动力学习题

墨蓝 2022-12-08 14:21 179阅读 0赞

# 动力学习题 #

6.1 求一均质的、坐标原点建立在其质心的刚性圆柱体的惯性张量。

解：假设圆柱体的半径为 R R R，高为 h h h，密度为 ρ \\rho ρ，则其总质量 M = 1 2 π R 2 h ρ M=\\frac\{1\}\{2\}\\pi R^2h\\rho M=21πR2hρ。

建立柱坐标系，则有  
 x = r c o s θ y = r s i n θ z = z d v = r d θ d r d z x=rcos\\theta\\\\ y=rsin\\theta \\\\ z=z\\\\ dv=rd\\theta drdz x=rcosθy=rsinθz=zdv=rdθdrdz

I x x = ∫ ∫ ∫ V ( z 2 + y 2 ) ρ d v = ∫ 0 2 π ∫ 0 R ∫ − h / 2 h / 2 ( z 2 + r 2 sin ⁡ 2 θ ) ρ d z d r d θ = ∫ 0 2 π ∫ 0 R ( r 3 h sin ⁡ 2 θ + 1 12 h 3 r ) ρ d r d θ = ∫ 0 2 π ( 1 4 R 4 h sin ⁡ 2 θ + 1 24 R 2 h 3 ) ρ d θ = 1 8 R 4 h ρ ∫ 0 2 π ( 1 − cos ⁡ 2 θ ) d θ + ∫ 0 2 π 1 24 R 2 h 3 d θ = 1 8 π R 4 h ρ + 1 12 π R 2 h 3 ρ = 1 4 M R 2 + 1 12 M h 2 I\_\{xx\}=\\int \\int \\int\_V (z^2+y^2)\\rho dv\\\\ =\\int\_0^\{2\\pi\}\\int\_0^R\\int\_\{-h/2\}^\{h/2\}(z^2+r^2\\sin^2\\theta)\\rho dzdrd\\theta\\\\ =\\int\_0^\{2\\pi\}\\int\_0^R(r^3h\\sin^2\\theta+\\frac\{1\}\{12\}h^3r )\\rho drd\\theta \\\\ =\\int\_0^\{2\\pi\}(\\frac\{1\}\{4\}R^4h\\sin^2\\theta+\\frac\{1\}\{24\}R^2h^3)\\rho d\\theta \\\\ =\\frac\{1\}\{8\}R^4h\\rho \\int\_0^\{2\\pi\}(1-\\cos2\\theta)d\\theta+\\int\_0^\{2\\pi\}\\frac\{1\}\{24\}R^2h^3d\\theta\\\\ =\\frac\{1\}\{8\}\\pi R^4h\\rho+\\frac\{1\}\{12\}\\pi R^2h^3\\rho\\\\ =\\frac\{1\}\{4\}MR^2+\\frac\{1\}\{12\}Mh^2 Ixx=∫∫∫V(z2\+y2)ρdv=∫02π∫0R∫−h/2h/2(z2\+r2sin2θ)ρdzdrdθ=∫02π∫0R(r3hsin2θ\+121h3r)ρdrdθ=∫02π(41R4hsin2θ\+241R2h3)ρdθ=81R4hρ∫02π(1−cos2θ)dθ\+∫02π241R2h3dθ=81πR4hρ\+121πR2h3ρ=41MR2\+121Mh2

同理，  
 I y y = 1 4 M R 2 + 1 12 M h 2 I\_\{yy\}=\\frac\{1\}\{4\}MR^2+\\frac\{1\}\{12\}Mh^2 Iyy=41MR2\+121Mh2

I z z = ∫ ∫ ∫ V ( x 2 + y 2 ) ρ d v = ∫ ∫ ∫ V r 2 ρ d v = 1 12 M R 2 I\_\{zz\}=\\int \\int\\int\_V(x^2+y^2)\\rho dv=\\int \\int\\int\_Vr^2\\rho dv=\\frac\{1\}\{12\}MR^2 Izz=∫∫∫V(x2\+y2)ρdv=∫∫∫Vr2ρdv=121MR2

由于对称性，或者通过积分计算可得  
 I x y = I x z = I y z = 0 I\_\{xy\}=I\_\{xz\}=I\_\{yz\}=0 Ixy=Ixz=Iyz=0  
因此，惯性张量为  
 C I = \[ 1 4 M R 2 + 1 12 M h 2 0 0 0 1 4 M R 2 + 1 12 M h 2 0 0 0 1 12 M R 2 \] ^CI= \\left\[ \\begin\{matrix\} \\frac\{1\}\{4\}MR^2+\\frac\{1\}\{12\}Mh^2 &0 &0\\\\ 0& \\frac\{1\}\{4\}MR^2+\\frac\{1\}\{12\}Mh^2& 0\\\\ 0&0&\\frac\{1\}\{12\}MR^2 \\end\{matrix\}\\right\] CI=⎣⎡41MR2\+121Mh200041MR2\+121Mh2000121MR2⎦⎤  
6.2 建立6.7节中二连杆操作臂（如下图所示）的动力学方程。将每个连杆看作为一个均质矩形刚体。各连杆的尺寸为 l i , w i , h i l\_i,w\_i,h\_i li,wi,hi，总质量为 m i m\_i mi。

