Anonymous answered this 221 answers a. area of wheel = pi*R² area of 6 cutouts = 6*pi*(R/5)² = 0.24*pi*R² So, percentage reduction of weight = (0.24)*100% = 24% b. 18.24% (using negative mass for cutouts and parallel axis theorem) c. 27.95% (using negative mass for cutouts and parallel axis theorem)

Anonymous answered this 221 answers a. area of wheel = piR² area of 6 cutouts = 6pi(R/5)² = 0.24piR² So, percentage reduction of weight = (0.24)100% = 24% b. 18.24% (using negative mass for cutouts and parallel axis theorem) c. 27.95% (using negative mass for cutouts and parallel axis theorem)

Precision Machining Technology (MindTap Course List)

2nd Edition

ISBN:9781285444543

Author:Peter J. Hoffman, Eric S. Hopewell, Brian Janes

Publisher:Peter J. Hoffman, Eric S. Hopewell, Brian Janes

Chapter5: Turning

Section5.5: Taper Turning

Problem 5RQ: What are the corresponding centerline and included angles to 5/8" TPF?

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