Fisica 2

Renata Vieira.\nA:\nT = 2.5 \\sqrt{\\frac{l}{Mgd}} \\implies I = \\left(\\frac{f}{2\\pi}\\right)^{2}.MgD\nB:\nMédia . Erros\nE para\n1.002 + 1.003 + 0.973 + 0.964 + 0.977 + 0.977 + 0.935 + 0.982 + 0.992 + 0.995 + 1.000 + 0.935 + 0.932 + 1.002 + 0.996 + 0.988 + 1.003 + 0.979 + 0.986 + 0.997\n= 20\n\\frac{19.884}{20} - 0.9942s \ \ t = 0.9942s\nC: d = 0.165 \ \ M = 0.0684 \ \ t = 0.5942s\nI = \\left(\\frac{t}{2\\pi}\\right)^{2}MgD = \\left(0.9942\\right)^{2} \\cdot 0.0684 \\cdot 0.165\n\\approx 0.00277 \\ kg.m^{2}\nD = I_{CM} = \\frac{1}{3}M\\left(a^{2} + b^{2}\\right)\nI_{p} = Mh^{2} + I_{CM} E = I_{CM} = \\frac{1}{3} M\\left(a^{2}+b^{2}\\right)\nI_{CM} = 0.0684 \\cdot \\left(0.4^{2}+0.0.345^{2}\\right) = 0.000919 \\ kg.m^{2}\nI_{p} = Mh^{2} + I_{CM}\nI_{p} = 0.0684 \\cdot (0.165)^{2} + 0.000919 = 0.002781 \\ kg.m^{2}\n\\% \\; \\text{Erro} = \\frac{I_{P} - I_{1}}{I_{I}} \cdot 100\n\\% = \\frac{0.002781 - 0.00277}{0.00277} \\cdot 100 = 0.395\\%\nA diferença entre o valor experimental I e o valor teórico I_{p} pode derivar de fatores externos com a força de resistência do ar, pode ocorrer devido ao atrito entre a barra suspensa e o eixo.