hw9 solutions - homework 09 – VENNES, ROSS – Due: Nov 2...

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 DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: homework 09 – VENNES, ROSS – Due: Nov 2 2007, 1:00 am 1 Question 1, chap 12, sect 2. part 1 of 2 10 points A record has an angular speed of 44 . 2 rev / min. What is its angular speed? Correct answer: 4 . 62862 rad / s (tolerance ± 1 %). Explanation: 1 rev = 6 . 28319 rad, and 1 min = 60 s , Therefore, ω 1 = (44 . 2 rev / min) parenleftbigg 2 π rad rev parenrightbiggparenleftbigg 1 min 60 s parenrightbigg = 4 . 62862 rad / s . Question 2, chap 12, sect 2. part 2 of 2 10 points Through what angle, in radians, does it rotate in 1 . 41 s? Correct answer: 6 . 52635 rad (tolerance ± 1 %). Explanation: θ = ω t = (4 . 62862 rad / s) (1 . 41 s) = 6 . 52635 rad . Question 3, chap 12, sect 2. part 1 of 2 10 points A racing car travels on a circular track of radius 335 m. The car moves with a constant linear speed of 42 . 4 m / s. Find its angular speed. Correct answer: 0 . 126567 rad / s (tolerance ± 1 %). Explanation: The linear speed v and the angular speed ω are related by, v = R ω ⇒ ω = v R . Question 4, chap 12, sect 2. part 2 of 2 10 points Find the magnitude of its acceleration. Correct answer: 5 . 36645 m / s 2 (tolerance ± 1 %). Explanation: If the car is moving at a constant speed, there is no tangential acceleration, thus the acceleration is purely radial, a r = v 2 R . Question 5, chap 12, sect 2. part 1 of 1 10 points A woman passes through a revolving door with a tangential speed of 1.8 m/s. If she is 0.78 m from the center of the door, what is the door’s angular speed? Correct answer: 2 . 30769 rad / s (tolerance ± 1 %). Explanation: Basic Concept: v t = rω Given: v t = 1 . 8 m / s r = 0 . 78 m Solution: ω = v t r = 1 . 8 m / s . 78 m = 2 . 30769 rad / s Question 6, chap 12, sect 2. part 1 of 1 10 points The speed of a moving bullet can be deter- mined by allowing the bullet to pass through two rotating paper disks mounted a distance 64 . 5 cm apart on the same axle. From the homework 09 – VENNES, ROSS – Due: Nov 2 2007, 1:00 am 2 angular displacement 48 ◦ of the two bullet holes in the disks and the rotational speed 866 rev / min of the disks, we can determine the speed v of the bullet. 48 ◦ v 866 rpm 64 . 5 cm What is the speed of the bullet? Correct answer: 69 . 8212 m / s (tolerance ± 1 %). Explanation: Let : ω = 866 rev / min , d = 64 . 5 cm , and θ = 48 ◦ . From θ = ω t the time to pass through an angle θ is t = θ ω = (48 ◦ ) (90 . 6873 rad / s) π rad 180 ◦ = 0 . 00923788 s . Then the speed of the bullet is v = d t = (64 . 5 cm) (0 . 01 m / cm) . 00923788 s = 69 . 8212 m / s . Question 7, chap 12, sect 2. part 1 of 3 10 points A car accelerates uniformly from rest and covers a distance of 53 m in 6 . 9 s....
View Full Document

This homework help was uploaded on 04/18/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

Page1 / 11

hw9 solutions - homework 09 – VENNES, ROSS – Due: Nov 2...

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