A very light rigid rod with a length of 0403 m extends strai

A very light rigid rod with a length of 0.403 m extends straight out from one end of a meter stick. The combination is suspended from a pivot at the upper end of the rod as shown in the following figure. The combination is then pulled out by a small angle and released. Determine the period of oscillation of the system. ls Your response differs from the correct answer by more than 10%. Double check your calculations, By what percentage does the period differ from the period of a simple pendulum 0.903 m long? Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully.%

Solution

given

L = 1 m

length of rod l = 0.403 m

d = L/2 + l = 0.5+0.403 =0.903 m

the equilvalent length

Leq = I / m*d

where I = moment of inetia of rod = Icm + m*d^2 = m*L^2 / 12 + m*d^2

Leq = m*(L^2/12 + d^2) / m*d = (L^2/12 + d^2) / d

Leq = (1/12 + (0.903)^2) / 0.903 = 0.995 m

T = 2*pi*sqrt (Leq / g) = 2*3.14*sqrt (0..995m / 9.8)

T = 2 s

(b)

given that

L = 0.903 m

T\' = 2*3.14 *sqrt (0.903 / 9.8)

T = 1.90 s

% change in period = (T/T\'-1)*100

% change = ((2-1.90) / 1.90) *100

% change = 5.26%