Lee Ein's Blog: Đề cương ôn tập cơ lý thuyết

18/5/12

Đề cương ôn tập cơ lý thuyết

1. Thiết lập công thức tính vận tốc, gia tốc

a. Tọa độ cực:

- Phương trình chuyển động: $\left\{ {\begin{array}{ccccccccccccccc} {r = r\left( t \right)}\\ {\varphi = \varphi \left( t \right)} \end{array}} \right.$

với $\left\{ {\begin{array}{ccccccccccccccc} {x = r\cos \varphi }\\ {y = r\sin \varphi } \end{array}} \right.$ ( $\begin{array}{l} 0 \le r = OM < \infty \\ 0 \le \varphi \le 2\pi \end{array}$ )

- Chọn ${q_1} = r$ , ${{q_1}}\limits^ \circ = r\limits^ \circ$ , ${{e_1}}\limits^ \to = {{e_r}}\limits^ \to$ , ${h_1} = {h_r}$

${q_2} = \varphi$ , ${{q_2}}\limits^ \circ = \varphi \limits^ \circ$ , ${{e_2}}\limits^ \to = {{e_\varphi }}\limits^ \to$ , ${h_2} = {h_\varphi }$

${h_1} = {h_r} = \sqrt {{{\left( {\frac{{\partial x}}{{\partial r}}} \right)}^2} + {{\left( {\frac{{\partial y}}{{\partial r}}} \right)}^2} + {0^2}} = \sqrt {{{\cos }^2}x + {{\sin }^2}x} = 1$

${h_2} = {h_\varphi } = \sqrt {{{\left( {\frac{{\partial x}}{{\partial \varphi }}} \right)}^2} + {{\left( {\frac{{\partial y}}{{\partial \varphi }}} \right)}^2} + {0^2}}$ $= \sqrt {{{\left( { - r\sin \varphi } \right)}^2} + {{\left( {r\cos \varphi } \right)}^2}} = r$

+ $v\limits^ \to = \sum\limits_i {{h_i} {{q_i}}\limits^ \circ } {{e_i}}\limits^ \to =$ ${h_r} {{q_r}}\limits^ \circ {{e_r}}\limits^ \to + {h_\varphi } {{q_\varphi }}\limits^ \circ {{e_\varphi }}\limits^ \to$

Vậy: $v\limits^ \to = r\limits^ \circ {{e_r}}\limits^ \to + {r\varphi }\limits^ \circ {{e_\varphi }}\limits^ \to$ hay ${v^2} = { r\limits^ \circ ^2} + {{r^2}{\varphi ^2}}\limits^{{\rm{ }} \circ } {{e_\varphi }}\limits^ \to$

+ $a\limits^ \to = \sum\limits_i {{a_i} {{e_{_i}}}\limits^ \to } = {a_r} {{e_{_r}}}\limits^ \to + {a_\varphi } {{e_\varphi }}\limits^ \to$

* ${a_r} = \frac{1}{{2{h_r}}}\left[ {\frac{d}{{dt}}\left( {\frac{{\partial {v^2}}}{{\partial r\limits^ \circ }}} \right) - \frac{{\partial {v^2}}}{{\partial r}}} \right]$ $= \frac{1}{2}\left[ {\frac{d}{{dt}}\left( {2 r\limits^ \circ } \right) - 2r{{ \varphi \limits^ \circ }^2}} \right]$

Vậy: ${a_r} = r\limits^{ \circ \circ } - r{ \varphi \limits^ \circ ^2}$

* ${a_\varphi } = \frac{1}{{2{h_\varphi }}}\left[ {\frac{d}{{dt}}\left( {\frac{{\partial {v^2}}}{{\partial \varphi \limits^ \circ }}} \right) - \frac{{\partial {v^2}}}{{\partial \varphi }}} \right]$ $= \frac{1}{{2r}}\left[ {\frac{d}{{dt}}\left( {2{r^2} r\limits^ \circ } \right) - 0} \right]$

$= \frac{1}{r}\frac{d}{{dt}}\left( {{r^2} r\limits^ \circ } \right) = \frac{1}{r}\left( {2r r\limits^ \circ \varphi \limits^ \circ + {r^2} \varphi \limits^{ \circ \circ } } \right)$

Vậy: ${a_\varphi } = 2 r\limits^ \circ \varphi \limits^ \circ + r \varphi \limits^{ \circ \circ }$

Lee Ein's Blog

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18/5/12

Đề cương ôn tập cơ lý thuyết

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