This preview shows pages 1–2. 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: Nguyen, Thanh – Homework 5 – Due: Sep 18 2007, 7:00 pm – Inst: D Weathers 1 This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A ball on the end of a string is whirled around in a horizontal circle of radius 0 . 206 m. The plane of the circle is 1 . 82 m above the ground. The string breaks and the ball lands 2 . 4 m away from the point on the ground directly beneath the ball’s location when the string breaks. The acceleration of gravity is 9 . 8 m / s 2 . Find the centripetal acceleration of the ball during its circular motion. Correct answer: 75 . 2801 m / s 2 . Explanation: In order to find the centripetal acceleration of the ball, we need to find the initial velocity of the ball. Let y be the distance above the ground. After the string breaks, the ball has no initial velocity in the vertical direction, so the time spent in the air may be deduced from the kinematic equation, y = 1 2 g t 2 . Solving for t , ⇒ t = r 2 y g . Let d be the distance traveled by the ball. Then v x = d t = d r 2 y g . Hence, the centripetal acceleration of the ball during its circular motion is a c = v 2 x r = d 2 g 2 y r = 75 . 2801 m / s 2 . keywords: 002 (part 1 of 1) 10 points Before throwing a 0 . 566 kg discus, an ath- lete rotates it along a circular path of radius 1 . 03 m. The maximum speed of the discus is 30 . 1 m / s. Determine the magnitude of its maximum radial acceleration. Correct answer: 879 . 621 m / s 2 . Explanation: The maximum radial acceleration is a r = v 2 r = (30 . 1 m / s) 2 1 . 03 m = 879 . 621 m / s 2 ....
View Full Document
This note was uploaded on 10/24/2011 for the course PHYS 1710 taught by Professor Weathers during the Winter '08 term at North Texas.
- Winter '08