Parallel and Perpendicular Axis Theorem

MOMENT OF INERTIA FOR DIFFERENT BODIES:

I) FOR THIN UNIFORM ROD:

a) About an axis passing through the centre of mass and perpendicular to its length:
Consider a thin uniform rod of length  and mass M. Let AB be the axis of rotation passing through its centre of mass, C and perpendicular to the length of the rod. Its moment of inertia about the given axis is 
                 
               
  Radius of gyration about this axis.
                K= \( \frac{l} {{\sqrt {12} }} \)  

b) About an axis passing through one end of the rod and perpendicular to its length.
Let A'B' be the axis of rotation passing through one end of the rod and perpendicular to its length. As we know that the moment of inertia of the rod about a parallel axis through its centre of mass,\( I_c = \frac{{Ml^2 }} {{12}} \) we can apply parallel axes theorem.
Moment of inertia of the rod about the axis A'B' Moment of inertia of the rod about AB + \( Mr^2 \)
         
       
            = \( \frac{{Ml^2 }} {{12}} + M\left( {\frac{l} {2}} \right)^2 \left( {\because r = \frac{l} {2}} \right) \)
         \( I = \frac{{Ml^2 }} {{12}} + \frac{{Ml^2 }} {4} = Ml^2 \left( {\frac{1} {{12}} + \frac{1} {4}} \right) = \frac{{Ml^2 }} {3} \)  
Radius of gyration,  
                       \( K=\sqrt {\frac{1} {M}} = \sqrt {\frac{{Ml^2 }} {{3M}}} = \frac{l} {{\sqrt 3 }} \)

Parallel and Perpendicular Axis

II) FOR A THIN UNIFORM CIRCULAR

a) About an axis passing through its centre and perpendicular to its plane or about a transverse axis
Consider a circular ring of mass M and radius R. Its moment of inerta about an axis passing through its centre and perpendicular to its plane is MR²
                   
   
                  \( \therefore I = MR^2 \)
        Radius of gyration K = R

b)About a tangent in the perpendicular Plane Consider a tangent in the perpendicular plane to the circular ring as the axis of rotation. The moment of inertia of the circular ring about a parallel axis through the centre of mass i.e., centre of the ring \( I_c = MR^2 \)
                      

By parallel axes theorem, moment of inertia about a tangent perpendicular to the plane of the ring is  \( I = I_c + MR^2 \)
         \( \therefore I_c = MR^2 \)                I=MR²+MR²
(r= distance between the tangent and centre = R)
            I=2MR²
Radius of gyration,
            K= \( \sqrt {\frac{1} {M}} = \sqrt {\frac{{2MR^2 }} {M}} = \sqrt 2 R \)  

c)About a Diameter
The moment of inertia of a circular ring about any diameter is same. Consider two diameters  and YY' of the circular ring. These two diameters are perpendicular to each other and intersect at the centre 'O'.
Moment of inertia of the circular ring about the diameter \( XX^1 \) = I_x
Moment of inertia of the circular ring about the diameter YY'= I_y
But  I_x=I_y=I due to symmetry.
By perpendicular axes theorem, the sum of I_x and I_y is equal to the moment of inertia of the circular ring about the axis ZZ' passing through the intersecting point O and perpendicular to its plane,  I_z=MR² .
            I_x+I_y = I_z
            I+I= MR²  
                          

           2I= \( MR^2 \Rightarrow I = \frac{{MR^2 }} {2} \)
        Radius of gyration,K= \( \sqrt {\frac{1} {M}} = \sqrt {\frac{{MR^2 }} {{2M}}} = \frac{R} {{\sqrt 2 }} \)

d) About a Tangent in its Plane
The tangent in the plane of the circular ring is parallel to the diameter. As we know that the moment of inertia of the circular ring about a diameter is
                         
