This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: arellano (aa39398) Practice Homework Chapter 10 Solutions weathers (17101) 1 This printout should have 53 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points A girl sitting on a merrygoround moves counterclockwise through an arc length of 2.57 m. If the girls angular displacement is 2.27 rad, how far is she from the center of the merrygoround? Correct answer: 1 . 13216 m. Explanation: Let : s = 2 . 57 m and = 2 . 27 rad . s = r r = s = 2 . 57 m 2 . 27 rad = 1 . 13216 m . keywords: 002 10.0 points A car tire rotates with an average angular speed of 29 rad/s. In what time interval will the tire rotate 3 . 9 times? Correct answer: 0 . 84498 s. Explanation: Let : avg = 29 rad / s and = 3 . 9 rev . avg = t t = avg = 3 . 9 rev 29 rad / s 2 rad 1 rev = . 84498 s . 003 (part 1 of 2) 10.0 points A wheel starts from rest and rotates with constant angular acceleration to an angular speed of 15 rad / s in 3 . 05 s. Find the magnitude of the angular acceler ation of the wheel. Correct answer: 4 . 91803 rad / s 2 . Explanation: Let : = 15 rad / s and t = 3 . 05 s . Angular acceleration is = w t = 15 rad / s 3 . 05 s = 4 . 91803 rad / s 2 . 004 (part 2 of 2) 10.0 points Find the angle in radians through which it rotates in this time. Correct answer: 22 . 875 rad. Explanation: The angle through which the wheel rotates during this time interval is = 1 2 t 2 = 1 2 (4 . 91803 rad / s 2 ) (3 . 05 s) 2 = 22 . 875 rad . keywords: 005 (part 1 of 2) 10.0 points A grinding wheel, initially at rest, is ro tated with constant angular acceleration of 5 . 4 rad / s 2 for 3 . 9 s. The wheel is then brought to rest with uniform deceleration in 8 . 75 rev. arellano (aa39398) Practice Homework Chapter 10 Solutions weathers (17101) 2 Find the angular acceleration required to bring the wheel to rest. Note that an in crease in angular velocity is consistent with a positive angular acceleration. Correct answer: 4 . 03365 rad / s 2 . Explanation: Let : = 0 rad / s , 1 = 5 . 4 rad / s 2 , t 1 = 3 . 9 s , and 2 = 8 . 75 rev . From kinematics f = + t. First find the speed attained before the wheel begins to slow down. Since = 0 rad / s for the acceleration, the final speed 1 = 1 t 1 for the acceleration is also the initial speed for the deceleration. Considering the wheel as it comes to rest, 2 2 = 1 2 + 2 2 2 = 0 2 = 2 1 2 2 = 2 1 t 2 1 2 2 = (5 . 4 rad / s 2 ) 2 (3 . 9 s) 2 2 (8 . 75 rev) 1 rev 2 rad = 4 . 03365 rad / s 2 . 006 (part 2 of 2) 10.0 points Determine the time needed to bring the wheel to rest....
View Full
Document
 Fall '10
 Weathers
 Work

Click to edit the document details