Determine o momento angular do disco do problema 18.5 sobre o ponto A

Passo 1

Velocidade angular do disco ω = ω 1 j + ω 2 k = ( 5 r a d / s ) j + ( 15 r a d / s ) k

Momento centroidal de inércia

I - z ' = I - y ' = 1 4 m r 2 = 1 4 10 32.2 6 12 f t 2 = 0.019410 f t ⋅ l b ⋅ s 2

I z ' = 1 2 m r 2 = 0.038820 f t ⋅ l b ⋅ s 2

Passo 2

Momento angular sobre ponto C

H C = I → x ω x i + J - y ⋅ ω y ' j + I z ' ω z k

= 0 + ( 0.019410 ) 5 j + ( 0.038820 ) 15 k = ( 0.09705 f t ⋅ l b ⋅ s ) j + ( 0.58230 f t ⋅ l b ⋅ s ) k

H C = ( 0.0970 f t ⋅ l b ⋅ s ) j + ( 0.582 f t ⋅ l b ⋅ s ) k

Localização do centro de massa

r C / A = ( 18  in.  ) i + ( 9  in  ) j = ( 1.5 f t ) i + ( 0.75 f t ) j

Velocidade do centro de massa

v C = ω 1 × r C / A = 5 j × ( 1.5 i + 0.75 j ) = - ( 7.5 f t / s ) k

Passo 3

Momento angular sobre ponto A

H A = H C + r C / A × m v C

H A = 0.09705 j + 0.58230 k + ( 1.5 i + 0.75 j ) × - ( 10 ) ( 7.5 ) 32.2 k = 0.09705 j + 0.058230 k + 3.4938 j - 1.74689 i

H A = - ( 1.747 l b ⋅ s ⋅ f t ) i + ( 3.59 l b ⋅ s ⋅ f t ) j + ( 0.0582 l b ⋅ s ⋅ f t ) k

Resposta

H A = - ( 1.747 l b ⋅ s ⋅ f t ) i + ( 3.59 l b ⋅ s ⋅ f t ) j + ( 0.0582 l b ⋅ s ⋅ f t ) k

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