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解：由于连杆看成均质矩形，因此在连杆坐标系下，惯性张量各分量为  
 1 I 1 x x = ∫ ∫ S y 2 ρ d s = ∫ 0 l 1 ∫ − h 1 / 2 h 1 / 2 y 2 ρ d y d x = 1 12 m 1 h 1 2 ^1I\_\{1xx\}=\\int\\int\_S y^2\\rho ds=\\int^\{l\_1\}\_0\\int^\{h\_1/2\}\_\{-h\_1/2\}y^2\\rho dydx=\\frac\{1\}\{12\}m\_1h\_1^2 1I1xx=∫∫Sy2ρds=∫0l1∫−h1/2h1/2y2ρdydx=121m1h12

1 I 1 y y = ∫ ∫ S x 2 ρ d s = ∫ 0 l 1 ∫ − h 1 / 2 h 1 / 2 x 2 ρ d y d x = 1 3 m 1 l 1 2 ^1I\_\{1yy\}=\\int\\int\_S x^2\\rho ds=\\int^\{l\_1\}\_0\\int^\{h\_1/2\}\_\{-h\_1/2\}x^2\\rho dydx=\\frac\{1\}\{3\}m\_1l\_1^2 1I1yy=∫∫Sx2ρds=∫0l1∫−h1/2h1/2x2ρdydx=31m1l12

1 I 1 z z = ∫ ∫ S ( x 2 + y 2 ) ρ d s = ∫ 0 l 1 ∫ − h 1 / 2 h 1 / 2 ( x 2 + y 2 ) ρ d y d x = 1 12 m 1 h 1 2 + 1 3 m 1 l 1 2 ^1I\_\{1zz\}=\\int\\int\_S (x^2+y^2)\\rho ds=\\int^\{l\_1\}\_0\\int^\{h\_1/2\}\_\{-h\_1/2\}(x^2+y^2)\\rho dydx=\\frac\{1\}\{12\}m\_1h\_1^2+\\frac\{1\}\{3\}m\_1l\_1^2 1I1zz=∫∫S(x2\+y2)ρds=∫0l1∫−h1/2h1/2(x2\+y2)ρdydx=121m1h12\+31m1l12

1 I 1 x y = 0 1 I 1 x z = 0 1 I 1 y z = 0 ^1I\_\{1xy\}=0 \\\\^1I\_\{1xz\}=0\\\\^1I\_\{1yz\}=0 1I1xy=01I1xz=01I1yz=0

因此连杆i的惯性张量为  
 1 I 1 = \[ 1 12 m 1 h 1 2 0 0 0 1 3 m 1 l 1 2 0 0 0 1 12 m 1 h 1 2 + 1 3 m 1 l 1 2 \] ^1I\_1=\\left\[ \\begin\{matrix\} \\frac\{1\}\{12\}m\_1h\_1^2& 0&0\\\\ 0&\\frac\{1\}\{3\}m\_1l\_1^2 &0\\\\ 0&0 & \\frac\{1\}\{12\}m\_1h\_1^2+\\frac\{1\}\{3\}m\_1l\_1^2 \\end\{matrix\}\\right\] 1I1=⎣⎡121m1h1200031m1l12000121m1h12\+31m1l12⎦⎤  
同理可得  
 2 I 2 = \[ 1 12 m 2 h 2 2 0 0 0 1 3 m 2 l 2 2 0 0 0 1 12 m 2 h 2 2 + 1 3 m 2 l 2 2 \] ^2I\_2=\\left\[ \\begin\{matrix\} \\frac\{1\}\{12\}m\_2h\_2^2& 0&0\\\\ 0&\\frac\{1\}\{3\}m\_2l\_2^2 &0\\\\ 0&0 & \\frac\{1\}\{12\}m\_2h\_2^2+\\frac\{1\}\{3\}m\_2l\_2^2 \\end\{matrix\}\\right\] 2I2=⎣⎡121m2h2200031m2l22000121m2h22\+31m2l22⎦⎤  
每个连杆的质心相对于本身的连杆坐标系的位置矢量为  
 1 P C 1 = l 1   1 X ^ 1 = \[ l 1 2 , 0 , 0 \] T 2 P C 2 = l 2   2 X ^ 2 = \[ l 2 2 , 0 , 0 \] T ^1P\_\{C1\}=l\_1 \\ ^1\\widehat X\_1=\[\\frac\{l\_1\}\{2\},0,0\]^T \\\\ ^2P\_\{C2\}=l\_2 \\ ^2\\widehat X\_2=\[\\frac\{l\_2\}\{2\},0,0\]^T \\\\ 1PC1=l1 1X1=\[2l1,0,0\]T2PC2=l2 2X2=\[2l2,0,0\]T

利用平行移轴定理的矢量-矩阵形式  
 C 1 I 1 = 1 I 1 − m i ( i P c 1 T ( i P c 1 ) I 3 − P c 1 P c 1 T ) ^\{C1\}I\_1=^1I\_1-m\_i(^i P^\{T \}\_\{c1\}( ^i P\_\{c1\}) I\_3-P\_\{c1\}P\_\{c1\}^T) C1I1=1I1−mi(iPc1T(iPc1)I3−Pc1Pc1T)

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