\frac{1}{2}m_{2}\Biggl(\frac{d(2l\cos\theta_{1}+l_{2}\cos\theta_{2})}{d t}\Biggr)^{2}简化
时间: 2023-09-02 19:07:27 浏览: 37
将分式展开,得到:
$$
\frac{1}{2}m_{2}\Biggl(\frac{d(2l\cos\theta_{1}+l_{2}\cos\theta_{2})}{d t}\Biggr)^{2}
$$
然后对括号内的式子进行求导,得到:
$$
\frac{1}{2}m_{2}\Bigl(-2l\sin\theta_{1}\frac{d\theta_{1}}{dt}-l_{2}\sin\theta_{2}\frac{d\theta_{2}}{dt}\Bigr)^{2}
$$
将括号内的式子展开,得到:
$$
\frac{1}{2}m_{2}\Bigl(2l\sin\theta_{1}\Bigr)^{2}\Bigl(\frac{d\theta_{1}}{dt}\Bigr)^{2}+m_{2}l_{2}l\sin\theta_{1}\sin\theta_{2}\Bigl(\frac{d\theta_{1}}{dt}\Bigr)\Bigl(\frac{d\theta_{2}}{dt}\Bigr)+\frac{1}{2}m_{2}\Bigl(l_{2}\sin\theta_{2}\Bigr)^{2}\Bigl(\frac{d\theta_{2}}{dt}\Bigr)^{2}
$$
化简后得到:
$$
\frac{1}{2}m_{2}\Bigl(4l^{2}\sin^{2}\theta_{1}\Bigr)\Bigl(\frac{d\theta_{1}}{dt}\Bigr)^{2}+m_{2}l_{2}l\sin\theta_{1}\sin\theta_{2}\Bigl(\frac{d\theta_{1}}{dt}\Bigr)\Bigl(\frac{d\theta_{2}}{dt}\Bigr)+\frac{1}{2}m_{2}\Bigl(l_{2}\sin\theta_{2}\Bigr)^{2}\Bigl(\frac{d\theta_{2}}{dt}\Bigr)^{2}
$$