Two skaters, each of mass 65 kg , approach each other along parallel paths separ

Question

Two skaters, each of mass 65 kg, approach each other along parallel paths separated by 6.9 m. They have equal and opposite velocities of 1.2 m/s. The first skater carries one end of a long pole with negligible mass, and the second skater grabs the other end of it as she passes; see the figure. Assume frictionless ice. Describe quantitatively the motion of the skaters after they have become connected by the pole. What is their angular speed? By pulling on the pole, the skaters reduce their separation to 1.0 m. What is their angular speed then? Calculate the ratio of the final kinetic energy to the original kinetic energy.

Explanation / Answer

a) they revolve around their commom CM (center of mass) which is at the center of the 3.8 m pole.
b)The total angular momentum about their CM is the sum of their individual L's. L=(Sum mvr). Which in turn equals the moment of inertia times the angular speed. L=(I)(omega)
c)The KE = (1/2)( I )(omega)^2
d)Since no external forces or torques are acting, angular momentum is conserved. So since the moment of inertia is reduced, the angular speed is increased.( L before)=(L after)
e)(KE after) =(1/2)( I after)(omega after)^2

f)internal energy of the skaters pulling on the pole changing the moment of inertia

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.