{"id":4227,"date":"2021-09-09T17:38:08","date_gmt":"2021-09-09T12:08:08","guid":{"rendered":"https:\/\/acejee.com\/blog\/?p=4227"},"modified":"2024-09-30T08:10:40","modified_gmt":"2024-09-30T02:40:40","slug":"moment-of-inertia-formula-for-different-shapes-jee-main","status":"publish","type":"post","link":"https:\/\/acejee.com\/blog\/moment-of-inertia-formula-for-different-shapes-jee-main\/","title":{"rendered":"Moment of Inertia formula for different shapes | JEE Main"},"content":{"rendered":"\t\t
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Moment of inertia or mass moment of inertia is the resistance of a rigid body to change in its angular velocity or we can say, resistance to angular acceleration, when a net external torque acts on it (similar to resistance offered by mass of a particle to acceleration, when a net force acts on it). In this blog we will explore moment of inertia formula for different shapes (ring<\/a>, disc<\/a>, hollow cylinder \/ annular disc<\/a>, ring\/disc (about an inclined axis)<\/a>, solid cylinder<\/a>, rod<\/a>, rod (about an inclined axis)<\/a>, sphere<\/a>, hollow sphere<\/a>, rectangle \/ rectangular plate<\/a>, square plate<\/a>, cube<\/a>, triangle<\/a>, cone<\/a><\/p>\n

For a rigid body made up of discrete particles (for example a pendulum with bob of mass $m$ and massless rod of length $l$), moment of inertia can be calculated using the formula $I = \\sum_i m_i r_i^2$ <\/p>\n

For a continuous rigid body (for example a uniform solid sphere or a uniform rod etc.), $I = \\int r^2 dm$<\/p>\n

Unit of moment of inertia $I$ is $Kg \\ m^2$<\/p>\n

Dimensional Formula = $[M^1L^2T^0]$<\/p>\n

Now, let’s look at the moment of inertia for different shapes, but before we do that, let’s review the parallel axis and perpendicular axis theorem<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Parallel Axis Theorem<\/h2>\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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$I = I_{CM} + Mh^2$<\/p>

Derivation<\/b>
Moment of Inertia about $AB$, $I_{CM}$ $=\\int r^2dm$<\/p>

Moment of inertia about $CD$ (parallel to $AB$), $I =\\int (h-r)^2 dm$ $=\\int (h^2 + r^2 -2hr)dm$ $= \\int r^2dm +$ $h^2 \\int dm -$ $2h \\int rdm$<\/p>

Now, $\\int r dm = M r_{CM} = 0$ as the axis $AB$ is passing through the center of mass (ping us if this didn’t make sense)<\/p>

Or, $I = I_{CM} + Mh^2$<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Perpendicular Axis Theorem (valid ONLY for planar objects)<\/h2>\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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\n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\"Perpendicular\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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$I_z = I_x + I_y$<\/p>

Derivation<\/b>
$I_z = \\sum_i m_i r_i^2$ $= \\sum_i m_i (a_i^2 + b_i^2)$ $=\\sum_i m_i a_i^2 + $ $\\sum_i m_i + b_i^2$ $=I_x + I_y$\u00a0<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Non – Planar Objects:
Note perpendicular axis theorem is NOT valid for non planar or 3 dimensional objects like a cube.<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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$I_z \\neq I_x + I_y$<\/p>

Reason<\/b>
$I_z = \\sum_i m_i (a_i^2 + b_i^2$
$I_x = \\sum_i m_i (b_i^2 + h_i^2)$
$I_y = \\sum_i m_i (b_i^2 + h_i^2)$<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Moment of Inertia of Ring<\/h2>\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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\n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\"Moment\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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About the central axis (passing through its center and perpendicular to the plane of the ring):
$I = MR^2$<\/p>

About the diameter:
$I = \\cfrac{1}{2} MR^2$<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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About the central axis (passing through the center of the ring and perpendicular to the plane of the ring):
$I = \\int R^2 dm$ $= R^2 \\int dm$ $= MR^2$<\/p>

About its diameter:
Using perpendicular axis theorem $I_z = I_x + I_y$, we will get $I = \\cfrac{1}{2} MR^2$<\/p><\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Moment of Inertia of Disc<\/h2>\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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\n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\"Moment\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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About the central axis (axis passing through the center and perpendicular to the plane of the disc)
$I = \\cfrac{1}{2}MR^2$<\/p>

