這裡記錄了我對書上知識的整理，以及對一些例題/習題的解法，這些解法不完全遵循書上的解法，而是結合了我自己的思考。希望對同樣學習該科目的人有幫助。
理論力學w，運動學w，動力學w
1D r = r ( t ) 2D cartesian r → = r → ( t ) = x i → + y j → = x ( t ) i → + y ( t ) j → polar r → = r → ( t ) = r e r → = r ( i → cos θ + j → sin θ ) = r ( t ) e r → ( θ ( t ) ) = r ( t ) ( i → cos θ ( t ) + j → sin θ ( t ) ) 3D cartesian r → = r → ( t ) = x i → + y j → + z k → = x ( t ) i → + y ( t ) j → + z ( t ) k → cylindrical r → = r → ( t ) = ρ e ρ → + z k → = ρ ( i → cos φ + j → sin φ ) + z k → spherical r → = r → ( t ) = r e r → = r ( i → sin θ cos φ + j → sin θ sin φ + k → cos θ ) {\displaystyle {\begin{alignedat}{4}&{\text{1D }}\ &&&&r=\,r(t)\\&{\text{2D}}&&{\text{cartesian}}&&{\vec {r}}={\vec {r}}(t)=x{\vec {i}}+y{\vec {j}}&&=x(t){\vec {i}}+y(t){\vec {j}}\\&&&{\text{polar}}&&{\vec {r}}={\vec {r}}(t)=r{\vec {e_{r}}}\qquad \ =r\left({\vec {i}}\cos \theta +{\vec {j}}\sin \theta \right)&&=r(t){\vec {e_{r}}}(\theta (t))=r(t)\left({\vec {i}}\cos \theta (t)+{\vec {j}}\sin \theta (t)\right)\\&{\text{3D}}&&{\text{cartesian}}&&{\vec {r}}={\vec {r}}(t)=x{\vec {i}}+y{\vec {j}}+z{\vec {k}}&&=x(t){\vec {i}}+y(t){\vec {j}}+z(t){\vec {k}}\\&&&{\text{cylindrical}}\quad &&{\vec {r}}={\vec {r}}(t)=\rho {\vec {e_{\rho }}}+z{\vec {k}}=\rho \left({\vec {i}}\cos \varphi +{\vec {j}}\sin \varphi \right)+z{\vec {k}}\\&&&{\text{spherical}}&&{\vec {r}}={\vec {r}}(t)=r{\vec {e_{r}}}=r\left({\vec {i}}\sin \theta \cos \varphi +{\vec {j}}\sin \theta \sin \varphi +{\vec {k}}\cos \theta \right)\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\end{alignedat}}}
1D v = r ˙ 2D cartesian r → = x i → + y j → r = | r → | = x 2 + y 2 v → = r → ˙ = x ˙ i → + y ˙ j → = v x i → + v y j → v = | v → | = x ˙ 2 + y ˙ 2 = v x 2 + v y 2 a → = v → ˙ = x ¨ i → + y ¨ j → = a x i → + a y j → a = | a → | = x ¨ 2 + y ¨ 2 = a x 2 + a y 2 polar r → = ( r − r θ ˙ 2 ) ( i → cos θ + j → sin θ ) = r r e r → v → = r → ˙ = ( r ˙ − r θ ˙ 2 ) ( i → cos θ + j → sin θ ) + ( 2 r θ ˙ + r θ ¨ ) ( − i → sin θ + j → cos θ ) = v r e r → + v θ e θ → v = | v → | = r ˙ 2 + ( r θ ˙ ) 2 = v r 2 + v θ 2 a → = v → ˙ = ( r ¨ − r θ ˙ 2 ) ( i → cos θ + j → sin θ ) + ( 2 r ˙ θ ˙ + r θ ¨ ) ( − i → sin θ + j → cos θ ) = a r e r → + a θ e θ → a = | a → | = ( r ¨ − r θ ˙ 2 ) 2 + ( 2 r ˙ θ ˙ + r θ ¨ ) 2 = a r 2 + a θ 2 local v → = ( v − r θ ˙ 2 ) ( i → cos θ + j → sin θ ) = v t e t → a → = v → ˙ = ( v ˙ − r θ ˙ 2 ) ( i → cos θ + j → sin θ ) + ( 2 v θ ˙ + r θ ¨ ) ( − i → sin θ + j → cos θ ) = a t e t → + a n e n → a = | a → | = v ˙ 2 + ( v θ ˙ ) 2 = a t 2 + a n 2 ρ = d s d θ = s ˙ θ ˙ = v θ ˙ = v 2 v θ ˙ = v 2 a n 3D cartesian r → = x i → + y j → + z k → r = | r → | = x 2 + y 2 + z 2 v → = r → ˙ = x ˙ i → + y ˙ j → + z ˙ k → = v x i → + v y j → + v z k → v = | v → | = x ˙ 2 + y ˙ 2 + z ˙ 2 = v x 2 + v y 2 + v z 2 a → = v → ˙ = x ¨ i → + y ¨ j → + z ¨ k → = a x i → + a y j → + a z k → a = | a → | = x ¨ 2 + y ¨ 2 + z ¨ 2 = a x 2 + a y 2 + a z 2 {\displaystyle {\begin{alignedat}{6}&{\text{1D }}\ &&&&v&&={\dot {r}}\\&{\text{2D}}&&{\text{cartesian}}\quad &&{\vec {r}}&&&&=x{\vec {i}}+y{\vec {j}}\\&&&&&r&&=|{\vec {r}}|&&={\sqrt {x^{2}+y^{2}}}\\&&&&&{\vec {v}}&&={\dot {\vec {r}}}&&={\dot {x}}{\vec {i}}+{\dot {y}}{\vec {j}}&&=v_{x}{\vec {i}}+v_{y}{\vec {j}}\\&&&&&v&&=|{\vec {v}}|&&={\sqrt {{\dot {x}}^{2}+{\dot {y}}^{2}}}&&={\sqrt {v_{x}^{2}+v_{y}^{2}}}\\&&&&&{\vec {a}}&&={\dot {\vec {v}}}&&={\ddot {x}}{\vec {i}}+{\ddot {y}}{\vec {j}}&&=a_{x}{\vec {i}}+a_{y}{\vec {j}}\\&&&&&a&&=|{\vec {a}}|&&={\sqrt {{\ddot {x}}^{2}+{\ddot {y}}^{2}}}&&={\sqrt {a_{x}^{2}+a_{y}^{2}}}\\&&&{\text{polar}}&&{\vec {r}}&&&&={\hphantom {\Big (}}r{\hphantom {{}-r{\dot {\theta }}^{2}{\Big )}}}\left({\vec {i}}\cos \theta +{\vec {j}}\sin \theta \right)&&=r_{\phantom {r}}{\vec {e_{r}}}\\&&&&&{\vec {v}}&&={\dot {\vec {r}}}&&={\hphantom {\Big (}}{\dot {r}}{\hphantom {{}-r{\dot {\theta }}^{2}{\Big )}}}\left({\vec {i}}\cos \theta +{\vec {j}}\sin \theta \right)+{\hphantom {{\Big (}2}}r{\dot {\theta }}{\hphantom {{}+r{\ddot {\theta }}{\Big )}}}\left(-{\vec {i}}\sin \theta +{\vec {j}}\cos \theta \right)&&=v_{r}{\vec {e_{r}}}+v_{\theta }{\vec {e_{\theta }}}\\&&&&&v&&=|{\vec {v}}|&&={\sqrt {{\dot {r}}^{2}+\left(r{\dot {\theta }}\right)^{2}}}&&={\sqrt {v_{r}^{2}+v_{\theta }^{2}}}\\&&&&&{\vec {a}}&&={\dot {\vec {v}}}&&=\left({\ddot {r}}-r{\dot {\theta }}^{2}\right)\left({\vec {i}}\cos \theta +{\vec {j}}\sin \theta \right)+\left(2{\dot {r}}{\dot {\theta }}+r{\ddot {\theta }}\right)\left(-{\vec {i}}\sin \theta +{\vec {j}}\cos \theta \right)&&=a_{r}{\vec {e_{r}}}+a_{\theta }{\vec {e_{\theta }}}\\&&&&&a&&=|{\vec {a}}|&&={\sqrt {\left({\ddot {r}}-r{\dot {\theta }}^{2}\right)^{2}+\left(2{\dot {r}}{\dot {\theta }}+r{\ddot {\theta }}\right)^{2}}}&&={\sqrt {a_{r}^{2}+a_{\theta }^{2}}}\\&&&{\text{local}}&&{\vec {v}}&&&&={\hphantom {\Big (}}v{\hphantom {{}-r{\dot {\theta }}^{2}{\Big )}}}\left({\vec {i}}\cos \theta +{\vec {j}}\sin \theta \right)&&=v_{\phantom {t}}{\vec {e_{t}}}\\&&&&&{\vec {a}}&&={\dot {\vec {v}}}&&={\hphantom {\Big (}}{\dot {v}}{\hphantom {{}-r{\dot {\theta }}^{2}{\Big )}}}\left({\vec {i}}\cos \theta +{\vec {j}}\sin \theta \right)+{\hphantom {{\Big (}2}}v{\dot {\theta }}{\hphantom {{}+r{\ddot {\theta }}{\Big )}}}\left(-{\vec {i}}\sin \theta +{\vec {j}}\cos \theta \right)&&=a_{t}{\vec {e_{t}}}+a_{n}{\vec {e_{n}}}\\&&&&&a&&=|{\vec {a}}|&&={\sqrt {{\dot {v}}^{2}+\left(v{\dot {\theta }}\right)^{2}}}&&={\sqrt {a_{t}^{2}+a_{n}^{2}}}\\&&&&&\rho &&={ds \over d\theta }&&={\frac {\dot {s}}{\dot {\theta }}}={\frac {v}{\dot {\theta }}}={\frac {v^{2}}{v{\dot {\theta }}}}&&={\frac {v^{2}}{a_{n}}}\\&{\text{3D}}&&{\text{cartesian}}&&{\vec {r}}&&&&=x{\vec {i}}+y{\vec {j}}+z{\vec {k}}\\&&&&&r&&=|{\vec {r}}|&&={\sqrt {x^{2}+y^{2}+z^{2}}}\\&&&&&{\vec {v}}&&={\dot {\vec {r}}}&&={\dot {x}}{\vec {i}}+{\dot {y}}{\vec {j}}+{\dot {z}}{\vec {k}}&&=v_{x}{\vec {i}}+v_{y}{\vec {j}}+v_{z}{\vec {k}}\\&&&&&v&&=|{\vec {v}}|&&={\sqrt {{\dot {x}}^{2}+{\dot {y}}^{2}+{\dot {z}}^{2}}}&&={\sqrt {v_{x}^{2}+v_{y}^{2}+v_{z}^{2}}}\\&&&&&{\vec {a}}&&={\dot {\vec {v}}}&&={\ddot {x}}{\vec {i}}+{\ddot {y}}{\vec {j}}+{\ddot {z}}{\vec {k}}&&=a_{x}{\vec {i}}+a_{y}{\vec {j}}+a_{z}{\vec {k}}\\&&&&&a&&=|{\vec {a}}|&&={\sqrt {{\ddot {x}}^{2}+{\ddot {y}}^{2}+{\ddot {z}}^{2}}}&&={\sqrt {a_{x}^{2}+a_{y}^{2}+a_{z}^{2}}}\end{alignedat}}}
