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Motion of System of Particles and Rigid Body - Theorems of Parallel and Perpendicular Axes

Grade 11ICSEPhysics

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

🔑Concepts

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A rigid body is a system of particles in which the distance between any two constituent particles remains constant under the influence of external forces.

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The Moment of Inertia (II) of a rigid body about a given axis is the sum of the products of the masses of its particles and the squares of their respective distances from the axis of rotation.

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The Theorem of Parallel Axes states that the moment of inertia (II) 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.

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The Theorem of Perpendicular Axes is applicable only to planar (laminar) bodies. It states that the moment of inertia (IzI_z) of a plane lamina about an axis perpendicular to its plane is equal to the sum of the moments of inertia about two mutually perpendicular axes (IxI_x and IyI_y) lying in its plane and intersecting at the point where the perpendicular axis passes through the lamina.

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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 can be assumed to be concentrated such that the moment of inertia remains the same: I=Mk2I = Mk^2.

📐Formulae

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

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

Iz=Ix+IyI_z = I_x + I_y

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

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

Iring,center=MR2I_{ring, center} = 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 its diameter is Idiameter=14MR2I_{diameter} = \frac{1}{4}MR^2.
  2. Using the Theorem of Parallel Axes: Itangent=Idiameter+Md2I_{tangent} = I_{diameter} + Md^2.
  3. Here, the distance between the center of mass (diameter) and the tangent is d=Rd = R.
  4. Itangent=14MR2+M(R)2=54MR2I_{tangent} = \frac{1}{4}MR^2 + M(R)^2 = \frac{5}{4}MR^2.

Explanation:

We first identify the moment of inertia about a central axis parallel to the required axis (the diameter) and then apply the parallel axis theorem where the shift distance is the radius RR.

Problem 2:

Given that the moment of inertia of a thin circular ring of mass MM and radius RR about an axis passing through its center and perpendicular to its plane is MR2MR^2, find its moment of inertia about its diameter.

Solution:

  1. Let Iz=MR2I_z = MR^2 be the M.I. about the axis perpendicular to the plane.
  2. Let IxI_x and IyI_y be the moments of inertia about two perpendicular diameters. By symmetry, Ix=Iy=IdI_x = I_y = I_d.
  3. According to the Theorem of Perpendicular Axes: Iz=Ix+IyI_z = I_x + I_y.
  4. MR2=Id+Id=2IdMR^2 = I_d + I_d = 2I_d.
  5. Therefore, Id=12MR2I_d = \frac{1}{2}MR^2.

Explanation:

Since a ring is a planar object, we can use the perpendicular axis theorem. Symmetry dictates that the resistance to rotation is identical for any diameter in the plane.

Theorems of Parallel and Perpendicular Axes Revision - Class 11 Physics ICSE