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Parallel axis theorem is use to calculate moment of inertia of 

Parallel axis theorem is use to calculate moment of inertia of  Correct Answer Both (a) & (b).

Correct option:3

Concept:-

Theorems on Moment of Inertia

There are two important theorems on moment of inertia, which enable the moment of inertia of a body to be determined about any general axis.

1. Theorem of Parallel Axes.

2. Theorem of Perpendicular Axes.

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  1. Theorem of Parallel Axes- It is a very useful theorem to relate the moment of inertia of a rigid body (either two or three dimensional) about two parallel axes, in which one passes through the centre of mass.
  • Let us consider two such axes are shown in figure for a body of mass M.

 

  • If the whole mass of a rigid body is kept at same distance x or R from the axis, then moment of inertia is mx2 or mR2, where m is the mass of whole body.

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  • If the whole mass of a rigid body is kept over the axis then, moment of inertia is zero. For example, moment of inertia of a thin rod about an axis passing through the rod is zero.

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Related Questions

According to parallel axis theorem, the moment of inertia of a section about an axis parallel to the axis through center of gravity (i.e. $$I$$P) is given by (where, A = Area of the section, $$I$$G = Moment of inertia of the section about an axis passing through its C.G. and h = Distance between C.G. and the parallel axis.)
According to parallel axis theorem, the moment of inertia of a section about an axis parallel to the axis through center of gravity (i.e. $${I_{\text{P}}}$$) is given by (where, A = Area of the section, $${I_{\text{G}}}$$ = Moment of inertia of the section about an axis passing through its C.G. and h = Distance between C.G. and the parallel axis.)
The parallel axis theorem can add any angle varied moment of inertias to give the perpendicular moment of inertia and it can be used in the determination of the moment of inertia about inclined axis.
The moment of inertia of a uniform sphere of radius, r about an axis passing through its centre is given by $$\frac{2}{5}\left( {\frac{{4\pi }}{3}{r^5}\rho } \right).$$    A rigid sphere of uniform mass density $$\rho $$ and radius R has two smaller spheres of radii $$\frac{R}{2}$$ hollowed out of it as shown in the figure given below.
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