2 . After 6 . 8 s, what is its angular velocity? Correct answer: 32 . 64 rad / s. Explanation: Let : ω = 0 , α = 4 . 8 rad / s 2 , and t = 6 . 8 s . The angular velocity is ω = ω + αt = αt = (4 . 8 rad / s 2 ) (6 . 8 s) = 32 . 64 rad / s . 003 (part 2 of 5) 10.0 points Through what angle did the wheel turn? Correct answer: 110 . 976 rad. Explanation: Δ θ = ω Δ t + 1 2 α (Δ t ) 2 = 1 2 α (Δ t ) 2 = 1 2 (4 . 8 rad / s 2 ) (6 . 8 s) 2 = 110 . 976 rad . 004 (part 3 of 5) 10.0 points Through how many revolutions did the wheel turn? Correct answer: 17 . 6624 rev. Explanation: Δ θ = (110 . 976 rad) parenleftbigg 1 rev 2 π rad parenrightbigg = 17 . 6624 rev . 005 (part 4 of 5) 10.0 points What is the speed of a point 0 . 6 m from the axis of rotation? Correct answer: 19 . 584 m / s. Explanation: Let : r = 0 . 6 m . The tangential speed is v = r ω = (0 . 6 m) (32 . 64 rad / s) = 17 . 6624 rev . 006 (part 5 of 5) 10.0 points What is the acceleration of a point 0 . 6 m from the axis of rotation? Correct answer: 639 . 228 m / s 2 . Explanation: Let : r = 0 . 6 m . lee (cyl327) – HW09 – gentle – (56245) 2 The acceleration has two components, the tangential and the centripetal acceleration, which act perpendicular to each other: a = radicalBig a 2 t + a 2 c = radicalBig ( r α ) 2 + ( r ω 2 ) 2 = r radicalbig α 2 + ω 4 = (0 . 6 m) × radicalBig (4 . 8 rad / s 2 ) 2 + (32 . 64 rad / s) 4 = 19 . 584 m / s . keywords: 007 (part 1 of 3) 10.0 points A car accelerates uniformly from rest and covers a distance of 61 m in 8 . 3 s. If the diameter of a tire is 49 cm, find the angular acceleration of the wheel. Correct answer: 7 . 22832 rad / s 2 . Explanation: Let : Δ x = 61 m , Δ t = 8 . 3 s , and R = 24 . 5 cm = 0 . 245 m . From kinematics, Δ x = 1 2 a Δ t 2 a = 2 Δ x Δ t 2 , so a = Rα α = a R = 2 Δ x R (Δ t ) 2 = 2 (61 m) (0 . 245 m) (8 . 3 s) 2 = 7 . 22832 rad / s 2 . 008 (part 2 of 3) 10.0 points Find the final angular velocity of one of the car’s wheels....
View Full Document

{[ snackBarMessage ]}

Page1 / 12

HW09-solutions - lee(cyl327 – HW09 – gentle –(56245 1...

This preview shows document pages 1 - 3. Sign up to view the full document.

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