解析力学案例

XantC

与速度垂直的力只改变其速度方向

\[ 令\boldsymbol{v}_0=(v_{0x}, v_{0y}),\;\boldsymbol{v}=(\varphi, \psi) \]

\[ 则 \boldsymbol{F}=k(-\psi,\varphi),其中k为随t变化的常数 \]

\[ \boldsymbol{v^{'}}=\boldsymbol{a}=\frac{k}{m}(-\psi,\varphi)=(\varphi^{'}, \psi^{'}) \]

\[ \Rightarrow \begin{cases} \displaystyle\varphi^{'}=-\frac{k}{m}\psi\\ \displaystyle\psi^{'}=\frac{k}{m}\varphi\\ \end{cases}\Rightarrow\begin{cases} \displaystyle\varphi=-A\sin\frac{k}{m}t+B\cos\frac{k}{m}t\\ \displaystyle\psi=A\cos\frac{k}{m}t+B\sin\frac{k}{m}t \end{cases} \]

\[ \because v\, \bigg |_{t=0}=(v_{0x}, v_{0y}) \]

\[ \therefore A=v_{0y},\;B=v_{0x} \]

\[ \therefore |\boldsymbol{v}|=\sqrt{\varphi^2+\psi^2}=\sqrt{A^2+B^2}=\sqrt{\boldsymbol{v}_{0x}^2+\boldsymbol{v}_{0y}^2}=|\boldsymbol{v}_0| \]

计算向心加速度大小

\[ \boldsymbol{s}=(r\cos\omega t, r\sin\omega t) \]

\[ \Rightarrow \boldsymbol{a}=(-r\omega^2\cos\omega t, -r\omega^2\sin\omega t) \]

\[ \Rightarrow |\boldsymbol{a}|=r\omega^2 \]