Esercizi e temi d'esame svolti di Meccanica Applicata alle Macchine

Esercizi Meccanica

Cinematica

Astro guidato agli estremi

V_a = 5 m/s a_a = 0,3 m/s² AB = l = 0,7 m

c = a + b ce^{i γ t} = ae^{i α t} + be^{i β t}

• Lungo x: c cos γ = a + b cos β
• Lungo y: -c sen γ = b sen β

Velocità: ce^{i γ t} = ae^{i α t} + be^{i β (t - π/2)}

• cos γ = λ + b cos (β + π/2)
• c sen γ = b sen β

a = c cos γ + bβ sen β 0 = c sen γ - bβ cos β det (A) = -(0,6791)

c = -β cos β + λ β = 0; det (A) = -γ

• a = cos γ - b cos β
• c = b sen β = 0,433 m sen γ
• β = sen γ - a

c = -b cos β = 4,48 m/s

β = -sen β - a = 5,23 rad/s

Accelerazione

cëᵧ = âe^i(β+π) + bβ²e^i(β+π)

• cosᵧ â + bβ³cos(β + ^π/₂) = â - bβ³cos(β + π)
• c(x)senᵧ = -bβsenβ + -bβ²senβ + bβ³senβ(β + π)
• cosᵧ â - bβ³cos^{= â - bβ3cosβ new}

cosᵧ senᵧ - â ß β ß cosβ senᵧ - ß

det(A)= -0.6761

• c senᵧ -b²cosβ -ß β β cosβ det(Â) = â - bβ cosβ â

= [ -bcosβ ß senß β 1 â - bβ cosβ ] Â = ß cosᵧ 1 ß cël ß ß

= â cosᵧ (-bβ²senβ) det(A)

Accelerazione:

ċ e^i(γ+ξ) + ċ ẏ e^i(δ+ξ) + c² i (1+γ+τ) e^i(1+γ+τ) + ä ẋ e^iξ + 2 ä e^iγ+τ + b e^iβ+ξ + ḃ β e^iβ+ξ + (b + ḃγ)e^i(β+ξ) + ḃβe^i(β+ξ) + ḃ e^i(β+ξ)

ċ e^iδ-1 + ċ ẏ e^i(γ+δ) + ä ẋ e^iδ-1 + ä e^iδ-1 + b e^iβ+1 - zç c j e^iδ-1 + c j e^iβ-1

[c cosγ - ẏ c sengγ] = ( - ä seng c - 2 ẋ cos x + ḃ cos x - 2 ḃ seng β - b1 seng β + x − ḃ j cos β ) - b β j cos γ + 2 ç j seng γ + c j cos γ - b3

| c seng γ ^{ẏ c cos ẏ} = ä^2 cosγ - 2 ä ẋ seng d + b seng β + 2 β j cos α β + b β ç cos b + 2 β seng β + b β seng β + 2 ç j sengγ + z ç j cosγ +

| c seng γ + 2 ç seng β + b j cosγ + c j seng β + b4

|[ cosγ c sengγ b3 ] | ẏ = 1 [ c o s ẏ seng γ b3 ]

|[ ẏ [ c ] = c [ 0 - seng j + b3] [ b4 ]

ċ = ^{cos γ b3 + c seng γ b4} / c

ẏ = ^{- seng γ b3 + cos γ b4} / c

Manovella con cuscino

OA = a = 0.4 m

AB = 1.4 m

AC = b = 0.6 m

α = 45°

α' = 25 rad/s

α'' = 0

β = 350°

ω_A, v_B, a_B?

c̅ = a̅ + b̅

c·e^{i γ} = a·e^{i α} + b·e^{i β}

a, c, γ fissi

b, α, β variano

1. c·cos γ = a·cos α + b·cos β
2. c·sen γ = a·sen α + b·sen β

γ = arc tan | a·sen α + b·sen β a·cos α + b·cos β

c = √(a·cos α + b·cos β)² + (a·sen α + b·sen β)² = 0,89 m

Note:

a·e^{(i α + π/2)} + b·e^{i β} + b·e^{i (β + π/2)} = 0

1. b·cos β - b·β·sen β = a·sen α
2. b·sen β + b·β·cos β = a·cos α

b = | b·cos β - b·β·sen β | = 7.07

| - b·sen β + b·β·cos β | = 7.07

b = 0.59 - 0.1

0.1 0.98

β = 0.77

5 = 0.59·(-7.07) + 0.1·7.07 = -8.13 m/s

β = 0.77·(-7.07) + 0.98·7.07 = 9.54 rad/s = ω_AB

DINAMICA

ES 1

α = cost (^dα⁄_dt = ^d2α⁄_dt² = 0)

Effettuare il sistema sia in equilibrio?

