《理论力学》课程教学资源（PPT课件，英文）Chapter 12 Theorem of moment of momentum

12.2 The moment of inertia of a rigid body against an axis2.3. The moment of inertia of a thin circular plate1The z-axis is perpendicular totheplane of theplatep= M/πR?dr :dm = p.2rdrxI,=[r'dm= f"r2.2=二元pR2To.rdr:mass:M , radius:R2↑z=MR?Let's take its center2axis,z2.4、The moment of inertia of a cylinderm,)R2=m;R2=三(=>MR22mass: M , radius:R

x y r R dr 2.3、The moment of inertia of a thin circular plate 2 dm    2rdr   M R 4 0 2 2 2 1 J r dm r 2 rdr R R z          12.2 The moment of inertia of a rigid body against an axis The z-axis is perpendicular to the plane of the plate mass：M ，radius：R dr : 2 2 1 Jz  MR 2.4、The moment of inertia of a cylinder z Let's take its center axis, z 2 2 2 2 1 ( ) 2 1 2 1 Jz   miR   mi R  MR mass：M ，radius：R

12.2 The moment of inertia of a rigid body against an axistZ2.5、 The moment of inertia of a thin planeLet me draw my coordinate axis. Take any microsurfaceyelement AS, and its mass is m, , thenJASxJ, =Em,y?J,=Zm,x,J, =Em,r? =Em,(x? +y?)=Em,x? +Em,yThus, the relation between the moment of inertia of the thin plate on the threecoordinateaxesisobtainedasfollows:J.=Jx+JFor a thin circular plate, notice its symmetry with respect to diameter, there is1MR2J.-24

2.5、The moment of inertia of a thin plane Let me draw my coordinate axis. Take any microsurface element Si and its mass is , then mi Si x y z O i x i y ir 2 y i i J   m x 2 x i i J   m y 2 2 2 2 2 ( ) z i i i i i i i i i J   m r   m x  y   m x   m y Thus, the relation between the moment of inertia of the thin plate on the three coordinate axes is obtained as follows: z x y J  J  J For a thin circular plate, notice its symmetry with respect to diameter, there is 2 4 1 2 1 J x  J y  J z  MR 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 axisty2.6, The moment of inertia of a rectangular thin plate%Let the quality of the board be M , thenx%1Maaa1212221Mb?Jin a similar wayx12And its moment of inertia for the centroid perpendicular to the plane of theplate isM(a? + b?)J=+12

x y 2 a 2 a 2 b 2 b 2.6、The moment of inertia of a rectangular thin plate Let the quality of the board be M , then 2 2 2 12 1 ( ) 12 1 12 1 J m a m a Ma y   i   i  in a similar way 2 12 1 J x  Mb And its moment of inertia for the centroid perpendicular to the plane of the plate is ( ) 12 1 2 2 J z  J x  J y  M a  b 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 axis3, Parallelaxistheorem for moment ofinertiaTheorem:the momentof inertia oftherigid bodyfor any axis is equal tothemoment of inertia oftherigid body for theaxis parallel totheaxis passing throughthe center of mass plus theproduct of themassof therigid body and the square ofthe distance between the two axes, namelyJ, = Jc + Md2It can be seen from the theorem that the moment of inertia of a rigidbody for all parallel axes is the smallest over the centroid axis

3、Parallel axis theorem for moment of inertia Theorem: the moment of inertia of the rigid body for any axis is equal to the moment of inertia of the rigid body for the axis parallel to the axis passing through the center of mass plus the product of the mass of the rigid body and the square of the distance between the two axes, namely 2 J z  J zC  Md 12.2 The moment of inertia of a rigid body against an axis It can be seen from the theorem that the moment of inertia of a rigid body for all parallel axes is the smallest over the centroid axis

12.2 The moment of inertia of a rigid body against an axisProve: as shown in the figure, make rectangularZZcoordinate system,thenJec =Em,ri =Zm,(x, + yi)rMdJ, =Em,r? =Em,(x? +y)O=Xdue to x, = Xi, y, = Yi; + d, soVyJ, = Em,| x +(yi, +d)?=Zm,(xi, + yi)+2dEm,yi, +d?mEm,y; , When the origin is atFrom the centroid coordinates formula yc =Mthe center of mass CJc =0Zm,=MZm,yl, = 0J, = Jc + Md?

x 1 x ( )1 y y z 1 z 1 x  x y 1 y 1 z  z 1r r d Mi O C Prove: as shown in the figure, make rectangular coordinate system, then 2 2 2 1 1 1 ( ) zC i i i i i J   m r   m x  y 2 2 2 ( ) z i i i i i J   m r   m x  y due to ， ，so i 1i x  x i 1i y  y  d 2 2 1 1 2 2 2 1 1 1 ( ) ( ) 2 z i i i i i i i i i J m x y d m x y d m y d m                12.2 The moment of inertia of a rigid body against an axis From the centroid coordinates formula ，When the origin is at the center of mass i i C m y y M   mi  M 2 J z  J zC  Md C  0 Cy 1 0 mi i  y 