using System; using L=Science.Physics.GeneralPhysics; namespace Serway.Chapter10 { /// /// Example08: Applying the Parallel-Axis Theorem /// Consider once again the uniform rigid rod of mass M and /// length L shown in Figure 10.10. Find the moment of inertia /// of the rod about an axis perpendicular to the rod through /// one end (the y's axis in Fig.10.10). /// I = 1/3 ML^2 /// public class Example08 { public Example08() { } private string result; public string Result { get{return result;} } public void Compute() { L.Mass m = new L.Mass(); m.kg = 10.0; L.RigidBody rod = new L.RigidBody(m); L.Length l = new L.Length(); l.m = 5.0; rod.SetRodAboutCenter(l); L.MomentOfInertia I = rod.RigidBodyMomentOfInertia; L.Length lh = new L.Length(); lh.m = l.m/2.0; L.MomentOfInertia.ParallelAxisTheorem(I,L.UnitVector.j,m,lh); result+=I.ToString(); } } } /* 20.8333333333333 +/- 0 +0 +/- 0 +0 +/- 0 0 +/- 0 +83.3333333333333 +/- 0 +0 +/- 0 0 +/- 0 +0 +/- 0 +0 +/- 0 (kgm^2) */