📝 题目
5．设螺旋形弹簧一圈的方程为 $x=a \cos t, y=a \sin t, z=k t$ ，其中 $0 \leqslant t \leqslant 2 \pi$ ，它的线密度 $\rho(x, y, z)= x^{2}+y^{2}+z^{2}$ ．求： （1）它关于 $z$ 轴的转动惯量 $I_{z}$ ； （2）它的质心．
💡 答案与解析
[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 $$ ds = \sqrt{\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}+\left(\frac{dz}{dt}\right)^{2}}\,dt = \sqrt{(-a\sin t)^{2}+(a\cos t)^{2}+k^{2}}\,dt = \sqrt{a^{2}+k^{2}}\,dt $$
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### （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