剛體的轉(zhuǎn)動(dòng)慣量專題

1、剛體的轉(zhuǎn)動(dòng)慣量專題1 .剛體的轉(zhuǎn)動(dòng)慣量的三要素剛體對(duì)某軸的轉(zhuǎn)動(dòng)慣量,是描述剛體在繞該軸的轉(zhuǎn)動(dòng)過(guò)程中轉(zhuǎn)動(dòng)慣性的物理量.有轉(zhuǎn)動(dòng)慣量的定義式I*可看出,剛體的轉(zhuǎn)動(dòng)慣量是與以下三個(gè)因素有關(guān)的.1與剛體的質(zhì)量有關(guān).例如半徑相同的兩個(gè)圓柱體,而它們的質(zhì)量不同,顯然,對(duì)于相應(yīng)的轉(zhuǎn)軸,質(zhì)量大的轉(zhuǎn)動(dòng)慣量也較大.2在質(zhì)量一定的情況下,與質(zhì)量的分布有關(guān).例如,質(zhì)量相同、半徑也相同的圓盤與圓環(huán),二者的質(zhì)量分布不同,圓環(huán)的質(zhì)量集中分布在邊緣,而圓盤的質(zhì)量分布在整個(gè)圓面上,所以,圓環(huán)的轉(zhuǎn)動(dòng)慣量較大.3還與給定轉(zhuǎn)軸的位置有關(guān),即同一剛體對(duì)于不同的轉(zhuǎn)軸,其轉(zhuǎn)動(dòng)慣量的大小也是不等的.例如,同一細(xì)長(zhǎng)桿,對(duì)通過(guò)其質(zhì)心且垂直于桿的轉(zhuǎn)

2、軸和通過(guò)其一端且垂直于桿的轉(zhuǎn)軸,二者的轉(zhuǎn)動(dòng)慣量不相同,且后者較大.這是由于轉(zhuǎn)軸的位置不同,從而也就影響了轉(zhuǎn)動(dòng)慣量的大小.剛體的轉(zhuǎn)動(dòng)慣量的三要素：剛體的總質(zhì)量、剛體的質(zhì)量分布情況、轉(zhuǎn)軸的位置.2 .轉(zhuǎn)動(dòng)慣量的普遍公式(1)轉(zhuǎn)動(dòng)慣量的定義式Inr2可知,對(duì)于形狀規(guī)那么、質(zhì)量均勻分布的連續(xù)剛體,其對(duì)特殊軸的轉(zhuǎn)動(dòng)慣量的計(jì)算可借助于定積分.這是,可設(shè)想將剛體分成許多小線元、面元、體元.dmdxdmdSdmdV于是Ir2dmr2dxlIr2dmSr2dS22IrdmVrdV一般說(shuō)來(lái),這是個(gè)三重的體積分,但對(duì)于有一定對(duì)稱性的物體,積分的重?cái)?shù)可以減少,甚至不需要積分.(2)剛體對(duì)某軸的轉(zhuǎn)動(dòng)慣量2a剛體對(duì)z軸的

3、轉(zhuǎn)動(dòng)慣量Izr2z2dmx2y2dm剛體對(duì)x軸的轉(zhuǎn)動(dòng)慣量,22,22,Ixrxdmyzdm2c剛體對(duì)y軸的轉(zhuǎn)動(dòng)慣量Iyr2y2dmx2z2dm仿照剛體對(duì)某軸的轉(zhuǎn)動(dòng)慣量來(lái)定義剛體對(duì)于某點(diǎn)的轉(zhuǎn)動(dòng)慣量：剛體中各質(zhì)點(diǎn)的質(zhì)量各自與其至某(參考)點(diǎn)的距離的平方的乘積,所得總和稱為剛體對(duì)該點(diǎn)的轉(zhuǎn)動(dòng)慣量.(3)剛體對(duì)某點(diǎn)的轉(zhuǎn)動(dòng)慣量剛體對(duì)坐標(biāo)原點(diǎn)o的轉(zhuǎn)動(dòng)慣量可表示為IoU-.由式、,得:IIIIBLBBIUH1|222.xyzdm1IO2IxIyIz即,質(zhì)點(diǎn)系剛體對(duì)于坐標(biāo)原點(diǎn)的轉(zhuǎn)動(dòng)慣量或極轉(zhuǎn)動(dòng)慣量等于它對(duì)于三個(gè)坐標(biāo)軸的轉(zhuǎn)動(dòng)慣量之和的一半.3 .剛體的平行軸定理許泰乃爾定理IImanaanunmiiM|2I1cm

4、d即,剛體對(duì)于任何一軸的轉(zhuǎn)動(dòng)慣量,等于剛體對(duì)于通過(guò)它的質(zhì)心并與該軸平行的轉(zhuǎn)動(dòng)慣量,加上剛體的質(zhì)量與兩軸間距離平方的乘積.注意：平行軸定理與剛體對(duì)質(zhì)心軸的轉(zhuǎn)動(dòng)慣量緊密聯(lián)系在一起,應(yīng)用此定理的參考點(diǎn)是剛體對(duì)質(zhì)心軸的轉(zhuǎn)動(dòng)慣量.根據(jù)平行軸定理,可得到如下關(guān)系：(1)剛體繞通過(guò)質(zhì)心的軸的轉(zhuǎn)動(dòng)慣量小于繞另一平行軸的轉(zhuǎn)動(dòng)慣量,二者之差為md2.(2)設(shè)有兩條平行軸PP'與QQ,均不通過(guò)質(zhì)心C.如果pp'比QQ'靠近C,那么剛體繞PP,軸的轉(zhuǎn)動(dòng)慣量小于繞QQ,軸的轉(zhuǎn)動(dòng)慣量如圖7.52(a)所示).(a)圖7.52平行軸定理的應(yīng)用a在不同圓上；b同一圓上(3)如果有一簇與質(zhì)心c的距離相等

