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Added sections about parallel and perpendicular axis theorems and some examples
The '''moment of [[inertia]]''' is a measure of a [[mass]] and its distance [[perpendicular]] to an [[axis]] of rotation. It describes an objects resistance to change in rotation.
The '''moment of [[inertia]]''' is a measure of a [[mass]] and its distance [[perpendicular]] to an [[axis]] of rotation.<ref>[https://www.britannica.com/science/moment-of-inertia Britannica.com]</ref> It describes an objects resistance to change in rotation.
The formula to calculate a moment of inertia between a point and a point mass is:
The formula to calculate a moment of inertia between a point and a point mass is:
where <math>r</math> is the displacement between the two points, <math>m</math> is the mass of the point mass and <math>I</math> is the moment of inertia.
where <math>r</math> is the displacement between the two points, <math>m</math> is the mass of the point mass and <math>I</math> is the moment of inertia.
If you have a finite set of point mass's and want to find their moment of inertia around a point you simply add the moment of inertia of each object to get the following:
If you have a finite set of point mass's and want to find their moment of inertia around a point you simply add the moment of inertia of each object to get the following:<ref>[http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html#mix Moment of inertia calculations on hyperphysics]</ref>
:<big><big><math>I = \sum_{n=1}^{n} m_{n} r_{n}^2</math></big></big>
:<big><big><math>I = \sum_{i=1}^{n} m_{i} r_{i}^2</math></big></big>
It is important to note that you can not simply take the centre of mass of an object and use it as a point mass. Instead you need to use integration:
It is important to note that you can not simply take the centre of mass of an object and use it as a point mass. Instead you need to use [[integration]]:
Mathematically, the moment of inertia <math>I</math> for a mass <math>m</math> that is moving a perpendicular distance of <math>r</math> from an axis is expressed as:
Mathematically, the moment of inertia <math>I</math> for a mass <math>m</math> that is moving a perpendicular distance of <math>r</math> from an axis is expressed as:
The formula for a moment of inertia is the result of combining the equation for [[kinetic energy]] with the equation for angular velocity.
The formula for a moment of inertia is the result of combining the equation for [[kinetic energy]] with the equation for angular velocity.
==Parallel Axis Theorem==
The [[parallel]] axis theorem states that the moment of inertia of a body is smallest when the axis of rotation passes through the centre of mass of that body.<ref>[http://physics.bu.edu/~redner/211-sp06/class-rigid-body/parallelaxis.html physics.bu.edu]</ref> Mathematically, it can be stated as:
:<math>
I = I_{cm} + Mh^2
</math>
where I<sub>cm</sub> is the moment of inertia for an axis through the centre of mass, M is the [[mass (science)|mass]] of the body and h is the perpendicular distance from the centre of mass to an axis '''parallel''' to the one through the centre of mass. This theorem is useful as once one moment of inertia is calculated, any other moment of inertia can be calculated easily (so long as the axes are parallel).
==Perpendicular Axis Theorem==
The [[perpendicular]] axis theorem states that the moment of inertia around an axis at right angles to the plane of a planar body, I<sub>z</sub> is equal to the sum of any two moments of inertia in the plane of the body so long as these two moments of inertia are perpendicular to each other.<ref>[http://hyperphysics.phy-astr.gsu.edu/hbase/perpx.html#ppx Perpendicular axis theorem on hyperphysics]</ref> This can be written as:
:<math>
I_z = I_x+I_y
</math>
where I<sub>x</sub> and I<sub>y</sub> are in the plane of the object. This theorem is particularly useful as means three dimensional objects can be broken down into a sum of two dimensional objects.
==Examples==
===Discrete example===
Suppose a massless string of length L has two particles, both of mass m, attached to it. One is attached halfway along the string, the other at the end (opposite end from the pivot). The string is now swung around so that it is taut. As there are a finite number of discrete masses, we must use the sum formula:
:<math>
I=\sum m_i r^2_i = m \left( \frac{L}{2}\right)^2 + mL^2 = \frac{5mL^2}{4}
</math>
===Continious Example===
Suppose a rod of mass m and length L is rotating about one end. As the rod is continuous, we use the integration form. Therefore in terms of differentials, dI=r<sup>2</sup>dm. Defining the [[density]] of the rod as λ=m/L, dI can be written as dI=r<sup>2</sup>λdr. Integrating gives moment of inertia:
:<math>
I=\int dI = \int^L_0 \lambda r^2 \, dr
</math>
This is integrated between 0 and L as the rod pivots about r=0 and has length L. If it pivoted about the middle, the boundaries would be -L/2 and +L/2. Performing the integration and substituting back in for λ:
:<math>
I = \frac{m}{3L} \left( L^3 - 0\right) = \frac{mL^2}{3}
</math>
==References==
{{reflist}}
==See also==
*[[Rotational mechanics]]
[[Category:Physics]]
[[Category:Physics]]
[[Category:Mechanics]]
[[Category:Calculus]]
[[Category:Calculus]]