Resolução da 1 Lista de Física 2

LISTA OSCILA\u00c7\u00d5ES\n 0) f = 0,254hz \u00bb \u03c9 = 2\u03c0f = \u03c9 = 12,56rad/s\n 1.0 \u03c6(0) = 0,37\n \u03d5(0) = 0\n c) x(t) = A cos(\u03c6) \u00bb x(0) = A cos(\u03c6) = 0,37\n x(t) = -A\u03c9 sin(\u03c9t + \u03c6) \u00bb x(0) = -A\u03c9 sin \u03c6 = 0\n \u00bb sin\u03c6 = 0 \u00bb \u03c6 = 0\n \n x(0) = A cos 0 = A = 0,37 \u00bb A = 0,37cm\n \n x(t) = -0,37 cos(\u03c9t) [cm]\n 1) \u03c9 = 2\u03c0 = 2\u03c0 \u00bb \u03c9 = \u03c0 rad/s \n 2) T = 4\n \n b) \u03c9 = 2\u03c0f = 2\u03c0/4 \n \u03c9 = \u03c0/2\n \n \u03d5(t) = -A\u03c9 sin(\u03c9t + \u03c6) = -0,37\u03c0 sin(\u03c0/2 t) [cm/s]\n \u03b1(t) = 0,58 sin(\u03c81) [cm/s]\n \n f) -Amax = 0,58sin(\u03c0/2 t)\n g) \u03b1(t) = -A\u03c9 cos(\u03b1t) = -0,34 \u03c0/4 cos(\u03c0/2 t)\n \n \u03b1max = 0,91cm/s\n \n h) x(1) = 0,39 cos(\u03c0/3) \u00bb x(1) = 0\n \n i) x(3) = 0,58 sin(\u03c1/3) \u00bb x(3) = 0,58 02) t = 0 \n \u03b8(t) = L0 + \u0394L + \u03b8(t) \n \u0398(t) = A cos(\u03c9t + \u03c6) \n \u03b8(0) = (0) = L0 \n \n \u03b4(t) = 0 \n \u03c9 = \sqrt{k/m} \n y(t) = -mg = L0 + \u0394x \n F = mg = \frac {L0}{ m g} \n \n \u03b8(t) = L0 \cos\u03c6(t)\n \n \u03b8(t) = \frac{L0 - L \cos(t)}{kL} = \frac{-mg}. \n \n L0 = -\frac{mgL0}{k} = \frac{mgL}{k} \n \n L0 = k \sqrt{\frac{mg}{k}} = 0 04) \n cos\u03b8 = \frac{d}{L + \u0394L}\n sin\u03b8 = \frac{y}{L + \u0394L}\n \n x << d \to sin\u03b8 \sim \frac{y}{d}\n L + \deltaL \sim d \n \n F = -k\u0394l sin\u03b8 \n \u0394l + L = d \rightarrow \u0394l = d \cos! \n F2 = -k \frac{(d \cos\u03b8 - L)}{y} \rightarrow L ≈ k \to \DeltaL \sim d\cos\n x << d\ddot{h} = \frac{d^2L + \deltaL}{L \sin\u03b8} = 0 \n \n m d^2x dt^2 + (d - L) = 0 d^2x dt^2 + (d - L = \frac[]{\sqrt{d-L}}{md} 05)\n I = ml²\n \n parâmetros solicitados: O \n \n - sen θ ≈ θ\n \n logue: S = F.S\n \n s.d. dist. ponto de ap. da força F em um rotativo.\n \n J = F.L = mg.sen θ L = N - mgL B\n \n S = (F/2) = -k L/2 = -k( sen θ ) L/2 = -k L²/4\n \n \n J = L² ∂²θ/∂t²\n \n -MgL sin θ = k L²/4 - Ml² ∂θ/∂t²\n \n - ∂²θ/∂t² + (g/L + k/4m)θ = 0\n \n ω² = \u221A(g/k) (2 + 4m) 08)\n I = I cm + m s² (teorema dos eixos)\n \n I cm = (1/12) m r² (m r² - m GM)\n \n I o = m L²/12 + m r²/4 = m( L²/12 + 3m²)/3 \n \n \n 2) T = 2π I/mgs = 2π (m r²/3)/(mgs) = 2π(z/3) (L/m)\n \n b) pêndulo simples equivalente: T² = 2π(z/g)\n \n T = 2π √(L/3g)\n \n c)\n S = zL/3 - L/2 = z(4/6) - L/6\n \n I = I cm + m*L²/8 = mL²/(12*8) => [I = m L²/9]\n \n T = 2π √(L/3g) 08)d)\n Tmax = (T o)max + m(a c)max)\n \n a c = r α² = (r/ R)² *[ d(θR)/dt]² = R * [d(θ)/dt]² → (dθ/dt) = R.