Question

2. Two men of equal mass M are skating in a circle on a perfectly frictionless pond. They are each holding onto a rope of length L. What happens to the size of the momentum p of each man if they both pull on the rope, "hand over hand", and shorten the distance between them to L/2? Assume the men again move in a circle and that the magnitude of their momenta are equal. Hint: what type of force act here? Does it exert a torque?

Answer 1

lar momentum will remain conserved. A= mul A = m vx / An. v.y=whvik (V=2.0] individual momentum p=mov, (p=2. m.vll momentum becomes double.