Resolução Lista 1 Cinematica 1

1) "Velocidade" = frequência = 1800 rpm = \( \frac{1800}{60} = 30 Hz \) \( \text{A veloc. angular} \Rightarrow \omega = 2 \pi f = 2 \pi (30) = 188,49 \text{ rad/s} \) Como \( \alpha \) é cte: \( \omega^2 = \omega_0^2 + 2 \alpha \Delta \Theta \) \( \omega^2 = 188,49^2 + 2\alpha \Delta \Theta \) \( = 65...(2 \pi) \) \( \Delta \Theta = [408,40] \) \( \alpha = -43,49 \text{ rad/s}^2 \) 2) "Velocidade Nominal" (a de funcionamento) \( \int \text{frequência} = 1800 \text{ rpm} = 30f \) \( \omega = 2 \pi f = 2 \pi (30) = 188,49 \text{ rad/s} \) a) Ao ligar: \( \omega = \omega_0 + \alpha t \) \( 188,49 = 0 + 2 \alpha(5) \) \( \alpha = -37,698 \text{ rad/s}^2 \) Ao desligar: \( \omega = \omega_0 + \alpha t \) \( 0 = 188,49 + \alpha(20) \) \( \alpha = -2,094 \text{ rad/s}^2 \) b) Ao ligar: \( \omega^2 = \omega_0^2 + 2\alpha \Delta \Theta \) \( (188,49)^2 = 0 + 2(37,698)\Delta \Theta \) \( \Delta \Theta = 952,47 \text{ rad} \) \( \approx 15,16 \text{ voltas} \) Ao desligar: \( \omega^2 = \omega_0^2 + 2\alpha \Delta \Theta \) \( 0 = (188,49)^2 + 2(-2,094)\Delta \Theta \) \( \Delta \Theta = 16,266,80 \) \( \approx 270,71 \text{ voltas} \) 3) \( \alpha = 0,5 \text{ rad/s}^2 \) \( t = 0_s \) \( \omega = \omega_0 + \alpha t \) \( \omega = 0 \) \( a_T = \alpha_T R \) \( a_n = 0 \) \( a_{res} = 1,10^{-2} \text{ m/s}^2 \) \( t = 2 \text{s} \) \( \omega = 0 + \alpha (2) \) \( \omega = 1 \text{ rad/s} \) \( a_{\rm test} \) até \( t=0_s \) \( a_T = 1,10^{-2} \text{ m/s}^2 (gol)” \) \( \omega^2 R = 1^3 R 0,02^2 = 0,02 \text{ m}^g \) \( a_n = \omega^2 R = 1,0^2 \times 0,02 \text{ m/s}^2 = 2,0 \) m/s} Ainda \( t = 2_s \) \( \alpha_T^2 = (1,10^{-2})^2 + (2,0^2) \) \( \alpha_{res} = 2,23 \times 10^{-2} \text{ m/s}^2 \) \( t = 5_s \) \( \omega = \omega_0 + \alpha t \Rightarrow \omega = 0,5(5) \) \( \omega = 2,5 \text{ rad/s} \) \( a_T = 1,10^{-2} \text{ m/s}^2 \) \( a_n = \omega^2 R = (2,5)^2(0,025) \) \( a_n = 6,25 \times 0,02 \) \( a_n \approx 1,25 \times 0^{-2} \text{ m/s}^2 \) \( a^2_{res} = a_T^2 + a_n^2 \) \( a^2_{res} = (1,10^{-2})^2 + (1,25 \times 10^{-2}) \) \( a^2_{res} = 12,54 \times 10^{-2} \text{ m/s}^2 \) 4) \( \nabla v_{ftb} = \nabla v_A = w_B \cdot r_B \) "nos pontos de contato" \( w_A \cdot r_A \) Para encontrar \( \alpha_A \) e \( \alpha_B \) Antes Deveréis: \( w_A = \frac{V_A}{R_A} = 0^6 \over \0,019 } \) \approx \) 31/41 \( w_B = \frac{Ub}{Rb} = 18,06 \over \0,03} \approx 18,86 \) Depois \( w_A = \0 & V_A & 15 &} & = 78,525 \) \( R^0_{B^2} {33,14, \15,05 RMS} & { 19, .^5} \) \( V^0, {00,02 } \right \) B \}} \ { V_{O & 0,02 \left \} α_A = Δw/Δt = 78,52 - 31,41/3 α_A = 15,70 rad/s² n° de voltas w² = w0² + 2.α.ΔΘ (78,52)² = (31,41)² + 2.(15,7).ΔΘ 6165,4 = 986,58 + 31.4.ΔΘ ΔΘ = 1649,93 rad ≈ 26 voltas α_B = 47,15 - 18,86/3 α_B = 9,43 rad/s² n° de voltas w² = w0² + 2.α.ΔΘ (47,15)² = (18,86)² + 2.(9,43).ΔΘ 2223,1 = 355,69 + 18.86.ΔΘ ΔΘ = 99,01 rad ≈ 15,76 voltas 5) Relação entre os raios: 1/5 Tambor: 25 para 45 rpm → w inicial = 2,61 rad/s 47,1 (2,61) w² = w0² + 2.α.ΔΘ (4,71)² = (2,61)² + 2.α.(21π) 22,20 = 6,81 + α.150,79 α = 0,1 rad/s² Dados do rodízio α = 0,5 rad/s² 5 vezes maior. b) w = w0 + α.t 4,71 = 2,61 + 0,1.t t = 21 seg. P/o tambor 6. Θ = 20t + 4t² w = dΘ/dt = 20 + 8t α = dw/dt = 8 Quando t = 20: Θ = 20.