若\cos \theta _j=\frac{L_{4}^{2}+L_{3}^{2}-L_{AC}^{2}}{2L_4L_3} 则theta等于什么
时间: 2023-11-01 18:29:02 浏览: 32
根据余弦定理,有
$$
\cos \theta_j = \frac{L_4^2 + L_3^2 - L_{AC}^2}{2L_4L_3} = \cos(\pi - \theta_j) = -\cos(\theta_j)
$$
移项可得
$$
\cos \theta_j + \cos \theta_j = 0
$$
即
$$
2\cos \theta_j = 0
$$
因此
$$
\cos \theta_j = 0
$$
解得
$$
\theta_j = \frac{\pi}{2} + k\pi, \quad k \in \mathbb{Z}
$$
注意到$\theta_j$的定义域应该为$[0,\pi]$,因此实际上只有$k=1$时有解,即
$$
\theta_j = \frac{3\pi}{2}
$$
相关问题
\frac{1}{2}m_{2}\Biggl(\frac{d(2l\cos\theta_{1}+l_{2}\cos\theta_{2})}{d t}\Biggr)^{2}简化
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:
$$
\frac{1}{2}m_{2}\Bigl(4l^{2}\sin^{2}\theta_{1}\Bigr)\Bigl(\frac{d\theta_{1}}{dt}\Bigr)^{2}+m_{2}l_{2}l\sin\theta_{1}\sin\theta_{2}\Bigl(\frac{d\theta_{1}}{dt}\Bigr)\Bigl(\frac{d\theta_{2}}{dt}\Bigr)+\frac{1}{2}m_{2}\Bigl(l_{2}\sin\theta_{2}\Bigr)^{2}\Bigl(\frac{d\theta_{2}}{dt}\Bigr)^{2}
$$
$$ \begin{aligned} & \frac{d}{dt}\left( m_{2}l_{2}^{2}\dot{\theta}_{2}+4m_{2}l_{1}l_{2}\dot{\theta}_{1}\cos(\theta_{2}-\theta_{1}) \right)
\end{aligned} $$
根据链式法则,我们有:
$$ \begin{aligned} & \frac{d}{dt}\left( m_{2}l_{2}^{2}\dot{\theta}_{2}+4m_{2}l_{1}l_{2}\dot{\theta}_{1}\cos(\theta_{2}-\theta_{1}) \right) \\ =& m_{2}l_{2}^{2}\ddot{\theta}_{2}+4m_{2}l_{1}l_{2}\left(-\dot{\theta}_{1}\dot{\theta}_{2}\sin(\theta_{2}-\theta_{1})+\ddot{\theta}_{1}\cos(\theta_{2}-\theta_{1})-\dot{\theta}_{1}\dot{\theta}_{2}\sin(\theta_{2}-\theta_{1})-\dot{\theta}_{1}^{2}\sin(\theta_{2}-\theta_{1})\right) \\ =& m_{2}l_{2}^{2}\ddot{\theta}_{2}+4m_{2}l_{1}l_{2}\left(\ddot{\theta}_{1}\cos(\theta_{2}-\theta_{1})-\dot{\theta}_{1}^{2}\sin(\theta_{2}-\theta_{1})-\dot{\theta}_{1}\dot{\theta}_{2}\sin(\theta_{2}-\theta_{1})\right) \end{aligned} $$
其中,$m_2$ 是挂在杆上的质量,$l_1$ 和 $l_2$ 分别是两个杆的长度,$\theta_1$ 和 $\theta_2$ 分别是两个杆的角度,$\dot{\theta}_1$ 和 $\dot{\theta}_2$ 分别是两个杆的角速度,$\ddot{\theta}_1$ 和 $\ddot{\theta}_2$ 分别是两个杆的角加速度。