r = e c t , r ˙ = c e c t , r ¨ = c 2 e c t {\displaystyle {\color {Sepia}r=e^{ct}},\quad {\color {RoyalBlue}{\dot {\color {black}r}}}={\color {RoyalBlue}c}e^{ct},\quad {\color {RoyalBlue}{\ddot {\color {black}r}}}=c^{\color {RoyalBlue}2}e^{ct}}
e r → = ( − i → cos ( b t ) + j → sin ( b t ) e r → ˙ = b ( − i → sin ( b t ) + j → cos ( b t ) ) = − b e θ → e r → ¨ = b 2 ( − i → cos ( b t ) − j → sin ( b t ) ) = − b 2 e r → {\displaystyle {\begin{alignedat}{3}&\color {Sepia}{\vec {e_{r}}}=&&\color {Sepia}{\hphantom {{\Big (}-}}{\vec {i}}\cos(bt)+{\vec {j}}\sin(bt)\\&{\color {RoyalBlue}{\dot {\color {black}{\vec {e_{r}}}}}}=\color {RoyalBlue}b&&\left({\color {RoyalBlue}-}{\vec {i}}{{}\color {RoyalBlue}\sin }(bt)+{\vec {j}}{{}\color {RoyalBlue}\cos }(bt)\right)={\phantom {-}}b&{\vec {e_{\theta }}}\\&{\color {RoyalBlue}{\ddot {\color {black}{\vec {e_{r}}}}}}=b^{\color {RoyalBlue}2}&&\left(-{\vec {i}}{{}\color {RoyalBlue}\cos }(bt){{}\color {RoyalBlue}-{}}{\vec {j}}{{}\color {RoyalBlue}\sin }(bt)\right)=-b^{2}&{\vec {e_{r}}}\end{alignedat}}}
r → = r e r → v → = r → ˙ = r ˙ e r → + r e r → ˙ = c e c t e r → + b e c t e θ → a → = r → ¨ = r ¨ e r → + 2 r ˙ e r → ˙ + r e r → ¨ = ( c 2 − b 2 ) e c t e r → + 2 b c e c t e θ → {\displaystyle {\begin{alignedat}{5}&\color {Sepia}{\vec {r}}&&\color {Sepia}{}=r{\vec {e_{r}}}\\&{\color {darkgreen}{\vec {v}}}=\color {RoyalBlue}{\dot {\color {black}{\vec {r}}}}&&={\color {RoyalBlue}{\dot {\color {black}r}}}{\vec {e_{r}}}+r\color {RoyalBlue}{\dot {\color {black}{\vec {e_{r}}}}}&&\color {darkgreen}=&\color {darkgreen}ce^{ct}{\vec {e_{r}}}+\,&&\color {darkgreen}be^{ct}{\vec {e_{\theta }}}\\&{\color {darkgreen}{\vec {a}}}=\color {RoyalBlue}{\ddot {\color {black}{\vec {r}}}}&&={\color {RoyalBlue}{\ddot {\color {black}r}}}{\vec {e_{r}}}+{\color {RoyalBlue}2{\dot {\color {black}r}}{\dot {\color {black}{\vec {e_{r}}}}}}+r\color {RoyalBlue}{\ddot {\color {black}{\vec {e_{r}}}}}&&\color {darkgreen}=\,&\color {darkgreen}\left(c^{2}-b^{2}\right)e^{ct}{\vec {e_{r}}}+\,&&\color {darkgreen}2bce^{ct}{\vec {e_{\theta }}}\end{alignedat}}}
s = − 4 b sin θ , θ ˙ = ω v = s ˙ = − 4 b ω cos θ v ˙ = − 4 b ω 2 sin θ {\displaystyle {\begin{alignedat}{4}&\color {Sepia}s&&\color {Sepia}{}={\phantom {-}}4b&\color {Sepia}\sin \theta &,\quad \color {Sepia}{\dot {\theta }}=\omega \\&v=\color {RoyalBlue}{\dot {\color {black}s}}&&={\phantom {-}}4b\color {RoyalBlue}\omega &{\color {RoyalBlue}\cos {}}\theta \\&\color {RoyalBlue}{\dot {\color {black}v}}&&={\color {RoyalBlue}-}4b\omega ^{\color {RoyalBlue}2}&{\color {RoyalBlue}\sin {}}\theta \end{alignedat}}}
e v → = ω ( − i → cos θ + j → sin θ e v → ˙ = ω ( − i → sin θ + j → cos θ ) = ω e θ → {\displaystyle {\begin{aligned}&\color {Sepia}{\vec {e_{v}}}={\hphantom {\omega {\Big (}-}}\,{\vec {i}}\cos \theta +{\vec {j}}\sin \theta \\&{\color {RoyalBlue}{\dot {\color {black}{\vec {e_{v}}}}}}={\color {RoyalBlue}\omega }\left({\color {RoyalBlue}-}{\vec {i}}{{}\color {RoyalBlue}\sin {}}\theta +{\vec {j}}{{}\color {RoyalBlue}\cos {}}\theta \right)=\omega {\vec {e_{\theta }}}\end{aligned}}}
v → = v e v → a → = v → ˙ = v ˙ e v → + v e v → ˙ = − 4 b ω 2 sin θ e v → + 4 b ω 2 cos θ e θ → a = | a → | = 4 | b | ω 2 {\displaystyle {\begin{alignedat}{4}&\color {Sepia}{\vec {v}}&&\color {Sepia}{}=v{\vec {e_{v}}}\\&{\vec {a}}=\color {RoyalBlue}{\dot {\color {black}{\vec {v}}}}&&={\color {RoyalBlue}{\dot {\color {black}v}}}{\vec {e_{v}}}+v\color {RoyalBlue}{\dot {\color {black}{\vec {e_{v}}}}}&&=-4b\omega ^{2}&\sin \theta {\vec {e_{v}}}+4b\omega ^{2}\cos \theta {\vec {e_{\theta }}}\\&{\color {darkgreen}a}={\color {RoyalBlue}|}{\vec {a}}\color {RoyalBlue}|\!\!\!\!&&&&{}\color {darkgreen}=\ 4{\color {RoyalBlue}|}b{\color {RoyalBlue}|}\omega ^{2}\end{alignedat}}}
r → = 2 ( i → sin ( 4 t ) + j → cos ( 4 t ) ) + 4 t k → v → = r → ˙ = 8 ( i → cos ( 4 t ) − j → sin ( 4 t ) ) + 4 k → v 2 = v → 2 = 80 a → = v → ˙ = − 32 ( i → sin ( 4 t ) + j → cos ( 4 t ) ) ⊥ v → a n = | a → | = 32 ρ = v 2 a n = 2.5 {\displaystyle {\begin{alignedat}{8}&\color {Sepia}{\vec {r}}&&&{}\color {Sepia}={}&&\color {Sepia}2&\color {Sepia}\left({\vec {i}}\sin(4t)+{\vec {j}}\cos(4t)\right)&&{}\color {Sepia}+4t&\color {Sepia}{\vec {k}}\\&\color {darkgreen}{\vec {v}}&&=\color {RoyalBlue}{\dot {\color {black}{\vec {r}}}}&\color {darkgreen}={}&&\color {RoyalBlue}8&\color {darkgreen}\left({\vec {i}}{{}\color {RoyalBlue}\cos }(4t){{}\color {RoyalBlue}-{}}{\vec {j}}{{}\color {RoyalBlue}\sin }(4t)\right)&&\color {darkgreen}{}+4&\color {darkgreen}{\vec {k}}\\&v^{2}&&={\vec {v}}^{\color {RoyalBlue}2}&={}&&\color {RoyalBlue}80&\\&\color {darkgreen}{\vec {a}}&&=\color {RoyalBlue}{\dot {\color {black}{\vec {v}}}}&\color {darkgreen}={}&&\color {RoyalBlue}{-32}&\color {darkgreen}\left({\vec {i}}{{}\color {RoyalBlue}\sin }(4t)+{\vec {j}}{{}\color {RoyalBlue}\cos }(4t)\right)&&&&\perp {\vec {v}}\\&a_{n}&&={\color {RoyalBlue}|}{\vec {a}}\color {RoyalBlue}|&={}&&32\\&\color {darkgreen}\rho &&=\color {RoyalBlue}{{\color {black}v^{2}} \over \color {black}a_{n}}&\color {darkgreen}={}&&\color {RoyalBlue}2.5\end{alignedat}}}
