同济高数 第11章 第11-1-5题

[AI解答]

[AI解答]
我们已知螺旋弹簧一圈的参数方程：
$$
x = a\cos t,\quad y = a\sin t,\quad z = kt,\quad 0\le t\le 2\pi
$$
线密度为
$$
\rho(x,y,z) = x^{2}+y^{2}+z^{2} = a^{2} + k^{2}t^{2}
$$
弧长微元为
$$
ds = \sqrt{\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}+\left(\frac{dz}{dt}\right)^{2}}\,dt
= \sqrt{(-a\sin t)^{2}+(a\cos t)^{2}+k^{2}}\,dt = \sqrt{a^{2}+k^{2}}\,dt
$$

---

（1）求转动惯量 $I_z$

关于 $z$ 轴的转动惯量公式为
$$
I_z = \int_{L} (x^{2}+y^{2})\,\rho\,ds
$$
由于 $x^{2}+y^{2} = a^{2}$，代入得
$$
I_z = \int_{0}^{2\pi} a^{2} \cdot (a^{2}+k^{2}t^{2}) \cdot \sqrt{a^{2}+k^{2}}\,dt
$$
提出常数：
$$
I_z = a^{2}\sqrt{a^{2}+k^{2}} \int_{0}^{2\pi} (a^{2}+k^{2}t^{2})\,dt
$$
计算积分：
$$
\int_{0}^{2\pi} a^{2}\,dt = 2\pi a^{2},\quad
\int_{0}^{2\pi} k^{2}t^{2}\,dt = k^{2}\cdot\frac{(2\pi)^{3}}{3} = \frac{8\pi^{3}k^{2}}{3}
$$
因此
$$
I_z = a^{2}\sqrt{a^{2}+k^{2}}\left(2\pi a^{2} + \frac{8\pi^{3}k^{2}}{3}\right)
= 2\pi a^{2}\sqrt{a^{2}+k^{2}}\left(a^{2} + \frac{4\pi^{2}k^{2}}{3}\right)
$$

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（2）求质心 $(\bar{x},\bar{y},\bar{z})$

总质量
$$
M = \int_{L} \rho\,ds = \sqrt{a^{2}+k^{2}}\int_{0}^{2\pi} (a^{2}+k^{2}t^{2})\,dt
= \sqrt{a^{2}+k^{2}}\left(2\pi a^{2} + \frac{8\pi^{3}k^{2}}{3}\right)
$$

质心坐标公式：
$$
\bar{x} = \frac{1}{M}\int_{L} x\,\rho\,ds,\quad
\bar{y} = \frac{1}{M}\int_{L} y\,\rho\,ds,\quad
\bar{z} = \frac{1}{M}\int_{L} z\,\rho\,ds
$$

