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Motion of System of Particles and Rigid Body - Moment of Inertia

Grade 11ICSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Moment of Inertia (II) is the rotational analogue of mass in linear motion, representing the measure of rotational inertia of a body.

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For a system of nn discrete particles, the moment of inertia is the sum of the products of the mass of each particle and the square of its perpendicular distance from the axis of rotation: I=∑miri2I = \sum m_i r_i^2.

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For a continuous rigid body, the moment of inertia is calculated using integration: I=∫r2dmI = \int r^2 dm.

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The Radius of Gyration (kk) is the radial distance from the axis of rotation to a point where the entire mass of the body could be concentrated without changing its moment of inertia: I=Mk2I = Mk^2.

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Theorem of Parallel Axes: The moment of inertia of a body about any axis is equal to the sum of its moment of inertia about a parallel axis passing through its center of mass (IcmI_{cm}) and the product of its mass (MM) and the square of the distance (dd) between the two axes: I=Icm+Md2I = I_{cm} + Md^2.

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Theorem of Perpendicular Axes: For a planar body (lamina), the moment of inertia about an axis perpendicular to its plane (IzI_z) is equal to the sum of its moments of inertia about two mutually perpendicular axes (IxI_x and IyI_y) in its plane, intersecting at the point where the perpendicular axis passes through: Iz=Ix+IyI_z = I_x + I_y.

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Factors affecting Moment of Inertia: Mass of the body, distribution of mass about the axis, and the orientation/position of the axis of rotation.

📐Formulae

I=∑i=1nmiri2I = \sum_{i=1}^{n} m_i r_i^2

I=∫r2dmI = \int r^2 dm

k=IMk = \sqrt{\frac{I}{M}}

I=Icm+Md2I = I_{cm} + Md^2

Iz=Ix+IyI_z = I_x + I_y

Iring, center=MR2I_{\text{ring, center}} = MR^2

Idisc, center=12MR2I_{\text{disc, center}} = \frac{1}{2}MR^2

Irod, center=112ML2I_{\text{rod, center}} = \frac{1}{12}ML^2

Isolid sphere=25MR2I_{\text{solid sphere}} = \frac{2}{5}MR^2

Ihollow sphere=23MR2I_{\text{hollow sphere}} = \frac{2}{3}MR^2

💡Examples

Problem 1:

Calculate the moment of inertia of a uniform rod of mass MM and length LL about an axis passing through one of its ends and perpendicular to its length.

Solution:

We know that the moment of inertia of a rod about its center is Icm=112ML2I_{cm} = \frac{1}{12}ML^2. The distance from the center to one end is d=L2d = \frac{L}{2}. Using the Parallel Axis Theorem: I=Icm+Md2=112ML2+M(L2)2=112ML2+14ML2=1+312ML2=13ML2I = I_{cm} + Md^2 = \frac{1}{12}ML^2 + M(\frac{L}{2})^2 = \frac{1}{12}ML^2 + \frac{1}{4}ML^2 = \frac{1+3}{12}ML^2 = \frac{1}{3}ML^2.

Explanation:

The Theorem of Parallel Axes is used to shift the axis of rotation from the center of mass to the end of the rod.

Problem 2:

A thin uniform disc has a mass MM and radius RR. Find its moment of inertia about a diameter.

Solution:

Let the disc lie in the XYXY plane. Due to symmetry, Ix=Iy=IdI_x = I_y = I_d (moment of inertia about a diameter). From the Theorem of Perpendicular Axes: Iz=Ix+IyI_z = I_x + I_y. We know IzI_z (about the central axis) is 12MR2\frac{1}{2}MR^2. Therefore, 12MR2=Id+Id=2Id  ⟹  Id=14MR2\frac{1}{2}MR^2 = I_d + I_d = 2I_d \implies I_d = \frac{1}{4}MR^2.

Explanation:

The Perpendicular Axis Theorem is applied to a 2D object (disc) where the ZZ-axis is the symmetry axis through the center.

Problem 3:

Find the radius of gyration of a solid sphere of radius RR about its diameter.

Solution:

The moment of inertia of a solid sphere about its diameter is I=25MR2I = \frac{2}{5}MR^2. We also have I=Mk2I = Mk^2. Equating the two: Mk2=25MR2  ⟹  k2=25R2  ⟹  k=25RMk^2 = \frac{2}{5}MR^2 \implies k^2 = \frac{2}{5}R^2 \implies k = \sqrt{\frac{2}{5}}R.

Explanation:

The radius of gyration is found by setting the standard formula for II equal to Mk2Mk^2 and solving for kk.

Moment of Inertia - Revision Notes & Key Formulas | ICSE Class 11 Physics