In Fig. 11-50, two skaters, each of mass 50 kg, approach each other along parallel paths separated by 3.0 m. They have opposite velocities of 1.4 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 around 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 pull along the pole until they are separated by 1.0 m. What then are (d) their angular speed and (e) the kinetic energy of the system? (f) What provided the energy for the increased kinetic energy?
The Correct Answer and Explanation is :
Solution
Given:
- Mass of each skater, ( m = 50 \, \text{kg} )
- Initial separation between skaters, ( d = 3.0 \, \text{m} )
- Initial velocity of each skater, ( v = 1.4 \, \text{m/s} )
- Negligible friction between skates and ice.
(a) Radius of the circle:
The center of mass (CM) lies at the midpoint between the two skaters. Since they are separated by ( 3.0 \, \text{m} ), the radius ( r ) of the circle traced by each skater is:
[
r = \frac{d}{2} = \frac{3.0}{2} = 1.5 \, \text{m}.
]
(b) Angular speed:
The total angular momentum ( L ) of the system is conserved because there is no external torque. The initial angular momentum can be written as:
[
L = 2 \cdot m \cdot v \cdot r,
]
where ( r = 1.5 \, \text{m} ) is the distance of each skater from the CM. Substituting values:
[
L = 2 \cdot 50 \cdot 1.4 \cdot 1.5 = 210 \, \text{kg·m}^2/\text{s}.
]
The moment of inertia of the system about the CM is:
[
I = 2 \cdot m \cdot r^2 = 2 \cdot 50 \cdot (1.5)^2 = 225 \, \text{kg·m}^2.
]
The angular speed ( \omega ) is:
[
\omega = \frac{L}{I} = \frac{210}{225} = 0.933 \, \text{rad/s}.
]
(c) Initial kinetic energy:
The total kinetic energy is:
[
KE = \frac{1}{2} I \omega^2 = \frac{1}{2} \cdot 225 \cdot (0.933)^2 = 97.8 \, \text{J}.
]
(d) New angular speed:
When the skaters pull themselves closer, the separation reduces to ( 1.0 \, \text{m} ), so the new radius ( r’ = 0.5 \, \text{m} ). The new moment of inertia is:
[
I’ = 2 \cdot m \cdot (r’)^2 = 2 \cdot 50 \cdot (0.5)^2 = 25 \, \text{kg·m}^2.
]
Angular momentum is conserved:
[
L = I \omega = I’ \omega’,
]
so:
[
\omega’ = \frac{L}{I’} = \frac{210}{25} = 8.4 \, \text{rad/s}.
]
(e) New kinetic energy:
The new kinetic energy is:
[
KE’ = \frac{1}{2} I’ (\omega’)^2 = \frac{1}{2} \cdot 25 \cdot (8.4)^2 = 882 \, \text{J}.
]
(f) Source of energy:
The increase in kinetic energy comes from the work done by the skaters in pulling themselves closer along the pole. This work increases the system’s rotational kinetic energy.
Summary of Results:
- Radius of circle: ( 1.5 \, \text{m} )
- Angular speed (initial): ( 0.933 \, \text{rad/s} )
- Kinetic energy (initial): ( 97.8 \, \text{J} )
- Angular speed (new): ( 8.4 \, \text{rad/s} )
- Kinetic energy (new): ( 882 \, \text{J} )
- Source of energy: Work done by skaters pulling along the pole.