r A ← S → = r S ′ ← S → + r A ← S ′ → v A ← S → = r A ← S → ˙ = r S ′ ← S → ˙ + r A ← S ′ → ˙ = v S ′ ← S → + v A ← S ′ → a A ← S → = v A ← S → ˙ = v S ′ ← S → ˙ + v A ← S ′ → ˙ = a S ′ ← S → + a A ← S ′ → {\displaystyle {\begin{alignedat}{3}&{\overrightarrow {r_{A\leftarrow S}}}&&={\overrightarrow {r_{S'\leftarrow S}}}+{\overrightarrow {r_{A\leftarrow S'}}}\\&{\overrightarrow {v_{A\leftarrow S}}}={\dot {\overrightarrow {r_{A\leftarrow S}}}}&&={\dot {\overrightarrow {r_{S'\leftarrow S}}}}+{\dot {\overrightarrow {r_{A\leftarrow S'}}}}&&={\overrightarrow {v_{S'\leftarrow S}}}+{\overrightarrow {v_{A\leftarrow S'}}}\\&{\overrightarrow {a_{A\leftarrow S}}}={\dot {\overrightarrow {v_{A\leftarrow S}}}}&&={\dot {\overrightarrow {v_{S'\leftarrow S}}}}+{\dot {\overrightarrow {v_{A\leftarrow S'}}}}&&={\overrightarrow {a_{S'\leftarrow S}}}+{\overrightarrow {a_{A\leftarrow S'}}}\end{alignedat}}}
d d φ ln tan φ 2 = 1 tan φ 2 d d φ sin φ 2 cos φ 2 = cos φ 2 sin φ 2 d sin φ 2 d φ cos φ 2 − sin φ 2 d cos φ 2 d φ cos 2 φ 2 = cos φ 2 sin φ 2 cos 2 φ 2 + sin 2 φ 2 cos 2 φ 2 1 2 = 1 sin φ {\displaystyle {d \over d\varphi }\ln \tan {\frac {\varphi }{2}}={\color {RoyalBlue}{\frac {1}{\color {black}\tan {\frac {\varphi }{2}}}}}{d \over d\varphi }{\color {RoyalBlue}{\frac {\sin \color {black}{\frac {\varphi }{2}}}{\cos \color {black}{\frac {\varphi }{2}}}}}={\frac {{\color {RoyalBlue}\cos {}}{\frac {\varphi }{2}}}{{\color {RoyalBlue}\sin {}}{\frac {\varphi }{2}}}}{\frac {\color {RoyalBlue}{d\color {black}\sin {\frac {\varphi }{2}} \over d\varphi }{\color {black}{}\cos {\frac {\varphi }{2}}}-{\color {black}\sin {\frac {\varphi }{2}}}{d\color {black}\cos {\frac {\varphi }{2}} \over d\varphi }}{\cos ^{\color {RoyalBlue}2}{\frac {\varphi }{2}}}}={\frac {\cos {\frac {\varphi }{2}}}{\sin {\frac {\varphi }{2}}}}{\frac {\cos ^{\color {RoyalBlue}2}{\frac {\varphi }{2}}\color {RoyalBlue}+\color {black}\sin ^{\color {RoyalBlue}2}{\frac {\varphi }{2}}}{\cos ^{2}{\frac {\varphi }{2}}}}{\color {RoyalBlue}{\frac {1}{2}}}={\frac {1}{\color {RoyalBlue}\sin \varphi }}}
r ˙ = − v rope r φ ˙ = − v water sin φ , v water ≠ 0 , assuming r > 0 , 0 < φ < π d ln r d φ = 1 r d r d φ = r ˙ r φ ˙ = v rope v water 1 sin φ = v rope v water d d φ ln tan φ 2 ln r r 0 = v rope v water d d φ ln tan φ 2 tan φ 0 2 r = r 0 ( tan φ 2 tan φ 0 2 ) v rope v water {\displaystyle {\begin{alignedat}{2}&\color {Sepia}{\dot {r}}&&\color {Sepia}{}=-\,v_{\text{rope}}\\&\color {Sepia}r{\dot {\varphi }}&&\color {Sepia}{}=-v_{\text{water}}\,\sin \varphi ,\quad v_{\text{water}}\neq 0,\quad {\text{assuming }}\,r>0,\quad 0<\varphi <\pi \\&{d{\color {RoyalBlue}{}\ln r} \over d\varphi }={\frac {1}{r}}{\color {RoyalBlue}{dr \over d\varphi }}=\color {RoyalBlue}{\frac {\color {black}{\dot {r}}}{\color {black}r{\dot {\varphi }}}}&&=\ \,\,\color {RoyalBlue}{\frac {\color {black}v_{\text{rope}}}{\color {black}v_{\text{water}}}}{\frac {1}{\color {black}\sin \varphi }}\\&&&=\ \,\,{\frac {v_{\text{rope}}}{v_{\text{water}}}}\color {RoyalBlue}{d \over d\varphi }\ln \,\,\tan {\frac {\varphi }{2}}\\&\ln \color {RoyalBlue}{\frac {\color {black}r}{r_{0}}}&&=\ \,\,{\frac {v_{\text{rope}}}{v_{\text{water}}}}{\hphantom {d \over d\varphi }}\ln \color {RoyalBlue}{\frac {\color {black}\tan {\frac {\varphi }{2}}}{\tan {\frac {\varphi _{0}}{2}}}}\\&\color {darkgreen}r&&\color {darkgreen}{}=\ \qquad \qquad {\color {RoyalBlue}r_{0}}{\left({\frac {\tan {\frac {\varphi }{2}}}{\tan {\frac {\varphi _{0}}{2}}}}\right)\!\!}^{\color {RoyalBlue}{\frac {v_{\text{rope}}}{v_{\text{water}}}}}\end{alignedat}}}
F → = m a → F B ← A → = − F A ← B → {\displaystyle {\begin{alignedat}{2}&{\vec {F}}&&=m{\vec {a}}\\&{\overrightarrow {F_{B\leftarrow A}}}&&=-{\overrightarrow {F_{A\leftarrow B}}}\end{alignedat}}}
F = − e E 0 cos ( ω t + θ ) a = F m = − e E 0 m cos ( ω t + θ ) v = v 0 + ∫ 0 t a d t = v 0 + e E 0 m ω sin θ − e E 0 m ω sin ( ω t + θ ) x = x 0 + ∫ 0 t v d t = x 0 + ( v 0 + e E 0 m ω sin θ ) t + e E 0 m ω 2 ( cos ( ω t + θ ) − cos θ ) {\displaystyle {\begin{alignedat}{6}&\color {Sepia}F&&\color {Sepia}=&&&&\color {Sepia}{}-\,eE_{0}&\color {Sepia}\cos(\omega t+\theta )\\&a=\color {RoyalBlue}{\frac {\color {black}F}{m}}&&=&&&&-\color {RoyalBlue}{\frac {\color {black}eE_{0}}{m}}&\cos(\omega t+\theta )\\&v=\color {RoyalBlue}v_{0}+\int _{0}^{t}{\color {black}a}dt&&=&&\color {RoyalBlue}v_{0}+{\frac {eE_{0}}{m\omega }}\sin \theta &&-{\frac {eE_{0}}{m\color {RoyalBlue}\omega }}&{\color {RoyalBlue}\sin }(\omega t+\theta )\\&{\color {darkgreen}x}=\color {RoyalBlue}x_{0}+\int _{0}^{t}{\color {black}v}dt&&\color {darkgreen}{}={\color {RoyalBlue}x_{0}}+{\bigg (}&&\color {darkgreen}v_{0}+{\frac {eE_{0}}{m\omega }}\sin \theta {\bigg )}\color {RoyalBlue}t&&\color {RoyalBlue}{}+\color {darkgreen}{\frac {eE_{0}}{m\omega \color {RoyalBlue}^{2}}}&\color {darkgreen}(\cos(\omega t+\theta )&\color {RoyalBlue}{}-\cos \theta \color {darkgreen})\end{alignedat}}}
