Question

A boxer receives a horizontal blow to the head that topples him over. The force, of magnitude 1.40 103 N, is applied for 1.3 10-2 s at a point 1.71 m above the floor. The boxer has a moment of inertia of 73.8 kg · m2 for rotation about an axis at his feet. (a) Find the boxer's final angular velocity. (b) How long would it take him to hit the floor if he continued with the same angular velocity? Why does his angular velocity increase as he falls?

Answer 1

a)

L = I *ω and τ = dL /dt

τ = dL /dt or L/t

F*r = L/t

L = F*t*r = I *ω

1.40*10^3*1.3*10^-2*1.71 = 73.8*ω

ω = ( 1.40*10^3*1.3*10^-2*1.71)/73.8 = 0.421707317

ω =  0.4217 rad/s

b)

the total angular displacement of his body as he falls on the ground

= pi /2 rad or 90 degree  

θ =ω *t

t =θ/ ω = ( pi /2) /0.421707317 = 3.7248 sec

C )

his center of mass gets closer and closer as he falls on the ground and because of this his angular velocity increases

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