About its diameter:
$I = \\cfrac{1}{4}MR^2$<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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About its central axis:\n$dI = r^2 dm$\n$dm = \\cfrac{M}{\\pi R^2} 2 \\pi r dr$\n$\\implies$ $dI = \\cfrac{2M}{R^2} r^3 dr$\n$I = \\int dI = \\cfrac{1}{2} MR^2$\n\nAbout its diameter:\nUsing the perpendicular axis theorem $I_z = I_x + I_y$, we will get $I = \\cfrac{1}{4}MR^2$<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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Top<\/a><\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Moment of Inertia of Ring \/ Disc About an Inclined Axis<\/h2>\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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\n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\"Moment\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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Moment of inertia of a ring, of mass $M$ and radius $R$, about an axis inclined at an angle $\\theta$ with it’s plane is
$= \\cfrac{M R^2}{2} (1 + \\sin^2 \\theta)$<\/p>

Moment of inertia of disc, of mass $M$ and radius $R$, about an axis inclined at an angle $\\theta$ with its plane is
$= \\cfrac{MR^2}{4} (1 + \\sin^2 \\theta)$\u00a0<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Ring:
$dm = \\cfrac{M}{2 \\pi r} r d \\alpha$
$dI = dm r^2 (\\sin^2 \\alpha + \\cos^2 \\alpha \\sin^2 \\theta)$
$= \\cfrac{M}{2 \\pi r} r^3 (\\sin^2 \\alpha + \\cos^2 \\alpha \\sin^2 \\theta)d \\alpha $<\/p>

$I = \\int dI$ $= \\cfrac{M}{2 \\pi} r^2 \\int_{\\alpha = 0}^{\\alpha = 2 \\pi}(\\sin^2 \\alpha + \\cos^2 \\alpha \\sin^2 \\theta)d \\alpha$
$=\\cfrac{Mr^2}{2} (1 + \\sin^2 \\theta)$<\/p>

Disc:
$dm = \\cfrac{M}{\\pi R^2} 2 \\pi r dr$
$dI = \\cfrac{dm r^2}{2} (1 + \\sin^2 \\theta)$
$I = \\int dI$ $= \\cfrac{M}{\\pi R^2} 2 \\pi \\int_0^R \\cfrac{r^3}{2} (1 + \\sin^2 \\theta) dr$
$= \\cfrac{MR^2}{4} (1 + \\sin^2 \\theta)$\u00a0<\/p><\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Moment of Inertia of Hollow Cylinder | Annular Disc<\/h2>\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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\n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\"Moment\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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About the central axis:
$I = \\cfrac{1}{2} M (R_1^2 + R_2^2)$<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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About its central axis:
$dI = r^2 dm$
$dm = \\cfrac{M}{\\pi (R_2^2 – R_1^2)} 2 \\pi r dr$
$\\implies$ $dI=$ $\\int_{r_1}^{r_2} r^2 dm$ $\\cfrac{2 M}{R_2^2 – R_1^2} \\int_{r_1}^{r_2} r^3 dr$
Or
$I = \\cfrac{1}{2} M (R_1^2 + R_2^2)$<\/p><\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Top<\/a><\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Moment of Inertia of Solid Cylinder<\/h2>\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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About the axis perpendicular to the length of the rod and passing through its center:
$I = \\cfrac{1}{4} MR^2 + \\cfrac{1}{12} ML^2$

About the central axis:
$I = \\cfrac{1}{2} MR^2$<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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About the axis perpendicular to its length and passing through its center:
$dI = \\cfrac{1}{4} R^2 dm + dm x^2$
$dm = \\cfrac{M}{L} dx$
$I = \\int dI = \\cfrac{1}{4} R^2 \\int dm + \\cfrac{M}{L} \\int_{-L\/2}^{L\/2} x^2 dx$
$I = \\cfrac{1}{4} MR^2 + \\cfrac{1}{12} ML^2$<\/p>

About the central axis:
Same as that of disc about its central axis<\/p><\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Top<\/a><\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Moment of Inertia of a Rod<\/h2>\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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About the axis perpendicular to its length and passing through the center:<\/p>

