Question
Download Solution PDFMoment of inertia of a thin spherical shell of mass M and radius R about a diameter is
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Moment of inertia:
A 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.
\({\rm{I}} = \frac{2}{3}{\rm{M}}{{\rm{R}}^2}\)
- For a rigid body system, the moment of inertia is the sum of the moments of inertia of all its particles taken about the same axis.
\(I=\sum m_{i}{r_{i}}^{2}\)
where I is the Moment of Inertia, m is point mass, r is the perpendicular distance from the axis of rotation.
The moment of inertia of different bodies is given in the below table:
| Shape | Axis of rotation | Moment of inertia |
| Ring | axis passing through the center perpendicular to the plane of the ring | \(I = mr^2\) |
| Ring | axis passing through the diameter of the ring | \(I = {1 \over 2}mr^2\) |
| Solid Cylinder | axis passing through the center perpendicular to the plane of the ring | \(I = {1 \over 2}mr^2\) |
| Solid sphere | through center | \(I = {2 \over 5}mr^2\) |
| Hollow sphere | through center | \(I = {2 \over 3}mr^2\) |
| Rod | through midpoint perpendicular to the rod | \(I = {1 \over 12}ml^2\) |
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