   
We can find the moment of inertia about a tangent in the plane of the ring by parallel axes. theorem. Moment of inertia about A'B' = Moment of inertia about AB through its centre of mass +\( MR^2 \)
                   I=\( \frac{{MR^2 }} {2} \)+\( MR^2 \) (as r=R)

          \( I = \frac{{3MR^2 }} {2} \)      
Radius of gyration,
            K= \( \sqrt {\frac{1} {M}} = \sqrt {\frac{{3MR^2 }} {{2M}}} = \sqrt {\frac{3} {2}} R \)                  

Parallel and Perpendicular Axis Theorem

III) FOR THIN UNIFORM DISC

a) About an axis passing through its centre and perpendicular to its plane or about a transverse axis.
Consider a uniform thin circular disc of mass M and radius R. Its moment of inertia about an axis passing through its centre and perpendicular MR2
to its plane is  \( \frac{{MR^2 }} {2} \)
                    \( \therefore I = \frac{{MR^2 }} {2} \)
Radius of gyration,about this axis is K= \( \frac{R} {{\sqrt 2 }} \)

b) About a tangent perpendicular to its plane Consider a tangent in a perpendicular plane to the circular disc as the axis of rotation. The moment of inertia of the circular disc about a parallel axis through the centre of mass i.e., centre of the disc                 
                     \( I_c = \frac{{MR^2 }} {2} \)
By parallel axes theorem, moment of inertia about a tangent perpendicular to its plane is 
            I= \( I_c + MR^2 \)
            I= \( MR^2 + \frac{{MR^2 }} {2} = \frac{{3MR^2 }} {2} \)
i.e.,            I= \( \frac{{3MR^2 }} {2} \)
(r distance between the two parallel axes = R)
Radius of gyration,
                \( K = \sqrt {\frac{I} {M}} = \sqrt {\frac{{3MR^2 }} {{2M}}} = \sqrt {\frac{3} {2}R} \)  

c) About a diameter
Moment of inertia of a circular disc about any diameter is same. Consider two diameters XX' and YY' of the circular disc. These two diameters are perpendicular to each other and intersect at the centre 'O'.
    Moment of intertia of the circular disc about XX' = I_x
    Moment of inertia of the circular disc about  YY' = I_y
    But   I_x= I_y = I due to symmetry.
By perpendicular axes theorem, the sum of  I_x, and I_y  is equal to the moment of inertia of the circular disc about the axis ZZ' passing through the intersecting point 'O' and perpendicular to its plane is,  \( I_z = \frac{{MR^2 }} {2} \)   
                
              
              \( I_x + I_y = I_z \)
                I+1= \( \frac{{MR^2 }} {2} \)
                2I= \( \frac{{MR^2 }} {2} \)
                I= \( \frac{{MR^2 }} {4} \)
Radius of gyration,  \( K = \sqrt {\frac{I} {M}} = \sqrt {\frac{{MR^2 }} {{4M}}} = \frac{R} {2} \)

d) About a tangent in its plane
The tangent in the plane of the circular disc is parallel to the diameter. As we know that the moment of inertia of the circular disc about the diameter is  \( \frac{{MR^2 }} {4} \), we can find the moment of inertia about a tangent in the plane of the disc by parallel axes theorem.
            I=\( I_c + MR^2 \)(r=R)
            =\( \frac{{MR^2 }} {4} \)+\( MR^2 \)=\( \frac{{5MR^2 }} {4} \)  i.e.,I=\( \frac{{5MR^2 }} {4} \)
Radius of gyration,  \( K = \sqrt {\frac{I} {M}} = \sqrt {\frac{{5MR^2 }} {{4M}}} = \frac{{\sqrt {5R} }} {2} \)  

