Question

(a) Prove the theorem of perpendicular axes. (Hint: Square of the distance of a point (x, y) in the x–y plane from an axis through the origin perpendicular to the plane is x2 + y2). (b) Prove the theorem of parallel axes. (Hint: If the centre of mass is chosen to be the origin ). question7.26 System Of Particles And Rotational Motion

Answer 1

perpendicular axis theorem:-acording to this theorem, the moment of inertia of a plane lamina about any axis perpendicular to the plane of the lamina is equal to sum of the moments of lamina about any two mutual perpendicular axes in the plane of the lamina ,meeting at a point where the given axis passes through the lamina .
  for a plane lamina, mathematically, it states that.
          Iz = Ix + Iy 

proof: let consider the lamina to made of the large number of particles each of mass m .let a consider of the particle at m in x-y plane.
      ∴ moment of inertia of this particle about OZ axis = mr²
     if Iz be the moment of inertia of the whole lamina about z -axis then,
          Iz = ∑mr²
              =∑m(x² + y²)
              =∑mx² +∑my² ________________(1)
also let Ix and Iy be the moment of inertia of the whole lamina about x and y axis respectively.
   Ix =∑my²
   Iy = ∑mx²

  Ix + Iy = ∑my² +∑mx²_______________________(2)
 from eqaution (1) and (2)
Iz = Ix +Iy proved 

parallel axis theorem:- according to this theorem, moment of inertia of a rigid body about any axis is equal  to moment of inertia of the body about another axis passing through centre of mass of the body in a direction parallel to the given axis, thus the product of total mass m of the body and square of the perpendicular distance between the two parallel axes .it states that 
            I = Ic + mh²
Ic = moment of inertia of the body about an axis through CG
I= moment of inertia parallel to CG
M = total mass of the body made by n particles =∑mi

proof :-  consider the particle located at a point P in the body at a distance ri from axis LL' passing through CG
 ∴ its distance from zz' axis = OP = OC + CP = h + r1 
if Ic = moment of inertia of the whole body about LL' axis 
then,   Ic = ∑miri²
if Ii be its moment of inertia about ZZ' axis then,
        Ii = mi(op)²
         I  = ∑Ii
            =∑mi(ri + h)²
           =∑miri² + ∑mih²+2h∑miri
           = Ic + Mh² + 0
           = Ic + Mh² proved