problem05_48

University Physics with Modern Physics with Mastering Physics (11th Edition)

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
5.48: a) This situation is equivalent to that of Example 5.23 and Problem 5.44, so Rg v 2 s = μ . Expressing v in terms of the period T , , 2 T R v π = so . 2 2 4 S g T R = A platform speed of 40.0 rev/min corresponds to a period of 1.50 s, so . 269 . 0 ) s m 80 . 9 ( s) (1.50 m) 150 . 0 ( 4 2 2 2 s = = μ b) For the same coefficient of static friction, the maximum radius is proportional to
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: the square of the period (longer periods mean slower speeds, so the button may be moved further out) and so is inversely proportional to the square of the speed. Thus, at the higher speed, the maximum radius is (0.150 m) ( 29 2 40 0 60 0 0 067 m . . = . ....
View Full Document

This document was uploaded on 02/04/2008.

Ask a homework question - tutors are online