Two skaters, each of mass 50 kg, approach each other along parallel paths separated by 3 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 the figure. Assume frictionless ice. Describe quantitatively the motion of the skaters after they have become connected by the pole.
What is the total kinetic energy of two-skater system?
By pulling on the pole, the skaters reduce their separation to 1.0 m. What is their angular speed then?
what is the new kinetic energy of the system after they have pulled themselves closer? explain the source of the increase kinetic energy
The Correct Answer and Explanation is :
Solution:
1. Total Initial Kinetic Energy
The total initial kinetic energy of the system is given by:
[
KE_{\text{initial}} = \frac{1}{2} m v_1^2 + \frac{1}{2} m v_2^2
]
Both skaters have the same mass (( m = 50 \, \text{kg} )) and speed (( v_1 = 1.4 \, \text{m/s} ) and ( v_2 = -1.4 \, \text{m/s} )):
[
KE_{\text{initial}} = \frac{1}{2} (50)(1.4)^2 + \frac{1}{2} (50)(1.4)^2 = 2 \times \frac{1}{2} (50)(1.96) = 98 \, \text{J}
]
Thus, the total initial kinetic energy is ( 98 \, \text{J} ).
2. Angular Speed After Reducing Separation
Initially, the skaters are ( r = 1.5 \, \text{m} ) away from the center of mass, as their paths are parallel and separated by ( 3.0 \, \text{m} ). When they reduce their separation to ( 1.0 \, \text{m} ), their distance from the center of mass becomes ( r’ = 0.5 \, \text{m} ).
Using conservation of angular momentum:
[
L_{\text{initial}} = L_{\text{final}}
]
Initially, angular momentum is:
[
L_{\text{initial}} = 2 \times m v r = 2 \times (50)(1.4)(1.5) = 210 \, \text{kg·m}^2/\text{s}
]
After reducing the separation, the angular velocity (( \omega’ )) relates to the new angular momentum:
[
L_{\text{final}} = 2 \times m r’^2 \omega’
]
Equating ( L_{\text{initial}} ) and ( L_{\text{final}} ):
[
210 = 2 \times (50)(0.5)^2 \omega’
]
[
\omega’ = \frac{210}{25} = 8.4 \, \text{rad/s}
]
Thus, the angular speed is ( 8.4 \, \text{rad/s} ).
3. New Kinetic Energy
Kinetic energy in rotational motion:
[
KE_{\text{final}} = \frac{1}{2} I \omega’^2
]
The moment of inertia (( I )) is:
[
I = 2 \times m r’^2 = 2 \times 50 \times (0.5)^2 = 25 \, \text{kg·m}^2
]
[
KE_{\text{final}} = \frac{1}{2} (25) (8.4)^2 = \frac{1}{2} (25)(70.56) = 882 \, \text{J}
]
Thus, the new kinetic energy is ( 882 \, \text{J} ).
4. Source of Increased Kinetic Energy
The increase in kinetic energy (( 882 – 98 = 784 \, \text{J} )) comes from work done by the skaters as they pull themselves closer along the pole. They exert forces, doing work against the tension in the pole, which transforms their internal energy (muscular energy) into rotational kinetic energy. This aligns with the principle of conservation of angular momentum but allows an increase in kinetic energy due to the external work they perform.