One ice skater of mass m moves with speed 2v to the right, while another of the same mass m moves with speed v toward the left. Their paths are separated by a distance b. At t = 0, when they are both at x = 0, they grasp a pole of length b and negligible mass. For t > 0, consider the system as a rigid body of two masses m separated by distance b, as shown in figure II. Which of the following is the correct formula for the motion after t = 0 of the skater initially at y = b/2?

(1)	x = 2vt, y = b/2
(2)	x = vt + 0.5b sin(3vt/b), y = 0.5b cos(3vt/b)
(3)	x = 0.5vt + 0.5b sin(3vt/b), y = 0.5b cos(3vt/b)
(4)	x = 0.5vt + 0.5b sin(6vt/b), y = 0.5b cos(6vt/b)

To analyze the motion of the two ice skaters after they grasp the pole, we need to consider the dynamics of the system as a rigid body. Initially, one skater is moving to the right with speed 2v, and the other is moving to the left with speed v. When they grab the pole, they effectively become a single system with a fixed distance between them, which is crucial for understanding their subsequent motion.

Understanding the System Dynamics

When the skaters grasp the pole, they will start to rotate around the center of mass of the system. The center of mass (CM) for two objects of equal mass can be found at the midpoint between them. Given that their initial positions are separated by a distance b, the center of mass will be located at:

• Position of skater 1 (moving right): 0
• Position of skater 2 (moving left): -b
• Center of mass (CM): (0 + (-b))/2 = -b/2

However, since they are moving and will start to rotate about this center of mass, we need to consider the angular motion that results from their initial linear velocities.

Analyzing the Motion After Grasping the Pole

After t = 0, the skaters will rotate around the center of mass. The angular velocity (ω) of the system can be derived from their initial linear velocities. The skater moving to the right has a linear velocity of 2v, while the one moving to the left has a velocity of v. The difference in their speeds will create a rotational motion.

The angular velocity can be calculated as:

• Relative speed = 2v + v = 3v
• Distance from CM to skater 1 = b/2
• Angular velocity (ω) = relative speed / distance = (3v) / (b/2) = (6v/b)

Now, we can express the positions of the skaters in terms of time (t) using the equations of circular motion. The skater initially at y = b/2 will move in a circular path around the center of mass.

Position Equations for the Skaters

The position of the skater initially at y = b/2 can be described using the following parametric equations:

• x(t) = (b/2) * sin(ωt) + (initial x position)
• y(t) = (b/2) * (1 - cos(ωt)) + (initial y position)

Substituting ω = 6v/b into these equations gives:

• x(t) = (b/2) * sin(6vt/b)
• y(t) = (b/2) * (1 - cos(6vt/b)) + b/2

However, we need to adjust these equations to account for the initial positions and velocities of the skaters. The correct transformation leads us to the final equations for the skater initially at y = b/2:

• x(t) = 0.5vt + 0.5b sin(6vt/b)
• y(t) = 0.5b cos(6vt/b)

Final Result

Thus, the correct formula for the motion after t = 0 of the skater initially at y = b/2 is:

(4) x = 0.5vt + 0.5b sin(6vt/b), y = 0.5b cos(6vt/b)

This equation captures the combined linear and rotational motion of the skaters as they move together around their center of mass after they grasp the pole.