$I = \\cfrac{1}{12} ML^2$<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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$dI = \\int_{-L\/2}^{L\/2} x^2 dm$\u00a0
$dm = \\cfrac{M}{L} dx$
$I = \\cfrac{1}{12}ML^2$<\/p><\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Top<\/a><\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Moment of Inertia of a Rod About an Inclined Axis<\/h2>\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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Moment of inertia of rod, of mass $M$ and length $L$, about an axis inclined at an angle $\\theta$ with the rod $= \\cfrac{ML^2}{12} \\sin^2 \\theta$\u00a0<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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$dm = \\cfrac{M}{L} dx$
$dI = dm x^2 \\sin^2 \\theta$
$I = \\int dI =\\cfrac{M}{L} \\int_{-L\/2}^{L\/2} x^2 \\sin^2 \\theta dx$\u00a0<\/p><\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Top<\/a><\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Moment of Inertia of Hollow Sphere<\/h2>\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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\n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\"Moment\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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About its diameter:<\/p>

$I = \\cfrac{2}{3} MR^2$<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Let’s consider the differential ring of mass $dm$ as shown:\u00a0<\/p>

$dm = \\cfrac{M}{4 \\pi R^2}2\\pi R \\sin \\theta . R d\\theta$ $=\\cfrac{1}{2} M \\sin \\theta d \\theta$<\/p>

$dI = R^2 \\sin^2 \\theta dm$ $=\\cfrac{1}{2} MR^2 \\sin^3 \\theta d\\theta$<\/p>

$I = \\int dI$ $= \\cfrac{1}{2} MR^2 \\int_0^{\\pi} \\sin^3 \\theta d\\theta$<\/p>

$I = \\cfrac{2}{3} MR^2$<\/p><\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Top<\/a><\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Moment of Inertia of Solid Sphere<\/h2>\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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\n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\"Moment\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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About its diameter:<\/p>

$I = \\cfrac{2}{5} MR^2$<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Consider a spherical shell or radius $r$ and thickness $dr$<\/p>

For this differential strip,
$dm = \\cfrac{M}{\\cfrac{4}{3} \\pi R^3} 4 \\pi r^2 dr$ $=\\cfrac{3M}{R^3} r^2dr$
$dI = \\cfrac{2}{3} dm r^2$ $=\\cfrac{2M}{R^3} r^4 dr$
$I = \\cfrac{2M}{R^3} \\int_0^R r^4 dr$ $=\\cfrac{2}{5} MR^2$<\/p><\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Top<\/a><\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Moment of Inertia of Rectangle | Rectangular Plate<\/h2>\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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About the axis passing through its center and perpendicular to its plane,<\/p>

$I = \\cfrac{M}{12} (a^2 + b^2)$<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Let’s consider differential rod like element of width $dx$,<\/p>

$dm = \\cfrac{M}{ab}adx$ $=\\cfrac{M}{b} dx$<\/p>

$dI = \\cfrac{1}{12} dm a^2 + dm (\\cfrac{b}{2} – x)^2$<\/p>

$I = \\int dI$ $=\\cfrac{1}{12} \\cfrac{Ma^2}{b} \\int_0^b dx + \\cfrac{M}{b} \\int_0^b (\\cfrac{b^2}{4} + x^2 – bx)dx$<\/p>

$I = \\cfrac{1}{12} M (a^2 + b^2)$<\/p><\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Top<\/a><\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Moment of Inertia of Square Plate<\/h2>\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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About the axis through its center and perpendicular to its plane:<\/p>

$I = \\cfrac{1}{6} Ma^2$<\/p>

About the axis passing through its center and in the plane of the square:<\/p>

$I = \\cfrac{1}{12} Ma^2$<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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About the axis through its center and perpendicular to its plane:
Derivation is same as that for rectangular plate<\/p>

About the axis passing through its center and in plane of the square ($I_x$ or $I_y$):
From perpendicular axis theorem: $I_z = I_x + I_y$
From Symmetry $I_x = I_y$<\/p>

$\\implies$ $I_x = I_y = \\cfrac{1}{12} Ma^2$<\/p><\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Top<\/a><\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Moment of Inertia of Cube<\/h2>\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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About an axis passing through its center and perpendicular to any two faces,<\/p>

$I = \\cfrac{1}{6} Ma^2$<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Consider the differential square plate element of thickness $dx$\u00a0<\/p>