Parallel and Perpendicular Axis Theorem

IV) FOR A THIN UNIFORM RECTANGULAR LAMINA

a) Moment of inertia of a rectangular about an axis passing through its centr perpendicular to its length is same as that of the moment of inertia of a thin uniform rod of length ! and an axis passing through its centre and perpendicular to its length i.e...
                                 \( I_y = \frac{{Ml^2 }} {{12}} \)
b) The moment of inertia of the thin rectangular lamina about an axis passing  through its centre and perpendicular to its breadth,
                               
     
c)Moment of inertia of the thin rectangular lamina about an axis of rotation passing through its centre and perpendicular to its plane can be obtained by using perpendicular axes theorem. The axes given in (a) and (b) above are in the plane of the lamina and perpendicular to each other intersecting at the centre. The moments of inertia about the two axes can be taken as
            \( I_x = \frac{{Mb^2 }} {{12}} \)  and \( I_y = \frac{{Ml^2 }} {{12}} \)
The axis passing through its centre and perpendicular to its plane can be taken as z-axis so that  
           \( I_z = I_x + I_y \) = \( M\left( {\frac{{l^2 + b^2 }} {{12}}} \right) \)  

Parallel and Perpendicular Axis Theorem

V) FOR A UNIFORM SOLID SPHERE

a) About an axis passing through its diameter 
Consider a solid sphere of mass M and radius R. Its moment of inertia about an axis of rotation passing through its diameter is
                  \( I = \frac{2} {5}MR^2 \)         
                      
           
b) About an axis passing through its tangent
Let \( A^1 B^1 \)  the tangent to the solid sphere. A parallel axis through its centre of mass is  AB. By parallel axes theorem,
Moment of inertia about the tangent = Moment of inertia about a diameter + \( MR^2 \)
                  \( \begin{gathered} I = \frac{2} {5}MR^2 + MR^2 \hfill \\ I = \frac{7} {5}MR^2 \hfill \\ \end{gathered} \)   
         

Parallel and Perpendicular Axis Theorem

VI) FOR A HOLLOW SPHERE(OR) SPHERICAL SHELL

a) Moment of inertia about an axis passing through the diameter of a hollow sphere of mass M and radius R is
                    \( I = \frac{2} {3}MR^2 \)        
                

b) Moment of inertia about an axis passing through its tangent can be obtained by applying parallel axes theorem. It is given by
                 \( \begin{gathered} I = I_c + MR^2 \hfill \\ I = \frac{2} {3}MR^2 + MR^2 \hfill \\ I = \frac{5} {3}MR^2 \hfill \\ \end{gathered} \)    
                           

Parallel and Perpendicular Axis Theorem

VII) FOR A UNIFORM SOLID CYLINDER

a) About its own axis i.e., axis of symmetry. A solid cyulinder of mass M can be supposed to be consisting of a large number of circular discs placed one over the other. The moment of inertia of each such disc of mass m is \( \frac{{MR^2 }} {2} \) and the moment of inertia of the solid cylinder is the sum of the moments of intertia of such discs.
.. Moment of inertia of a solid cylinder about its own axis =\( \sum m \frac{{R^2 }} {2} = \frac{{R^2 }} {2}\sum m \)
                  \( \therefore I = \frac{{MR^2 }} {2} \)  
            

It is same as that of a circular disc.

b) Moment of inertia about an axis perpendicular to its geometrical axis and passing through its centre of mass is
                                 \( I = M\left( {\frac{{l^2 }} {{12}} + \frac{{R^2 }} {4}} \right) \)   
                       

Parallel and Perpendicular

VIII)FOR A UNIFORM HOLLOW CYLINDER 

a) Its moment of inertia about its own axis is
                     \( I = MR^2 \)  
              

It is same as that of a circular ring.

b) Moment of inertia about an axis perpendicular to its geometrical axis and passing through its centre of mass is
                                   \( I = M\left( {\frac{{l^2 }} {{12}} + \frac{{R^2 }} {2}} \right) \)  
                                     

Parallel and Perpendicular Axis Theorem

MOMENT OF INERTIA OF SOME  RIGID BODIES

Parallel and Perpendicular Axis Theorem