Answer:
The period of the system of ladder and woman, T = 2.5 seconds
Explanation:
Mass of the ladder, [tex]m_1 = 6 kg[/tex]
Mass of the artiste, [tex]m_2 = 42.0 kg[/tex]
Length of the ladder, L = 42.0 kg
The total moment of inertia can be calculated using the equation:
[tex]I = \frac{1}{3} M_1 L^2 + m_2 (\frac{L}{2} )^2\\I = \frac{1}{3} *6*3^2 + 42* (\frac{3}{2} )^2\\I = 18 + 94.5\\I = 112.5 kg m^2[/tex]
D = L/2 = 3/2
D = 1.5 m
The frequency of the system of ladder and woman follows that of a physical pendulum which can be given by the equation:
[tex]f = \frac{1}{2\pi } \sqrt{\frac{mgD}{I} } \\f = \frac{1}{2\pi } \sqrt{\frac{48*9.8*1.5}{112.5} }\\f = 0.4[/tex]
The period of the system of ladder and woman is given by:
T = 1/f
T = 1/0.4
T = 2.5 seconds
