EXAMPLE: ROLLING CYLINDER
b) This portion of the problem is resolved with the parallel axis theorem: IOzz = ICzz + md2. Recalling that x2 + y2 = r2, and assuming that the plane of the example figure is the mid plane of the cylinder, the triple integral that describes the moment of inertia can be formed and solved as:
IOzz  =  (T)/(2) − (T)/(2)2π0R0(x2 + y2)rρ01 − (r)/(2R)drdθdz + R2ρ0dV  =  (T)/(2) − (T)/(2)2π0R0ρ0r31 − (r)/(2R)drdθdz + R2(T)/(2) − (T)/(2)2π0R0ρ0r − (r2)/(2R)drdθdz  =  T2πρ0R0r3 − (r4)/(2R)dr + R2ρ02πTR0r − (r2)/(2R)dr  =  T2πρ0(R4)/(4) − (R4)/(10) + (R4)/(2) − (R4)/(6)  =  R42πρ0T(3)/(4) − (3)/(30) − (5)/(30) = 2R4πρ0T(29)/(60) = (29)/(30)R4πρ0T
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