In the figure, two skaters, each of mass 54.9 kg, approach each other along para

Question

In the figure, two skaters, each of mass 54.9 kg, approach each other along parallel paths separated by 2.29 m. They have opposite velocities of 1.73 m/s each. One skater carries one end of a long pole of negligible mass, and the other skater grabs the other end as she passes. The skaters then rotate about the center of the pole. Assume that the friction between skates and ice is negligible. What are (a) the radius of the circle, (b) the angular speed of the skaters, and (c) the kinetic energy of the two-skater system? Next, the skaters each pull along the pole until they are separated by 0.91 m. What then are (d) their angular speed and (e) the kinetic energy of the system?

Explanation / Answer

This problem can be solved as follows:

(a)  Rotation is about their combined centre of mass. Being equal masses this is the mid-point of the bar,

r = 2.29/2 = 1.145 m

(b) For each skater, linear momentum (mv) transfers to angular mom (I) ..
I = moment of inertia as a point mass at radius r (I = mr²), = angular velocity

mv = mr². => = v/r² = 1.73m/s / (1.145m)² => = 1.32 rad/s

(c) Individual KE(rot) = ½I² = ½(mr²)²
= ½ x 54.9kg x (1.145m)² x 1.32² = 62.70 J

For both skaters, net KE = 2 x 62.70J = 125.40 J

(d) By conservation of mom (with no external force acting on the system)
Initial angular momentum (2 x Ii.i) = final mom (2 x If.f) ..
ri = 1.145 m, rf = 0.455 m

2 x m.ri².i = 2 x m.rf².f

f = ri².i / rf² = 1.145² x 1.32 / 0.455² => f = 8.36 rad/s

(e) KEf = 2(½.If.f²) = If.f² = m.rf² x 8.36²
= 54.9 kg x (0.455m)² x 8.36² = 794.34 J

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