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Revision:Moments Of Inertia

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TSR Wiki > Study Help > Subjects and Revision > Revision Notes > Mathematics > Moments of Inertia

These notes are based on the requirements of the M5 A Level mathematics module.

1 Moments of inertia
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Moments of inertia

Calculating moments of inertia

$\displaystyle M.I._{\mathsf{of\ a\ point}} = mr^2$

$\displaystyle M.I._{\mathsf{total}} = \sum_i m_i r^2_i$

To find moments of inertia:

Take a small part of length $\delta x$ which is at distance from axis, the mass of that small part is: M/[length or area of the body] . [length or area of the small part].

Substitute the mass of the small part in the equation of M.I., r will be x,

$\displaystyle \sum m_i r^2_i$ will contain $\delta x$ ,

So you let $\delta x \longrightarrow 0$ then:

$\displaystyle M.I. = \lim_{\delta x \rightarrow 0}\sum m_i r^2_i$ , so you integrate to get M.I.

eg.:

Show that the moments of inertia of a uniform rod f mass M and length 2a about an axis through its centre perp. to its length to be $\displaystyle \frac{1}{3}Ma^2$ .

Take a small part of length $\delta x$ , which is at distance from the axis.

Mass of rod per length $\displaystyle = \frac{M}{2a}$

So mass of the small piece which is of length $\displaystyle \delta x = \frac{M}{2a} . \delta x$

So $\displaystyle M.I._{\mathsf{small\ piece}} = mr^2 = \frac{M}{2a} . \delta x . x^2$

So total $\displaystyle M.I. = lim_{\delta x \rightarrow 0} \sum_{x=-a}^{x=a} \frac{M}{2a} . \delta x . x^2$

Which means:

$\displaystyle M.I. = \int_{-a}^a \frac{M}{2a} . x^2 \delta x$

$\displaystyle = \frac{M}{2a} \int_{-a}^{a} x^2 \delta x$

$\displaystyle = \frac{M}{2a} [\frac{x^3}{3}]_{-a}^{a}$

$\displaystyle = \frac{M}{2a} [ \frac{a^3}{3} - \frac{(-a)^3}{3}] = \frac{M}{2a} (\frac{2a^3}{3}) = \frac{1}{3} M a^2$ .

Standard moments of inertia that are given in the formula sheet

For uniform bodies of mass m:

Thin rod, length 2l, about perpendicular axis through centre: $\displaystyle \frac{1}{3} ml^2$

Rectangular lamina about axis in plane bisecting edges of length 2l: $\displaystyle \frac{1}{3} ml^2$

Thin rod, length 2l, about perpendicular axis through end: $\displaystyle \frac{4}{3} ml^2$

Rectangular lamina about edge perp. to edges of length 2l: $\displaystyle \frac{4}{3} ml^2$

Rectangular lamina, sides 2a and 2b, about perpendicular axis through centre: $\displaystyle \frac{1}{3} m(a^2 + b^2)$

Hoop or cylindrical shell of radius r about axis through centre: $\displaystyle mr^2$

Hoop of radius r about a diameter: $\displaystyle \frac{1}{2}mr^2$

Disc or solid cylinder of radius r about axis through centre: $\displaystyle \frac{1}{2} mr^2$

Disc of radius r about a diameter: $\displaystyle \frac{1}{4} mr^2$

Solid sphere, radius r, about diameter: $\displaystyle \frac{2}{5} mr^2$

Spherical shell of radius r about a diameter: $\displaystyle \frac{2}{3} mr^2$

Additive rule

If two bodies have moments of inertia I₁ and I₂ about the same axis, then the moments of inertia of the composite body about that axis is I₁ + I₂.

eg. A uniform rod of mass 2m and length a, has a particle of mass m at distance 2a/3 from one of its ends. Find M.I. of the system.

$\displaystyle I_{\mathsf{rod}} = \frac{1}{3} (2m) (\frac{a}{2})^2 = \frac{1}{6} . ma^2$

$\displaystyle I_{\mathsf{particle}} = m (\frac{1}{3} a)^2 = \frac{1}{9} . ma^2$

$\displaystyle I_{\mathsf{system}} = ma^2 ( \frac{1}{6} + \frac{1}{9}) = \frac{5}{18} ma^2$

Stretching rule

If one body can be obtained from another body by stretching parallel to the axis without altering the distribution of mass relative to the axis, then the moments of inertia of the two bodies about the axis is the same.

For example: a solid cylinder is a disc stretched parallel to the axis through its centre perp. to it. A uniform rectangular lamina is a rod stretched parallel to the axis through its centre perp. to it.

Radius of Gyration

$\displaystyle I = mk^2$

Or $\displaystyle k = \sqrt{\frac{I}{m}}$

Where $\displaystyle k$ is the radius of gyration.

Parallel axis theorem

If a body of mass m has moments of inertia I about the axis through the centre of mass, then the moments of inertia of that body about an axis PARALLEL to and is at distance d from the first axis is $\displaystyle I_{CM} + md^2$ .

Perpendicular axes theorem

If a lamina lies on the plane xy, where Ox and Oy are perp., and has moments of inertia I_x and I_y about Ox and Oy respectively, and Oz is an axis perp to Ox, Oy and the lamina, then moments of inertia about Oz is I_z, where I_z = I_x + I _y.