# 问题及解答

## 求螺旋形弹簧的质量

Posted by haifeng on 2024-05-18 22:40:16 last update 2024-05-18 22:40:16 | Edit | Answers (1)

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

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

1

Posted by haifeng on 2024-05-19 09:08:22

(1) 质量为

$\begin{split} \int_{\Gamma}\rho(x,y,z)\mathrm{d}s&=\int_{0}^{2\pi}(x^2+y^2+z^2)\sqrt{(x'(t))^2+(y'(t))^2+(z'(t))^2}\mathrm{d}t\\ &=\int_{0}^{2\pi}(a^2\cos^2 t+a^2\sin^2 t+k^2 t^2)\sqrt{(-a\sin t)^2+(a\cos t)^2+k^2}\mathrm{d}t\\ &=\int_{0}^{2\pi}(a^2+k^2 t^2)\sqrt{a^2+k^2}\mathrm{d}t\\ &=\sqrt{a^2+k^2}\cdot(a^2 t+\frac{1}{3}k^2 t^3)\biggr|_{0}^{2\pi}\\ &=\sqrt{a^2+k^2}\cdot(2\pi a^2+\frac{8}{3}k^2\pi^3)\\ &=\frac{2\pi}{3}\sqrt{a^2+k^2}(3a^2+4\pi^2k^2). \end{split}$

(2)  对 $z$ 轴的转动惯量为

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

(3)

$\bar{x}=\frac{\int_{\Gamma}x\rho(x,y,z)\mathrm{d}s}{\int_{\Gamma}\rho(x,y,z)\mathrm{d}s}$

$\begin{split} \int_{\Gamma}x\rho(x,y,z)\mathrm{d}s&=\int_{0}^{2\pi}a\cos t\cdot(a^2+k^2 t^2)\sqrt{a^2+k^2}\mathrm{d}t\\ &=\sqrt{a^2+k^2}\cdot\biggl[a^3\int_{0}^{2\pi}\cos t\mathrm{d}t+ak^2\int_{0}^{2\pi}t^2\cos t\mathrm{d}t\biggr]\\ &=\sqrt{a^2+k^2}\cdot\biggl[a^3\cdot\sin t\biggr|_{0}^{2\pi}+ak^2\int_{0}^{2\pi}t^2\mathrm{d}\sin t\biggr]\\ &=\sqrt{a^2+k^2}\cdot\biggl[0+ak^2\cdot\Bigl(t^2\sin t\biggr|_{0}^{2\pi}-\int_{0}^{2\pi}\sin t\mathrm{d}t^2\Bigr)\biggr] \end{split}$

$\begin{split} -\int_{0}^{2\pi}2t\sin t\mathrm{d}t&=2\int_{0}^{2\pi}t\mathrm{d}\cos t=2\biggl[t\cos t\biggr|_{0}^{2\pi}-\int_{0}^{2\pi}\cos t\mathrm{d}t\biggr]\\ &=2\Bigl[(2\pi\cos 2\pi-0)-0\Bigr]\\ &=4\pi. \end{split}$

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

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

$\bar{y}=\frac{\int_{\Gamma}y\rho(x,y,z)\mathrm{d}s}{\int_{\Gamma}\rho(x,y,z)\mathrm{d}s}$

$\begin{split} \int_{\Gamma}y\rho(x,y,z)\mathrm{d}s&=\int_{0}^{2\pi}a\sin t\cdot(a^2+k^2 t^2)\sqrt{a^2+k^2}\mathrm{d}t\\ &=\sqrt{a^2+k^2}\cdot\biggl[a^3\int_{0}^{2\pi}\sin t\mathrm{d}t+ak^2\int_{0}^{2\pi}t^2\sin t\mathrm{d}t\biggr]\\ &=\sqrt{a^2+k^2}\cdot\biggl[a^3\cdot(-\cos t)\biggr|_{0}^{2\pi}-ak^2\int_{0}^{2\pi}t^2\mathrm{d}\cos t\biggr]\\ &=\sqrt{a^2+k^2}\cdot\biggl[0-ak^2\cdot\Bigl(t^2\cos t\biggr|_{0}^{2\pi}-\int_{0}^{2\pi}\cos t\mathrm{d}t^2\Bigr)\biggr] \end{split}$

$\begin{split} \int_{0}^{2\pi}2t\cos t\mathrm{d}t&=2\int_{0}^{2\pi}t\mathrm{d}\sin t=2\biggl[t\sin t\biggr|_{0}^{2\pi}-\int_{0}^{2\pi}\sin t\mathrm{d}t\biggr]\\ &=2\Bigl[0+0\Bigr]\\ &=0. \end{split}$

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

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

$\bar{z}=\frac{\int_{\Gamma}z\rho(x,y,z)\mathrm{d}s}{\int_{\Gamma}\rho(x,y,z)\mathrm{d}s}$

$\begin{split} \int_{\Gamma}z\rho(x,y,z)\mathrm{d}s&=\int_{0}^{2\pi}kt\cdot(a^2+k^2 t^2)\sqrt{a^2+k^2}\mathrm{d}t\\ &=\sqrt{a^2+k^2}\cdot\biggl[a^2 k\int_{0}^{2\pi}t\mathrm{d}t+k^3\int_{0}^{2\pi}t^3\mathrm{d}t\biggr]\\ &=\sqrt{a^2+k^2}\cdot\biggl[a^2 k\cdot(\frac{1}{2}t^2)\biggr|_{0}^{2\pi}+k^3\cdot(\frac{1}{4}t^4)\biggr|_{0}^{2\pi}\biggr]\\ &=\sqrt{a^2+k^2}\cdot\biggl[a^2 k 2\pi^2+k^3 4\pi^4\biggr]\\ &=\sqrt{a^2+k^2}\cdot 2\pi^2 k(a^2+2\pi^2 k^2). \end{split}$

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