Two skaters, each of mass 70 kg, skate at speeds of 4 m/s in opposite directions

Question

Two skaters, each of mass 70 kg, skate at speeds of 4 m/s in opposite directions along parallel lines 1.5 m apart. As they are about to pass one another they join hands and go into circular paths about their common center of mass. (a) What is their total angular momentum? (b) A third skater is skating at 2 m/s along a line parallel to the initial directions of the other two and 6 m o to the side of the track of the nearer one. From his standpoint, what is the total angular momentum of the other two skaters as they rotate?

Explanation / Answer

A) Here the linear mumentum of the of each skater is p = m v

which is 70 * 4 = 280 kgm/s

When the two skaters join hand and start rotating in a circle the diameter of the circle will be equal to the distance between the two lines.

Angular mumentum of one skater is the multiplication of linear mumentum, radius of the circle and sin of angle between radius and linear mumentum. Since here radius is perpendicular to the linear mumentum so angle will be 90 degree and sine of 90 is 1. Our angular mumentum of one skater will be given by 280 * 0.75 = 210 so their total angular mumentum will be = 210+210 = 420

b) Angular mumentum from the point of view of third skater will be given by

Total angular mumentum from third skater's point of view = Angular mumentum of center of mass about third skater + angular mumentum of system about the center of mass

here angular mumentum about center of mass we have already calculated in part a

To calculate angular mumentum of center of mass about third skater = distance of third skater from center of mass * total mass * velocity of third skater

= (6+0.75)*140*2

= 1890

so we get

Total angular mumentum from third skater's point of view = Angular mumentum of center of mass about third skater + angular mumentum of system about the center of mass

= 1890+420

= 2310

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