Esercizi Meccanica Razionale - Dinamica

determinare ω per cui la configurazione è stabile

la configurazione è stabile per

• sen ψ ≠ 0
• ω ² > ³/₂ k/m

z₁ = l/2 cos ψ

z₂ = l cos ψ - l/2 cos ψ = l/2 cos ψ

X_B = l sen ψ

U_p = -mg z₁ - mg z₂

= -mg l/2 cos ψ + mg l/2 cos ψ = 0

U_K = 1/2 k (X_B)² = 1/2 k l² sen² ψ

U_c = 1/2 1/2 ω² d² Σ 1/3 m l²

= U_c = 1/2 I ω², I = ml²

U_c = 1/2 Σ 1/3 m l² sen ψ il³/(3 l)

- 1/2 m ω² sen ϕ z₁ IL

U_tot = -{m l ω² sen ψ - k l² sen ψ

U_tot(de) = {sen ψ/2 m l² ω² cos ψ - k l² cos ψ} = 0

{cos ψ/1/3 ml² ω² - k l²} = 0

{sen ψ cos ψ (2 ml² ω² k l²)} = 0

sen ψ cos ψ = 0

ω² = 3/2 ke K

U_tot = cos ψ = 0 π/2, 3 π/2

U_{tot π/2} = 3/2 ω² ml² + K k² = 0

ω² - 3/2 k m

tot (π/2) = 2/3 ml ω² - ke K

ω² - 3/2 m > k e

U_{tot 3/2 π} = 3/2 l ω² - ke K

ω² - 3/2 k m > 0

AB = l

BC = 3l

Determinate w t. c. n. ha configurazione stabile

U_k = 1/2 k x_A²

= 1/2 k x_C²

U_c = 1/2 ω²ρ ds (s senφ)² + ∫ 1/2 ω²ρ ds′ (s senφ)² =

= ∫ 1/2 ω²ρ sen²φ s³/3₀ + ∫ 1/2 ω²ρ sen²φ s′³/3_l =

= 1/6 ω²ρ sen²φ + 4/3 ω²ρ sen²φ l³ =

= 5/6 ω²ρ sen²φ l³ - 3 ω² sen²φ ml²/2

U_p = -mg z₁ + l 1/2 mg z₂, z₁ = l/2 cosφ, z₂ = -l cosφ

U_p = -mg 2 cosφ + mg l cosφ - l/2 mg cosφ

U_tot = 1/2 k l² sen²φ - 1/2 k 4 l² sen²φ + 3 ω² sen²φ ml² + 3 l mg cosφ/2

dU_tot/dφ = -1/2 kl²senφ cosφ - 2 kl²senφ cosφ + 3/2 ω²senφ cosφ ml² - 3/2 l mg senφ

senφ (5 k²l² + 3 ω²cosφ ml² - 3 mg)/2 = 0 ⇒ senφ = 0 = φ = 0

⇒ φ = π

cosφ (-5 k² l² + 3 ω²m l²)/2 + 3/2 l mg

cosφ = -3/10 k²l²subg/9 + 6 ω²m l = -3 m g/10k l² + 6 ω²m l

asta - l, m

corpo - P, m

U_a = m g z_a

U_a = m g l cos ψ

z_a = - ^l⁄₂ cos ψ

U_p = m g z_o

U_p = m g l cos ψ

z_o = - l cos ψ

U_cp = ¹⁄₂ m ω² r² sen² ψ

U_ca = ¹⁄₂ ω² ρ sen ψ dsi = ...

U_tot = m g l cos ψ ...

... = ³⁄₂ m g l cos ψ + ²⁄₃ m ω² l² sen ψ cos ψ

... sen ψ = ...

cos ψ = ⁹⁄₈ ω² > 1

cos θ = ...

U₍₀₎ = ...

Determinare ω tale che la configurazione di equilibrio è stabile.

Molle collegate in serie ⇒ \( \frac{1}{K_{eq}} = \frac{1}{K_1} + \frac{1}{K_2} \), ma \( K_1 = K_2 \) ⇒ \( \frac{1}{K_{eq}} = \frac{2}{K} \)

\( U_k = \frac{1}{2} K \frac{X^2}{4} \) \( X = 2\ell \sin \varphi \)

⇒ \( U_k = \frac{1}{4} K 4 \ell^2 \sin^2 \varphi \)

\( U_A = -mg \cdot 2o \) \( 2o = \frac{\ell}{2} \sin \varphi \)

\( U_c = \frac{1}{2} \int^1_0 ω^2 \rho d s \int^s_0 \rho (\ell \sin \varphi + s \sin \varphi )^2= \)

\( = \frac{ω^2}{2} \rho \ell \sin^2 \varphi \frac{\ell^3}{3} \cdot \frac{3}{2} \int^1_0 ω^2 d s \rho (\ell^2 \sin^2 \varphi + 3s \sin^2 \varphi + 2 \ell \sin \varphi s \sin \varphi ) = \)

\(= \frac{1}{2} ω \rho \ell \sin \varphi \ell^2 + \frac{1}{4} ω^2 \rho \ell \sin \varphi + \frac{1}{3} ω^2 \rho \ell \sin \varphi \ell^2 =\)

\(= \frac{2}{3} ω m \cdot \sin \varphi (\ell \sin \varphi X) + \frac{\ell^2}{3} \sin \varphi + 2 \int s \sin \varphi \frac{\ell^2}{2} =\)

\(= \frac{ω}{2} m \ell \sin \varphi \frac{\ell^2}{3} - \frac{ω m \ell \ell \sin \varphi}{4} + \frac{\omega^2 m}{2} \varphi \ell \frac{\ell^2}{3} = \)

\(= \frac{3}{6} ω m \ell \sin \varphi + \frac{1}{2} ω m \ell \sin \varphi + \frac{1}{4} ω m \ell \sin \varphi \frac{\ell^2}{3} =\)

\(= \frac{1}{3} ω \ell \sin \varphi \ell \ell + ω \cdot m \ell \sin \varphi \ell \)

\(- ω \sin \varphi \cdot m \ell \ell + \frac{\ell \cdot m \ell \sin \varphi \right = \frac{4}{3} m ω^2 \ell \sin \varphi \)

\(U_{tot} = -k \ell^2 \sin^2 \varphi + m g \ell \sin \varphi + \frac{4}{3} m ω^2 \ell \sin \varphi \ell^2 \)

\( \frac{dU_{tot}}{d\varphi}= -k \ell^2 \cos \varphi \sin \varphi + mg \ell \cos \varphi + \frac{4}{3} m ω^2 \sin^2 \varphi \cos \varphi \ell^2 =\)

\(\cos \varphi (- 2 k \ell^2 \sin \varphi + mg \ell \cos \varphi + \frac{4}{3} m ω^2 2 \sin \varphi \cos \varphi \ell^2) = 0\)

\(\cos \varphi (- 2 k \ell^2 \sin \varphi + mg \ell \cos \varphi + \frac{8}{3} m ω^2 \ell^2) = -mg \ell \)

\(\cdot \sin \varphi = - ( - 2 k \ell^2 \cos \varphi + \frac{8}{3} m ω^2 \ell^2) \)

\(\cos \varphi = 0 = \frac{\pi}{2\cdot \frac{3 \cdot \pi}{2}} \)

\(\cdot \sin \varphi = \frac{mg}{2k \cdot \ell \frac{8}{3} m ω^2 \right ) \cdot > 1 \, ; = 1 \, < 1