先计算 $\int x\rho\,ds$：
$$
\int_{0}^{2\pi} a\cos t \cdot (a^{2}+k^{2}t^{2})\cdot\sqrt{a^{2}+k^{2}}\,dt
= a\sqrt{a^{2}+k^{2}}\int_{0}^{2\pi} (a^{2}+k^{2}t^{2})\cos t\,dt
$$
由于 $\cos t$ 在整周期积分为零，且 $t^{2}\cos t$ 是偶函数在对称区间？这里区间是 $[0,2\pi]$，不是对称于0，但可直接计算：
$$
\int_{0}^{2\pi} t^{2}\cos t\,dt
$$
用分部积分：
令 $u=t^{2}, dv=\cos t\,dt$，则 $du=2t\,dt, v=\sin t$
$$
\int t^{2}\cos t\,dt = t^{2}\sin t - 2\int t\sin t\,dt
$$
再对 $\int t\sin t\,dt$ 分部：$u=t, dv=\sin t\,dt$，得 $-t\cos t + \int \cos t\,dt = -t\cos t + \sin t$
所以
$$
\int t^{2}\cos t\,dt = t^{2}\sin t - 2(-t\cos t + \sin t) = t^{2}\sin t + 2t\cos t - 2\sin t
$$
代入上下限 $0$ 到 $2\pi$：
在 $t=2\pi$：$ (4\pi^{2})(0) + 2(2\pi)(1) - 2(0) = 4\pi$
在 $t=0$：$0+0-0=0$
所以 $\int_{0}^{2\pi} t^{2}\cos t\,dt = 4\pi$
因此
$$
\int x\rho\,ds = a\sqrt{a^{2}+k^{2}}\left(a^{2}\cdot 0 + k^{2}\cdot 4\pi\right) = 4\pi a k^{2}\sqrt{a^{2}+k^{2}}
$$
于是
$$
\bar{x} = \frac{4\pi a k^{2}\sqrt{a^{2}+k^{2}}}{M}
$$
而 $M = \sqrt{a^{2}+k^{2}}\left(2\pi a^{2} + \frac{8\pi^{3}k^{2}}{3}\right)$，约去 $\sqrt{a^{2}+k^{2}}$：
$$
\bar{x} = \frac{4\pi a k^{2}}{2\pi a^{2} + \frac{8\pi^{3}k^{2}}{3}}
= \frac{4\pi a k^{2}}{2\pi\left(a^{2} + \frac{4\pi^{2}k^{2}}{3}\right)}
= \frac{2a k^{2}}{a^{2} + \frac{4\pi^{2}k^{2}}{3}}
$$

同理，$\bar{y}$ 计算：
$$
\int y\rho\,ds = a\sqrt{a^{2}+k^{2}}\int_{0}^{2\pi} (a^{2}+k^{2}t^{2})\sin t\,dt
$$
$\int_{0}^{2\pi} \sin t\,dt = 0$，$\int_{0}^{2\pi} t^{2}\sin t\,dt$ 计算如下：
分部：$u=t^{2}, dv=\sin t\,dt$，得
$$
\int t^{2}\sin t\,dt = -t^{2}\cos t + 2\int t\cos t\,dt
$$
而 $\int t\cos t\,dt = t\sin t + \cos t$，所以
$$
\int t^{2}\sin t\,dt = -t^{2}\cos t + 2(t\sin t + \cos t) = -t^{2}\cos t + 2t\sin t + 2\cos t
$$
代入上下限：
$t=2\pi$：$-4\pi^{2}(1) + 0 + 2(1) = -4\pi^{2}+2$
$t=0$：$0+0+2 = 2$
差值为 $(-4\pi^{2}+2)-2 = -4\pi^{2}$
因此
$$
\int y\rho\,ds = a\sqrt{a^{2}+k^{2}} \cdot k^{2}(-4\pi^{2}) = -4\pi^{2} a k^{2}\sqrt{a^{2}+k^{2}}
$$
所以
$$
\bar{y} = \frac{-4\pi^{2} a k^{2}}{2\pi a^{2} + \frac{8\pi^{3}k^{2}}{3}}
= \frac{-4\pi^{2} a k^{2}}{2\pi\left(a^{2} + \frac{4\pi^{2}k^{2}}{3}\right)}
= \frac{-2\pi a k^{2}}{a^{2} + \frac{4\pi^{2}k^{2}}{3}}
$$

最后 $\bar{z}$：
$$
\int z\rho\,ds = \sqrt{a^{2}+k^{2}}\int_{0}^{2\pi} kt (a^{2}+k^{2}t^{2})\,dt
= k\sqrt{a^{2}+k^{2}}\int_{0}^{2\pi} (a^{2}t + k^{2}t^{3})\,dt
$$
计算积分：
$$
\int_{0}^{2\pi} a^{2}t\,dt = a^{2}\cdot\frac{(2\pi)^{2}}{2} = 2\pi^{2}a^{2}
$$
$$
\int_{0}^{2\pi} k^{2}t^{3}\,dt = k^{2}\cdot\frac{(2\pi)^{4}}{4} = k^{2}\cdot\frac{16\pi^{4}}{4} = 4\pi^{4}k^{2}
$$
所以
$$
\int z\rho\,ds = k\sqrt{a^{2}+k^{2}}\left