• ΣF_x = 0
• ΣF_y = 0
• ΣM_B = 0

H_B - F = 0

V_A - mg = 0

V_Alcosα + Flsenα - mg l⁄₂ cosα = 0

F = H_B

V_A - mg

mg lcosα - mg l⁄₂ cosα = Flsenα

F = ¹⁄₂mg cotα

V₀ = ċ î

Ȧ₀ = ċ î

G₂ V_G2 = V' + ω̄ Λ (c_x2 - B) = ċ î + βＫΛ (-^b⁄₂ cos β î + ^b⁄₂ senβ ĵ ) =

= ċ î - β² ^b⁄₂ cos β î - β^{2 b}⁄₂ senβ î

X_xG2 = 2,16 ^m⁄_s

Y_u2 = -0,75 ^m⁄_s

|V_u2| = 2,286 ^m⁄_s

Ȧ_G2 = Ȧ + βȦ + ω̄Λ ( c_x2 - B ) = ω̄Λ ( ω̄Λ ( c_x2 - B ) - ċ î + βΛΛ (-^b⁄₂ cosβ î + ^b⁄₂ senβ ĵ ) +

+ βΛΛ ( βＫΛ (-^b⁄₂ cosβ î + ^b⁄₂ senβ î ) -

+ βＫΛ (-^b⁄₂ cosβ î ) - ċ î + β^{2 b}⁄₂ cosβ - β^{2 b}⁄₂ senβ î

⁄_{- β^{2 b}⁄2 senβ î + β î = ċ î + β^{2 b}⁄2 cosβ î - β^{2 b}⁄2 senβ î +}

+^b⁄₂ cosβ î - β^{2 b}⁄₂ senβ î

{ ^Ẋ⁄_xG2 = -9,75 ^m⁄_s²

Ẏ_u2 = 0,43 ^m⁄_s² }

Θ̇ = ^ċ⁄_R = (-3,46) = 34,6 rad

Θ̈ = ^ċ⁄_R = -(-18 ) = 180 rad

@₁,₀ @,1

Incognito C_x H_o V_u H_w V_w

J_o = MR² / 2

0.1 kg m²

X_o: X - Rsenα

Y_o: Rcosα

Y_r: Y_o + Rcosα

Y_s: Y_m + b

X_c: Θ/3

ΣF_y = V_c - Mg - μ_g - m_j = 0

V_c = Mg + μ (per ψ')

ΣF = FRcosα + V_cRsenα - M_x = JΘ̈

F = V_cRsenα + M_xRcosα + M_rRsenα - JΘ̈

ΣM_o = H_o b / 2 (μ_j + m_j) Θ/3 - (M_x + F)V_θ + (V_c - M_g)X_c - JΘ̈ = 0

H_o = (μ_j + m_j) Θ/3 + (M_x + F)V_θ + (V_c - M_g)X + JΘ̇ = 1.41 N

ΣF_x = H_s + H_o - M_x - F = 0

H_B = M_x + F - H_o = 3.88 N

Pistone ideale

È Ă Ā: ¿Ỹ αє: © ?

p (pressione per garantire il moto con e senza attrito?)

V̅_n = V̅_B + ω̅ ∧ (A̅ - B̅)

V̅_x = αb senα

• V̅_x = 0,017 m/s
• α = 0,053 rad/s

Derivo le espressioni per trovare l’accelerazione

⍺_n = α ̇b senα + α²b cosα

• ⍺_n = 0,0021 m/s²
• α ̇ = α² tanα̇ = 0,0046 rad/s²

V̇_C = V̅_B + ω̅ ∧ (C̅ - B̅)

V̅_x = -2b senα̇ = -0,0085 m/s

V̅_y = -ȳ ı̇ + α̇b cosα̇ = 0,045 m/s

|V̅| = 0,046 m/s