Moment of inertia of a thin spherical shell of mass M and radius R, about its diameter is

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UPPSC AE Mechanical 2019 Official Paper I (Held on 13 Dec 2020)
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  1. \(MR^2\)
  2. \(\frac{1}{2}MR^2\)
  3. \(\frac{2}{5}M{R^2}\)
  4. \(\frac{2}{3}M{R^2}\)

Answer (Detailed Solution Below)

Option 4 : \(\frac{2}{3}M{R^2}\)
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Detailed Solution

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Explanation:

Moment of inertia:

Moment of inertia is a measure of the resistance of a body to angular acceleration about a given axis that is equal to the sum of the products of each element of mass in the body and the square of the element’s distance from the axis.

I = ∑( m1r12 + m2r22 + m3r32 +m4r42 + …….. + mnrn2)

Moment of inertia of a thin spherical shell of mass M and radius R about its diameter.

\({\rm{I}} = \frac{2}{3}{\rm{M}}{{\rm{R}}^2}\)

Additional Information

Moment of inertia of some important shapes:

Body

Axis of Rotation

Moment of inertia

Uniform circular ring of radius R

Perpendicular to its plane and through the center

MR2

Uniform circular ring of radius R

About diameter

\(\frac{MR^2}{2}\)

Uniform circular disc of radius R Perpendicular to its plane and through the center \(\frac{MR^2}{2}\)
Uniform circular disc of radius R About diameter \(\frac{MR^2}{4}\)

A solid cylinder of radius R

Axis of the cylinder

\(\frac{MR^2}{2}\)

A hollow cylinder of radius R Axis of cylinder MR2
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