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    University of Melbourne BMEN30005 Introduction To Biomechanics SEMESTER 1 - 2020 Tutorial 4 - Solutions

    Uploaded by

    Juan Felipe Uribe Cifuentes
    This document contains solutions to tutorial questions about biomechanics. It defines key formulas for calculating mass moment of inertia, including the formulas for a point mass and using the parallel axis theorem. It then works through examples of applying these formulas to calculate the mass moments of inertia for various body parts and shapes, such as a rod. The solutions integrate the point mass formula to find the mass moments of inertia of continuous objects.

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    University of Melbourne BMEN30005 Introduction To Biomechanics SEMESTER 1 - 2020 Tutorial 4 - Solutions

    Uploaded by

    Juan Felipe Uribe Cifuentes
    21 views4 pages
    This document contains solutions to tutorial questions about biomechanics. It defines key formulas for calculating mass moment of inertia, including the formulas for a point mass and using the parallel axis theorem. It then works through examples of applying these formulas to calculate the mass moments of inertia for various body parts and shapes, such as a rod. The solutions integrate the point mass formula to find the mass moments of inertia of continuous objects.

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    This document contains solutions to tutorial questions about biomechanics. It defines key formulas for calculating mass moment of inertia, including the formulas for a point mass and using the parallel axis theorem. It then works through examples of applying these formulas to calculate the mass moments of inertia for various body parts and shapes, such as a rod. The solutions integrate the point mass formula to find the mass moments of inertia of continuous objects.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    21 views4 pages

    University of Melbourne BMEN30005 Introduction To Biomechanics SEMESTER 1 - 2020 Tutorial 4 - Solutions

    Uploaded by

    Juan Felipe Uribe Cifuentes
    This document contains solutions to tutorial questions about biomechanics. It defines key formulas for calculating mass moment of inertia, including the formulas for a point mass and using the parallel axis theorem. It then works through examples of applying these formulas to calculate the mass moments of inertia for various body parts and shapes, such as a rod. The solutions integrate the point mass formula to find the mass moments of inertia of continuous objects.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    of 4

    UNIVERSITY OF MELBOURNE

    BMEN30005
    INTRODUCTION TO BIOMECHANICS
    SEMESTER 1 – 2020

    TUTORIAL 4 – SOLUTIONS

    Question 1

    The mass moment of inertia of a point mass is 𝐼 = 𝑚𝑟 2 where m is the total mass of the
    point and r is the perpendicular distance between the mass and the axis you’d like to find the
    mass moment of inertia about. This is a very important formula to memorize for this topic.

    𝐼𝑦 = 𝑚𝑟𝑦2

    𝐼𝑦 = 𝑚(𝐿𝑐𝑜𝑠𝜃)2

    𝐼𝑦 = (5)(2 ∗ 𝑐𝑜𝑠25°)2

    𝑰𝒚 = 𝟏𝟔. 𝟒 kg-m2

    𝐼𝑥 = 𝑚𝑟𝑥2

    𝐼𝑥 = 𝑚(𝐿𝑠𝑖𝑛𝜃)2
    𝐼𝑥 = (5)(2 ∗ 𝑠𝑖𝑛25°)2
    𝑰𝒙 = 𝟑. 𝟔 kg-m2

    1
    Question 2

    The KJC (or the axis that the knee rotates about) is the Z-axis shown in the diagram
    below. It is directed out of the page.
    The radius of gyration (https://en.wikipedia.org/wiki/Radius_of_gyration) is a method
    of communicating the total mass moment of inertia of some body about an axis as if it
    were a point mass. For instance, if a given body has a total mass m and total mass
    moment of inertia about some axis I, then its radius of gyration k will be the distance
    you’d need to place a point mass of mass m from the same axis to also give it a mass
    moment of inertia I. So a radius of gyration can tell us the mass moment of inertia of a
    body using our point mass formula (𝐼 = 𝑚𝑘 2).
    The parallel axis theorem (https://en.wikipedia.org/wiki/Parallel_axis_theorem) is a
    method for determining the mass moment of inertia of an object about a new axis, if you
    already know the mass moment of inertia about an axis that passes through the center
    of mass of your body and is parallel to your new axis. We find it using the following
    formula: 𝐼 ′ = 𝐼𝐶𝑂𝑀 + 𝑚𝑦 2 , where 𝐼𝐶𝑂𝑀 is the mass moment of inertia about the COM, 𝐼 ′ is the
    mass moment of inertia about the new axis, 𝑚 is the total mass of the object, and 𝑦 is the
    distance between the two axes. It’s important to remember that 𝐼𝐶𝑂𝑀 always starts on the right
    hand side of the equation, that way moving the axis further away from the COM will always
    result in a greater mass moment of inertia.

    𝐼𝑥 ′ = 𝐼𝑐 = 𝑚𝑘𝑐 2

    = 2.5 × 0.1222 = 0.04𝑘𝑔𝑚2


    𝐼𝑥 = 𝐼𝑥′ + 𝑚𝑦 2

    = 0.04 + 2.5 × 0.152 = 𝟎. 𝟏𝟎𝒌𝒈𝒎𝟐

    2
    𝐼𝑥 = 𝐼𝑥′ + 𝑚𝑦 2

    = 0.04 + 2.5 × (0.15 + 0.38)2 = 𝟎. 𝟕𝟒𝒌𝒈𝒎𝟐

    3
    Question 3

    For a point mass we know that: 𝐼 = 𝑚𝑟 2 . Using integration, we can take advantage of that
    formula to find the mass moment of inertia of any shape we want to by summing up all of the
    mass moment of inertias of an infinite amount of infinitely small masses that make up some
    larger shape.
    Not that this bar is unusual in that it only exists in one dimension (i.e., it only has a length in the
    x-direction, no thickness in y or z).

    Integrating our point mass equation across the bar gives:


    𝑚
    𝐼𝑦1 = ∫𝑚 𝑥 2 𝑑𝑚 𝑤ℎ𝑒𝑟𝑒 𝑑𝑚 = 𝑑𝑥
    𝐿
    𝐿/2
    𝑚
    𝐼𝑦1 = ∫ 𝑥 2 𝑑𝑥
    𝐿
    −𝐿/2

    𝐿/2
    𝑚 𝑥3
    𝐼𝑦1 = [ ]
    𝐿 3 −𝐿/2

    𝑚 𝐿3 𝐿3
    𝐼𝑦1 = ( + )
    3 8𝐿 8𝐿
    𝒎𝑳𝟐
    𝑰𝒚𝟏 =
    𝟏𝟐

    For the axis at the end of the rod only the bounds change:
    𝐿
    𝑚 𝑥3
    𝐼𝑦2 = [ ]
    𝐿 3 0

    𝒎𝑳𝟐
    𝑰𝒚𝟐 =
    𝟑

    For common shapes we can check our answer against online or textbook tables of common mass
    moment of inertias (https://en.wikipedia.org/wiki/List_of_moments_of_inertia).

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