Help finding angular momentum and kinetic energy?

Question

Two ice skaters hold hands and rotate, making one revolution in 1.5 s. Their masses are 40 kg and 70 kg, and they are separated by 1.7 m.

(a) Find the angular momentum of the system about their center of mass.

____ J·s

(b) Find the total kinetic energy of the system.

____ J

Answer 1

r7=Dm/(M+m)

r4=DM/(M+m)

Answer 2

If D = 1.7 m, then r7 = (m/(m + M) D/2 = (40/110) D/2 the rotation axis from the 70 kg skater and r4 = (M/(m + M) D/2 = (70/110) D/2 from the 40 kg skater.  D = r7 + r4 and the two r's are the distances from the respective skaters to CM, the center of mass.

Total angular momentum is the sum of the two momenta L = L7 + L4 = k(M(r7)^2 + m(r4)^2)w = kwD/2 (1/(m + M))(Mm + mM) = kwD(Mm)/(m + M); look up k for point mass on the end of each rotation arm around CM and plug in the numbers.  w = 2 pi/1.5 radians per second, m = 40 kg, M = 70 kg, D = 1.7 m.  (k = 1 if the two skaters are treated as mass on the end of an axis.)

Kinetic energy is KE = E7 + E4 = 1/2 Iw^2; where I = I7 + I4 = kD(Mm)/(m + M) from the previous problem.  So KE = 1/2 kD(Mm)w^2/(M + m); you can plug the numbers.