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Ch. 11 - Angular Momentum; General Rotation
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook
Chapter 11, Problem 49c

Two ice skaters, both of mass 68 kg, approach on parallel paths 1.6 m apart. Both are moving at 3.5 m/s with their arms outstretched. They join hands as they pass, still maintaining their 1.6-m separation, and begin rotating about one another. Treat the skaters as particles with regard to their rotational inertia. If they now pull on each other’s hands, reducing their radius to half its original value, what is their common angular speed after reducing their radius?

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First, recognize that this is a conservation of angular momentum problem. Angular momentum is conserved because there are no external torques acting on the system. The initial angular momentum of the system must equal the final angular momentum.
The angular momentum of a system is given by the formula: \( L = I \omega \), where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. For point masses, the moment of inertia is \( I = m r^2 \), where \( m \) is the mass and \( r \) is the radius of rotation.
Calculate the initial moment of inertia of the system. Since there are two skaters, each of mass \( m = 68 \ \text{kg} \), and the radius of rotation is half the distance between them (\( r = 1.6 \ \text{m} / 2 = 0.8 \ \text{m} \)), the total initial moment of inertia is \( I_\text{initial} = 2 m r^2 \).
Determine the initial angular velocity \( \omega_\text{initial} \). The skaters are initially moving in straight lines with a speed of \( v = 3.5 \ \text{m/s} \). The relationship between linear velocity and angular velocity is \( v = r \omega \), so \( \omega_\text{initial} = v / r \).
Now, calculate the final angular velocity \( \omega_\text{final} \) after the skaters reduce their radius to half its original value (\( r_\text{final} = 0.8 \ \text{m} / 2 = 0.4 \ \text{m} \)). Using conservation of angular momentum, \( I_\text{initial} \omega_\text{initial} = I_\text{final} \omega_\text{final} \). Substitute \( I_\text{final} = 2 m r_\text{final}^2 \) and solve for \( \omega_\text{final} \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Conservation of Angular Momentum

The principle of conservation of angular momentum states that if no external torque acts on a system, the total angular momentum of that system remains constant. In this scenario, the two skaters initially have a certain angular momentum due to their rotation. When they pull on each other's hands and reduce their radius, the angular momentum must be conserved, allowing us to relate their initial and final angular speeds.
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Moment of Inertia

Moment of inertia is a measure of an object's resistance to changes in its rotational motion, depending on the mass distribution relative to the axis of rotation. For the skaters, as they pull closer together, their moment of inertia decreases because the mass is concentrated closer to the axis of rotation. This change in moment of inertia is crucial for calculating the new angular speed after they reduce their radius.
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Angular Speed

Angular speed is the rate at which an object rotates around an axis, typically measured in radians per second. In this problem, we need to determine the common angular speed of the skaters after they reduce their separation. By applying the conservation of angular momentum and the relationship between moment of inertia and angular speed, we can find the new angular speed after the skaters pull on each other's hands.
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Related Practice
Textbook Question

Two ice skaters, both of mass 68 kg, approach on parallel paths 1.6 m apart. Both are moving at 3.5 m/s with their arms outstretched. They join hands as they pass, still maintaining their 1.6-m separation, and begin rotating about one another. Treat the skaters as particles with regard to their rotational inertia. Calculate the change in kinetic energy for this process.

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Textbook Question

Suppose a 5.2 x 10¹⁰kg meteorite struck the Earth at the equator with a speed v = 2.2 x 10⁴ m/s, as shown in Fig. 11–38 and remained stuck. By what factor would this affect the rotational frequency of the Earth (1 rev/day)?

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Textbook Question

A particle is at the position (x, y, z) = (1.0, 2.0, 3.0)m. It is traveling with a vector velocity (-5.0 ,+ 2.8, -3.1)m/s. Its mass is 4.3 kg. What is its vector angular momentum about the origin?

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Textbook Question

Two lightweight rods 24 cm in length are mounted perpendicular to an axle and at 180° to each other (Fig. 11–35). At the end of each rod is a 480-g mass. The rods are spaced 42 cm apart along the axle. The axle rotates at 4.5 rad/s.

(a) What is the component of the total angular momentum along the axle?

(b) What angle does the vector angular momentum make with the axle? [Hint: Remember that the vector angular momentum must be calculated about the same point for both masses, which could be the cm.]

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Textbook Question

Two ice skaters, both of mass 68 kg, approach on parallel paths 1.6 m apart. Both are moving at 3.5 m/s with their arms outstretched. They join hands as they pass, still maintaining their 1.6-m separation, and begin rotating about one another. Treat the skaters as particles with regard to their rotational inertia. They now pull on each other’s hands, reducing their radius to half its original value. Calculate the change in kinetic energy for this process.

2
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Textbook Question

On a level billiards table a cue ball, initially at rest at point O on the table, is struck so that it leaves the cue stick with a center-of-mass speed v₀ and ω₀ a “reverse” spin of angular speed (see Fig. 11–41). A kinetic friction force acts on the ball as it initially skids across the table. Using conservation of angular momentum, find the critical angular speed ωC such that, if ω₀=ωC, kinetic friction will bring the ball to a complete (as opposed to momentary) stop.

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