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what is moment of inertia in engineering mechanics?

even i know the answer i'm not convinced, because it's confusing, it says moment of inertia is sum of masses of object x square of the distance from some reference axis, but how this square of distnce comes?!!!!! i dont know

becasue we can understand force x per dist from axis of application is moment it is simple and understandable, but

how this square of distance come in moment of inertia comes i don't know ? please help me to understand this anybody

3 Answers

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  • Kes
    Lv 7
    2 decades ago
    Favorite Answer

    Knowing how the formula for Moment of Inertia is derived may help you understand why it takes the form it does and why it is correct. It will also relate to other (linear) formulas that you know. Unlocking the secrets of formulas you use will take you far beyond those who blindly apply them. Good luck.

  • 2 decades ago

    Understanding WHY moment of inertia is decribed as I=mr^2 is more of a physics question than an engineering question.

    For simplicity, let us consider a point particle with mass m rotating about a given axis. The system can be generalized as the sum over all point particles to understand the system. This particle rotates about the axis at a radius of r with a frequency w. It is known that the rotational kinetic energy of this particle is

    T(rot) = .5*m*(w x r)^2

    [for w x r as vector cross product]. By a mathematical identity, this is

    T(rot) = .5*m*{w^2*r^2 - (w . r)^2}

    [where w . r is the dot product]. This can be expressed as in component form to give you a 3 by 3 matrix for I [in 3 dimensional space] called the inertia tensor.

    T(rot) = .5 * sum(ij)[ I(ij) * w(i) * w(j) ]

    I(ij) = m * { delta(ij)sum(k)[x(k)^2] - x(i)*x(j) }

    [where delta(ij) is the Kronicker delta: is 1 when i=j, else 0]

    For an elementary treatment when the particle is always only moving in a direction that is perpendicular to the axis of rotation, this inertia tensor will only have diagonal elements and a scalar I is described where I = sum(ij)[I(ij)] which is the moment of inertia you are familiar with. Now,

    T(rot)=.5 * I * w^2

    I = m * r^2

    Essentially, the square of the length term comes from the fact that a rotating particle takes more energy to rotate it based on how far away it is from the axis as well as how quickly it tries to rotate (which is also proportional to the radius).

    Source(s): Classical Dynamics of Particles and Systems, Marion Thornton
  • 2 decades ago

    moment of inertia is taken for rigid bodies about a center

    to consider the whole body we consider

    its revolution about the axis about which inertia is to be calculated

    and this revolution of the elementary part of the body leads to insertion of term x square

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