Two skaters, each of mass 35 kg, approach each other along parallel paths separated by 3.6 m. They have equal and opposite velocities of 1.4 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 Fig. 12-42. Assume frictionless ice.Figure 12-42(a) Describe quantitatively the motion of the skaters after they have become connected by the pole. m (radius) rad/s (angular speed)(b) By pulling on the pole, the skaters reduce their separation to 1.0 m. What is their angular speed then? rad/s(c) Calculate the kinetic energy of the system in (a) and (b). J (energy for system a) J (energy for system b)
Question
Two skaters, each of mass 35 kg, approach each other along parallel paths separated by 3.6 m. They have equal and opposite velocities of 1.4 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 Fig. 12-42. Assume frictionless ice.Figure 12-42(a) Describe quantitatively the motion of the skaters after they have become connected by the pole. m (radius) rad/s (angular speed)(b) By pulling on the pole, the skaters reduce their separation to 1.0 m. What is their angular speed then? rad/s(c) Calculate the kinetic energy of the system in (a) and (b). J (energy for system a) J (energy for system b)
Solution
(a) The skaters' motion after they have become connected by the pole can be described using the concept of conservation of angular momentum. The initial angular momentum of the system is given by the sum of the individual angular momenta of the skaters, which is the product of their mass, velocity, and distance from the center of mass (which is halfway between them, so 1.8m).
L_initial = 2 * (mass * velocity * distance) = 2 * (35 kg * 1.4 m/s * 1.8 m) = 176.4 kg*m^2/s
After they grab the pole, they will start rotating around their center of mass. The final angular momentum is given by the moment of inertia times the angular speed. The moment of inertia for two point masses rotating around a common axis is given by I = 2 * m * r^2, where m is the mass of one skater and r is the distance from the center of mass to each skater (1.8m).
I = 2 * (35 kg * (1.8 m)^2) = 226.8 kg*m^2
By conservation of angular momentum, L_initial = L_final, so
176.4 kgm^2/s = 226.8 kgm^2 * w
Solving for w (angular speed), we get w = 0.78 rad/s.
(b) If the skaters reduce their separation to 1.0 m, the distance from the center of mass to each skater becomes 0.5m. The moment of inertia becomes I = 2 * m * r^2 = 2 * (35 kg * (0.5 m)^2) = 35 kg*m^2.
Using conservation of angular momentum again, we have
176.4 kgm^2/s = 35 kgm^2 * w
Solving for w, we get w = 5.04 rad/s.
(c) The kinetic energy of the system in both cases is given by K = 1/2 * I * w^2.
For (a), K = 1/2 * 226.8 kg*m^2 * (0.78 rad/s)^2 = 69.3 J.
For (b), K = 1/2 * 35 kg*m^2 * (5.04 rad/s)^2 = 446.4 J.
So, the kinetic energy increases when the skaters pull themselves closer together.
Similar Questions
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