Question

Find the period of the leg of a man who is 1.83 m in height with a mass of 67 kg. The moment of inertia of a cylinder rotating about a perpendicular axis at one end is ml2/3. Write your answer with one decimal place.

Answer 1

2.2 s

Step-by-step explanation:

Using the equation for the period of a physical pendulum, T = 2π√(I/mgh) where I = moment of inertia of leg about perpendicular axis at one point = mL²/3 where m = mass of man = 67 kg and L = height of man = 1.83 m, g = acceleration due to gravity = 9.8 m/s² and h = distance of leg from center of gravity of man = L/2 (center of gravity of a cylinder)

So, T = 2π√(I/mgh)

T = 2π√(mL²/3 /mgL/2)

T = 2π√(2L/3g)

substituting the values of the variables into the equation, we have

T = 2π√(2L/3g)

T = 2π√(2 × 1.83 m/(3 × 9.8 m/s² ))

T = 2π√(3.66 m/(29.4 m/s² ))

T = 2π√(0.1245 s² ))

T = 2π(0.353 s)

T = 2.22 s

T ≅ 2.2 s

So, the period of the man's leg is 2.2 s