0 → = F → + b e b t / m v → + m g j → , b > 0 0 → = 0 → m = a → + b m e b t / m v → + m g j → 0 → = e b t / m 0 → = e b t / m a → + b m e b t / m v → + m g e b t / m j → 0 → = ∫ 0 t 0 → d t = e b t / m v → − v 0 → − m g b j → + m g b e b t / m j → v → = ( v 0 → + m g b j → ) ( e − b t / m − m g b e b t / m j → → t → + ∞ − m g b j → r → = r 0 → + ∫ 0 t v → d t = r 0 → − m b ( v 0 → + m g b j → ) ( e − b t / m − 1 ) − m g b t j → → t → + ∞ ( x 0 + m v x 0 b ) i → − ∞ j → {\displaystyle {\begin{alignedat}{7}&\color {Sepia}{\vec {0}}&&&&\color {Sepia}{}={}&\color {Sepia}{\vec {F}}+{}&\ \,\color {Sepia}b\ \,{\hphantom {e^{bt/m}}}{\vec {v}}&&\color {Sepia}{}+\,mg&&\color {Sepia}{\vec {j}},\quad b>0\\&{\vec {0}}&&=\color {RoyalBlue}{\frac {\color {black}{\vec {0}}}{m}}&&=&{\color {RoyalBlue}{\vec {a}}}+{}&{\color {RoyalBlue}{\frac {\color {black}b}{m}}}{\hphantom {e^{bt/m}}}{\vec {v}}&&{}+\,{\phantom {m}}g&&{\vec {j}}\\&{\vec {0}}&&={\color {RoyalBlue}e^{bt/m}}{\vec {0}}&&=&{\color {RoyalBlue}e^{bt/m}}{\vec {a}}+{}&{\frac {b}{m}}{\color {RoyalBlue}e^{bt/m}}{\vec {v}}&&+\,{\phantom {m}}g\,{\color {RoyalBlue}e^{bt/m}}&&{\vec {j}}\\&{\vec {0}}&&=\color {RoyalBlue}\int _{0}^{t}{\color {black}{\vec {0}}}dt&&=&e^{bt/m}\color {RoyalBlue}{\vec {v}}-{}&\color {RoyalBlue}{\vec {v_{0}}}-{\frac {mg}{b}}{\vec {j}}&&+{\color {RoyalBlue}{\frac {m\color {black}g}{b}}}e^{bt/m}&&{\vec {j}}\\&\color {darkgreen}{\vec {v}}&&&&=&{\bigg (}&{\vec {v_{0}}}+{\frac {mg}{b}}{\vec {j}}{\bigg )}\color {RoyalBlue}{\phantom {\Big (}}e^{-bt/m}&&-{\frac {mg}{b}}{\phantom {e^{bt/m}}}&&{\vec {j}}\quad \color {darkgreen}\xrightarrow {t\to +\infty } -{\frac {mg}{b}}{\vec {j}}\\&\color {darkgreen}{\vec {r}}&&=\color {RoyalBlue}{\vec {r_{0}}}+\int _{0}^{t}{\color {black}{\vec {v}}}dt&&\color {darkgreen}{}={}&\color {RoyalBlue}{\vec {r_{0}}}-{\frac {m}{b}}\color {darkgreen}{\bigg (}&\color {darkgreen}{\vec {v_{0}}}+{\frac {mg}{b}}{\vec {j}}{\bigg )}{\Big (}e^{-bt/m}{\color {RoyalBlue}{}-1}{\Big )}&&\color {darkgreen}{}-{\frac {mg}{b}}\color {RoyalBlue}t&&\color {darkgreen}{\vec {j}}\quad \xrightarrow {t\to +\infty } \left(\color {RoyalBlue}x_{0}+{\frac {mv_{x0}}{b}}\right){\vec {i}}-\infty {\vec {j}}\end{alignedat}}}
0 = F + k x , k > 0 0 = 0 m = a + k m x = ( d 2 d t 2 + k m ) x = ( d d t − i k m ) ( d d t + i k m ) x = ( d d t − i k m ) ( v + i k m x ) 0 = e − i k / m t 0 = ( e − i k / m t d d t − i k m e − i k / m t ) ( v + i k m x ) = d d t ( e − i k / m t ( v + i k m x ) ) {\displaystyle {\begin{alignedat}{2}&\color {Sepia}0&&\color {Sepia}{}=F+kx,\quad k>0\\&0=\color {RoyalBlue}{\frac {\color {black}0}{m}}&&={\color {RoyalBlue}a}+{\color {RoyalBlue}{\frac {\color {black}k}{m}}}x=\left({\color {RoyalBlue}{d^{2} \over dt^{2}}}+{k \over m}\right)x=\left({d \over dt}-i{\sqrt {k \over m}}\right)\left({d \over dt}+i{\sqrt {k \over m}}\right)x=\left({d \over dt}-i{\sqrt {k \over m}}\right)\left({\color {RoyalBlue}v}+i{\sqrt {k \over m}}x\right)\\&0={\color {RoyalBlue}e^{-i{\sqrt {k/mt}}}}0&&=\left({\color {RoyalBlue}e^{-i{\sqrt {k/m}}t}}{d \over dt}-i{\sqrt {k \over m}}\color {RoyalBlue}e^{-i{\sqrt {k/m}}t}\right)\left(v+i{\sqrt {k \over m}}x\right)={\color {RoyalBlue}{d \over dt}}\left(e^{-i{\sqrt {k/m}}t}\left(v+i{\sqrt {k \over m}}x\right)\right)\end{alignedat}}}
e − i k / m t ( v + i k m x ) = v 0 + i k m x 0 {\displaystyle e^{-i{\sqrt {k/m}}t}\left(v+i{\sqrt {k \over m}}x\right)=\color {RoyalBlue}v_{0}+i{\sqrt {k \over m}}x_{0}}
v + i k m x = ( v 0 + i k m x 0 ) e i k / m t = ( v 0 + i k m x 0 ) ( cos ( k m t ) + i sin ( k m t ) ) = ( v 0 cos ( k m t ) − k m x 0 sin ( k m t ) ) + i k m ( m k v 0 sin ( k m t ) + x 0 cos ( k m t ) ) v = ( v 0 cos ( k m t ) − k m x 0 sin ( k m t ) , x = m k v 0 sin ( k m t ) + x 0 cos ( k m t ) {\displaystyle {\begin{alignedat}{2}&v+i{\sqrt {k \over m}}x&&=\left(v_{0}+i{\sqrt {k \over m}}x_{0}\right){\color {RoyalBlue}e^{i{\sqrt {k/m}}t}}=\left(v_{0}+i{\sqrt {k \over m}}x_{0}\right)\left({\color {RoyalBlue}\cos }\left({\sqrt {k \over m}}t\right){\color {RoyalBlue}\!{}+i\sin }\left({\sqrt {k \over m}}t\right)\right)\\&&&=\left(v_{0}\cos \left({\sqrt {k \over m}}t\right)-{\sqrt {k \over m}}x_{0}\sin \left({\sqrt {k \over m}}t\right)\right)+i{\color {RoyalBlue}{\sqrt {k \over m}}}\left({\color {RoyalBlue}{\sqrt {m \over k}}}v_{0}\sin \left({\sqrt {k \over m}}t\right)+x_{0}\cos \left({\sqrt {k \over m}}t\right)\right)\\&v&&={\hphantom {\bigg (}}v_{0}\cos \left({\sqrt {k \over m}}t\right)-{\sqrt {k \over m}}x_{0}\sin \left({\sqrt {k \over m}}t\right),\qquad \quad \,\color {darkgreen}x={\sqrt {m \over k}}v_{0}\sin \left({\sqrt {k \over m}}t\right)+x_{0}\cos \left({\sqrt {k \over m}}t\right)\end{alignedat}}}
0 = F + m k e k t v + m g , k > 0 , v 0 = 0 , x 0 = h 0 = 0 m = a + m k e k t v + g 0 = e k t 0 = e k t a + m k e k t v + g e k t 0 = ∫ 0 t 0 d t = e k t v − g k + g k e k t v = + g k ( e − k t − g k → t → + ∞ − g k x = h + ∫ 0 t v d t = h − g k 2 ( e − k t − 1 ) − g k t {\displaystyle {\begin{alignedat}{9}&\color {Sepia}{\overset {}{0}}&&\color {Sepia}=&&\!\color {Sepia}F&&\color {Sepia}{}+mk&&{\hphantom {e^{kt}}}\color {Sepia}v&&\color {Sepia}{}+{}&\color {Sepia}mg\,&\!\color {Sepia},&&\ \color {Sepia}k>0,\quad v_{0}=0,\quad x_{0}=h\\&0=\color {RoyalBlue}{\frac {\color {black}0}{m}}&&=&&\color {RoyalBlue}a&&+{\phantom {m}}k&&{\hphantom {e^{kt}}}v&&{}+{}&g\,\\&0={\color {RoyalBlue}e^{kt}}0&&=\color {RoyalBlue}e^{kt}&&a&&+{\phantom {m}}k&&{\color {RoyalBlue}e^{kt}}v&&{}+{}&g\,&\color {RoyalBlue}e^{kt}\\&0=\color {RoyalBlue}\int _{0}^{t}{\color {black}0}dt&&=e^{kt}&&\color {RoyalBlue}v&&\color {RoyalBlue}{}-\ {\frac {g}{k}}&&&&{}+{}&{\color {RoyalBlue}{\frac {\color {black}g}{k}}}&e^{kt}\\&\color {darkgreen}v&&=&&&&{\phantom {{}+{}}}\ {\frac {g}{k}}&&{\hphantom {(}}\color {RoyalBlue}e^{-kt}&&{}-{}&{\frac {g}{k}}&&&\color {darkgreen}\xrightarrow {t\to +\infty } -{\frac {g}{k}}\\&{\color {darkgreen}x}=\color {RoyalBlue}h+\int _{0}^{t}{\color {black}v}dt&&\color {darkgreen}=&&\color {RoyalBlue}h&&\color {RoyalBlue}{}-\color {darkgreen}{\frac {g}{k\color {RoyalBlue}^{2}}}&&\color {darkgreen}\left(e^{-kt}\color {RoyalBlue}{}-1\right)&&\color {darkgreen}{}-{}&\color {darkgreen}{\frac {g}{k}}&\color {RoyalBlue}t\end{alignedat}}}
