问题及解答

设螺旋形弹簧一周的参数方程为

$$\{\begin{array}{l}x=a\cos t,\\ y=a\sin t,\\ z=kt,\end{array}$$

这里 $t\in [0,2\pi ]$. 不妨记此曲线为 $\mathrm{\Gamma }$, 设其线密度函数为 $\rho (x,y,z)=x^2+y^2+z^2$, 求:

(1)  $\mathrm{\Gamma }$ 的质量;
(2)  $\mathrm{\Gamma }$ 关于 $z$ 轴的转动惯量;
(3)  $\mathrm{\Gamma }$ 的质心.

(1) 质量为

$$\begin{array}{rl}{\int }_{\mathrm{\Gamma }}\rho (x,y,z)\mathrm{d}s& ={\int }_0^{2\pi }(x^2+y^2+z^2)\sqrt{(x^{\prime }(t){)}^2+(y^{\prime }(t){)}^2+(z^{\prime }(t){)}^2}\mathrm{d}t\\ & ={\int }_0^{2\pi }(a^2{\cos }^{2}t+a^2{\sin }^{2}t+k^2{t}^2)\sqrt{(-a\sin t{)}^2+(a\cos t{)}^2+k^2}\mathrm{d}t\\ & ={\int }_0^{2\pi }(a^2+k^2{t}^2)\sqrt{a^2+k^2}\mathrm{d}t\\ & =\sqrt{a^2+k^2}\cdot (a^{2}t+\frac{1}{3}{k}^2{t}^3){|}_0^{2\pi }\\ & =\sqrt{a^2+k^2}\cdot (2\pi a^2+\frac{8}{3}{k}^2{\pi }^3)\\ & =\frac{2\pi }{3}\sqrt{a^2+k^2}(3a^2+4{\pi }^2{k}^2).\end{array}$$

(2)  对 $z$ 轴的转动惯量为

$${\int }_{\mathrm{\Gamma }}(x^2+y^2)\rho (x,y,z)\mathrm{d}s={\int }_{\mathrm{\Gamma }}{a}^2\rho (x,y,z)\mathrm{d}s=a^2\sqrt{a^2+k^2}\cdot (2\pi a^2+\frac{8}{3}{k}^2{\pi }^3)=\frac{2\pi a^2}{3}\sqrt{a^2+k^2}(3a^2+4{\pi }^2{k}^2).$$

(3)

$$\overline{x}=\frac{{\int }_{\mathrm{\Gamma }}x\rho (x,y,z)\mathrm{d}s}{{\int }_{\mathrm{\Gamma }}\rho (x,y,z)\mathrm{d}s}$$

其中

$$\begin{array}{rl}{\int }_{\mathrm{\Gamma }}x\rho (x,y,z)\mathrm{d}s& ={\int }_0^{2\pi }a\cos t\cdot (a^2+k^2{t}^2)\sqrt{a^2+k^2}\mathrm{d}t\\ & =\sqrt{a^2+k^2}\cdot [a^3{\int }_0^{2\pi }\cos t\mathrm{d}t+a{k}^2{\int }_0^{2\pi }{t}^2\cos t\mathrm{d}t]\\ & =\sqrt{a^2+k^2}\cdot [a^3\cdot \sin t{|}_0^{2\pi }+a{k}^2{\int }_0^{2\pi }{t}^2\mathrm{d}\sin t]\\ & =\sqrt{a^2+k^2}\cdot [0+a{k}^2\cdot (t^2\sin t{|}_0^{2\pi }-{\int }_0^{2\pi }\sin t\mathrm{d}{t}^2)]\end{array}$$

其中

$$\begin{array}{rl}-{\int }_0^{2\pi }2t\sin t\mathrm{d}t& =2{\int }_0^{2\pi }t\mathrm{d}\cos t=2[t\cos t{|}_0^{2\pi }-{\int }_0^{2\pi }\cos t\mathrm{d}t]\\ & =2[(2\pi \cos 2\pi -0)-0]\\ & =4\pi .\end{array}$$

因此, 

$${\int }_{\mathrm{\Gamma }}x\rho (x,y,z)\mathrm{d}s=\sqrt{a^2+k^2}a{k}^2\cdot 4\pi .$$

于是

$$\overline{x}=\frac{{\int }_{\mathrm{\Gamma }}x\rho (x,y,z)\mathrm{d}s}{{\int }_{\mathrm{\Gamma }}\rho (x,y,z)\mathrm{d}s}=\frac{\sqrt{a^2+k^2}a{k}^2\cdot 4\pi }{\frac{2\pi }{3}\sqrt{a^2+k^2}(3a^2+4{\pi }^2{k}^2)}=\frac{6a{k}^2}{3a^2+4{\pi }^2{k}^2}.$$

