This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Gozick, Brandon – Homework 5 – Due: Sep 22 2006, 5:00 pm – Inst: D Weathers 1 This printout should have 9 questions. Multiplechoice 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 . 207 m. The plane of the circle is 1 . 03 m above the ground. The string breaks and the ball lands 2 . 37 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: 129 . 088 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 = 129 . 088 m / s 2 . keywords: 002 (part 1 of 1) 10 points Before throwing a 0 . 48 kg discus, an ath lete rotates it along a circular path of radius 1 . 03 m. The maximum speed of the discus is 8 . 16 m / s. Determine the magnitude of its maximum radial acceleration....
View
Full
Document
This note was uploaded on 09/12/2008 for the course PHYS 1710 taught by Professor Weathers during the Fall '08 term at North Texas.
 Fall '08
 Weathers
 mechanics, Work

Click to edit the document details