2.4.8 Ache as componentes cilíndricas circulares da velocidade e da aceleração de uma partícula em movimento,
\begin{array}{ll} v_{\rho}=\dot{\rho}, & a_{\rho}=\ddot{\rho}-\rho \dot{\varphi}^{2} \\\ v_{\varphi}=\rho \dot{\varphi}, & a_{\varphi}=\rho \ddot{\varphi}+2 \dot{\rho} \dot{\varphi} \\\ v_{z}=\dot{z}, & a_{z}=\ddot{z} \end{array}
Sugestão:
\begin{aligned} \mathbf{r}(t) & =\hat{\boldsymbol{\rho}}(t) \rho(t)+\hat{\mathbf{z}} z(t) \\\ & =[\hat{\mathbf{x}} \cos \varphi(t)+\hat{\mathbf{y}} \operatorname{sen} \varphi(t)] \rho(t)+\hat{\mathbf{z}} z(t) \end{aligned}
Nota: \dot{\rho}=d \rho / d t, \ddot{\rho}=d^{2} \rho / d t^{2}, e assim por diante.

Passo 1

O enunciado pede para obter as componentes cilíndricas da velocidade e da aceleração de uma partícula cuja posição é dada por
\mathbf{r}(t)=\hat{\boldsymbol{\rho}}\,\rho(t)+\hat{\mathbf{z}}\,z(t)
com \hat{\boldsymbol{\rho}}=\hat{\mathbf{x}}\cos\varphi(t)+\hat{\mathbf{y}}\sin\varphi(t) e \hat{\boldsymbol{\varphi}}=-\hat{\mathbf{x}}\sin\varphi(t)+\hat{\mathbf{y}}\cos\varphi(t). O conteúdo é derivação de vetores em coordenadas cilíndricas. Vamos precisar das regras de derivação dos versores \hat{\boldsymbol{\rho}},\hat{\boldsymbol{\varphi}},\hat{\mathbf{z}}.

Passo 2

Primeiro anotamos as derivadas dos versores cilíndricos em função de \dot\varphi=d\varphi/dt:
\frac{d}{dt}\,\hat{\boldsymbol{\rho}}=\dot\varphi\,\hat{\boldsymbol{\varphi}}, \qquad \frac{d}{dt}\,\hat{\boldsymbol{\varphi}}=-\dot\varphi\,\hat{\boldsymbol{\rho}}, \qquad \frac{d}{dt}\,\hat{\mathbf{z}}=\mathbf{0}.
Essas fórmulas vêm de derivar \cos\varphi e \sin\varphi em função do tempo.

Passo 3

Agora calculamos a velocidade
\mathbf{v}=\frac{d\mathbf{r}}{dt} =\frac{d}{dt}(\rho\,\hat{\boldsymbol{\rho}})+\frac{d}{dt}(z\,\hat{\mathbf{z}}) =\dot\rho\,\hat{\boldsymbol{\rho}}+\rho\,\frac{d\hat{\boldsymbol{\rho}}}{dt}+\dot z\,\hat{\mathbf{z}}.
Usando \tfrac{d\hat{\boldsymbol{\rho}}}{dt}=\dot\varphi\,\hat{\boldsymbol{\varphi}}, obtemos
\mathbf{v} =\dot\rho\,\hat{\boldsymbol{\rho}}+\rho\,\dot\varphi\,\hat{\boldsymbol{\varphi}}+\dot z\,\hat{\mathbf{z}}.
Logo as componentes são
v_{\rho}=\dot\rho,\quad v_{\varphi}=\rho\,\dot\varphi,\quad v_{z}=\dot z.

Passo 4

Para a aceleração diferenciamos \mathbf{v}:
\mathbf{a}=\frac{d\mathbf{v}}{dt} =\frac{d}{dt}(\dot\rho\,\hat{\boldsymbol{\rho}}) +\frac{d}{dt}(\rho\,\dot\varphi\,\hat{\boldsymbol{\varphi}}) +\frac{d}{dt}(\dot z\,\hat{\mathbf{z}}).
Calculemos cada termo:
  1. \displaystyle\frac{d}{dt}(\dot\rho\,\hat{\boldsymbol{\rho}}) =\ddot\rho\,\hat{\boldsymbol{\rho}}+\dot\rho\,\dot\varphi\,\hat{\boldsymbol{\varphi}}.
  2. \displaystyle\frac{d}{dt}(\rho\,\dot\varphi\,\hat{\boldsymbol{\varphi}}) =\bigl(\dot\rho\,\dot\varphi+\rho\,\ddot\varphi\bigr)\hat{\boldsymbol{\varphi}} +\rho\,\dot\varphi\,\frac{d\hat{\boldsymbol{\varphi}}}{dt} =\bigl(\dot\rho\,\dot\varphi+\rho\,\ddot\varphi\bigr)\hat{\boldsymbol{\varphi}} -\rho\,\dot\varphi^{2}\,\hat{\boldsymbol{\rho}}.
  3. \displaystyle\frac{d}{dt}(\dot z\,\hat{\mathbf{z}}) =\ddot z\,\hat{\mathbf{z}}.
Somando e agrupando em \hat{\boldsymbol{\rho}},\hat{\boldsymbol{\varphi}},\hat{\mathbf{z}},
\mathbf{a} =\bigl(\ddot\rho-\rho\,\dot\varphi^{2}\bigr)\hat{\boldsymbol{\rho}} +\bigl(\rho\,\ddot\varphi+2\,\dot\rho\,\dot\varphi\bigr)\hat{\boldsymbol{\varphi}} +\ddot z\,\hat{\mathbf{z}}.

Resposta

v_{\rho}=\dot{\rho},\quad v_{\varphi}=\rho\,\dot{\varphi},\quad v_{z}=\dot{z}
a_{\rho}=\ddot{\rho}-\rho\,\dot{\varphi}^{2},\quad a_{\varphi}=\rho\,\ddot{\varphi}+2\,\dot{\rho}\,\dot{\varphi},\quad a_{z}=\ddot{z}

Exercícios de Livros Relacionados