Two friends hold on to a rope, one at each end, on a smooth, frictionless ice su

Question

Two friends hold on to a rope, one at each end, on a smooth, frictionless ice surface. They skate in a circle about an axis through the center of the rope and perpendicular to the ice. The mass of one friend is 78.0 kg and the other has a mass of 125.0 kg. The rope of negligible mass is 3.0 m long and they move at a speed of 5.00 m/s. (a) What is the magnitude of the angular momentum of the system comprised of the two friends? 1522.5 Correct: Your answer is correct. kg · m2/s (b) They now pull on the rope and move closer to each other so that the rope between them is now half as long. Determine the speed with which they move now. Incorrect: Your answer is incorrect. Consider the angular momentum principle. Note that there is no external torque acting on the system. m/s (c) The two friends have to do work in order to move closer to each other. How much work did they do?

Explanation / Answer

a) Initail angular momentum = m1*v*r + m2*v*r

= 78*5*1.5 + 125*5*1.5

= 1522.5 kg.m^2

b) Let v' is the speed of each person when the come closer.

Apply conservation of angular momentum

final angular momentum = initial angular momentum

(m1*v'*(r/2) + m2*v'*(r/2) = 1522.5

v'*(m1*r + m2*r)/2 = 1522.5

v'*(78*1.5 + 125*1.5)/2 = 1522.5

v' = 1522.5*2/(78*1.5 + 125*1.5)

= 10 m/s

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

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