5、的平行軸,那么,剛體繞這些軸的轉(zhuǎn)動(dòng)慣量均相等(如圖7.52(b)所示).4.剛體的垂直軸定理(正交軸定理、薄片定理)設(shè)想剛體為平面薄片,即厚度可以略去不計(jì),因而剛體為平面圖形.iIzIxIy!(6即,平面圖形對(duì)于圖形內(nèi)的兩條正交軸的轉(zhuǎn)動(dòng)慣量之和,等于這個(gè)圖形對(duì)過(guò)二軸交點(diǎn)且垂直于圖形平面的那條轉(zhuǎn)軸的轉(zhuǎn)動(dòng)慣量.注意：正交軸定理對(duì)于有限厚度的板不成立.5.轉(zhuǎn)動(dòng)慣量的疊加原理實(shí)際上,有些物體是由幾種形狀不同的剛體的組合.它對(duì)于某軸的轉(zhuǎn)動(dòng)慣量,可視為各局部對(duì)于同一轉(zhuǎn)軸的轉(zhuǎn)動(dòng)慣量之和,因而,HIlI2I3L(7即,由幾個(gè)局部組成的剛體對(duì)某軸的轉(zhuǎn)動(dòng)慣量,等于各局部對(duì)同軸的轉(zhuǎn)動(dòng)慣量之和.此即轉(zhuǎn)動(dòng)慣量的疊加原理

6、.疊加原理是根據(jù)加法的組合定那么,把屬于各局部的項(xiàng)分別相加,然后求和而得.同理,設(shè)有一物體挖去假設(shè)干局部,那么剩余局部的轉(zhuǎn)動(dòng)慣量,等于原物體的轉(zhuǎn)動(dòng)慣量,減去挖去局部的轉(zhuǎn)動(dòng)慣量.例題1在質(zhì)量為m,半徑為R的勻質(zhì)圓盤上挖出半徑為的兩個(gè)圓孔,圓孔中央在半徑R的中點(diǎn),求剩余局部對(duì)過(guò)大圓盤中央且與盤面垂直的軸線的轉(zhuǎn)動(dòng)慣量圖7.53轉(zhuǎn)動(dòng)慣量的疊加原理的應(yīng)用解大圓盤對(duì)過(guò)圓盤中央o且與盤面垂直的軸線以下簡(jiǎn)稱軸的轉(zhuǎn)動(dòng)慣量為I1mR2.由于對(duì)稱放置,兩個(gè)小圓盤對(duì)O軸的轉(zhuǎn)動(dòng)慣量相等,設(shè)為,圓盤質(zhì)量的面密度,根據(jù)平行軸定理,有I'1,r224mr2R212一mr4設(shè)挖去兩個(gè)小圓盤后,剩余局部對(duì)O軸的轉(zhuǎn)動(dòng)慣量為

7、I''2I1:mR24rm2R212-mr22r4R26.轉(zhuǎn)動(dòng)慣量的標(biāo)度變換法轉(zhuǎn)動(dòng)慣量的標(biāo)度變換法是計(jì)算轉(zhuǎn)動(dòng)慣量的一種簡(jiǎn)便的方法由于在幾何上具有相似性的均勻物體,它們對(duì)相應(yīng)轉(zhuǎn)軸的轉(zhuǎn)動(dòng)慣量的表達(dá)式也具有相似性,在根據(jù)轉(zhuǎn)動(dòng)慣量的平行軸定理、疊加原理等,確定彼此關(guān)系,比擬系數(shù),從而獲得物體對(duì)該軸的轉(zhuǎn)動(dòng)慣量.故這種方法可以不用積分即能求得某些特殊形狀的物體的轉(zhuǎn)動(dòng)慣量.例題2求均勻立方體繞通過(guò)面心的中央軸的轉(zhuǎn)動(dòng)慣量IC.圖7.54標(biāo)度變換法用于計(jì)算立方體對(duì)通過(guò)面心的中央軸的轉(zhuǎn)動(dòng)慣量解令立方體的總質(zhì)量為m,邊長(zhǎng)為l,設(shè)均勻立方體繞通過(guò)面心的中央軸的轉(zhuǎn)動(dòng)慣量:為1ckml2c其中,系數(shù)k是無(wú)

8、量綱的量.由于一切立方體在幾何上都是相似的,它們應(yīng)該具有同樣的k.中央軸到棱邊的距離為d罵2根據(jù)平行軸定理,立方體繞棱邊的轉(zhuǎn)動(dòng)慣量為2x2212IDkmlmlkml22現(xiàn)將立方體等分為8個(gè)小立方體,每個(gè)小立方體的質(zhì)量為,邊8長(zhǎng)為；,繞棱邊的轉(zhuǎn)動(dòng)慣量為ID'132ml28個(gè)立方體繞棱邊的轉(zhuǎn)動(dòng)慣量之和應(yīng)等于大立方體繞中央軸的轉(zhuǎn)動(dòng)慣量,即比擬系數(shù),得于是,求得所以,IC8ID-1k工322IC工ml6下面介紹利用定積分法計(jì)算質(zhì)量均勻分布、圖形具有對(duì)稱性的剛體對(duì)于一些特殊的轉(zhuǎn)軸的轉(zhuǎn)動(dòng)慣量.勻質(zhì)細(xì)桿例題3質(zhì)量為m、長(zhǎng)為l的勻質(zhì)細(xì)桿,繞其質(zhì)心且垂直于桿的軸旋轉(zhuǎn),桿的轉(zhuǎn)動(dòng)慣量是多少？解設(shè)桿的線密度為

