p4轉(zhuǎn)動(dòng)慣量表要點(diǎn)

圓環(huán)(轉(zhuǎn)軸沿幾何軸)R圓環(huán)(轉(zhuǎn)軸沿直徑)圓柱細(xì)棒圓盤和圓柱圓盤(轉(zhuǎn)軸沿直徑)(轉(zhuǎn)軸過中心與幾何軸線垂直)(轉(zhuǎn)軸過中心與棒垂直)(轉(zhuǎn)軸過端點(diǎn)與棒垂直)球殼(轉(zhuǎn)軸沿直徑)球體(轉(zhuǎn)軸沿直徑)RRLLLL2R2RR2R1圓圈和圓筒R2R1R常見物體的轉(zhuǎn)動(dòng)慣量(轉(zhuǎn)軸沿幾何軸)(轉(zhuǎn)軸沿幾何軸)細(xì)棒LL(轉(zhuǎn)軸過中心與板面垂直)及薄板薄板N

試求質(zhì)量為M,長和寬分別為L和N的勻質(zhì)矩形的轉(zhuǎn)動(dòng)慣量,其轉(zhuǎn)軸是L邊的中垂線.[解]矩形的面積為S=LN,取一矩形面積元dS=Ndx,矩形元的轉(zhuǎn)動(dòng)慣量為

dJ=x2dm=Nσx2dx,質(zhì)量的面密度為σ=M/S=M/LN.其質(zhì)量為dm=σdS=Nσdx,矩形的轉(zhuǎn)動(dòng)慣量為M探討:(1)轉(zhuǎn)動(dòng)慣量與N邊的大小無關(guān),當(dāng)N邊很小的時(shí)候,矩形就變成細(xì)棒,轉(zhuǎn)動(dòng)慣量不變.LNyodSxdx(2)假如轉(zhuǎn)軸是N邊的中垂線,則轉(zhuǎn)動(dòng)軸量為L

試求質(zhì)量為M,長和寬分別為L和N的勻質(zhì)矩形的轉(zhuǎn)動(dòng)慣量,其轉(zhuǎn)軸通過中心且與板面垂線.[解]矩形的面積為S=LN,質(zhì)量的面密度為σ=M/S=M/LN.

在矩形上取一矩形面積元dS=dxdy,質(zhì)量元的轉(zhuǎn)動(dòng)慣量為dJ=r2dm=σ(x2+y2)dxdy,其質(zhì)量為dm=σdS=σdxdy,矩形繞z軸的轉(zhuǎn)動(dòng)慣量為yoMdSLNxr

方法一:質(zhì)點(diǎn)法.z

在矩形上取一質(zhì)量元dm,繞z軸的轉(zhuǎn)動(dòng)慣量為dJ=r2dm=(x2+y2)dm=x2dm+y2dm,矩形繞z軸的轉(zhuǎn)動(dòng)慣量為yoMdmLNxr

方法二:正交軸定理.z其中x2dm是dm繞y軸的轉(zhuǎn)動(dòng)慣量dJy,y2dm是dm繞x軸的轉(zhuǎn)動(dòng)慣量dJx,所以dJ=dJy+dJx,而矩形繞x軸的轉(zhuǎn)動(dòng)慣量為Jx=MN2/12,因此J=Jx+Jy,矩形繞y軸的轉(zhuǎn)動(dòng)慣量為Jy=ML2/12,

在矩形上取一長為N,寬為dx的面積元dm,yoMdSLNx

方法三:平行軸定理.z依據(jù)平行軸定理,面積元繞z軸的轉(zhuǎn)動(dòng)慣量為矩形繞z軸的轉(zhuǎn)動(dòng)慣量為繞z'軸的轉(zhuǎn)動(dòng)慣量為其面積為dS=Ndx,其質(zhì)量為dm=σdS,z'試求質(zhì)量為M,半徑為R的勻質(zhì)圓環(huán)的轉(zhuǎn)動(dòng)慣量,其轉(zhuǎn)軸沿著直徑.[解]方法一:質(zhì)點(diǎn)法.圓環(huán)的周長為C=2πR,在圓環(huán)上取一弧元,長度為ds=Rdθ,弧元的轉(zhuǎn)動(dòng)慣量為dJ=D2dm=R2cos2θλRdθ質(zhì)量的線密度為

λ=M/C=M/2πR;其質(zhì)量為

dm=λds=λRdθ,整個(gè)圓環(huán)繞直徑的轉(zhuǎn)動(dòng)慣量為RDoMds弧元到軸線的距離為D=Rcosθ,=λR3cos2θdθ,θ