[(d(θ)/dt)²]\n \n T = mg cos θ + mR (d²θ/dt²)\n \n Tmax = mg + mR(d²θ/dt²) = 9.3 + 0.3 + 0.8 x 0.9 (17²)\n \n => Tmax = 3,152 N L = 80cm = 0.8m\nm = 0.3kg\n\nt = 0\n\\( \\theta(0) = 15^\\circ = \\frac{\\pi}{12} \\) \n\\( \\frac{d\\theta(0)}{dt} = 0 \\)\n\n\\( \\omega = \\sqrt{ \\frac{9.8}{0.8} } = \\sqrt{12.25} \\) \n\\( \\omega = 3.5 rad/s \\) \nT = \\frac{2\\pi}{\\omega} = \\frac{2\\pi}{3.5} \\) \nT = 1.795s\n\n\\( \\theta(t) = \\Theta_{max} \\cos(\\omega t + \\phi) \\) \n\\( \\frac{d\\theta(t)}{dt} = \\omega_{max} \\sin(\\omega t + \\phi) \\) \n\\( \\theta(0) = \\Theta_{max} \\cos \\phi = \\frac{\\pi}{12} \\) \n\\( \\frac{d\\theta(0)}{dt} = -\\omega_{max} \\sin \\phi = 0 \\) \n\\( \\Rightarrow \\sin \\phi = 0 \\Rightarrow \\phi = 0 \\)\n\n\\( \\theta_{max} \\cos 0 = \\frac{\\pi}{12} \\Rightarrow \\Theta_{max} = \\frac{\\pi}{12} \\)\n\\( \\theta(t) = \\frac{\\pi}{12} \\cos(3.5t) \\) \n\n\\( \\frac{d\\theta_{max}}{dt} = -\\omega_{max} = 3.5 \\times 0.262 \\Rightarrow \\frac{d\\theta_{max}}{dt} = 0.947 rad/s \\) I = \\int_0 ^{}\\theta\\, dt\ndF_{x} = -KDR^2 - k\\theta = I\\frac{d^2\\theta}{dt^2} \nS = I\\alpha = -KDR^2 - k\\theta\n\\int\\frac{d^2\\theta}{dt^2} + (KR + k)\\theta = 0\nI\\frac{d^2\\theta}{dt^2} + (K R^2 + k)\\theta = 0\n\\Rightarrow \\omega = \\sqrt{\\frac{KR^2 + k}{I}} B_0 = 0\nJ = -K\\theta\nM = 0.08kg\nL = 30cm = 0.3m\nK = 2\\times10^3.\\frac{N m^2}{kg/m^2}\n\na) T = 2\\pi\\sqrt{\\frac{I}{K}}\nI = I_1 + I_2 = M(\\frac{L}{2})^2 + M(\\frac{L}{2})^2 = \\frac{M L^2}{2} = 0.08\\times 0.09 = 3.6\\times10^{-3}\nT = 2\\pi\\sqrt{\\frac{3.6 \\times 10^{-3}}{2\\times10^3}} = 2.666\\n\n\nb) t = 0 \\frac{d\\theta(0)}{dt} = 0\nE = U_{max} = \\frac{1}{2} K x_{max}^{2} - \\frac{1}{2} m v_{max}^{2} = \\frac{1}{2} mR^{2} \\frac{d^2\\theta_{max}}{dt^{2}}\n\\theta(t) = \\Theta_{max} \\cos(\\omega t + \\phi) \\Rightarrow \\theta(0) = \\Theta_{max} \\cos \\phi = 0.1 rad\n\\frac{d\\theta(0)}{dt} = -\\omega \\Theta_{max} \\sin(\\omega t + \\phi)|_{t=0} = \\frac{d\\theta(0)}{dt} = \\Theta_{max} \\omega_{max} = 0\n\\Rightarrow \\cos \\phi = 0 \\Rightarrow \\phi = 0\n\\Rightarrow \\Theta_{max} \\cos 0 = \\Theta_{max} \\Rightarrow \\Theta_{max} = 0.1 rad\nE = \\frac{1}{2} K x_{max}^{2} = \\frac{1}{2} \\times 2K \\times 10^{-2}\\times 10^{-6}\n\\Rightarrow E = A \\cdot 10^{4} J c) \\( \\alpha = R B(t) \\Rightarrow \\dot{\\alpha} = R \\frac{dB(t)}{dt} \\quad \\kappa = \\frac{1}{2} \\quad \\dot{s}_{max} = \\frac{R \\left( \\frac{d\\alpha}{dt} \\right)_{max}}{g} \\dot{\\alpha} = \\underline{\\theta}_{max} \\omega = \\theta'_{max} \\Rightarrow \\frac{2 \\pi}{T} = 0.1 \\times \\frac{2 \\pi}{7.666} \\quad \\underline{s}_{max} = 0.15 \\times 0.236 = \\quad \\underline{\\beta_{max}} = 7.84 \\times 10^{-3} \\text{ m/s}^2