In the figure, two skaters, each of mass 58.0 kg, approach each other along parallel paths separated by 2.79 m. They have opposite velocities of 1.50 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 1.25 m. What then are (d) their angular speed and (e) the kinetic energy of the system

67

To solve this problem, we can use the principle of conservation of angular momentum and the formula for rotational kinetic energy. Let's break down the problem step-by-step:

(a) The radius of the circle:
When the skaters grab the pole, they rotate about its center. The moment of inertia of a system of two particles with masses m1 and m2, separated by a distance d, rotating about an axis perpendicular to the line connecting them is given by: I = m1 * d^2 + m2 * d^2.
In this case, both skaters have the same mass m = 58.0 kg and the distance between them is d = 2.79 m.
So, the moment of inertia I = 2 * m * d^2 = 2 * 58.0 kg * (2.79 m)^2.

(b) The angular speed of the skaters:
The principle of conservation of angular momentum states that the initial angular momentum is equal to the final angular momentum.
The initial angular momentum is zero since the skaters have opposite velocities.
The final angular momentum is given by: L = I * ω, where ω is the angular speed.
So, the angular speed ω = 0 / I = 0 rad/s.

(c) The kinetic energy of the two-skater system:
The kinetic energy of a rotating body is given by: KE_rot = (1/2) * I * ω^2.
In this case, the kinetic energy of the two-skater system is KE_rot = (1/2) * I * 0^2 = 0 Joules.

(d) Their angular speed when they are separated by 1.25 m:
Now the skaters are pulling along the pole, so the moment of inertia changes.
The new moment of inertia I' can be calculated using: I' = m * d'^2 + m * d'^2, where d' is the new distance between the skaters (1.25 m).
So, I' = 2 * 58.0 kg * (1.25 m)^2.

(e) The kinetic energy of the system when they are separated by 1.25 m:
Similarly, the new kinetic energy KE_rot' can be calculated using: KE_rot' = (1/2) * I' * ω'^2, where ω' is the new angular speed.
So, KE_rot' = (1/2) * I' * ω'^2.

To find ω', we can apply the principle of conservation of angular momentum again. The initial angular momentum is zero since the skaters have stopped rotating. The final angular momentum is given by: L' = I' * ω'.
Since the initial and final angular momenta are equal, we can set them equal to each other and solve for ω'.

I * ω = I' * ω'
(2 * 58.0 kg * (2.79 m)^2) * 0 = (2 * 58.0 kg * (1.25 m)^2) * ω'
Solving for ω', we get: ω' = 0 rad/s.

Therefore, when they are separated by 1.25 m, the angular speed of the skaters is still zero and the kinetic energy of the system remains zero.

To find the answers to these questions, we can use the principles of conservation of angular momentum and conservation of kinetic energy.

(a) To determine the radius of the circle, we can use the conservation of angular momentum. The initial angular momentum of the system is equal to zero since the skaters have opposite velocities. When they grab the pole and start rotating, the angular momentum is conserved. We can write this as:

initial angular momentum = final angular momentum

Since the moment of inertia for the skaters is negligible compared to the moment of inertia of the pole, we can focus on the pole. The moment of inertia of a uniform rod rotating about its center is given by (1/12) * m * L^2, where m is the mass of the rod and L is its length. In this case, the length is the separation between the skaters, which is 2.79 m.

The initial angular momentum is zero because the skaters have opposite velocities. The final angular momentum is given by the moment of inertia of the pole times the angular speed of rotation, which we'll call ω. Therefore:

0 = (1/12) * m * L^2 * ω

Rearranging the equation, we can solve for ω:

ω = 0

Since ω = 0, the angular speed is zero, and the skaters are not rotating. Therefore, the radius of the circle is also zero.

(b) The angular speed of the skaters is zero, as explained in part (a).

(c) The kinetic energy of the two-skater system is given by the sum of their individual kinetic energies. The kinetic energy for each skater is given by (1/2) * m * v^2, where m is the mass and v is the velocity. Both skaters have the same mass (58.0 kg) and velocity (1.50 m/s). Calculating the individual kinetic energies and summing them up gives:

Kinetic energy = (1/2) * m * v^2 + (1/2) * m * v^2
= (1/2) * (58.0 kg) * (1.50 m/s)^2 + (1/2) * (58.0 kg) * (1.50 m/s)^2
= 87 J + 87 J
= 174 J

Therefore, the kinetic energy of the two-skater system is 174 Joules.

(d) When the skaters pull along the pole until they are separated by 1.25 m, the conservation of angular momentum still applies. Since there is no external torque acting on the system, the angular momentum will remain constant. We can write this as:

initial angular momentum = final angular momentum

Again, the initial angular momentum is zero since the skaters have opposite velocities. The final angular momentum is given by the moment of inertia of the pole times the new angular speed, which we'll call ω'.

0 = (1/12) * m * L^2 * ω'

Solving for ω', we get:

ω' = 0

Similar to part (b), the angular speed of the skaters is zero when they are pulling along the pole.

(e) When the skaters are pulling along the pole, they are not rotating, so the kinetic energy of the system will only depend on their individual velocities. Each skater has a velocity of 1.50 m/s, and their masses remain the same (58.0 kg). Therefore, the kinetic energy of the system is given by:

Kinetic energy = (1/2) * m * v^2 + (1/2) * m * v^2
= (1/2) * (58.0 kg) * (1.50 m/s)^2 + (1/2) * (58.0 kg) * (1.50 m/s)^2
= 87 J + 87 J
= 174 J

Hence, the kinetic energy of the system remains at 174 Joules.

Jobby