9、,那么m-選擇如下圖的坐標(biāo)軸,桿的質(zhì)心位于原點(diǎn),取一個(gè)長(zhǎng)度為dx、與質(zhì)心的距離為x的微元,那么1dxOxx圖7.55勻質(zhì)細(xì)桿對(duì)質(zhì)心軸的轉(zhuǎn)動(dòng)慣量2 ,2,dIxdmxdxl/221312IOxdxlml1/21212根據(jù)平行軸定理,桿對(duì)通過(guò)其一端且垂直于桿的軸的轉(zhuǎn)動(dòng)慣量為21121212IIOmmlmlml21243當(dāng)然用定積分也可得相同的結(jié)果lx2dx1l3-ml2033勻質(zhì)正方形薄板例題4求質(zhì)量為m、邊長(zhǎng)為a的勻質(zhì)正方形薄板對(duì)其邊為軸的轉(zhuǎn)動(dòng)慣量.解勻質(zhì)薄板可視為細(xì)長(zhǎng)條的組合.根據(jù)疊加原理可得對(duì)一邊的轉(zhuǎn)動(dòng)慣量.Odx圖7.56勻質(zhì)正方形薄板對(duì)一邊為軸的轉(zhuǎn)動(dòng)慣量Ix1mia21ma233同理,可

10、得1212-mia-ma33或利用定積分,其中,鳥(niǎo)為面密度.'a對(duì)z軸的轉(zhuǎn)動(dòng)慣量對(duì)質(zhì)心軸的轉(zhuǎn)動(dòng)慣量ax2adx10312一ma322IzIxIy3ma一2II-22212121cIzmama-ma-ma2326對(duì)以對(duì)角線為軸的轉(zhuǎn)動(dòng)慣量IxIy'1Ic11ma21ma222612當(dāng)然,對(duì)z軸的轉(zhuǎn)動(dòng)慣量也可用二重積分計(jì)算得到22,1xydxdya0dyax2dx0aa2dxydyoo22一ma3勻質(zhì)矩形薄板例題5求質(zhì)量為m、長(zhǎng)和寬分別為a和b的勻質(zhì)矩形薄板對(duì)其邊為軸的轉(zhuǎn)動(dòng)慣量.解方法同上,不難得到a圖7.57勻質(zhì)矩形薄板對(duì)一邊為軸的轉(zhuǎn)動(dòng)慣量,1.2.12Ixmb,Ivmaxy3 3

11、由垂直軸定理,可以進(jìn)一步求得矩形薄板對(duì)通過(guò)頂點(diǎn)且垂直于板平面的軸的轉(zhuǎn)動(dòng)慣量如圖7.57為12,2IzIxIymab3當(dāng)然,對(duì)z軸的轉(zhuǎn)動(dòng)慣量也可用二重積分計(jì)算得到Iz22,xydxdy,a2,dyxdxodxby2dy1a4b4ma2b2033矩形薄板對(duì)通過(guò)質(zhì)心且垂直于板平面的軸的轉(zhuǎn)動(dòng)慣量為J<1III111<1IIIII11JJBIII11JJBIII11JIIIIBJHI222122,ab122I'mabmmab3212b/2b/2O1O.O2b/2b/2圖7.58勻質(zhì)矩形薄板對(duì)過(guò)中央且垂直于板面的軸的轉(zhuǎn)動(dòng)慣量另解：從量綱上考慮,所求的轉(zhuǎn)動(dòng)慣量可表示為ilII!1ilI1

12、«!I!:II22IOc1mac2mabQmb其中,ci1,2,3為待定系數(shù).將a和b轉(zhuǎn)置后,IOc1mb202mbaQma?但I(xiàn)O不會(huì)由于a和b轉(zhuǎn)置而發(fā)生變化,比擬系數(shù),有0102IOc1ma2b2c2mba7.54利用勻質(zhì)矩形板可等分為兩個(gè)小勻質(zhì)矩形板的特點(diǎn),如圖所示,有2m2bmbIOiIO2Ci二a二J二a二2222mbOIO124IO22b2IOim422bGma一2bQma一2比擬系數(shù),有得,因而,C1117c1,二C2c24162Ci112C201一m122,2ab勻質(zhì)長(zhǎng)方體例題6求質(zhì)量為m、長(zhǎng)、寬和高分別為a、b和c的勻質(zhì)長(zhǎng)方體對(duì)其棱邊為軸的轉(zhuǎn)動(dòng)慣量.P圖7.59勻質(zhì)

13、長(zhǎng)方體對(duì)其棱邊為軸的轉(zhuǎn)動(dòng)慣量解由疊加原理,不難得到以棱邊c為軸的轉(zhuǎn)動(dòng)慣量Iz12,212,2-miab-mab33同理可得,以棱邊a為軸的轉(zhuǎn)動(dòng)慣量Ix1,221,22-mibc-mbc33以棱邊b為軸的轉(zhuǎn)動(dòng)慣量122122-miacmac33當(dāng)然,對(duì)z軸的轉(zhuǎn)動(dòng)慣量也可用三重積分計(jì)算得到Iz22xydxdydzcba2cab2dzdyxdxdzdxydy0000001313abcabc33121,2一ma-mb33122mab3對(duì)x軸的轉(zhuǎn)動(dòng)慣量也可用三重積分計(jì)算得到Ix22yzdxdydzacb2abc2dxdzydydxdyzdz0000001,22mbc3對(duì)x軸的轉(zhuǎn)動(dòng)慣量也可用三重積分計(jì)算得

14、到Iy22xzdxdydzcba2abc2dzdyxdxdxdyzdz000000122-mac3根據(jù)平行軸定理,對(duì)通過(guò)長(zhǎng)方體面心為軸的轉(zhuǎn)動(dòng)慣量paiBBii1blb=4：:1I：janmsahhbh:awbi:;a;g:.2ab1221221221Pp,Izm-mabmabmab23412如果將上述長(zhǎng)方體換成邊長(zhǎng)為a的立方體,那么繞其棱邊的轉(zhuǎn)動(dòng)慣量均相等,且I2ma23對(duì)通過(guò)正方體面心為軸的轉(zhuǎn)動(dòng)慣量Ipp'1ma2PP6余此類推.對(duì)于特殊剛體,勻質(zhì)細(xì)圓環(huán)例題7求質(zhì)量為m、半徑為R的勻質(zhì)細(xì)圓環(huán)對(duì)通過(guò)中央并與環(huán)面垂直的軸的轉(zhuǎn)動(dòng)慣量.圖7.60勻質(zhì)細(xì)圓環(huán)對(duì)通過(guò)中央并與環(huán)面垂直的軸的轉(zhuǎn)動(dòng)慣量