r → = x | t 1 i → + 1 4 b x 2 j → , x 0 = 2 b , x 1 = 0 , x ˙ | t = 0 = 0 , x ˙ | t 1 < 0 v → = r → ˙ = x ˙ | t 1 i → + 1 2 b x x ˙ j → v 0 → = 0 → , v 1 → = x ˙ | t 1 i → 0 = v → ⋅ F r → = 2 v → ⋅ ( F → + m g j → ) 0 = 2 m 0 = 2 v → ⋅ ( a → + m g j → ) = 2 v → ⋅ a → + g 2 b 2 x x ˙ = d d t ( v → 2 + g 2 b x 2 ) 0 = ∫ 0 t 0 d t = ( v 1 → 2 − g 2 b x 0 2 = ( x ˙ | t 1 ) 2 − 2 b g x ˙ | t 1 = − 2 b g , v 1 → = − 2 b g i → a → = v → ˙ = x ¨ | t 1 i → + 1 2 b ( x ˙ 2 + x x ¨ ) j → a 1 → = x ¨ | t 1 i → + g j → 0 = v 1 → ⋅ ( a 1 → + g j → ) = − 2 b g x ¨ | t 1 x ¨ | t 1 = 0 , a 1 → = g j → F 1 → = m a 1 → = m g j → F r 1 → = F 1 → + m g j → = 2 m g j → {\displaystyle {\begin{alignedat}{3}&\color {Sepia}{\vec {r}}&&&&\color {Sepia}{}=x{\hphantom {|_{t_{1}}}}{\vec {i}}+{\frac {1}{4b}}x^{2}\,{\vec {j}},\quad x_{0}=2b,\quad x_{1}=0,\quad {\dot {x}}|_{t=0}=0,\quad {\dot {x}}|_{t_{1}}<0\\&{\vec {v}}&&=\color {RoyalBlue}{\dot {\color {black}{\vec {r}}}}&&={\color {RoyalBlue}{\dot {\color {black}x}}}{\hphantom {|_{t_{1}}}}{\vec {i}}+{\frac {1}{{\color {RoyalBlue}2}b}}x{\color {RoyalBlue}{\dot {\color {black}x}}}{\vec {j}}\\&{\vec {v\color {RoyalBlue}_{0}}}&&=\color {RoyalBlue}{\vec {0}}\color {black},\qquad \quad {\vec {v\color {RoyalBlue}_{1}}}&&={\dot {x}}{\color {RoyalBlue}|_{t_{1}}}{\vec {i}}\\&\color {Sepia}0&&\color {Sepia}{}={\vec {v}}\cdot {\vec {F_{r}}}&&\color {Sepia}{}={\phantom {2}}{\vec {v}}\cdot \left({\vec {F}}+mg{\vec {j}}\right)\\&0&&={\color {RoyalBlue}{\frac {2}{m}}}0&&={\color {RoyalBlue}2}{\vec {v}}\cdot \left({\color {RoyalBlue}{\vec {a}}}\ +{\phantom {m}}g{\vec {j}}\right)=2{\vec {v}}\cdot {\vec {a}}+{\color {RoyalBlue}{\color {black}g \over 2b}}2{\color {RoyalBlue}x{\dot {x}}}={\color {RoyalBlue}{d \over dt}}\left({\vec {v}}{\color {RoyalBlue}^{2}}+{g \over 2b}x\color {RoyalBlue}^{2}\right)\\&0&&=\color {RoyalBlue}\int _{0}^{t}{\color {black}0}dt&&={\hphantom {\Big (}}{\vec {v\color {RoyalBlue}_{1}}}^{2}\quad \color {RoyalBlue}\,-\color {black}{\frac {g}{2b}}x_{\color {RoyalBlue}0}^{2}\\&&&&&=\left(\color {RoyalBlue}{\dot {x}}|_{t_{1}}\right)^{2}-{\color {RoyalBlue}2b}g\\&{\dot {x}}|_{t_{1}}&&={\color {RoyalBlue}-{\sqrt {\color {black}2bg}}},\quad \color {darkgreen}{\vec {v_{1}}}&&\color {darkgreen}{}=\qquad \ \ \,{-{\sqrt {2bg}}}\,\color {RoyalBlue}{\vec {i}}\\&{\vec {a}}&&=\color {RoyalBlue}{\dot {\color {black}{\vec {v}}}}&&={\color {RoyalBlue}{\ddot {\color {black}x}}}{\hphantom {|_{t_{1}}}}{\vec {i}}+{\frac {1}{2b}}\left({\dot {x}}^{\color {RoyalBlue}2}+x\color {RoyalBlue}{\ddot {\color {black}x}}\right){\vec {j}}\\&{\vec {a\color {RoyalBlue}_{1}}}&&&&={\ddot {x}}{\color {RoyalBlue}|_{t_{1}}}{\vec {i}}+\qquad \qquad \quad \ \,{\color {RoyalBlue}g}{\vec {j}}\\&0&&={\vec {v\color {RoyalBlue}_{1}}}\cdot \left({\vec {a\color {RoyalBlue}_{1}}}+g{\vec {j}}\right)&&{}={\color {RoyalBlue}\qquad \ \ \,{-{\sqrt {2bg}}}\,{\ddot {x}}|_{t_{1}}}\\&{\ddot {x}}|_{t_{1}}&&=0,\qquad \quad {\vec {a_{1}}}&&=\qquad \qquad \qquad \qquad \ \ \,g{\vec {j}}\\&{\vec {F_{1}}}&&={\color {RoyalBlue}m}{\vec {a_{1}}}&&=\qquad \qquad \qquad \qquad \!{\color {RoyalBlue}m}g{\vec {j}}\\&\color {darkgreen}{\vec {F_{r\color {RoyalBlue}1}}}&&={\vec {F\color {RoyalBlue}_{1}}}+mg{\vec {j}}&&\color {darkgreen}{}=\qquad \qquad \qquad \quad \,\,{\color {RoyalBlue}2}mg{\vec {j}}\end{alignedat}}}
a A ← S → = m A a S ′ ← S → + m A a A ← S ′ → F A → = m A a A ← S → = m A a S ′ ← S → + m A a A ← S ′ → F A → − m A a S ′ ← S → = m A a S ′ ← S → + m A a A ← S ′ → {\displaystyle {\begin{alignedat}{2}&{\overrightarrow {a_{A\leftarrow S}}}&&={\hphantom {m_{A}}}{\overrightarrow {a_{S'\leftarrow S}}}+{\hphantom {m_{A}}}{\overrightarrow {a_{A\leftarrow S'}}}\\&{\vec {F_{A}}}=m_{A}{\overrightarrow {a_{A\leftarrow S}}}&&=m_{A}{\overrightarrow {a_{S'\leftarrow S}}}+m_{A}{\overrightarrow {a_{A\leftarrow S'}}}\\&{\vec {F_{A}}}-m_{A}{\overrightarrow {a_{S'\leftarrow S}}}&&={\hphantom {m_{A}{\overrightarrow {a_{S'\leftarrow S}}}+{}}}m_{A}{\overrightarrow {a_{A\leftarrow S'}}}\end{alignedat}}}
W = ∫ F → ⋅ d r → P = W ˙ = ∫ F → ⋅ d r → ˙ = F → ⋅ v → {\displaystyle {\begin{alignedat}{2}&W&&=\int {\vec {F}}\cdot d{\vec {r}}\\&P={\dot {W}}&&={\hphantom {\int {}}}{\vec {F}}\cdot {\phantom {d}}{\dot {\vec {r}}}={\vec {F}}\cdot {\vec {v}}\end{alignedat}}}
if 0 → = ∇ × F → = | i → j → k → ∂ ∂ x ∂ ∂ y ∂ ∂ z F x F y F z | = ( ∂ F z ∂ y − ∂ F y ∂ z ) i → + ( ∂ F x ∂ z − ∂ F z ∂ x ) j → + ( ∂ F y ∂ x − ∂ F x ∂ y ) k → then F → = − ∇ V = − ∂ V ∂ x i → − ∂ V ∂ y j → − ∂ V ∂ z k → d W = F → ⋅ d r → = − ∂ V ∂ x d x − ∂ V ∂ y d y − ∂ V ∂ z d z = − d V {\displaystyle {\begin{aligned}&{\text{if }}\,\qquad {\vec {0}}=\nabla \times {\vec {F}}={\begin{vmatrix}{\vec {i}}&{\vec {j}}&{\vec {k}}\\[5mu]{\dfrac {\partial }{\partial x}}&{\dfrac {\partial }{\partial y}}&{\dfrac {\partial }{\partial z}}\\[5mu]F_{x}&F_{y}&F_{z}\end{vmatrix}}=\left({\partial F_{z} \over \partial y}-{\partial F_{y} \over \partial z}\right){\vec {i}}+\left({\partial F_{x} \over \partial z}-{\partial F_{z} \over \partial x}\right){\vec {j}}+\left({\partial F_{y} \over \partial x}-{\partial F_{x} \over \partial y}\right){\vec {k}}\\[5mu]&\!{\begin{alignedat}{6}{\text{then}}\quad &{\vec {F}}&&=-\nabla V&&=-{\partial V \over \partial x}{\vec {i}}&&-{\partial V \over \partial y}{\vec {j}}&&-{\partial V \over \partial z}{\vec {k}}\\&dW&&={\vec {F}}\cdot d{\vec {r}}&&=-{\partial V \over \partial x}dx&&-{\partial V \over \partial y}dy&&-{\partial V \over \partial z}dz=-dV\end{alignedat}}\end{aligned}}}