方法二:正交軸定理.由于對稱,圓環(huán)繞x軸和繞y軸的轉(zhuǎn)動(dòng)慣量相等,即Jx=Jy,其中x2dm是dm繞y軸的轉(zhuǎn)動(dòng)慣量dJy,y2dm是dm繞x軸的轉(zhuǎn)動(dòng)慣量dJx,所以dJz=dJy+dJx,而圓環(huán)繞z軸的轉(zhuǎn)動(dòng)慣量為Jz=MR2,在圓環(huán)上取一質(zhì)量元dm,繞z軸的轉(zhuǎn)動(dòng)慣量為dJz=R2dm

,由于R2=x2+y2,可得dJz=(x2+y2)dm=x2dm+y2dm,所以圓環(huán)繞x軸或y軸的轉(zhuǎn)動(dòng)慣量為這就是圓環(huán)繞直徑的轉(zhuǎn)動(dòng)慣量.RoMdmxyz因此Jz=Jx+Jy=2Jx=2Jy,試求質(zhì)量為M,半徑為R的勻整圓盤的轉(zhuǎn)動(dòng)慣量,其轉(zhuǎn)軸沿著直徑.[解]圓盤的面積為S=πR2,

方法一:質(zhì)點(diǎn)法.在圓盤上取一面元,面積為dS=rdθdr,質(zhì)量元的轉(zhuǎn)動(dòng)慣量為dJ=D2dm=σcos2θdθr3dr質(zhì)量的面密度為σ=M/S=M/πR2.其質(zhì)量為dm=σdS=σrdθdr,整個(gè)圓盤繞直徑的轉(zhuǎn)動(dòng)慣量為DoMdS質(zhì)量元到軸線的距離為D=rcosθ,θr

方法二:圓環(huán)法.在圓盤上取一細(xì)圓環(huán),面積為dS=2πrdr,其質(zhì)量為dm=σdS=σ2πrdr,圓環(huán)的轉(zhuǎn)動(dòng)慣量為dJ=r2dm/2=σπrdr,圓盤的轉(zhuǎn)動(dòng)慣量為RoMdSr探討:在質(zhì)點(diǎn)法中,質(zhì)量元的轉(zhuǎn)動(dòng)慣量可表示為對圓括號中的θ從0到2π的積分值為π,方括號中就是圓環(huán)的面積dS,這時(shí)dJ表示圓環(huán)的轉(zhuǎn)動(dòng)慣量,可見:圓環(huán)法的基礎(chǔ)照舊是質(zhì)點(diǎn)法.R

方法三:細(xì)棒法之一.在圓盤上取一平行于軸的細(xì)棒,其長度為l=2Rsinθ,RloMdSθyx由于x=Rcosθ,其寬度為|dx|=Rsinθdθ,其質(zhì)量為dm=σdS=σ2R2sin2θdθ,細(xì)棒的轉(zhuǎn)動(dòng)慣量為圓盤繞直徑的轉(zhuǎn)動(dòng)慣量為其面積為

dS=l|dx|=2R2sin2θdθ,

方法四:細(xì)棒法之二.在圓盤上取一垂直于軸的細(xì)棒,其長度為l=2Rcosθ,其質(zhì)量為dm=σdS=σ2R2cos2θdθ,細(xì)棒的轉(zhuǎn)動(dòng)慣量為圓盤繞直徑的轉(zhuǎn)動(dòng)慣量為RloMdSθ由于y=Rsinθ,其寬度為dy=Rcosθdθ,y其面積為dS=ldy=2R2cos2θdθ,依據(jù)公式可得xn為偶數(shù).

方法五:正交軸定理.圓盤繞x軸和繞y軸的轉(zhuǎn)動(dòng)慣量相等Jx=Jy,可得dJz=(x2+y2)dm=dJy+dJx,因此得Jz=Jy+Jx=2Jx=2Jy,而圓盤繞z軸的轉(zhuǎn)動(dòng)慣量為在圓盤上取一質(zhì)量元dm,繞z軸的轉(zhuǎn)動(dòng)慣量為dJz=r2dm,y由于r2=x2+y2,xRoMrz所以圓盤繞x軸或y軸的轉(zhuǎn)動(dòng)慣量為這就是圓盤繞直徑的轉(zhuǎn)動(dòng)慣量.dm試求質(zhì)量為M,半徑為R的勻整球殼的轉(zhuǎn)動(dòng)慣量,其轉(zhuǎn)軸沿著直徑.[解]球殼的面積為S=4πR2,