15、解細(xì)圓環(huán)的質(zhì)量可以認(rèn)為全局部布在半徑為r的圓周上,即在距離中央小于或大于r的各處,質(zhì)量均為零,所以轉(zhuǎn)動(dòng)慣量為222：IzmRRmimR：zR2dmmR2又由垂直軸定理,可以得到其對(duì)直徑為轉(zhuǎn)軸的轉(zhuǎn)動(dòng)慣量為12IdmR22再利用平行軸定理,可得細(xì)圓環(huán)對(duì)其任意切線為轉(zhuǎn)軸的轉(zhuǎn)動(dòng)慣量為.»UB»»>11_2_23_2ItmRmR-mR22.!BIII*圖7.61勻質(zhì)細(xì)圓環(huán)對(duì)任意切線為軸的轉(zhuǎn)動(dòng)慣量m2水其中,為細(xì)圓環(huán)的線密度,那么dmRd細(xì)圓環(huán)對(duì)切線的轉(zhuǎn)動(dòng)慣量2TtRRcoso327cR3128soR32武冗Rd2cosd3uR32mR2勻質(zhì)中空薄圓盤例題8求質(zhì)量為m、

16、內(nèi)半徑為K、外半徑為R2的勻質(zhì)中空薄圓盤對(duì)通過(guò)中央并與盤面垂直的軸的轉(zhuǎn)動(dòng)慣量.圖7.62勻質(zhì)中空薄圓盤對(duì)通過(guò)中央并與盤面垂直的軸的轉(zhuǎn)動(dòng)慣解勻質(zhì)中空薄圓盤可視為無(wú)限多個(gè)同心的細(xì)圓環(huán)的組合,所以,根據(jù)疊加原理可以得到該中空薄圓盤對(duì)通過(guò)中央且垂直于盤面的轉(zhuǎn)軸的轉(zhuǎn)動(dòng)慣量.中空薄圓盤的質(zhì)量為22m兀R2R1其中,為中空薄圓盤的面密度,那么dm2：rdr中空薄圓盤對(duì)通過(guò)中央且垂直于盤面的轉(zhuǎn)軸的轉(zhuǎn)動(dòng)慣量22IOr2<drRR232九rdrR.冗R4R142122R2Ri2一九R2R22mR2R1iiii:»n:)當(dāng)然,中空薄圓盤對(duì)通過(guò)中央且垂直于盤面的轉(zhuǎn)軸的轉(zhuǎn)動(dòng)慣量也可用二重積分計(jì)算得到.2

17、21tR23R23122IOrrdrddrdr2冗rdr-mR2R10RR2勻質(zhì)薄圓盤例題9求質(zhì)量為m、半徑為R的勻質(zhì)薄圓盤對(duì)通過(guò)中央并與環(huán)面垂直的軸的轉(zhuǎn)動(dòng)慣量.圖7.63勻質(zhì)薄圓盤對(duì)通過(guò)中央并與環(huán)面垂直的軸的轉(zhuǎn)動(dòng)慣量解勻質(zhì)薄圓盤可視為無(wú)限多個(gè)同心的細(xì)圓環(huán)的組合,所以,根據(jù)疊加原理可以得到該厚圓環(huán)對(duì)通過(guò)中央且垂直于環(huán)面的轉(zhuǎn)軸的轉(zhuǎn)動(dòng)慣量.薄圓盤的質(zhì)量為m?R2其中,為薄圓盤的面密度,那么dm2：rdr薄圓盤對(duì)通過(guò)中央且垂直于盤面的轉(zhuǎn)軸的轉(zhuǎn)動(dòng)慣量R2_R31_41_2NOr2<dr2urdr-uR-mR：0022當(dāng)然,薄圓盤對(duì)通過(guò)中央且垂直于盤面的轉(zhuǎn)軸的轉(zhuǎn)動(dòng)慣量也可用二重積分計(jì)算得到.c27

18、tRcR1c233,12IOrrdrddrdr2冗rdr-mRO0002可見(jiàn),薄圓盤是中空?qǐng)A盤的特例同樣,根據(jù)垂直軸定理,得其對(duì)直徑為轉(zhuǎn)軸的轉(zhuǎn)動(dòng)慣量為PMMIII:III；ID-mR2：4II:II:再利用平行軸定理,可得其對(duì)切線為轉(zhuǎn)軸的轉(zhuǎn)動(dòng)慣量為122521tmRmRmR44勻質(zhì)薄壁圓筒例題10求質(zhì)量為m、半徑為R的勻質(zhì)薄壁圓筒對(duì)中央軸線的轉(zhuǎn)動(dòng)慣量.解勻質(zhì)薄壁圓筒可視為半徑相同,圓心在同一條直線上且各個(gè)環(huán)面均垂直于該直線的一系列細(xì)圓環(huán)的組合.根據(jù)疊加原理,由圓環(huán)對(duì)該直線的轉(zhuǎn)動(dòng)慣量較易求出此圓筒對(duì)該直線為轉(zhuǎn)軸的轉(zhuǎn)動(dòng)慣量：u(：j|MO甲R(shí)2mR2AR1i!II一O1UJ圖7.64勻質(zhì)薄壁圓筒對(duì)

