Two skaters, each of mass 40 kg, approach each other along parallel paths that a

Question

Two skaters, each of mass 40 kg, approach each other along parallel paths that are separated by a distance of 2 m. Both skaters have a speed of 10 m/s.
The first skater carries a 2-m pole that may be considered massless. As he passes the pole, the second skater catches hold of the end. The two skaters then go around in a circle about the center of the pole.
What is their combined angular momentum about the center of the pole?
I get 800 kg m^2/s but i need to double check with anyone who could answer. I will rate you well if you show all your work.

Explanation / Answer

so just before they grab on they both a velocity perpendicular to the vector from the middle of the rod to themselves. so we can use the L=m v r sin=40*10*1 sin(90), note r=1 since its measured from the middle of the pole

so each of them has an L=400 kg m^2/s

so the total L is 800 kg m^2/s

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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