$dm = \\cfrac{M}{a^3} a^2 dx$ $= \\cfrac{M}{a} dx$<\/p>

$dI = \\cfrac{1}{6} a^2 dm$<\/p>

$I = \\int dI = \\cfrac{1}{6} Ma \\int_0^a dx$<\/p>

$I = \\cfrac{1}{6} Ma^2$<\/p><\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Top<\/a><\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Moment of Inertia of Triangle<\/h2>\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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1) Moment of Inertia of Equilateral Triangle about centroid <\/h3>\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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Moment of inertia about an axis passing through the centroid and perpendicular to the triangle, $I = \\cfrac{1}{12} Ma^2$<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Let’s find the moment of inertia of an equilateral triangle of side $a$ indirectly by dividing the equilateral triangle into 4 equilateral triangles of slide $\\cfrac{a}{2}$ as shown.<\/p>\n

Now, the moment of inertia is proportional to mass and square of the side lengths, <\/p>\n

So moment of inertia of smaller equilateral triangles, $i$, (about perpendicular axis passing through their centroids) would be $\\cfrac{1}{16}$ of the moment of inertia $I$ of the full triangle<\/p>\n

So using parallel axis theorem, we can write<\/p>\n

$I = 4 \\times \\cfrac{I}{16} + 3 \\times \\cfrac{M}{4} \\times (\\cfrac{a}{2 \\sqrt{3}})^2$<\/p>\n

Or, $I = \\cfrac{1}{12} Ma^2$<\/p><\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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2) Moment of Inertia of Triangle (of base $a$ and height $h$) about centroid (with axis in-plane of triangle)<\/h3>\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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Moment of inertia about the axis passing through centroid and parallel to base:<\/p>

$I = \\cfrac{\\rho ah^3}{36} = \\cfrac{Mh^2}{18}$<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Consider the differential element $dm$ as shown\n\n$dm = \\cfrac{M}{\\cfrac{1}{2}ah} \\cfrac{a}{h} x dx$\n\n$dI = dm (\\cfrac{2}{3}h – x)^2$\n\n$I = \\int_0^h dI$<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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3) Moment of Inertia of Triangle (of base $a$ and height $h$) about its base<\/h3>\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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Moment of inertia about the base:<\/p>

$I = \\cfrac{\\rho ah^3}{12} = \\cfrac{Mh^2}{6}$<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Using parallel axis theorem:\u00a0<\/p>

$I_{base} = I_{centroid} + M(\\cfrac{h}{3})^2$ $=\\cfrac{Mh^2}{6}$<\/p><\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Top<\/a><\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Moment of Inertia of Solid Cone<\/h2>\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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Moment of inertia of solid cone about its central axis, $I$ $=\\cfrac{3}{10} MR^2$<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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\n\t\t\t\t\t\t\t\t\t\t\t\t\n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t<\/i><\/span>\n\t\t\t\t\t\t\t\t<\/i><\/span>\n\t\t\t\t\t\t\t\t\t\t\t\t\t<\/span>\n\t\t\t\t\t\t\t\t\t\t\t\tDerivation<\/a>\n\t\t\t\t\t<\/div>\n\n\t\t\t\t\t

Consider the differential disc element as shown<\/p>\n

$dm = \\cfrac{M}{\\cfrac{1}{3} \\pi R^2 h} \\pi (\\cfrac{R}{h} x)^2 dx$
$dI = \\cfrac{1}{2} (\\cfrac{R}{h} x)^2 dm$<\/p>\n

$I = \\int_0^h dI$<\/p><\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Top<\/a><\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t<\/div>\n\t\t","protected":false},"excerpt":{"rendered":"

