MTH622 Final Term Solved Papers by Stylish and Sahir

Solution:

Let  be the mass and
the radius of the ring. Then the mass per unit length will be  22

We regard this ring to be composed of small elements of mass ( ) each of length  ,

we can write it as 22
⟹  =  22


Moment of inertia of the element about an axis through center O and the perpendicular to the

plane of the ring equals 2 4


Therefore the M.I of the whole ring will be

22
4
  × 22
= 4


Hence is the required moment of inertia of the ring. 4
 the end point O. Let and  be the mass and the length of the specific element of the rod at a distance  from

Then

Then the moment of inertia of this element about the given axis is

Hence the moment of inertia of the whole rod will be 3 

= 1 3