类似地,

$$\overline{y}=\frac{{\int }_{\mathrm{\Gamma }}y\rho (x,y,z)\mathrm{d}s}{{\int }_{\mathrm{\Gamma }}\rho (x,y,z)\mathrm{d}s}$$

其中

$$\begin{array}{rl}{\int }_{\mathrm{\Gamma }}y\rho (x,y,z)\mathrm{d}s& ={\int }_0^{2\pi }a\sin t\cdot (a^2+k^2{t}^2)\sqrt{a^2+k^2}\mathrm{d}t\\ & =\sqrt{a^2+k^2}\cdot [a^3{\int }_0^{2\pi }\sin t\mathrm{d}t+a{k}^2{\int }_0^{2\pi }{t}^2\sin t\mathrm{d}t]\\ & =\sqrt{a^2+k^2}\cdot [a^3\cdot (-\cos t){|}_0^{2\pi }-a{k}^2{\int }_0^{2\pi }{t}^2\mathrm{d}\cos t]\\ & =\sqrt{a^2+k^2}\cdot [0-a{k}^2\cdot (t^2\cos t{|}_0^{2\pi }-{\int }_0^{2\pi }\cos t\mathrm{d}{t}^2)]\end{array}$$

其中

$$\begin{array}{rl}{\int }_0^{2\pi }2t\cos t\mathrm{d}t& =2{\int }_0^{2\pi }t\mathrm{d}\sin t=2[t\sin t{|}_0^{2\pi }-{\int }_0^{2\pi }\sin t\mathrm{d}t]\\ & =2[0+0]\\ & =0.\end{array}$$

因此, 

$${\int }_{\mathrm{\Gamma }}x\rho (x,y,z)\mathrm{d}s=-\sqrt{a^2+k^2}a{k}^2\cdot 4{\pi }^2.$$

于是

$$\overline{y}=\frac{{\int }_{\mathrm{\Gamma }}x\rho (x,y,z)\mathrm{d}s}{{\int }_{\mathrm{\Gamma }}\rho (x,y,z)\mathrm{d}s}=\frac{-\sqrt{a^2+k^2}a{k}^2\cdot 4{\pi }^2}{\frac{2\pi }{3}\sqrt{a^2+k^2}(3a^2+4{\pi }^2{k}^2)}=\frac{-6\pi a{k}^2}{3a^2+4{\pi }^2{k}^2}.$$

$$\overline{z}=\frac{{\int }_{\mathrm{\Gamma }}z\rho (x,y,z)\mathrm{d}s}{{\int }_{\mathrm{\Gamma }}\rho (x,y,z)\mathrm{d}s}$$

其中

$$\begin{array}{rl}{\int }_{\mathrm{\Gamma }}z\rho (x,y,z)\mathrm{d}s& ={\int }_0^{2\pi }kt\cdot (a^2+k^2{t}^2)\sqrt{a^2+k^2}\mathrm{d}t\\ & =\sqrt{a^2+k^2}\cdot [a^{2}k{\int }_0^{2\pi }t\mathrm{d}t+k^3{\int }_0^{2\pi }{t}^3\mathrm{d}t]\\ & =\sqrt{a^2+k^2}\cdot [a^{2}k\cdot (\frac{1}{2}{t}^2){|}_0^{2\pi }+k^3\cdot (\frac{1}{4}{t}^4){|}_0^{2\pi }]\\ & =\sqrt{a^2+k^2}\cdot [a^{2}k2{\pi }^2+k^{3}4{\pi }^4]\\ & =\sqrt{a^2+k^2}\cdot 2{\pi }^{2}k(a^2+2{\pi }^2{k}^2).\end{array}$$

于是

$$\overline{z}=\frac{{\int }_{\mathrm{\Gamma }}z\rho (x,y,z)\mathrm{d}s}{{\int }_{\mathrm{\Gamma }}\rho (x,y,z)\mathrm{d}s}=\frac{\sqrt{a^2+k^2}\cdot 2{\pi }^{2}k(a^2+2{\pi }^2{k}^2)}{\frac{2\pi }{3}\sqrt{a^2+k^2}(3a^2+4{\pi }^2{k}^2)}=\frac{3\pi k(a^2+2{\pi }^2{k}^2)}{3a^2+4{\pi }^2{k}^2}.$$