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Q7.26 (a)     Prove the theorem of perpendicular axes. (Hint: The square of the distance of a point (x,y)  in the x-y     plane from an axis through the origin and perpendicular to the plane is (x^{2}+y^{2}).

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Consider the figure given below:  

                                       Rotational motion, 20190

The moment of inertia about the x-axis is given by :

                                                   I_x\ =\ mx^2

And the moment of inertia about the y-axis is :

                                                   I_y\ =\ my^2

Now about z-axis : 

                                        I_z\ =\ m\left ( \sqrt{\left ( x^2\ +\ y^2 \right )} \right )^2

or                                    I_z\ =\ I_x\ +\ I_y

Posted by

Devendra Khairwa

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