19、中央軸線的轉(zhuǎn)動(dòng)慣量當(dāng)然,也可定積分法求解勻質(zhì)中空?qǐng)A柱體例題11求質(zhì)量為m、內(nèi)半徑為以、外半徑為R2的勻質(zhì)中空?qǐng)A柱體對(duì)中央軸線的轉(zhuǎn)動(dòng)慣量:圖7.65勻質(zhì)中空?qǐng)A柱體對(duì)中央軸線的轉(zhuǎn)動(dòng)慣量解勻質(zhì)中空?qǐng)A柱體可視圓心在同一條直線上且環(huán)面均垂直于該直線的一系列中空?qǐng)A盤的組合.根據(jù)疊加原理,由中空?qǐng)A盤對(duì)該直線的轉(zhuǎn)動(dòng)慣量較易求出此中空?qǐng)A柱體對(duì)該直線為轉(zhuǎn)軸的轉(zhuǎn)動(dòng)慣量paMBi11！I1IlaaiaBimiimIOmiR；R2mR；R2當(dāng)然,也可定積分法求解.m兀R；R；L其中,為體密度.dm2<LdrR22rRi2"Ldr2TLR23rdrRi晨21-tL21R24R4R22R2R22R222mR

20、2Ri勻質(zhì)實(shí)心圓柱體例題12求質(zhì)量為m、半徑為R的勻質(zhì)實(shí)心圓柱體對(duì)中央軸線的轉(zhuǎn)動(dòng)慣量.圖7.66勻質(zhì)實(shí)心圓柱體對(duì)中央軸線的轉(zhuǎn)動(dòng)慣量解勻質(zhì)實(shí)心圓柱體可視圓心在同一條直線上且圓面均垂直于該直線的一系列薄圓盤的組合.根據(jù)疊加原理,由薄圓盤對(duì)該直線的轉(zhuǎn)動(dòng)慣量較易求出此圓柱體對(duì)該直線為轉(zhuǎn)軸的轉(zhuǎn)動(dòng)慣212miRmR2當(dāng)然,也可定積分法求解tR2L其中,為體密度.dm2<Ldr2<Ldr2L3,1,_rdr一社R2R4Ri41mR22當(dāng)然,實(shí)心圓柱體對(duì)中央軸線的轉(zhuǎn)動(dòng)慣量可用三重積分計(jì)算得到.27tLRqR,10r2rdrddzddzrdr2社rdrmRo0oo2可見(jiàn),厚圓筒是實(shí)心圓柱體的特例.同

21、樣,根據(jù)垂直軸定理,得其對(duì)直徑為轉(zhuǎn)軸的轉(zhuǎn)動(dòng)慣量為ID-mR24«UII!勻質(zhì)實(shí)心圓柱體例題12求質(zhì)量為m、半徑為R的勻質(zhì)實(shí)心圓柱體對(duì)中央直徑(a)z.(b)圖7.67勻質(zhì)實(shí)心圓柱體對(duì)中央直徑的轉(zhuǎn)動(dòng)慣量解|設(shè)勻質(zhì)實(shí)心圓柱體由x2z2R2與y?y圍成.mtR2L其中,為體密度.dmdVdrddy繞Z軸的轉(zhuǎn)動(dòng)慣量為22xydV27rRL/222drdrrcos00L/227t2R3L/2cosdrdrdy00L/2R41cR22L3LL2tt4238y2dy27rRL/220d0rdrl/2ydytR4L1122,3tRL1mR2-1mL2412同理可得,繞x軸的轉(zhuǎn)動(dòng)慣量為Ixvy2dVr

22、drL/2r2cos2L/2y2dy2兀sinoR3L/2d0rdr_2dy27rRL/22od0rdrl/2ydyr4lLL4123tR2L3121mR2mL2412勻質(zhì)實(shí)心圓柱體例題13求質(zhì)量為m、半徑為R的勻質(zhì)實(shí)心圓柱體對(duì)端面直徑為軸的轉(zhuǎn)動(dòng)慣量.圖7.68勻質(zhì)實(shí)心圓柱體對(duì)端面直徑的轉(zhuǎn)動(dòng)慣量解設(shè)勻質(zhì)實(shí)心圓柱體由x2z2R2與y0、yL圍成.其中,為體密度.繞Z軸的轉(zhuǎn)動(dòng)慣量為dmdVdrddyIzx2y2dVVy2dy.L2.rdr0ydy27tRL22drdrrcos00027r2R3Lcosdrdrdy000R4R2L3冗L2冗4231tR4LmR2加2tR2L343同理可得,繞x軸的轉(zhuǎn)

23、動(dòng)慣量為IxVy22兀dz2dVRl22rdrrsin00027r2R3Locosd0rdr0dyr4r2L3冗L2九4231tR4L1tR2L3431mR2-mL2y2dy2兀doRL2°rdr°ydy當(dāng)然,利用平行軸定理也可得到相同的結(jié)果勻質(zhì)球殼例題14求質(zhì)量為m、半徑為R的勻質(zhì)球殼對(duì)球心的轉(zhuǎn)動(dòng)慣量、對(duì)任意直徑和切線的轉(zhuǎn)動(dòng)慣量.圖7.69勻質(zhì)球殼對(duì)球心、對(duì)任意直徑和切線為軸的轉(zhuǎn)動(dòng)慣量解由于在距離球心大于或小于r處,質(zhì)量均為零,而質(zhì)量m均勻分布于球殼上2m4R.解法一：根據(jù)剛體對(duì)坐標(biāo)原點(diǎn)的轉(zhuǎn)動(dòng)慣量的定義式,有OrnR2R2mmR2：ahijti或paMBMlJB(*；11