(20) + 4.(20)² = 34 200 rad = 5445 voltas. w = 20 + 8.(20) = 740 rad/s α = 8 rad/s = const. 7. α = (2t + 2t²) e Como α = dw/dt e dw = α dt ∫{w0}{w} dw = ∫{t=0}{t=10s} α dt similar, ∫{t=0}{t=10} dΘ = ∫{t=0}{t=10} w dt ∫{t=0}{t=10} dw = ∫{t=0}{t=10} α dt w - w0 = ∫{t=0}{t=10} (2t + 2t²) dt = [t² + 2/3 t³] (t=10) w - w0 = 10² + 2/3.10³ w = 466,66 rad/s α = 2t + 2t² p/t=10s α = 220 rad/s² ∫{t=0}{t=10} dΘ = ∫{t=0}{t=10} w dt ΔΘ = ∫{t=0}{t=10} (t² + 2/3 t²) dt = [t³/3 + t⁴/6] (t=10) ΔΘ = 1.999,99 ≈ 2000 rad 8 α = 10 . Θ^{1/3} Procurar uma relação entre \( \alpha, \omega, \Theta \) \( \alpha = \frac{d\omega}{dt} \, \rightarrow \, dt = \frac{d\omega}{\alpha} \, \quad \small\text{igualando} \, \frac{d\omega}{\alpha} = \frac{d\Theta}{\omega} \\ \omega = \frac{d\Theta}{dt} \, \rightarrow \, dt = \frac{d\Theta}{\omega} \\ \omega \, d\omega = \alpha \, d\Theta \\ \int_{\omega_0}^{\omega} \omega \, d\omega = \int_{\Theta_0}^{\Theta} \alpha \, d\Theta = \int_0^{\Theta} 10 . \Theta^{1/3} \, d\Theta \\ \left[ \frac{\omega^2}{2} \right]_{\omega_0}^{\omega} = \left[ \frac{10 . 3 . \Theta^{4/3}}{4} \right]_0^{\Theta} \, \omega_0 = 0 \\ \therefore \omega^2 = 2 . \frac{10 . 3 . \Theta^{4/3}}{4} = 15 . \Theta^{4/3} \, \rightarrow \, \omega = \left( 15 . \Theta^{4/3} \right)^{1/2} \omega = 3,87 . \Theta^{2/3} \\ Mas \omega = \frac{d\Theta}{dt} \, \rightarrow \, 3,87 . \Theta^{2/3} = \frac{d\Theta}{dt} \\ 3,87 dt = \frac{d\Theta}{\Theta^{2/3}} \\ \therefore \ \Theta^{-2/3} d\Theta = 3,87 dt \\ \text{Integrando} \\ \int_0^{\Theta} \Theta^{-2/3} d\Theta = \int_0^t 3,87 dt \\ \frac{\Theta^{1/3}}{1/3} = 3,87 . t \\ \Theta^{1/3} = (3,87 . t) / 3 \\ \therefore \ \Theta^{1/3(3)} = \left[ \frac{1}{3} . 3,87 . t \right]^3 \rightarrow \Theta = 2,14 . t^3 Quando \ t = 4 \, s \ \Theta = 2,14 (4)^3 = 137 \pi \\ \left[ \omega = \frac{d\Theta}{dt} = 2,14 . (3) . t^2 = 6,42 t^2 \right] \\ Quando \ t = 4 \, s \ \omega = 102 \pi \, rad/s \\ 9) \\ a) \vec{A} \times \vec{B} = \begin{vmatrix} \, \hat{i} \ & \hat{j} \ & \hat{k} \, \ \, 5 \ & 0 \ & 2 \, \ \, -20 \ & 2 \ & 5 \, \ \end{vmatrix} \\ = 294 \hat{i} - 790 \hat{j} + 500 \hat{k} \\ b) \vec{A} \times \vec{B} = \begin{vmatrix} \, \hat{j} \ & \hat{k} \, \ \, 0 \ & 2 \ & 2 \, \ \, -2 \ & 2 \ & 5 \, \ \end{vmatrix} \\ = 54 \hat{i} - 44 \hat{j} + 4 \hat{k} \\ c) \vec{A} \times \vec{B} = \begin{vmatrix} \, \hat{i} \ & \hat{j} \ & \hat{k} \, \ \, 5 \ & 2 \ & 2 \, \ \, 10 \ & 4 \ & 4 \, \ \end{vmatrix} \\ = 0 \hat{i} + 0 \hat{j} + 0 \hat{k} d) \( \vec{A} \times \vec{B} \): | \( \hat{i} \quad \hat{j} \quad \hat{k} \) | | \( 5 \quad 1 \quad 2 \) | | \( 1 \quad 2 \quad 15 \) | = 13\hat{i} - 73\hat{j} + 9\hat{k} e) \( \vec{A} \times \vec{B} \): | \( \hat{i} \quad \hat{j} \quad \hat{k} \) | | \( 8 \quad 2 \quad -12 \) | | \( 0 \quad 2 \quad 15 \) | = 114\hat{i} - 120\hat{j} + 16\hat{k} (10) Exemplo: Vetor \( \vec{r}_A \) = (A - O) \[ \frac{\hat{i} \quad \hat{j} \quad \hat{k}}{x_A - x_O, y_A - y_O, z_A - z_O } \] \( \vec{r}_A \) = 0\hat{i} + 200\hat{j} + 120\hat{k} \text{ mm} \] \( \vec{r}_B \) = 300\hat{i} + 200\hat{j} + 120\hat{k} \text{ mm} \] \( \vec{r}_C \) = 300\hat{i} + 0\hat{j} + 120\hat{k} \] \( \vec{r}_D \) = 300\hat{i} + 0\hat{j} + 0\hat{k}