Time period of pendulum is given by,
T = 2*pi*sqrt(I/(m*g*h))
here, I = moment of inertia of pendulum = 0.5*m*r^2 + m*(r+L)^2
m = mass of disk = 0.8261 kg
g = 9.81 m/s^2
h = distance of center of gravity of disk from hinge = r+L
L = length of rod = ??
T = time period = 1.592 sec.
r = radius of disk = 14.38 cm = 0.1438 m
So, 1.592 = 2*pi*sqrt[(0.5*0.8261*0.1438^2 + 0.8261*(0.1438+L)^2)/(0.8261*9.81*(0.1438 + L))]
1.592^2/(4*pi^2) = (0.5*0.8261*0.1438^2 + 0.8261*(0.1438+L)^2)/(0.8261*9.81*(0.1438 + L))
(0.5*0.8261*0.1438^2 + 0.8261*(0.1438+L)^2) = (1.592^2/(4*pi^2))*(0.8261*9.81*(0.1438 + L))
0.8261*L^2 + 2*0.1438*0.8261*L + 0.1438^2*0.8261 + 0.5*0.8261*0.1438^2 - (1.592^2/(4*pi^2))*0.8261*9.81*0.1438 -(1.592^2/(4*pi^2))*(0.8261*9.81*L = 0
0.8261*L^2 - 0.28268*L - 0.04919= 0
By solving above quadratic equatuion:
L = 0.4691 m
L = 46.91 cm
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