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System of Particles and Rotational Motion - Moment of Inertia and Theorems of Parallel and Perpendicular Axes

Grade 11CBSEPhysics

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

🔑Concepts

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Moment of Inertia (II): It is the rotational analogue of mass. It measures the resistance of a body to rotational motion about a specific axis. It depends on the mass of the body, its shape, size, and the distribution of mass relative to the axis of rotation.

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For a system of discrete particles, the moment of inertia is defined as I=∑miri2I = \sum m_i r_i^2, where rir_i is the perpendicular distance of the ithi^{th} particle from the axis.

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For a continuous body, I=∫r2dmI = \int r^2 dm.

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Radius of Gyration (kk): The 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. It is given by I=Mk2I = Mk^2.

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Theorem of Perpendicular Axes: Applicable to planar bodies (laminae). It states that the moment of inertia of a planar body 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) lying in its plane and intersecting at the same point: Iz=Ix+IyI_z = I_x + I_y.

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Theorem of Parallel Axes: Applicable to bodies of any shape. It states that the moment of inertia of a body about any axis (II) 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.

📐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}}

Iz=Ix+IyI_z = I_x + I_y

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

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

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

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

Isolid_sphere,center=25MR2I_{solid\_sphere, center} = \frac{2}{5}MR^2

Ihollow_sphere,center=23MR2I_{hollow\_sphere, center} = \frac{2}{3}MR^2

💡Examples

Problem 1:

Calculate the moment of inertia of a uniform disc of mass MM and radius RR about an axis passing through its tangent in the plane of the disc.

Solution:

  1. The moment of inertia of a disc about an axis passing through its center and perpendicular to its plane is Iz=12MR2I_z = \frac{1}{2}MR^2.
  2. Using the Perpendicular Axis Theorem, Iz=Ix+IyI_z = I_x + I_y. For a symmetric disc, Ix=Iy=IdiameterI_x = I_y = I_{diameter}. Thus, 12MR2=2Idiameter  ⟹  Idiameter=14MR2\frac{1}{2}MR^2 = 2I_{diameter} \implies I_{diameter} = \frac{1}{4}MR^2.
  3. Using the Parallel Axis Theorem, the moment of inertia about a tangent in the plane is Itangent=Idiameter+MR2I_{tangent} = I_{diameter} + M R^2.
  4. Itangent=14MR2+MR2=54MR2I_{tangent} = \frac{1}{4}MR^2 + MR^2 = \frac{5}{4}MR^2.

Explanation:

We first found the MOI about the diameter using the perpendicular axis theorem and then shifted that axis to the tangent using the parallel axis theorem.

Problem 2:

A thin rod of length LL and mass MM has a moment of inertia 112ML2\frac{1}{12}ML^2 about an axis passing through its center. Find the moment of inertia about an axis passing through one of its ends.

Solution:

  1. We use the Parallel Axis Theorem: I=Icm+Md2I = I_{cm} + Md^2.
  2. Here, Icm=112ML2I_{cm} = \frac{1}{12}ML^2.
  3. The distance dd between the center of the rod and one end is d=L2d = \frac{L}{2}.
  4. Iend=112ML2+M(L2)2=112ML2+14ML2I_{end} = \frac{1}{12}ML^2 + M(\frac{L}{2})^2 = \frac{1}{12}ML^2 + \frac{1}{4}ML^2.
  5. Iend=1+312ML2=412ML2=13ML2I_{end} = \frac{1+3}{12}ML^2 = \frac{4}{12}ML^2 = \frac{1}{3}ML^2.

Explanation:

The Parallel Axis Theorem allows us to find the MOI at the end by using the known value at the center of mass and the distance L2\frac{L}{2}.

Moment of Inertia and Theorems of Parallel and Perpendicular Axes Revision - Class 11 Physics CBSE