Moment of inertia of a rigid body is its resistance to change in its angular velocity. Let’s explore the moment of inertia of commonly encountered simple objects<\/p>\n","protected":false},"author":1,"featured_media":4230,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"ocean_post_layout":"","ocean_both_sidebars_style":"","ocean_both_sidebars_content_width":0,"ocean_both_sidebars_sidebars_width":0,"ocean_sidebar":"0","ocean_second_sidebar":"0","ocean_disable_margins":"enable","ocean_add_body_class":"","ocean_shortcode_before_top_bar":"","ocean_shortcode_after_top_bar":"","ocean_shortcode_before_header":"","ocean_shortcode_after_header":"","ocean_has_shortcode":"","ocean_shortcode_after_title":"","ocean_shortcode_before_footer_widgets":"","ocean_shortcode_after_footer_widgets":"","ocean_shortcode_before_footer_bottom":"","ocean_shortcode_after_footer_bottom":"","ocean_display_top_bar":"default","ocean_display_header":"default","ocean_header_style":"","ocean_center_header_left_menu":"0","ocean_custom_header_template":"0","ocean_custom_logo":0,"ocean_custom_retina_logo":0,"ocean_custom_logo_max_width":0,"ocean_custom_logo_tablet_max_width":0,"ocean_custom_logo_mobile_max_width":0,"ocean_custom_logo_max_height":0,"ocean_custom_logo_tablet_max_height":0,"ocean_custom_logo_mobile_max_height":0,"ocean_header_custom_menu":"0","ocean_menu_typo_font_family":"0","ocean_menu_typo_font_subset":"","ocean_menu_typo_font_size":0,"ocean_menu_typo_font_size_tablet":0,"ocean_menu_typo_font_size_mobile":0,"ocean_menu_typo_font_size_unit":"px","ocean_menu_typo_font_weight":"","ocean_menu_typo_font_weight_tablet":"","ocean_menu_typo_font_weight_mobile":"","ocean_menu_typo_transform":"","ocean_menu_typo_transform_tablet":"","ocean_menu_typo_transform_mobile":"","ocean_menu_typo_line_height":0,"ocean_menu_typo_line_height_tablet":0,"ocean_menu_typo_line_height_mobile":0,"ocean_menu_typo_line_height_unit":"","ocean_menu_typo_spacing":0,"ocean_menu_typo_spacing_tablet":0,"ocean_menu_typo_spacing_mobile":0,"ocean_menu_typo_spacing_unit":"","ocean_menu_link_color":"","ocean_menu_link_color_hover":"","ocean_menu_link_color_active":"","ocean_menu_link_background":"","ocean_menu_link_hover_background":"","ocean_menu_link_active_background":"","ocean_menu_social_links_bg":"","ocean_menu_social_hover_links_bg":"","ocean_menu_social_links_color":"","ocean_menu_social_hover_links_color":"","ocean_disable_title":"on","ocean_disable_heading":"default","ocean_post_title":"","ocean_post_subheading":"","ocean_post_title_style":"","ocean_post_title_background_color":"","ocean_post_title_background":0,"ocean_post_title_bg_image_position":"","ocean_post_title_bg_image_attachment":"","ocean_post_title_bg_image_repeat":"","ocean_post_title_bg_image_size":"","ocean_post_title_height":0,"ocean_post_title_bg_overlay":0.5,"ocean_post_title_bg_overlay_color":"","ocean_disable_breadcrumbs":"off","ocean_breadcrumbs_color":"","ocean_breadcrumbs_separator_color":"","ocean_breadcrumbs_links_color":"","ocean_breadcrumbs_links_hover_color":"","ocean_display_footer_widgets":"default","ocean_display_footer_bottom":"default","ocean_custom_footer_template":"0","cybocfi_hide_featured_image":"yes","ocean_post_oembed":"","ocean_post_self_hosted_media":"","ocean_post_video_embed":"","ocean_link_format":"","ocean_link_format_target":"self","ocean_quote_format":"","ocean_quote_format_link":"post","ocean_gallery_link_images":"off","ocean_gallery_id":[],"footnotes":""},"categories":[16],"tags":[],"class_list":["post-4227","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-iit-jee-physics-chapter-wise-approach","entry","has-media"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/acejee.com\/blog\/wp-json\/wp\/v2\/posts\/4227","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/acejee.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/acejee.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/acejee.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/acejee.com\/blog\/wp-json\/wp\/v2\/comments?post=4227"}],"version-history":[{"count":3,"href":"https:\/\/acejee.com\/blog\/wp-json\/wp\/v2\/posts\/4227\/revisions"}],"predecessor-version":[{"id":9016,"href":"https:\/\/acejee.com\/blog\/wp-json\/wp\/v2\/posts\/4227\/revisions\/9016"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/acejee.com\/blog\/wp-json\/wp\/v2\/media\/4230"}],"wp:attachment":[{"href":"https:\/\/acejee.com\/blog\/wp-json\/wp\/v2\/media?parent=4227"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/acejee.com\/blog\/wp-json\/wp\/v2\/categories?post=4227"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/acejee.com\/blog\/wp-json\/wp\/v2\/tags?post=4227"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}