F → = ( x + 2 y + z + 5 ) i → + ( 2 x + y + z ) j → + ( x + y + z − 6 ) k → r → = i → cos θ + j → sin θ + 7 θ k → , θ 0 = 0 , θ 1 = 2 π r 0 → = i → + r 1 → = i → + 14 π k → {\displaystyle {\begin{alignedat}{6}&\color {Sepia}{\vec {F}}&&\color {Sepia}{}={}&\color {Sepia}(x+2y+z+5){\vec {i}}+{}&&\color {Sepia}(2x+y+z){\vec {j}}+{}&&\color {Sepia}(x+y+z-6){\vec {k}}\\&\color {Sepia}{\vec {r}}&&\color {Sepia}=&\color {Sepia}{\vec {i}}\cos \theta +{}&&\color {Sepia}{\vec {j}}\sin \theta +{}&&\color {Sepia}7\theta {\vec {k}}&\color {Sepia},\quad \theta _{0}=0,\quad \theta _{1}=2\pi \\&{\vec {r\color {RoyalBlue}_{0}}}&&=&{\vec {i}}{\phantom {{}+{}}}\\&{\vec {r\color {RoyalBlue}_{1}}}&&=&{\vec {i}}+{}&&&&{\color {RoyalBlue}14\pi }{\vec {k}}\end{alignedat}}}
P = F → ⋅ v → = + 2 x x ˙ + 2 y x ˙ + z x ˙ + 5 x ˙ = + 2 x y ˙ + 2 y y ˙ + z y ˙ = + 2 x z ˙ + 2 y z ˙ + z z ˙ − 6 z ˙ = d d t ( 1 2 ( x 2 + y 2 + z 2 ) + 2 x y + x z + y z + 5 x − 6 z ) W = ∫ 0 t 1 P d t = d d t ( 1 2 ( x 2 + y 2 + z 2 ) + 2 x y + x z + y z + 5 x − 6 z ) | 1 , 0 , 0 1 , 0 , 14 π = 98 π 2 − 70 π {\displaystyle {\begin{alignedat}{3}&P&&={\vec {F}}\color {RoyalBlue}\cdot {\vec {v}}&&={\phantom {+2}}x{\color {RoyalBlue}{\dot {x}}}+2y{\color {RoyalBlue}{\dot {x}}}+z{\color {RoyalBlue}{\dot {x}}}+5\color {RoyalBlue}{\dot {x}}\\&&&&&{\phantom {=}}+2x{\color {RoyalBlue}{\dot {y}}}\,+{\phantom {2}}y{\color {RoyalBlue}{\dot {y}}}+z\color {RoyalBlue}{\dot {y}}\\&&&&&{\phantom {=}}+{\phantom {2}}x{\color {RoyalBlue}{\dot {z}}}\,+{\phantom {2}}y{\color {RoyalBlue}{\dot {z}}}+z{\color {RoyalBlue}{\dot {z}}}-6\color {RoyalBlue}{\dot {z}}\\&&&&&={\color {RoyalBlue}{d \over dt}}\left({\color {RoyalBlue}{\frac {1}{2}}}\left(x^{\color {RoyalBlue}2}+y^{\color {RoyalBlue}2}+z\color {RoyalBlue}^{2}\right)+2xy+xz+yz+5x-6z\right)\\&\color {darkgreen}W&&=\color {RoyalBlue}\int _{0}^{t_{1}}{\color {black}P}dt&&={\hphantom {d \over dt}}\,\color {RoyalBlue}\left.\color {black}\left({\frac {1}{2}}\left(x^{2}+y^{2}+z^{2}\right)+2xy+xz+yz+5x-6z\right)\right|_{1,\,0,\,0}^{1,\,0,\,14\pi }{\color {darkgreen}{}={}}98\pi ^{2}-70\pi \end{alignedat}}}
0 → = r → × F → 0 → = 0 → m = r → × a → = r → × a → + v → × v → = d d t ( r → × v → ) r → × v → = ( r 0 → × v 0 → 0 = ( r → × v → ) ⋅ r → = ( r 0 → × v 0 → ) ⋅ r → = J 0 → m ⋅ r → {\displaystyle {\begin{alignedat}{2}&\color {Sepia}{\vec {0}}&&\color {Sepia}{}={\vec {r}}\times {\vec {F}}\\&{\vec {0}}=\color {RoyalBlue}{\frac {\color {black}{\vec {0}}}{m}}&&={\vec {r}}\times {\color {RoyalBlue}{\vec {a}}}={\vec {r}}\times {\vec {a}}{\color {RoyalBlue}{}+{\vec {v}}\times {\vec {v}}}={\color {RoyalBlue}{d \over dt}}({\vec {r}}\times {\color {RoyalBlue}{\vec {v}}})\\&{\vec {r}}\times {\vec {v}}&&={\hphantom {\Big (}}\color {RoyalBlue}{\vec {r_{0}}}\times {\vec {v_{0}}}\\&{\color {darkgreen}0}=({\vec {r}}\times {\vec {v}})\color {RoyalBlue}\cdot {\vec {r}}&&\color {darkgreen}{}=\left({\vec {r_{0}}}\times {\vec {v_{0}}}\right){\color {RoyalBlue}\!{}\cdot {\vec {r}}}={\frac {\vec {J_{0}}}{m}}\cdot {\vec {r}}\end{alignedat}}}
r → = − l θ ˙ 2 ( i → sin θ − j → cos θ ) , v 0 = 0 , θ 1 = 0 v → = r → ˙ = − l θ ˙ 2 ( i → cos θ + j → sin θ ) a → = v → ˙ = − l θ ˙ 2 ( i → sin θ − j → cos θ ) + l θ ¨ ( i → cos θ + j → sin θ ) {\displaystyle {\begin{alignedat}{2}&\color {Sepia}{\vec {r}}&&\color {Sepia}{}={\phantom {-}}l{\phantom {{\dot {\theta }}^{2}}}\left({\vec {i}}\sin \theta -{\vec {j}}\cos \theta \right),\quad v_{0}=0,\quad \theta _{1}=0\\&{\vec {v}}=\color {RoyalBlue}{\dot {\color {black}{\vec {r}}}}&&={\phantom {-}}l{\color {RoyalBlue}{\dot {\theta }}}^{\phantom {2}}\left({\vec {i}}{\color {RoyalBlue}{}\cos {}}\theta {\color {RoyalBlue}{}+{}}{\vec {j}}{\color {RoyalBlue}{}\sin {}}\theta \right)\\&{\vec {a}}=\color {RoyalBlue}{\dot {\color {black}{\vec {v}}}}&&={\color {RoyalBlue}-}l{\dot {\theta }}^{\color {RoyalBlue}2}\left({\vec {i}}{\color {RoyalBlue}{}\sin {}}\theta -{\vec {j}}{\color {RoyalBlue}{}\cos {}}\theta \right)+l{\color {RoyalBlue}{\ddot {\color {black}\theta \,}}\!}\left({\vec {i}}\cos \theta +{\vec {j}}\sin \theta \right)\end{alignedat}}}
解法1：
0 = 2 ( F → + m g j → ) ⋅ v → 0 = 2 m 0 = 2 ( a → + m g j → ) ⋅ v → = 2 a → ⋅ v → + 2 l g θ ˙ sin θ = d d t ( v → 2 − 2 l g cos θ ) = d d t ( v 2 − 2 l g cos θ ) 0 = ∫ 0 t 1 0 d t = v 1 2 − 2 l g ( 1 − cos θ 0 ) v 1 = 2 l g ( 1 − cos θ 0 ) {\displaystyle {\begin{alignedat}{2}&\color {Sepia}0&&\color {Sepia}{}={\phantom {2}}\left({\vec {F}}+mg{\vec {j}}\right)\cdot {\vec {v}}\\&0={\color {RoyalBlue}{\frac {2}{m}}}0&&={\color {RoyalBlue}2}\left({\color {RoyalBlue}{\vec {a}}}\ +{\phantom {m}}g{\vec {j}}\right)\cdot {\vec {v}}=2{\vec {a}}\cdot {\vec {v}}+2{\color {RoyalBlue}l{\color {black}g}{\dot {\theta }}\sin \theta }={\color {RoyalBlue}{d \over dt}}\left({\vec {v}}\color {RoyalBlue}^{2}-{\color {black}2lg}\cos \color {black}\theta \right)={d \over dt}\left(v^{2}-2lg\cos \theta \right)\\&0=\color {RoyalBlue}\int _{0}^{t_{1}}{\color {black}0}dt&&=v_{\color {RoyalBlue}1}^{2}-2lg({\color {RoyalBlue}1}-\cos \theta _{\color {RoyalBlue}0})\\&\color {darkgreen}v_{1}&&\color {darkgreen}{}=\quad \,\color {RoyalBlue}{\sqrt {\color {darkgreen}2lg(1-\cos \theta _{0})}}\end{alignedat}}}
