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Theorems of Perpendicular and Parallel Axes

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Topics

Estimated time: 7 minutes
  • Theorem of Perpendicular Axes
  • Theorem of Parallel Axes
  • Application of perpendicular and parallel axes theorem on different regular bodies
Maharashtra State Board: Class 11

Theorem: Perpendicular Axis Theorem

Statement: The moment of inertia (Iz​) of a laminar object about an axis (z) perpendicular to its plane is equal to the sum of its moment of inertias about two mutually perpendicular axes (x and y) in its plane, all three axes being concurrent.

Iz = Ix + Iy
Maharashtra State Board: Class 11

Theorem: Parallel Axis Theorem

Statement: The moment of inertia (Io​) of an object about any axis is equal to the sum of its moment of inertia (Ic) about an axis parallel to the given axis and passing through the centre of mass, and the product of the mass of the object and the square of the distance between the two axes.

Io = Ic + Mh2

where M = mass of the object, h = distance between the two parallel axes.

Maharashtra State Board: Class 11

Key Points: M.I. of Symmetrical Objects

  • Thin ring / Hollow cylinder (Central axis) → I = MR2
  • Thin ring (Diameter) → I = \[\frac {1}{2}\]M R2
  • Annular ring / Thick hollow cylinder (Central) → I = \[\frac{1}{2}M (R_1^2+R_2^2)\]
  • Uniform disc / Solid cylinder (Central) → I = \[\frac {1}{2}\]M R2
  • Uniform disc (Diameter) → I = \[\frac {1}{4}\]M R2
  • Solid sphere (Central) → I = \[\frac {2}{5}\]M R2
  • Thin walled hollow sphere (Central) → I = \[\frac {2}{3}\]M R2
  • Thin rod (Centre, ⊥ to length) → I = \[\frac {1}{12}\]M L2
  • Thin rod (One end, ⊥ to length) → I = \[\frac {1}{3}\]M L2
  • Solid cone (Central) → I = \[\frac {3}{10}\]M R2, Hollow cone (Central) → I = \[\frac {1}{2}\]M R2

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Principle of parallel and perpendicular axes [00:15:57]
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