Para as condições descritas no Prob. 7/36, determine a velocidade v e a aceleração a do centro A da ferramenta esférica em termos de β .

Passo 1

Para resolvermos o problema, usaremos as relações de velocidade e aceleração de um ponto dadas no livro. Também converteremos os valores para deixar no sistema SI para facilitar nossas contas.

Passo 2

Vamos começar utilizando a equação para velocidade relativa:

v A = v 0 + Ω × r A O + v r e l

Com v 0 = - R ⋅ Ω ⋅ i ^ , Ω = Ω ⋅ k ^ , r A O = b ⋅ s e n β ⋅ j ^ + b ⋅ cos ⁡ β ⋅ k ^ e v r e l = b ⋅ β ˙ ⋅ cos ⁡ β ⋅ j ^ - s e n β ⋅ k ^ . Todos dados obtidos no exercício que é usado como referência (7.36)

Agora substituindo os valores obtidos na equação inicial, temos que:

v A = - R ⋅ Ω ⋅ i ^ + Ω ⋅ k ^ × b ⋅ s e n β ⋅ j ^ + cos ⁡ β ⋅ k ^ + b ⋅ β ˙ ⋅ cos ⁡ β ⋅ j ^ - s e n β ⋅ k ^

v A = - Ω ⋅ R + b ⋅ s e n β ⋅ i ^ + b ⋅ β ˙ ⋅ ( cos ⁡ β ⋅ j ^ - s e n β ⋅ k ^ )

Para encontrarmos a aceleração, usaremos outra relação dada que se encontra derivando a relação da velocidade. Temos então que:

a A = a 0 + Ω ˙ × r A O + Ω × Ω × r A O + 2 ⋅ Ω × v r e l + a r e l

Onde:

a 0 = - R ⋅ Ω 2 ⋅ j ^ , Ω ˙ = 0 .

Ω × Ω × r A O = - Ω 2 ⋅ b ⋅ s e n β ⋅ Ω ⋅ j ^

2 ⋅ Ω × v r e l = 2 ⋅ Ω ⋅ b ⋅ β ˙ ⋅ cos ⁡ β ⋅ i ^

a r e l = b ⋅ β ˙ 2 ( s e n β ⋅ j ^ + cos ⁡ β ⋅ k ^

Assim, temos que a aceleração final será:

a A = - 2 ⋅ b ⋅ Ω ⋅ β ˙ ⋅ cos ⁡ β ⋅ i ^ - Ω 2 ⋅ R + b ⋅ s e n β + b ⋅ β ˙ 2 ⋅ s e n β ⋅ j ^ - b ⋅ β ˙ 2 ⋅ cos ⁡ β ⋅ k ^

Resposta

Resposta: v A = - Ω ⋅ R + b ⋅ s e n β ⋅ i ^ + b ⋅ β ˙ ⋅ ( cos ⁡ β ⋅ j ^ - s e n β ⋅ k ^ )

a A = - 2 ⋅ b ⋅ Ω ⋅ β ˙ ⋅ cos ⁡ β ⋅ i ^ - Ω 2 ⋅ R + b ⋅ s e n β + b ⋅ β ˙ 2 ⋅ s e n β ⋅ j ^ - b ⋅ β ˙ 2 ⋅ cos ⁡ β ⋅ k ^

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