Steiner’s theorem and gyration radius. A metal tank wheel in the shape of a disc of radius R is lightened by cutting out six identical through- holes in the shape of discs, whose axes are parallel to the wheel’s axis of rotation, and whose radii are all r. The holes’ centers are positioned at the vertices of a regular hexagon of side A, with the wheel’s axis passing through the center of the hexagon. By what percentage is the wheel lightened, as a result of the cutouts, for r=R/5? Is the solution: area of wheel = pi*R? area of 6 cutouts = 6* pi*(R/5)2 = 0.24* pi*R2 So, percentage reduction of weight = (0.24)*100% = 24%?
Steiner’s theorem and gyration radius. A metal tank wheel in the shape of a disc of radius R is lightened by cutting out six identical through- holes in the shape of discs, whose axes are parallel to the wheel’s axis of rotation, and whose radii are all r. The holes’ centers are positioned at the vertices of a regular hexagon of side A, with the wheel’s axis passing through the center of the hexagon. By what percentage is the wheel lightened, as a result of the cutouts, for r=R/5? Is the solution: area of wheel = pi*R? area of 6 cutouts = 6* pi*(R/5)2 = 0.24* pi*R2 So, percentage reduction of weight = (0.24)*100% = 24%?
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter11: Gravity, Planetary Orbits, And The Hydrogen Atom
Section: Chapter Questions
Problem 46P
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Steiner’s theorem and gyration radius.
A metal tank wheel in the shape of a disc of radius R is lightened by cutting out six identical through- holes in the shape of discs, whose axes are parallel to the wheel’s axis of rotation, and whose radii are all r. The holes’ centers are positioned at the vertices of a regular hexagon of side A, with the wheel’s axis passing through the center of the hexagon.
By what percentage is the wheel lightened, as a result of the cutouts, for r=R/5?
Is the solution:
area of wheel = pi*R?
area of 6 cutouts = 6* pi*(R/5)2 = 0.24* pi*R2
So, percentage reduction of weight = (0.24)*100% = 24%?
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