方法一:質(zhì)點(diǎn)法.在球殼上取一面積元,其大小為質(zhì)量的面密度為σ=M/S=M/4πR2.oMdSθzyRxφ轉(zhuǎn)動(dòng)慣量為dJ=D2dm=σR4dφsin3θdθ,到轉(zhuǎn)動(dòng)軸z的距離為D=Rsinθ,球殼繞z軸的的轉(zhuǎn)動(dòng)慣量為dS=R2sinθdθdφ,其質(zhì)量為dm=σdS=σR2sinθdθdφ,D方法二:圓環(huán)法之一.在球殼上取一細(xì)圓環(huán),其軸線是球殼的轉(zhuǎn)動(dòng)軸.細(xì)環(huán)的轉(zhuǎn)動(dòng)慣量為dJ=r2dm=2πσR4sin3θdθ,其質(zhì)量為dm=σdS=2πσR2sinθdθ,球殼繞直徑的轉(zhuǎn)動(dòng)慣量為RroMdSdθ其弧寬為ds=Rdθ,其面積為dS=2πrds=2πR2sinθdθ,細(xì)環(huán)半徑為r=Rsinθ,dsθ方法三:圓環(huán)法之二.在球殼上取一細(xì)圓環(huán),其軸線與轉(zhuǎn)動(dòng)軸垂直.RroMdSθD細(xì)環(huán)繞過半徑的轉(zhuǎn)動(dòng)慣量為dJc=r2dm/2,依據(jù)平行軸定理得dJ=dJc+D2dm由于左右對稱,球殼繞直徑的轉(zhuǎn)動(dòng)慣量為其質(zhì)量為dm=σdS=2πσR2cosθdθ,其弧長為ds=Rdθ,其面積為dS=2πrds=2πR2cosθdθ,細(xì)環(huán)半徑為r=Rcosθ,細(xì)環(huán)到軸線的距離為D=Rsinθ,=πσR4cos3θdθ+2πσR4sin2θcosθdθ=πσR4(1+sin2θ)cosθdθ,

方法四:正交軸定理.在球殼上取一質(zhì)量元dm,繞過z軸的轉(zhuǎn)動(dòng)慣量為dJz=D2dm=(x2+y2)dm,即dJo=[dJz+dJy+dJx]/2,所以Jo=(Jx+Jy+Jz)/2.

由于dJo=(x2+y2+z2)dmoMdmzyRxD同理,繞過y軸的轉(zhuǎn)動(dòng)慣量為

dJy=(z2+x2)dm,繞過x軸的轉(zhuǎn)動(dòng)慣量為dJx=(y2+z2)dm.

該質(zhì)量元繞o點(diǎn)的轉(zhuǎn)動(dòng)慣量為

dJo=R2dm,積分得球殼繞o點(diǎn)的轉(zhuǎn)動(dòng)慣量為Jo=MR2.=[(x2+y2)dm+(y2+z2)dm+(z2+x2)dm]/2,

由對稱可得Jx=Jy=Jz,所以Jo=3Jz/2.因此Jz=2Jo/3,即這種方法避開了積分運(yùn)算.試求質(zhì)量為M,半徑為R的勻整球體的轉(zhuǎn)動(dòng)慣量,其轉(zhuǎn)軸沿著直徑.[解]球體的體積為V=4πR2/3,

方法一:質(zhì)點(diǎn)法.在球體中取一體積元,其體積為轉(zhuǎn)動(dòng)慣量為dJ=D2dm=ρdφsin3θdθr4dr,質(zhì)量的體密度為ρ=M/V.到轉(zhuǎn)動(dòng)軸z的距離為D=rsinθ,球體的轉(zhuǎn)動(dòng)慣量為oMdvθdv=r2sinθdθdrdφ,rzyRx其質(zhì)量為dm=ρdv=ρr2sinθdθdrdφ,φD方法二:圓盤法之一.在球體上取一薄圓盤,其軸線是球體的轉(zhuǎn)動(dòng)軸.圓盤的轉(zhuǎn)動(dòng)慣量為dJ=y2dm/2=πρR5sin5θdθ/2,其質(zhì)量為dm=ρdv=πρR3sin3θdθ,球體的轉(zhuǎn)動(dòng)慣量為oMdvθ由于z=Rcosθ,其厚度為|dz|=Rsinθdθ,其體積為dv=πy2|dz|=πR3sin3θdθ,圓盤的半徑為y=Rsinθ,yRzy設(shè)u=cosθ,則du=-sinθdθ,可得轉(zhuǎn)動(dòng)慣量注:利用積分公式可以干脆計(jì)算轉(zhuǎn)動(dòng)慣量:n是奇數(shù).

