Esercizi di Meccanica razionale

Esempi Note Voi

P=O= l(ũsinθ-ĵcosθ)

v̅ = lθ̇(ũcosθ+ĵsinθ)

a̅ = [lθ̈(ĩcosθ+ĵsinθ) + θ̇²(-ũsinθ+ĵcosθ)]

con Φ̅ = Φ(-ĩsinθ+ĵcosθ)

⇒ mȧ̅ = F̅ + Φ̅

x̄: mẋ = Φ̄x

ȳ: mÿ = Φ̄y - mg

z̄: ż = 0

Punto P: X: lsinθ Ẋ: lθ̇cosθ Ẍ: l[θ̈cosθ - θ̇²sinθ]

Y: lcosθ Ẏ: -lθ̇sinθ Ÿ: -l[θ̈sinθ + θ̇²cosθ]

⇒ ml[θ̈cosθ - θ̇²sinθ] = Φ̄cosθ

⇒ ml[θ̈sinθ + θ̇²cosθ] = Φ̄sinθ - mg

⇒ mlθ̈² = mgcosθ

⇒ θ̈ + g/l sinθ

Pro: moto pendolo

→reazione Φ̅ = con. iniziali: Θ(t=0) = Θ₀ Φ(t=0) = Φ₀ Θ̇(t=0) = 0

θ̈ + θ̇₀θ̇ sin θ

→

d/dt [ θ̇²/2 ] - d/dt [ cos θ ] = θ̇²/2

d/dt [ θ̇²/2 cos θ ] = θ̇²/l cos θ

(primo)

E = T + V = 1/2 mϕ̇² + mgγ = 1/2 l² θ̇

(rparo) l² θ̇² _سات - ϕ₀/l cos θ

l = l′

sommuso ϱ con inverso p

ψ²/2 - θ̇²/l cos θ = θ̇²/2 ϕ₀/l cos θ

→

θ̇² = ψ² - 2 ϕ₀/l cos θ + ϕ_{0/l cos θ}

= ψ²/2

= ϕ₀/l cos θ = θ̇²

(quindi siamo)

ϕ = m (lθ̇² + ϕ₀ cos θ)

= m [ z lθ̇ ₀/2 cos θ +ϕ₀ cos θ ] → ϕ = 3 lmg cos θ

Oscillatore armonico

T = 1/2 m ẋ²

V = 1/2 k x²

L = T - V = 1/2 m ẋ² - 1/2 k x²

d/dt (∂L/∂ẋ) - ∂L/∂x = 0

∂L/∂x = - kx

d/dt (mẋ) = mẍ

Quindi: mẍ + kx = 0

ẍ + (k/m)x = 0 ⇒ ẍ + ω²x = 0

Oscillatore armonico smorzato (m - k - c)

P = (m)

k > 0

F_A = - λ ẋ

F_A = (NC)

T = 1/2 m ẋ²

V = 1/2 k x²

L = T - V = 1/2 m ẋ² - 1/2 k x²

d/dt (∂L/∂ẋ) - ∂L/∂x = Q_k^(NC)

Con ∂L/∂x = - kx

d/dt (mẋ) = mẍ

ẍ = -kx

Q_k^(NC) = - λ ẋ = λ ẋ

⇒ mẍ - (-kx) = λ ẋ

⇒ mẍ + λ ẋ + kx = 0

L = T - V = \left(\frac{1}{2} ms^2 \dot{s}^2 + \frac{1}{2} m \dot{\theta}^2 q^2 + \frac{1}{2} m s^2 + m l \dot{\theta} s \cos \theta\right) - \left(- m g s \cos \theta + \frac{1}{2} ks^2\right)

\rightarrow L(s, \dot{s}) \rightarrow \frac{\partial L}{\partial s} \quad \frac{\partial L}{\partial \dot{s}}

\frac{\partial L}{\partial s} = 2 ms + ml \dot{\theta} \cos \theta \qquad \frac{\partial L}{\partial \dot{s}} = - k s

\rightarrow \text{eq. di Lagrange:} \quad \frac{d}{dt} \left(\frac{\partial L}{\partial \dot{s}}\right) - \frac{\partial L}{\partial s} = 0

\rightarrow 2 ms + ml (\ddot{\theta} \cos \theta - \dot{\theta}^2 \sin \theta) + k s = 0 \quad \text{1\textsuperscript{a} eq. di Lagrange}

\rightarrow L(\theta, \dot{\theta}) \rightarrow \frac{\partial L}{\partial \theta} \quad \frac{\partial L}{\partial \dot{\theta}}

\frac{\partial L}{\partial \theta} = m l \dot{s} \dot{\theta} + m g s \cos \theta \qquad \frac{\partial L}{\partial \dot{\theta}} = - m l s \sin \theta - m g l \sin \theta

\frac{d}{dt} \left(\frac{\partial L}{\partial \dot{\theta}}\right) = m \dot{s} \dot{\theta} - m l (\dot{s}\dot{\theta} \sin \theta + m g s \cos \theta) = m \ddot{s}\dot{\theta} + m l (\dot{s}\dot{\theta} \sin \theta - s \dot{\theta} \sin \theta) = 0

\rightarrow \text{eq. di Lagrange:} \quad \frac{d}{dt} \left(\frac{\partial L}{\partial \dot{\theta}}\right) - \frac{\partial L}{\partial \theta} = 0

\rightarrow m \dot{s^2}\ddot{\theta} + m l (\dot{s}\cos \theta - \dot{s} \sin \theta) + m l \dot{\theta} s \sin \theta + m g l \sin \theta = 0

\rightarrow m \dot{s^2} + m g s \cos \theta + m g l \sin \theta = 0

\text{Abbiamo trovato le 2 eq. di Lagrange ed avremo quindi $\frac{\partial R}{\partial \dot{\theta}} = 0$. Posso modificare il sistema trovando la mole:}

x \quad (UI \text{averto})

T = \frac{1}{2} m \dot{s}^2 + \frac{1}{2} m \dot{\theta}^2 r (\dot{s})

T = \frac{1}{2} m s^2 \dot{\theta}^2 + \frac{1}{2} m r (\dot{\theta}^2 + e^{-2}) s \cos \theta

V = - m g x + m g y \cos \theta - m g \cos \theta

L = T - V = \frac{1}{2} m s^2 \dot{s}^2 + \frac{1}{2} m q \dot{\theta}^2 r + \frac{1}{2} m g s cos \theta + m g l \sin \theta

\sin Q \p - (\dot{\theta}, \dot{s}, p, \dot{\theta})

σ = _m/^(ab) ⇒ m = δab

I_m = ∫∫_D (y² + z²) dS = δ∫^b/2_-b/2 ∫^a/2_-a/2 (y²+ z²) dx dy = δ∫_-b/2 ^b/2 dx ∫_-a/2^a/2 dy

= δ∫_-b/2^b/2 dx ∫_-a/2^{a/2 [y3/₃]}