解法2：
0 = ( F → + m g j → ) ⋅ ( i → cos θ + j → sin θ ) 0 = 0 m = ( a → + m g j → ) ⋅ ( i → cos θ + j → sin θ ) = 2 l 2 θ ˙ θ ¨ + 2 l g θ ˙ sin θ 0 = 2 l θ ˙ 0 = 2 l 2 θ ˙ θ ¨ + 2 l g θ ˙ sin θ = d d t ( l 2 θ ˙ 2 − 2 l g cos θ ) = d d t ( v 2 − 2 l g cos θ ) 0 = ∫ 0 t 1 0 d t = v 1 2 − 2 l g ( 1 − cos θ 0 ) v 1 = 2 l g ( 1 − cos θ 0 ) {\displaystyle {\begin{alignedat}{2}&\color {Sepia}0&&\color {Sepia}{}=\left({\vec {F}}+mg{\vec {j}}\right)\cdot \left({\vec {i}}\cos \theta +{\vec {j}}\sin \theta \right)\\&0=\color {RoyalBlue}{\frac {\color {black}0}{m}}&&=\left({\color {RoyalBlue}{\vec {a}}}\ +{\phantom {m}}g{\vec {j}}\right)\cdot \left({\vec {i}}\cos \theta +{\vec {j}}\sin \theta \right)\\&&&=\color {RoyalBlue}{\phantom {2}}l^{\phantom {2}}{\phantom {\dot {\theta }}}{\ddot {\theta }}\color {black}+{\phantom {2l}}g{\phantom {\dot {\theta }}}\sin \theta \\&0={\color {RoyalBlue}2l{\dot {\theta }}}0&&={\color {RoyalBlue}2{\color {black}l}^{2}{\dot {\theta }}}{\ddot {\theta }}+{\color {RoyalBlue}2l{\color {black}g}{\dot {\theta }}}\sin \theta ={\color {RoyalBlue}{d \over dt}}\left(l^{2}{\dot {\theta }}\color {RoyalBlue}^{2}-{\color {black}2lg}\cos \color {black}\theta \right)={d \over dt}\left({\color {RoyalBlue}v}^{2}-2lg\cos \theta \right)\\&0=\color {RoyalBlue}\int _{0}^{t_{1}}{\color {black}0}dt&&=v_{\color {RoyalBlue}1}^{2}-2lg({\color {RoyalBlue}1}-\cos \theta _{\color {RoyalBlue}0})\\&\color {darkgreen}v_{1}&&\color {darkgreen}{}=\quad \,\color {RoyalBlue}{\sqrt {\color {darkgreen}2lg(1-\cos \theta _{0})}}\end{alignedat}}}
0 = F x , v projectile x 0 = v cannon x 0 = v cannon y = 0 0 = ∫ 0 t 1 F x d t = m projectile v projectile x 1 + m cannon v cannon x 1 ( 1 ) v ′ cos α = m projectile v projectile x 1 − m cannon v cannon x 1 ( 2 ) m cannon v ′ cos α = ( m projectile + m cannon ) v projectile x 1 ( 1 ) + m cannon ( 2 ) v projectile x 1 = − m cannon m projectile + m cannon v ′ cos α < v ′ cos α v projectile y 1 = − m cannon m projectile + m cannon v ′ cos α i → + v ′ sin α v projectile 1 → = − m cannon m projectile + m cannon v ′ cos α i → + v ′ sin α j → tan θ = v projectile y 1 v projectile x 1 = − m projectile + m cannon m cannon v ′ tan α > tan α v cannon x 1 = v projectile x 1 − v ′ cos α = − m projectile m projectile + m cannon v ′ cos α v cannon 1 → = − m projectile m projectile + m cannon v ′ cos α i → {\displaystyle {\begin{alignedat}{4}&\color {Sepia}0&&&&\color {Sepia}{}=F_{x},\quad v_{{\text{projectile }}x0}=v_{{\text{cannon }}x0}=v_{{\text{cannon }}y}=0\\&0&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!=\color {RoyalBlue}\int _{0}^{t_{1}}{\color {black}F_{x}}dt&&=\color {RoyalBlue}m_{\text{projectile}}v_{{\text{projectile }}x1}+m_{\text{cannon}}v_{{\text{cannon }}x1}&&\color {CadetBlue}(1)\\&\color {Sepia}v'\cos \alpha &&&&\color {Sepia}{}={\phantom {m_{\text{projectile}}}}v_{{\text{projectile }}x1}-{\phantom {m_{\text{cannon}}}}v_{{\text{cannon }}x1}&&\color {CadetBlue}(2)\\&{\color {RoyalBlue}m_{\text{cannon}}}v'\cos \alpha &&&&=\ \,\left(m_{\text{projectile}}\color {RoyalBlue}+m_{\text{cannon}}\right)v_{{\text{projectile }}x1}&&\color {CadetBlue}(1)+m_{\text{cannon}}(2)\\&v_{{\text{projectile }}x1}&&&&={\phantom {-}}{\color {RoyalBlue}{\frac {\color {black}m_{\text{cannon}}}{m_{\text{projectile}}+m_{\text{cannon}}}}}v'\cos \alpha &&<v'\cos \alpha \\&\color {Sepia}v_{{\text{projectile }}y1}&&&&\color {Sepia}{}={\hphantom {-{\frac {m_{\text{cannon}}}{m_{\text{projectile}}+m_{\text{cannon}}}}v'\cos \alpha \,{\vec {i}}+{}}}v'\sin \alpha \\&\color {RoyalBlue}{\overrightarrow {\color {darkgreen}v_{{\text{projectile }}1}}}&&&&\color {darkgreen}{}={\phantom {-}}{\frac {m_{\text{cannon}}}{m_{\text{projectile}}+m_{\text{cannon}}}}v'\cos \alpha \,\color {RoyalBlue}{\vec {i}}+{\color {darkgreen}v'\sin \alpha }\,{\vec {j}}\\&\tan \theta &&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!=\color {RoyalBlue}{\frac {\color {black}v_{{\text{projectile }}y1}}{\color {black}v_{{\text{projectile }}x1}}}&&={\phantom {-}}\color {RoyalBlue}{\frac {\color {black}m_{\text{projectile}}+m_{\text{cannon}}}{\color {black}m_{\text{cannon}}}}{\phantom {v'}}\!\!\ \tan \color {black}\alpha &&>\tan \alpha \\&v_{{\text{cannon }}x1}&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!=v_{{\text{projectile }}x1}-v'\cos \alpha &&={\color {RoyalBlue}-}{\frac {\color {RoyalBlue}m_{\text{projectile}}}{m_{\text{projectile}}+m_{\text{cannon}}}}v'\cos \alpha \\&\color {RoyalBlue}{\overrightarrow {\color {darkgreen}v_{{\text{cannon }}1}}}&&&&\color {darkgreen}{}=-{\frac {m_{\text{projectile}}}{m_{\text{projectile}}+m_{\text{cannon}}}}v'\cos \alpha \,\color {RoyalBlue}{\vec {i}}\end{alignedat}}}
0 = F A rope − F B rope = m A ( a A + g ) − m B ( a B + g ) = m A ( 2 s A t 2 + g ) − m B ( 2 s B t 2 + g ) = 2 ( m A s A − m B s B ) t 2 + ( m A − m B ) g t = 2 ( m A s A − m B s B ) ( m B − m A ) g {\displaystyle {\begin{alignedat}{2}&\color {Sepia}0&&\color {Sepia}{}=F_{A{\text{ rope}}}-F_{B{\text{ rope}}}=m_{A}(a_{A}+g)-m_{B}(a_{B}+g)=m_{A}\left({\color {RoyalBlue}{\frac {2s_{A}}{t^{2}}}}+g\right)-m_{B}\left({\color {RoyalBlue}{\frac {2s_{B}}{t^{2}}}}+g\right)\\&&&=\quad {\frac {2(m_{A}s_{A}-m_{B}s_{B})}{t^{2}}}+(m_{A}-m_{B})g\\&\color {darkgreen}t&&\color {darkgreen}{}=\color {RoyalBlue}{\sqrt {\frac {\color {darkgreen}2(m_{A}s_{A}-m_{B}s_{B})}{(m_{B}-m_{A})g}}}\end{alignedat}}}
