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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.