24、0R2dmmR2當(dāng)然,極轉(zhuǎn)動(dòng)慣量也可用二重積分計(jì)算得到.22427t7t42IOR2R2smddR40d0s1nd4冗R4mR2根據(jù)關(guān)系式1IO2IxIyIz對(duì)于勻質(zhì)球殼,球心為坐標(biāo)原點(diǎn).根據(jù)對(duì)稱性,可知IxIyIz22IxIyIzmRy3amiiiinuum即為球殼對(duì)任意直徑的轉(zhuǎn)動(dòng)慣量.解法當(dāng)然,Ix、Iy和Iz也可利用二重積分計(jì)算得到22IzRsinRsindd22TtTt&Rdsind00R42tt-232mR23IxRsinsinRcos2.Rsin427t2R3Rsindsindoo4242R兀一2R2u-33-mR23R422cossind2IyRsincosRcos427

25、t2K3Rcosdsind00R4冗22R42冗2332.Rsin427tit2Rdcos00sind-mR23解法三:球殼可視為一系列薄圓環(huán)的組合.dIxr2dmr2dS其中,r表示薄圓環(huán)的半徑,面元.dm為薄圓環(huán)的元質(zhì)量,dS為薄圓環(huán)的dS冗Ix2冗x0rRcos2.2<Rd2tRcosd438422R4cos3d-uR4-mR233根據(jù)平行軸定理,可得球殼對(duì)任意切線為軸的轉(zhuǎn)動(dòng)慣量22252It-mRmRmR33|假設(shè)將該球殼切除一半,求剩余局部球冠對(duì)任一直徑的轉(zhuǎn)動(dòng)慣!i1i3.!根據(jù)剛體對(duì)坐標(biāo)原點(diǎn)的轉(zhuǎn)動(dòng)慣量的定義式,有：IOmiR2R2mimR2.Oii2212IORdmmR2當(dāng)然

26、,極轉(zhuǎn)動(dòng)慣量也可用二重積分計(jì)算得到_2_2.IORRsindd427t酒2Rdsind00_c41c22冗RmR2顯然,Ix、ly和Iz可利用二重積分計(jì)算得到22IzRsinRsindd427t,23Rdsind00-4c2R2九一3-mR23Ix2RsinsinRcos_2Rsindd.27rRsin0公23dsind0/2兀R4d0422cossindRY-mR23Rsincos2Rcos2_2RsinR42cos2d02sin3d427tRd01t2c2.,cossindo口42R冗一3R42l12mR3將該球殼局部切除,假設(shè)剩余局部球冠的高度為直徑的1/4,I求其對(duì)任一直徑的轉(zhuǎn)動(dòng)慣量.

27、:此時(shí)在球坐標(biāo)系中的極角極轉(zhuǎn)動(dòng)慣量可用二重積分計(jì)算得到.427tf34112R2R2sinddR4dsind2冗R4mR20024Ix、I,和iz可利用二重積分計(jì)算得到.Iz22RsinRsindd427ty33Rdsind00R42Tt2448mR2IxRsinsin222RcosRsindd427t2Rsindo7t33sind427r正32Rdcossind00"5-R冗R24生mR296Rsincos2Rcos_2Rsindd/2兀4Rcos0,33sindo:任32dcossin0這里已利用積分Tt3sino4r2962240sincossin1一cos3cos024I將該

28、球殼局部切除,假設(shè)剩余局部為原來(lái)的1/8,求其對(duì)任一直徑I的轉(zhuǎn)動(dòng)慣量.sind12mR8極轉(zhuǎn)動(dòng)慣量可用二重積分計(jì)算得到一2_2_4屈2IOR2R2sinddR40dIx、Iy和Iz可利用二重積分計(jì)算得到Iz2RsinIxRRRsinsinn22sindo冗24R434f2R4d0'R4-23mR2122Rcos7t23sindo冗123工20R2sinddsin3d22Rsindd4W2J22Rdcossindoo112mR2RsincosRcos_2RsinddR4"2cos2o23dsind0R4,2d*cos2oosind口4九2口4九1RR432312mR12勻質(zhì)實(shí)心

29、球體例題15求質(zhì)量為m、半徑為R的勻質(zhì)實(shí)心球體對(duì)球心的轉(zhuǎn)動(dòng)慣量、對(duì)任意直徑的轉(zhuǎn)動(dòng)慣量.z圖7.70勻質(zhì)球體對(duì)球心、對(duì)任意直徑和切線為軸的轉(zhuǎn)動(dòng)慣量解解法球體可視為球殼的組合,m#33其中,為體密度.根據(jù)剛體對(duì)坐標(biāo)原點(diǎn)的轉(zhuǎn)動(dòng)慣量的定義式,有pmkit-IrIe;a««aaaIsi<maj,R22,R4,43_2“Or4<dr4冗rdr-uR-mR；0055當(dāng)然,極轉(zhuǎn)動(dòng)慣量也可利用三重積分計(jì)算得到.2227r冗R4R432Iorrsindrdddsindrdr47rrdr-mRO00005根據(jù)關(guān)系式1IocIxIyIz2對(duì)于勻質(zhì)球體,球心為坐標(biāo)原點(diǎn).根據(jù)對(duì)稱性,可知I

30、xIyIz,2_2IxIyIzmR5即為球體對(duì)任意直徑的轉(zhuǎn)動(dòng)慣量.Ix、Iy和Iz也可利用三重積分計(jì)算得到.Izrsin2r2sindrdd27r兀oR/0d0sind0rdr.2.R52235-mR25rsinsinrcos27ro兀&sindsind007tRr4drr2sindrdd7t2cos0.,R4,sindrdr0解法3227t3imR2R55R5R5rsincos27r2cosdo2rcosr2sindrdd冗3R4sindrdroo冗2cos0.R4.sindrdr0R55ImR2R55球體可視為球殼的組合,根據(jù)疊加原理,也可較易求得其對(duì)直徑的轉(zhuǎn)動(dòng)慣量為Ia!3ILa