這道題似乎並不需要使用動量矩定理。書中認為內力矩互相抵消，即默認繩兩端拉力相等；且題目默認滑輪不轉動、不打滑。（？）
此題中各物理量均取絕對值。
F A = F B , v A 0 = v B 0 = 0 , r 1 = r 0 2 m A v A 1 = ∫ 0 t 1 F A d t = ∫ 0 t 1 F B d t = m B v B 1 m A 2 v A 1 2 − m B 2 v B 1 2 = 0 ( 1 ) m A a A = m B a B = k m A m B r 2 d d t ( m A v A 2 + m B v B 2 ) = 2 m A a A v A + 2 m B a B v B = 2 k m A m B r 2 ( v A + v B ) = 2 k m A m B r 2 ( − d r d t ) = d d t 2 k m A m B r m A v A 1 2 + m B v B 1 2 = 2 k m A m B ( 1 r 0 / 2 − 1 r 0 ) = 2 k m A m B r 0 ( 2 ) ( m A 2 + m A m B ) v A 1 2 = 2 k m A m B 2 r 0 ( 1 ) + m B ( 2 ) v A 1 = m B 2 k r 0 ( m A + m B ) v B 1 = m A m B v A 1 = m A 2 k r 0 ( m A + m B ) {\displaystyle {\begin{alignedat}{4}&\color {Sepia}F_{A}&&&&\color {Sepia}{}=F_{B},\quad v_{A0}=v_{B0}=0,\quad r_{1}={\frac {r_{0}}{2}}\\&\color {RoyalBlue}m_{A}v_{A1}&&=\color {RoyalBlue}\int _{0}^{t_{1}}{\color {black}F_{A}}dt\color {black}=\color {RoyalBlue}\int _{0}^{t_{1}}{\color {black}F_{B}}dt&&=\color {RoyalBlue}m_{B}v_{B1}\\&m_{A}^{\color {RoyalBlue}2}v_{A1}^{\color {RoyalBlue}2}\color {RoyalBlue}-\color {black}m_{B}^{\color {RoyalBlue}2}v_{B1}^{\color {RoyalBlue}2}&&&&=0&&\color {CadetBlue}(1)\\&\color {Sepia}m_{A}a_{A}=m_{B}a_{B}&&&&\color {Sepia}{}={\frac {km_{A}m_{B}}{r^{2}}}\\&{\color {RoyalBlue}{d \over dt}}\left(m_{A}v_{A}^{\color {RoyalBlue}2}+m_{B}v_{B}^{\color {RoyalBlue}2}\right)&&={\color {RoyalBlue}2}m_{A}a_{A}\color {RoyalBlue}v_{A}+2{\color {black}m_{B}a_{B}}v_{B}&&={\frac {{\color {RoyalBlue}2}km_{A}m_{B}}{r^{2}}}{\color {RoyalBlue}(v_{A}+v_{B})}={\frac {2km_{A}m_{B}}{r^{2}}}\left(\color {RoyalBlue}-{dr \over dt}\right)\\&&&&&={\color {RoyalBlue}{d \over dt}}{\frac {2km_{A}m_{B}}{r}}\\&m_{A}v_{A\color {RoyalBlue}1}^{2}+m_{B}v_{B\color {RoyalBlue}1}^{2}&&&&=2km_{A}m_{B}\left({\frac {1}{\color {RoyalBlue}r_{0}/2}}{\color {RoyalBlue}{}-{}}{\frac {1}{r\color {RoyalBlue}_{0}}}\right)={\frac {2km_{A}m_{B}}{r_{0}}}&&\color {CadetBlue}(2)\\&\left(m_{A}^{2}\color {RoyalBlue}+{\color {black}m_{A}}m_{B}\right)v_{A1}^{2}&&&&={\frac {2km_{A}m_{B}^{\color {RoyalBlue}2}}{r_{0}}}&&\color {CadetBlue}(1)+m_{B}(2)\\&\color {darkgreen}v_{A1}&&&&\color {darkgreen}{}=m_{B}\color {RoyalBlue}{\sqrt {\color {darkgreen}{\frac {2k}{r_{0}\color {RoyalBlue}(m_{A}+m_{B})}}}}\\&\color {darkgreen}v_{B1}&&={\color {RoyalBlue}{\frac {m_{A}}{m_{B}}}}v_{A1}&&\color {darkgreen}{}={\color {RoyalBlue}m_{A}}{\sqrt {\frac {2k}{r_{0}(m_{A}+m_{B})}}}\end{alignedat}}}
0 = F − m g = d d t ( m v ) − m g = d d t ( ( m 0 + λ t ) v ) − ( m 0 + 1 2 λ t ) g , v 0 = 0 , s 0 = 0 0 = ∫ 0 t 0 d t = d d t ( ( m 0 + λ t ) v ) − ( m 0 t + 1 2 λ t 2 ) g v = 2 m 0 t + λ t 2 2 ( m 0 + λ t ) g = ( m 0 + λ t ) ( t + m 0 λ ) − m 0 2 λ 2 ( m 0 + λ t ) g = 1 2 g t 2 + m 0 g 2 λ t − m 0 2 g 2 λ 1 m 0 + λ t = d d t ( 1 4 g t 2 + m 0 g 2 λ t − m 0 2 g 2 λ 2 ln ( m 0 + λ t ) ) s = ∫ 0 t v d t = 1 4 g t 2 + m 0 g 2 λ t − m 0 2 g 2 λ 2 ln m 0 + λ t m 0 {\displaystyle {\begin{alignedat}{3}&\color {Sepia}0&&\color {Sepia}{}=F-mg={d \over dt}(mv)-mg&&\color {Sepia}{}={d \over dt}((m_{0}+\lambda t)v)-\ \ (m_{0}\ +{\hphantom {\frac {1}{2}}}\lambda t)\quad g,\quad v_{0}=0,\quad s_{0}=0\\&0&&=\color {RoyalBlue}\int _{0}^{t}{\color {black}0}dt&&={\hphantom {{d \over dt}(}}(m_{0}+\lambda t)v{\hphantom {)}}-\left(m_{0}\color {RoyalBlue}t{\color {black}{}+{}}{\frac {1}{2}}{\color {black}\lambda t}^{2}\right)g\\&v&&&&={\color {RoyalBlue}{\frac {2\color {black}m_{0}t+\lambda t^{2}}{2(m_{0}+\lambda t)}}}g={\frac {(m_{0}+\lambda t)\color {RoyalBlue}\left(t+{\frac {m_{0}}{\lambda }}\right)-{\frac {m_{0}^{2}}{\lambda }}}{2(m_{0}+\lambda t)}}g\\&&&&&=\qquad \,{\frac {1}{2}}gt{\hphantom {^{2}}}+{\frac {m_{0}g}{2\lambda }}{\hphantom {t}}-{\frac {m_{0}^{2}g}{2\lambda }}\quad \ {\frac {1}{m_{0}+\lambda t}}\\&&&&&={\color {RoyalBlue}{d \over dt}}\left({\frac {1}{\color {RoyalBlue}4}}gt^{\color {RoyalBlue}2}+{\frac {m_{0}g}{2\lambda }}{\color {RoyalBlue}t}-{\frac {m_{0}^{2}g}{2\lambda \color {RoyalBlue}^{2}}}\color {RoyalBlue}\ln \color {black}(m_{0}+\lambda t)\right)\\&\color {darkgreen}s&&=\color {RoyalBlue}\int _{0}^{t}{\color {black}v}dt&&\color {darkgreen}{}=\qquad \,{\frac {1}{4}}gt^{2}+{\frac {m_{0}g}{2\lambda }}t-{\frac {m_{0}^{2}g}{2\lambda ^{2}}}\ln \color {RoyalBlue}{\frac {\color {darkgreen}m_{0}+\lambda t}{m_{0}}}\end{alignedat}}}
![{\displaystyle {\begin{aligned}&{\text{if }}\,\qquad {\vec {0}}=\nabla \times {\vec {F}}={\begin{vmatrix}{\vec {i}}&{\vec {j}}&{\vec {k}}\\[5mu]{\dfrac {\partial }{\partial x}}&{\dfrac {\partial }{\partial y}}&{\dfrac {\partial }{\partial z}}\\[5mu]F_{x}&F_{y}&F_{z}\end{vmatrix}}=\left({\partial F_{z} \over \partial y}-{\partial F_{y} \over \partial z}\right){\vec {i}}+\left({\partial F_{x} \over \partial z}-{\partial F_{z} \over \partial x}\right){\vec {j}}+\left({\partial F_{y} \over \partial x}-{\partial F_{x} \over \partial y}\right){\vec {k}}\\[5mu]&\!{\begin{alignedat}{6}{\text{then}}\quad &{\vec {F}}&&=-\nabla V&&=-{\partial V \over \partial x}{\vec {i}}&&-{\partial V \over \partial y}{\vec {j}}&&-{\partial V \over \partial z}{\vec {k}}\\&dW&&={\vec {F}}\cdot d{\vec {r}}&&=-{\partial V \over \partial x}dx&&-{\partial V \over \partial y}dy&&-{\partial V \over \partial z}dz=-dV\end{alignedat}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/53b5b536218cd10a7eba83e8c5880a3d762a5b05)
