In the figure below, two skaters, each of mass 55 kg, approach each other along
ID: 2035245 • Letter: I
Question
In the figure below, two skaters, each of mass 55 kg, approach each other along parallel paths separated by 4.0 m. They have opposite velocities of 1.3 m/s each. One skater carries one end of a long pole with negligible mass, and the other skater grabs the other end as she passes. The skaters then rotate around the center of the pole. Assume that the friction between skates and ice is negligible. (a) What is the radius of the circle? (b) What is the angular speed of the skaters? 0.65 rad/s (c) What is the kinetic energy of the two-skater system? 92.95 Next, the skaters pull along the pole until they are separated by 1.2 m. (d) What then is their angular speed? X rad/s 5.1 (e) What then is the kinetic energy of the system? (1) What provided the energy for the increased kinetic energy? (Select all that apply.) U centrifugal force O gravity O friction O internal energy of the skaters O centripetal forceExplanation / Answer
(a) The rotation is about the combined center of mass , since masses are equal,the radius is the mid-point of the bar
r =4/2 = 2.0 m
(b) For each skater,the linear momentum is transferred to the angular momentum
linear momentum = mv
angular momentum = m?r,
where ? = ang velocity
so
mv = mr.?
? = v/r = 1.3m/s / (2) = 0.65
thus ? = 0.65 rad/s
(c) Kinetic Energy = ½I?² = ½(mr²)?²
= ½ x 55kg x (2)² x (0.65)² = 46.475 J
For both the skaters, net Kinetic Energy = 2 x 46.475 J = 92.95 J
(d) from the conservation of momentum
Initial angular momentum (2 x Ii.?i) = final angular momentum (2 x If.?f)
ri = 2m, rf = 0.8m
thus
2 x m.ri².?i = 2 x m.rf².?f
?f = ri².?i / rf² = (2)² x 0.65 / (0.8)² = 4.0625 rad/s
e) Kinetic energy of the system = K.Ef = 2(½.If.?f²) = If.?f² = m.rf² x (4.0625)²
= 55kg x (0.8m)² x (4.0625)² = 580.94 J
f) The work done by both the skaters as they pull inwards (F) against the cemtripetal force (pulling themselves inward) to get closer (e)