31、a«r,:日ijbbbiua*a1海aiaaaaau.；«|22R2228R48522：IDrdm-r4<dr一冗rdruR-mR；30330155：,»3«»hua»a«:m>ii解法三：球體可視為一系列薄圓盤的組合.12,1221222dIX一rdm-r<dx冗Rxdx222其中,r表示薄圓盤的半徑,dm為薄圓盤的元質(zhì)量,dV為薄圓環(huán)的體元,x為薄圓盤到質(zhì)心軸的距離,R1222.R22IXR-RXdX%RXdx2為薄圓環(huán)的厚度.dxu-R5-mR2155根據(jù)平行軸定理,可得球體對(duì)任意切線為軸的轉(zhuǎn)動(dòng)慣量2

32、2272ItmR2mR2mR255annnbm0a假設(shè)將該球體切除一半,求剩余局部對(duì)任一直徑的轉(zhuǎn)動(dòng)慣量根據(jù)剛體轉(zhuǎn)動(dòng)慣量的疊加原理,有32%,nmR210:1:nn當(dāng)然,極轉(zhuǎn)動(dòng)慣量也可用三重積分計(jì)算得到.2221tM2R4R432IOrrsindrdddsindrdr2urdrmR000010顯然,Ix、ly和Iz也可利用三重積分計(jì)算得到.Izrsin2r2sindrdd27tt2QRd0d°sind0rdr92R52冗35-mR25rcossindrdd27rsind02R5*.3sinR4drdro27r獲22dcos00.,R4,sindrdr07t27t3R527t3R55mR

33、2rsincosrcos27r2cosd02 R5幾3 5兀23sindoR55r4drr2sindrdd27r拒22dcos00.,R4,sindrdr0R555mR2將該球殼局部切除,假設(shè)剩余局部的高度為直徑的1/4,求其對(duì)任I一直徑的轉(zhuǎn)動(dòng)慣量.此時(shí)在球坐標(biāo)系中的極角極轉(zhuǎn)動(dòng)慣量可用三重積分計(jì)算得到.2227r欠3R4R432IOrrsindrdd0dosind0rdr支0rdrmRIx、ly和Iz可利用三重積分計(jì)算得到.,.22Izrsinrsindrdd27t33aR.dsindrdr000c5R52冗245mR216Ixrsinsinrcos2.rsindrdd27r0sin2o3si

34、n5花24R5oR5Rr4dr32cossinoR4drdroTt2457小mR480rsincoscossindrdd27t2cos0一一3sinr4dr7t32cossinodRr4dro5花24R57R52冗245式mR2480將該球體局部切除,假設(shè)剩余局部為原來(lái)的1/8,求其對(duì)任一直徑的轉(zhuǎn)動(dòng)慣量.極轉(zhuǎn)動(dòng)慣量可用三重積分計(jì)算得到IOr2r2sindrdd12R4dsindrdr000Rr4dr32mR40Ix、Iy和Iz可利用三重積分計(jì)算得到22Izrsinrsindrdd諜評(píng)3R4dsindrdr0002R5235-mR220222rsinsinrcosrsindrdd下2122dco

35、s00R4sindrdr0不22疝23R4sindsindrdr0002t2RL1R543523520mR22221yrsincosrcosrsindrdd2223R4工2小2R4cosdsindrdrdcossindrdr0000002_r51R543523512mR20勻質(zhì)中空球體例題16求質(zhì)量為m、內(nèi)半徑為R、外半徑為R2的勻質(zhì)中空球體對(duì)球心的轉(zhuǎn)動(dòng)慣量、對(duì)任意直徑的轉(zhuǎn)動(dòng)慣量.圖7.71勻質(zhì)中空球體對(duì)球心、對(duì)任意直徑和切線為軸的轉(zhuǎn)動(dòng)慣量解中空球體可視為球殼的組合,433m冗R2R13其中,為體密度.根據(jù)剛體對(duì)坐標(biāo)原點(diǎn)的轉(zhuǎn)動(dòng)慣量的定義式,有,R222.R24.PoRir4/dr4冗J力5兀R

36、553r5Ri5R3c35R2R1當(dāng)然,極轉(zhuǎn)動(dòng)慣量也可利用三重積分計(jì)算得到IOr2r2sindrdd%而d,dr4兀RYdr00RiR5R；R3根據(jù)關(guān)系式1.IOcIxlyIz2對(duì)于勻質(zhì)球體,球心為坐標(biāo)原點(diǎn).根據(jù)對(duì)稱性,可知IxIyIzIIIIIIUMIIUIHMh.2R2R5!1yz看展r3：即為球殼對(duì)任意直徑的轉(zhuǎn)動(dòng)慣量.Ix、Iy和Iz也可利用三重積分計(jì)算得到.22Izrsinrsindrdd21t1t3R24dsindrdr00RiR5R52R5R5-m335R3R3Ixrsinsinrcossindrdd2兀sin兀3dsin0R24.rdr7t2cos0.,R24,sindrdrRR

37、5R5R15R5R5R15m-3R3R13rsin27r2cosR5R5cosrcossindrddsinR2dR1r4dr冗2cos0.,R24,sindrdr耳R5R52vR5R15r：R15另解:R5R15R3r3中空球體可視為球殼的組合,根據(jù)疊加原理,也可較易求得其對(duì)直徑的轉(zhuǎn)動(dòng)慣量為55R5R52R5R5m335R3r312r2dmR22r24ddr8兀Rydrg冗3Ri33Ri15練習(xí):1求質(zhì)量為m、邊長(zhǎng)為a的勻質(zhì)等邊三角形ABD對(duì)過(guò)頂點(diǎn)A且垂直于板面的軸的轉(zhuǎn)動(dòng)慣量圖7.72等邊三角形對(duì)過(guò)頂點(diǎn)且垂直于板面為軸的轉(zhuǎn)動(dòng)慣量解m存對(duì)x軸的轉(zhuǎn)動(dòng)慣量,2ycot60dy231.34a-a2Ix

