Question

(b) Given the moment of inertia of a disc of massĀ MĀ and radiusĀ RĀ about any of itsĀ diameters to beĀ $MR_{2}/4$, find its moment of inertia about an axis normal to theĀ disc and passing through a point on its edge.

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Updated on : 2022-09-05

Solution

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(a)Ā

The moment of inertia (M.I.) of a sphere about its diameter$=2MR_{2}/5$

According to the theorem of parallel axes, the moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes.

The M.I. about a tangent of the sphere =$2MR_{2}/5+MR_{2}=7MR_{2}/5$

(b)Ā

The moment of inertia of a disc about its diameter =$MR_{2}/4$

According to the theorem of perpendicular axis, the moment of inertia of a planar body (lamina) about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes concurrent with perpendicular axis and lying in the plane of the body.

The M.I. of the disc about an axis normal to the disc through the center $=MR_{2}/4+MR_{2}/4=MR_{2}/2$

Now, Applying the theorem of parallel axes:

The moment of inertia about an axis normal to the disc and passing through a point on its edgeĀ

$=MR_{2}/2+MR_{2}=3MR_{2}/2$

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