Figure 1 of 1 4 ft 1 ft SolutionGiven data Mass of the outer

Solution

Given data:

Mass of the outer rim =M=159 lb (r=4 ft)

Mass of each spoke = 24lb (l=3ft)

Mass of inner rim = m=23 lb (r=1ft)

From the figure we can conclude that the total moment of inertia of wheel about its centre will be

I total= 8(I spokes)+Mr²+mr²

But in question it is given that moment of inertia passing through A.Then changed moment of inertia will be

I total (A)=8(Ml²/3)+2Mr²+2mr²

  8*(24*9/3)+(2*159*16)+(2*23*1)

= 5710 lb -ft²