Question

Two ice skaters hold hands and rotate, making one revolution in 2.47 s. Their masses are 60.0 kg and 73.0 kg, and they are separated by 1.00 m. Find the angular momentum of the system about their center of mass. If the skaters pull towards each other and decrease their separation to 0.50 m, what is the new period of rotation about the center of mass?

Answer 1

let the center of mass be at a distnace x from the 60 kg person

x=73*1/(73+60)=0.55 m

distance of center of mass from 73 kg person=1-0.55=0.45m

mass moment of inertia = 60*0.55²+73*0.45² = 32.93 kg-m²

angular velocity=2*pi/2.47=2.54 rad/s

angular momentum = 32.93*2.54=83.77 kg-m²/s

if the separation is reduced to 0.5 m,let us determine the new center of mass location and mass moment of inertia

let the center of mass be at a distnace x from the 60 kg person

x=73*0.5/(73+60)=0.274 m

distance of center of mass from 73 kg person=0.5-0.274=0.226m

mass moment of inertia =73*0.226²+60*0.274² = 8.23 kg-m²

since the angular momentum will be conserved, new angular velocity = 83.77/8.23=10.17 rad/s

period of rotation = 2*pi/10.17=0.618 sec