38、Jyaa3a32a3,33.394aa386161232-ma-ma2812一ma8a2a22n.x一axcot30dx212一ma6對(duì)y軸的轉(zhuǎn)動(dòng)慣量,錯(cuò)誤的做法：正確的做法:.3一a2xcot30a22dx2x0Jaxcot302dx3a2xcot30dx3243321一ma612一ma81一ma24根據(jù)垂直軸定理,IzIx1一ma6Iz2OC,Ia2ACIaIz21AOma652一ma12|假設(shè)直接計(jì)算對(duì)經(jīng)過(guò)頂點(diǎn)A且垂直于三角形平面的軸的轉(zhuǎn)動(dòng)慣1-I量,取窄條后,那么窄條上各點(diǎn)到軸的距離并非處處相等,故此法|不可行！!2求質(zhì)量為m、底面為邊長(zhǎng)a的等邊三角形、高L的勻質(zhì)正三棱柱對(duì)以其高為軸的

39、轉(zhuǎn)動(dòng)慣量.解正三棱柱可視為由無(wú)限多個(gè)正三角形的組合,根據(jù)第1題的結(jié)論,利用轉(zhuǎn)動(dòng)慣量的疊加原理,有525252miaamima1212123求質(zhì)量為m、邊長(zhǎng)為a且一個(gè)頂角為60勻質(zhì)棱形對(duì)過(guò)頂點(diǎn)且垂直于板面的軸的轉(zhuǎn)動(dòng)慣量.圖7.73一個(gè)頂角為60的棱形對(duì)過(guò)頂點(diǎn)且垂直于板面為軸的轉(zhuǎn)動(dòng)慣量解根據(jù)第1題結(jié)論,利用轉(zhuǎn)動(dòng)慣量的疊加原理,有2IaId2IzIbIe2Iza121272mma-ma-ma23412321232132m-a-mama-ma234124求質(zhì)量為m、底面為邊長(zhǎng)a且一個(gè)頂角為60勻質(zhì)棱形、勻質(zhì)正四棱柱對(duì)以其高為軸的轉(zhuǎn)動(dòng)慣量.解該正四棱柱可視為由無(wú)限多個(gè)棱形的組合,根據(jù)第的結(jié)論,利用轉(zhuǎn)動(dòng)慣

40、量的疊加原理,有7272IAID-mia-ma,132132IBIEmiama12125求質(zhì)量為M、邊長(zhǎng)為a勻質(zhì)正六邊形對(duì)過(guò)頂點(diǎn)且垂直于板面的軸的轉(zhuǎn)動(dòng)慣量.圖7.74正六邊形對(duì)過(guò)頂點(diǎn)且垂直于板面為軸的轉(zhuǎn)動(dòng)慣量解根據(jù)第1題結(jié)論,利用轉(zhuǎn)動(dòng)慣量的疊加原理,有52217-2IaIbIdIeIfIgma6MaMa1212其中M6m.6求質(zhì)量為M、底面為邊長(zhǎng)a的正六邊形、高L的勻質(zhì)正六棱柱對(duì)以其高為軸的轉(zhuǎn)動(dòng)慣量.解該正六棱柱可視為由無(wú)限多個(gè)正六邊形的組合,根據(jù)第5題的結(jié)論,利用轉(zhuǎn)動(dòng)慣量的疊加原理,有IaIbIdIeIfIgmiaMa212表7.2一些簡(jiǎn)單幾何圖形的轉(zhuǎn)動(dòng)慣量剛體轉(zhuǎn)軸的位置轉(zhuǎn)動(dòng)慣量細(xì)桿通過(guò)中央

41、且垂直于桿12-ml12m,l通過(guò)桿端且垂直于桿-ml23通過(guò)中央且-2mR剛體轉(zhuǎn)軸的位置轉(zhuǎn)動(dòng)慣M細(xì)圓環(huán)m,R與環(huán)面垂直沿直徑12一mR2沿切線3D2一mR2中空?qǐng)A盤m,Ri,旦通過(guò)中央且與環(huán)面垂直mR12R22沿直徑1-22mR1R24加體轉(zhuǎn)軸的位置轉(zhuǎn)動(dòng)慣M薄圓盤m,R通過(guò)中央且與盤面垂直-mR22沿直徑-mR24沿切線5c2一mR4溥壁中空?qǐng)A筒通過(guò)中央軸mR2沿直徑12一mR2剛體轉(zhuǎn)軸的位置轉(zhuǎn)動(dòng)慣m,R中空?qǐng)A柱體m,Ri,旦通過(guò)中央軸-mR12R；2沿直徑工mR2R24通過(guò)中央軸-mR22沿直徑-mR24剛體轉(zhuǎn)軸的位置轉(zhuǎn)動(dòng)慣中實(shí)圓柱體m,R,L沿中央直徑-mR2mL2412沿端面直徑102

42、12一mR-mL43球殼m,R沿直徑21-12一mR3沿切線-mR23加體轉(zhuǎn)軸的位置轉(zhuǎn)動(dòng)慣中空球體m,Ri,R沿直徑2r5R5rm5r2R3球體m,R沿直徑202一mR5沿切線72一mR5第七章習(xí)題選講7.1.4半徑為0.1m的圓盤在鉛直平面內(nèi)轉(zhuǎn)動(dòng),在圓盤平面內(nèi)建立Oxy坐標(biāo)系,原點(diǎn)在軸上,x和y軸沿水平和鉛直向上的方向.邊緣上一點(diǎn)A當(dāng)t.時(shí)恰好在x軸上,該點(diǎn)的角坐標(biāo)滿足1.2tt2(的單位為rad,t的單位為s).(1)t0時(shí),(2)自to開(kāi)始轉(zhuǎn)45o時(shí),(3)轉(zhuǎn)過(guò)90o時(shí),a點(diǎn)的速度和加速度在x和y軸上的投影.解:dt1.22t,dt2.0(1)當(dāng)10時(shí),1.2,x0,axay(2)兀/4時(shí),1.2tt2anat1.2R1.2R2.00.10.12m/s20.144m/s20.10.10.2m/s2t0.47s2