Two skaters, each of mass 80 kg , approach each other along parallel paths separ

Question

Two skaters, each of mass 80 kg, approach each other along parallel paths separated by 6.1 m. They have equal and opposite velocities of 1.8 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?

Explanation / Answer

Data:
Mass of each skater, m = 80 kg

Velocity of each, v = 1.8 m/s

Radius of circular path, r = 6.1 m / 2
                                      = 3.05 m Solution:

Angular momentum of each, L = m v r
                                               = 439.2 kg m2 s-1

Moment of inertia of each, I = m r2
                                           = (80 kg)  (3.05 m)2
                                           = 744.2 kg m2

Angular momentum of both, L' = 2L
                                               = 878.4 kg m2 s-1

Moment of inertia of both, I' = 2 I
                                           = 1488.4 kg m2

(a)

L' = I'
Angular speed, = 878.4 kg m2 s-1 / 1488.4 kg m2                            = 0.59 rad/s

(b)

Radius of circular path, r = 1 m / 2
                                         = 0.5 m


Moment of inertia of each, I1= m r^2
                                              = (80 kg)(0.5 m )2
                                              = 20 kg m2


Moment of inertia of both, I1' = 2 I1
                                               = 40 kg m2

According to law of conservation of angular momentum,

L ( after ) = L ( before )

I1' ' = 878.4 kg m2 s-1

(40 kg m2 )( ' ) = 878.4 kg m2 s-1


'= 21.96 rad/s

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