12.2 The moment of inertia of a rigid body against an axisExamplel As shown in the figure, given the mass of thehomogeneous rod is Mand the moment of inertia oftheshaft z, is J , find the moment of inertia of the rodbapairJ2CSolution: J, =J,c + Md2, obtainJ, = Jzc + Ma?(1)J, = Jzc + Mb(2)(2) -(1)obtainJ, = J, + M(b? -a?
C 1z z 2 z a b Example1 As shown in the figure, given the mass of the homogeneous rod is , and the moment of inertia of the shaft is , find the moment of inertia of the rod pair . M 1z 1 J 2 J Solution： ，obtain 2 Jz JzC Md 2 J1 JzC Ma （1） 2 J2 JzC Mb （2） (2) (1)obtain ( ) 2 2 2 1 J J M b a 12.2 The moment of inertia of a rigid body against an axis
12.2 The moment of inertia of a rigid body against an axis4、The moment of inertia of the composite rigid bodyExample2 The dimensions of homogeneous right-angle0folding rod are shown in the figure, its mass is 3m, and21the moment of inertia of the rod against the axis O isAB21calculated.Jo= JoA + JAB =1(2m)(21)* + (2m)(V21)ml312= 5ml?AOBBA2121
2l O A B l 2l O A B l 2l Example2 The dimensions of homogeneous right-angle folding rod are shown in the figure, its mass is , and the moment of inertia of the rod against the axis is calculated. 3m O 2 2 2 2 5 (2 )(2 ) (2 )( 2 ) 12 1 3 1 ml ml m l m l J O JOA J AB O A B l 2l 4、The moment of inertia of the composite rigid body 12.2 The moment of inertia of a rigid body against an axis
12.2 The moment of inertia of a rigid body against an axisZExample3 As shown in the figure, the outer diameter R, andinner diameter R of the homogeneous hollow cylinder withmass is m , and the moment of inertia of the central axis zis calculated.Solution: The hollow cylinder can be regarded as composed oftwo solid cylinders. The moment of inertia of the outercylinder is Jin,andthe moment of inertia ofthe inner cylinderisJournegatively.2'RJ.=JJout - JiyLet, m, and m, respectively be the mass of the outer and inner cylinder, then1m,R?jm,R?1Jin :out221SO,mRm,R,T22
z 2R1 2R2 l Example3 As shown in the figure, the outer diameter and inner diameter of the homogeneous hollow cylinder with mass is , and the moment of inertia of the central axis is calculated. m z R1 R2 z in J J J out Solution: The hollow cylinder can be regarded as composed of two solid cylinders. The moment of inertia of the outer cylinder is , and the moment of inertia of the inner cylinder is negatively. Jout in J Let, m1 and m2 respectively be the mass of the outer and inner cylinder, then 2 out 1 1 2 1 J m R 2 2 2 i 2 1 J n m R so, 2 2 2 2 1 1 2 1 2 1 Jz m R m R 12.2 The moment of inertia of a rigid body against an axis
12.2 The moment of inertia of a rigid body against an axisLet the mass per unit volume be P, thenm = p元R,1m, = p元R;l:=-pa(R, -R:)Substitute into thepreceding formula and get1pd(R - R:)(R + R)noticepl(R2-R)=m , obtain1=m(R2 + R2)1二2
Let the mass per unit volume be , then m R l2 1 1 m R l2 2 2 Substitute into the preceding formula and get ( )( ) 2 1 ( ) 2 1 2 2 2 1 2 2 2 1 4 2 4 1 l R R R R Jz l R R l(R R ) m 2 2 2 1 notice ，obtain ( ) 2 1 2 2 2 J z m R1 R 12.2 The moment of inertia of a rigid body against an axis