方法三:圓盤法之二.在球體上取一薄圓盤,其軸線與球體的轉(zhuǎn)動(dòng)軸垂直.其質(zhì)量為dm=ρdv=πρR3cos3θdθ,由于對稱,可得球體的轉(zhuǎn)動(dòng)慣量為oMdvθ由于y=Rsinθ,其厚度為dy=Rcosθdθ,其體積為dv=πz2dy=πR3cos3θdθ,圓盤的半徑為z=Rcosθ,zRzyy圓盤繞半徑的轉(zhuǎn)動(dòng)慣量為dJc=z2dm/4,依據(jù)平行軸定理得圓盤的轉(zhuǎn)動(dòng)慣量dJ=dJc+y2dm,即:方法四:球殼法.在球體上取一半徑為r,厚度為dr薄球殼,其質(zhì)量為dm=ρdv=4πρr2dr,球體的轉(zhuǎn)動(dòng)慣量為oMdvRzy球殼繞半徑的轉(zhuǎn)動(dòng)慣量為其體積為dv=4πr2dr,r由此可見:用球殼法求球體的轉(zhuǎn)動(dòng)慣量最簡潔.

方法五:正交軸定理.在球體內(nèi)取一體積元dv,繞過z軸的轉(zhuǎn)動(dòng)慣量為

dJz=d2dm=(x2+y2)dm,即dJo=[dJz+dJy+dJx]/2,所以Jo=(Jx+Jy+Jz)/2.

由于dJo=(x2+y2+z2)dm=[(x2+y2)dm+(y2+z2)dm+(z2+x2)dm]/2,同理,繞過y軸的轉(zhuǎn)動(dòng)慣量為

dJy=(z2+x2)dm,繞過x軸的轉(zhuǎn)動(dòng)慣量為

dJx=(y2+z2)dm.

該質(zhì)量元繞o點(diǎn)的轉(zhuǎn)動(dòng)慣量為dJo=r2dm,積分得球殼繞o點(diǎn)的轉(zhuǎn)動(dòng)慣量為由對稱可得Jx=Jy=Jz,所以Jo=3Jz/2.因此Jz=2Jo/3,即這種方法避開了繁雜的積分運(yùn)算.oMdvθrzyRxφD即dJo=ρr4sinθdθdrdφ,其質(zhì)量為

dm=ρdv=ρr2sinθdθdrdφ,

試求質(zhì)量為M,半徑為R,長為L的勻質(zhì)圓柱體的轉(zhuǎn)動(dòng)慣量,其轉(zhuǎn)軸垂直軸線.[解]圓柱體的體積為V=πR2L,質(zhì)量的體密度為ρ=M/V=M/πR2L.

方法一:質(zhì)點(diǎn)法.在圓柱體內(nèi)取一厚度為dx的體積元,到y(tǒng)軸的距離為r,則r2=x2+z2;到x軸的距離為r',則z=r'cosθ.繞y軸的轉(zhuǎn)動(dòng)慣量為其質(zhì)量為

dm=ρdv=ρr'dθdr'dx,整個(gè)圓柱體繞y軸的轉(zhuǎn)動(dòng)慣量為其體積為dv=r'dθdr'dx,dJ=r2dm=ρ(x2+z2)dmRMLxrzyr'θo=ρ(x2+r'2cos2θ)r'dθdr'dx,

方法二:平板法.在圓柱上取一垂直于y軸的薄板,其長為L,寬為2z,厚為dy,其體積為dv=2Lzdy.繞y軸的轉(zhuǎn)動(dòng)慣量為其質(zhì)量為dm=ρdv=2ρLR2cos2θdθ,整個(gè)圓柱體繞y軸的轉(zhuǎn)動(dòng)慣量為利用積分公式RMLxzyo

由于y=Rsinθ,所以dy=Rcosθdθ;又由于z=Rcosθ,可得dv=2LR2cos2θdθ.yzoθn為偶數(shù).R過板的中心且垂直y軸有一軸,過此軸的轉(zhuǎn)動(dòng)慣量為dJc=L2dm/12,整個(gè)圓柱體繞y軸的轉(zhuǎn)動(dòng)慣量為依據(jù)平行軸定理,薄板繞y軸的轉(zhuǎn)動(dòng)慣量為

方法三:平行軸定理之一.在圓柱上取一垂直于z軸的薄板,其長為L,寬為2y,厚為dz,其體積為dv=2Lydz.其質(zhì)量為dm=ρdv