The period of the leg can be approximated by treating the leg as a physical pendulum, with a period of , where I is the moment of inertia, m is the mass, and h is the distance from the pivot point to the center of mass. The leg can be considered to be a right cylinder of constant density. For a man, the leg constitutes 16% of his total mass and 48% of his total height. Find the period of the leg of a man who is 1.85 m in height with a mass of 65 kg. The moment of inertia of a cylinder rotating about a perpendicular axis at one end is (m*l^2)/3 sec

This question is incomplete, the complete question is;

The period of the leg can be approximated by treating the leg as a physical pendulum, with a period of 2π√[ / mgh ] , where I is the moment of inertia, m is the mass, and h is the distance from the pivot point to the center of mass. The leg can be considered to be a right cylinder of constant density. For a man, the leg constitutes 16% of his total mass and 48% of his total height. Find the period of the leg of a man who is 1.85 m in height with a mass of 65 kg. The moment of inertia of a cylinder rotating about a perpendicular axis at one end is (m*^2)/3___________ sec.  

the period of the leg of the man is 1.5 sec

Explanation:

Given the data in the question;

Time period = 2π√[ / mgh ]

also, The moment of inertia of a cylinder rotating about a perpendicular axis at one end is ( m/3)

= m/3 and h = /2

hence;

Time period = 2π√[ m/3 / mgh ]

Time period = 2π√[ m / 3mgh ]

Time period = 2π√[ / 3g( /2) ]

Time period = 2π√[ 2 / 3g ]

Time period = 2π√[ 2 / 3g ]

now, = 48% of 1.85 = 0.48 × 1.85 = 0.888 m

acceleration due to gravity g = 9.8 m/s²

so we substitute these values into the equation;

Time period = 2π√[ (2×0.888 ) / (3×9.8) ]

Time period = 2π√[ 1.776 / 29.8 ]

Time period = 2π√[ 0.059597 ]

Time period = 2π × 0.24412

Time period = 1.5 sec

Therefore, the period of the leg of the man is 1.5 sec

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