A bicycle wheel of radius 0.3m has a rim of mass 1 kg and 50 spokes , each of mass 0.01kg. what is its moment of inertia about its axis.

							
hi pulkit 
            i think this might help u 
as we know that the moment of inertia of ring about the axis passing through its center and perpendicucal to the plane is MR2 and that of rod of length L about an aixs passing through its end and perpenducular to the length of rod is ML2/3
so by using these two bodies we can find the moment of inertia of this wheel ( considering rim as ring and spokes as rods )
we get
moment of inertia of rim as MR2 = 1*0.3*0.3 = 0.09 kg m2
and the moment of inertia of all spokes =50*ML2/3=50*0.01*0.3*0.3/3 =0.015 kg m2
so the total moment of inertia of the wheel about its axis is 0.09+0.015

						

							
WE MAY BE ABLE TO FIND THE MOMENT OF INERIA OF WHOLE OBJECT BY THE SUM OF MOMENT OF INERTIA OF THE RING AND THE SPOKES SEPERATELY:-
 
MOMENT OF INERTIA OF A RING =MR2= 1.00 X (0.3)2
                                                                                 = 0.09
 
MOMENT OF INERTIA OF THE SPOKES= MOMENT OF INERTIA OF A ROD WITH RADIUS 0.3 AND MASS 0.5(0.01 X 50)
                                                                   =MR2/3 = [0.5 X (0.3)2]/3
                                                                   = 0.015.
 
MOMENT OF INERTIA WHOLE BODY= 0.09+0.015 = 0.105KG-M