Explanation The moment of inertia of the tire is the sum of the moment of inertia of the side walls and the thread wall. The side walls can be considered as a uniform disc with a circular disc removed at the center. Write the expression for moment of inertia of a hollow cylinder for R 2 > R 1 . I = 1 2 M ( R 1 2 + R 2 2 ) (I) Here, I is moment of inertia, M is the mass of tire, R 1 is the radius of inner wall and R 2 is the radius of outer wall. Mass per unit area of the side wall will be given by the product of volume of unit area and density. Write the expression for the mass of a hollow cylinder of height. M = ρ V (II) Here, ρ is the density and V is the volume of the cylinder. Write the expression for the volume of the hollow cylinder. V = π ( R 2 2 − R 1 2 ) h Here, R 2 is the outer radius, R 1 is the inner radius and h is the height of cylinder. Substitute π ( R 2 2 − R 1 2 ) h for V in equation (II). M = ρ π ( R 2 2 − R 1 2 ) h Substitute ρ π ( R 2 2 − R 1 2 ) h for M in equation (I). I = 1 2 ( ρ π ( R 2 2 − R 1 2 ) h ) ( R 1 2 + R 2 2 ) Rearrange the above equation. I = 1 2 ρ π ( R 2 4 − R 1 4 ) h (III) The total moment of inertia is the sum of moment of inertia of two walls and tread which is equal to the moment of inertia of hollow cylinder. Write the expression for the moment of inertia of two side walls. I 1 = 2 ( I ) Here, I 1 is the moment of inertia due to inertia of two side walls. Substitute 1 2 ρ π ( R 2 4 − R 1 4 ) h for I in above equation. I 1 = 2 ( 1 2 ρ π ( R 2 4 − R 1 4 ) h ) (IV) Write the expression for the moment of inertia of tread wall. I 2 = I Here, I 2 is the moment of inertia due to inertia of tread wall. Substitute 1 2 ρ π ( R 2 4 − R 1 4 ) h for I 2 in above equation. I 2 = 1 2 ρ π ( R 2 4 − R 1 4 ) h (V) Write the expression for the moment of inertia of the tire about its perpendicular axis through center. I total = I 1 + I 2 (VI) Here, I total is the moment of inertia of the tire about itsperpendicular axis through center

Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University

9th Edition

ISBN: 9781305372337

Author: Raymond A. Serway | John W. Jewett

Publisher: Cengage Learning

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Chapter 10, Problem 41P

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The moment of inertia about an axis perpendicular through its center.

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Chapter 10 Solutions

Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University

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The moment of inertia about an axis perpendicular through its center.

Solution Summary: The author explains the moment of inertia of a tire, the sum of its side walls and thread walls.

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Chapter 10 Solutions