Question

(a) Prove the theorem of perpendicular axes.
(Hint: square of the distance of a point (x,y) in the xy plane from an
axis through the origin and perpendicular to the plane is \[{x^2} + {y^2}\]
(b) Prove the theorem of parallel axes.
(Hint: If the centre of mass is chosen to be the origin \[\sum {{m_i}{r_i} = 0} \]

Answer 1

Hint: From the moment of inertia of the particle about different axis in a plane, moment of inertia of the whole lamina can be found about the axis perpendicular to the plane.

Complete step by step answer:
(a) Let perpendicular axes x, y and z (which meet at origin o) so that the body lies in the xy plane and z-axis is perpendicular to the plane of the body.
Let \[{I_x},\,\,{I_y},\,\,{I_z}\] be moments of inertia about axis x, y, z respectively the perpendicular axis theorem states that
\[{I_z}\,\, = \,\,{I_x}\,\,\, + \,\,{I_y}\]
Proof:
\[{I_x}\,\, = \,\,\int {{y^2}dx} \] because the distance of point (x, y) from x-axis is y.
Similarly,
\[{I_y}\,\, = \,\,\int {{x^2}dm} \]
\[{I_z}\,\, = \,\,\int {\left( {{x^2} + {y^2}} \right)dm} \]
\[ \Rightarrow \,\,{I_z}\,\, = \,\,{I_x}\,\, + \,\,{I_y}\]

Additional Information:
The perpendicular axis theorem status that the moment of inertia of a planar lamina about an axis perpendicular to the plane of the lamina is equal to the sum of the moment of inertia of the lamina about any of the two mutually perpendicular axis in it’s own plane such that they intersect each other at a point from where perpendicular axes pass through the lamina.
(b) Suppose a body of mass m is made to rotate about an axis z passing through the body’s centre of gravity. The body has a moment of inertia \[{I_{cm}}\]with respect to this axis. The parallel axis theorem states that if the body is made to rotate instead about a new axis z, which is parallel to the first axis and displaced from it by a distance d, then the moment of inertia I with respect to the new axis related to \[{I_{cm}}\] by
\[{I_{cm}}\,\, = \,\,{I_{cm}} + m{d^2}\]
Proof:
\[{I_{cm}}\,\, = \,\,\int {{r^2}dm} \]
\[I\,\, = \,\,\int {{{\left( {r\,\, \pm \,\,d} \right)}^2}dm} \]
\[ = \,\,\int {{r^2}\,\,dm} \,\,\, + \,\,\, = \,\,\int {{d^2}\,\,dm} \,\, \pm \,\, = \,\,\int {2dr\,\,dm} \]
\[ = \,\,{l_{cm}}\,\, + \,\,m{d^2}\]
Last term it is zero since it gives a centre of mass.

Note:
The theorem of parallel axis states that the moment of inertia of a body about an axis parallel to an axis passing through the centre of mass is equal to the moment of inertia of body about centre of mass plus product of mass of body and square of perpendicular distance between the two parallel axes.