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  • Given the moment of inertia of a disc of mass M and radius R about any of its diameters to be MR^ 2 /4, find its moment of inertia about an axis normal to the disc and passing through a point on its edge.

Q 7.10(b)     Given the moment of inertia of a disc of mass M and radius R  about any of its
                    diameters to be MR^{2}/4 , find its moment of inertia about an axis normal to the
                   disc and passing through a point on its edge.

Answers (1)

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We know that moment of inertia of a disc about its diameter is :

                                                                                                          =\ \frac{1}{4}MR^2

Using perpendicular axes theorem we can write :

Moment of inertia of disc about its centre:-  

                                                                                      =\ \frac{1}{4}MR^2\ +\ \frac{1}{4}MR^2\ =\ \frac{1}{2}MR^2

Using parallel axes theorem we can find the required MI :

Moment of inertia about an axis normal to the disc and passing through a point on its edge is given by :

                                                                                       =\ \frac{1}{2}MR^2\ +\ MR^2\ =\ \frac{3}{2}MR^2           

Posted by

Devendra Khairwa

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