Question

Linear mass density of a thin rod of length 2L varies with distance from its centre as M= Mo| x| Moment of
inertia of the rod about a perpendicular axis through its centre is?

​

Answer 1

Answer:

Moment of Inertia  of a thin  rod about the axis perpendicular to its length and passing through its center of mass is  Ml^2/12    i.e  

I1 = Ml^2/12.........................................(a)

Now, rod of length ''l'' is formed into a circular ring say Radius ''R''

then  

2πR = l

R = l/2π

Moment of Inertia of the ring passing through the center of mass is MR^2   i.e

I2 = MR^2

(a)/(b) = I1/I2 = (Ml^2/12 )/Ml^2/4π^2  

                           = π^2/3

or 9.8/3 or 98/30 = 49/15  

For more approximation treat π^2 = 10  

thus, I1/I2 = 10/3 or 10 : 3