Ch. 11 - Angular Momentum; General Rotation

Giancoli Douglas - Physics for Scientists and Engineers 5th edition

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Giancoli Douglas 5th edition

Ch. 11 - Angular Momentum; General Rotation

Problem 53c

Chapter 11, Problem 53c

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. If ω₀ is 10% smaller than ω_C , i.e., ω₀ = 0.90ω_C, determine the ball’s cm velocity v_CM when it starts to roll without slipping.

Verified step by step guidance

Identify the key concepts: The problem involves rotational and translational motion, friction, and the condition for rolling without slipping. The cue ball starts with an initial velocity \(v_0\) and reverse angular velocity \(\omega_0\). The goal is to find the center-of-mass velocity \(v_{CM}\) when the ball transitions to rolling without slipping.

Understand the condition for rolling without slipping: Rolling without slipping occurs when the linear velocity \(v_{CM}\) of the center of mass and the angular velocity \(\omega\) are related by \(v_{CM} = R\omega\), where \(R\) is the radius of the ball.

Set up the equations of motion: The kinetic friction force \(f_k\) acts to reduce the linear velocity \(v\) and increase the angular velocity \(\omega\). Using Newton's second law for translation and rotation, we have: \(f_k = m \frac{dv}{dt}\) and \(f_k R = I \frac{d\omega}{dt}\), where \(I = \frac{2}{5}mR^2\) is the moment of inertia of the ball.

Relate the changes in \(v\) and \(\omega\): From the equations above, \(\frac{dv}{dt} = \frac{5}{2R} \frac{d\omega}{dt}\). Integrating this relationship over time, we find \(v - v_0 = -\frac{5}{2}R(\omega - \omega_0)\).

Substitute the given condition \(\omega_0 = 0.90\omega_C\): The critical angular velocity \(\omega_C\) is defined as \(\omega_C = \frac{v_0}{R}\). Substituting \(\omega_0 = 0.90\omega_C\) into the equation, solve for \(v_{CM}\) when \(v_{CM} = R\omega\). This will yield the final expression for the center-of-mass velocity when rolling without slipping begins.

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Key Concepts

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

Kinetic Friction

Kinetic friction is the force that opposes the motion of two surfaces sliding past each other. It is proportional to the normal force and is characterized by the coefficient of kinetic friction. In the context of the cue ball, this force acts to slow down the ball as it skids across the table, affecting its transition from sliding to rolling.

Rolling Without Slipping

Rolling without slipping occurs when the point of contact between a rolling object and the surface does not slide. This condition is met when the linear velocity of the center of mass (v_CM) is equal to the angular velocity (ω) multiplied by the radius (r) of the object. For the cue ball, achieving this state means that the ball will roll smoothly without any relative motion at the contact point.

Conservation of Angular Momentum

Conservation of angular momentum states that if no external torque acts on a system, the total angular momentum remains constant. In the case of the cue ball, as it transitions from sliding to rolling, the angular momentum associated with its spin and the linear momentum must be considered to determine the final velocity of the center of mass when it starts rolling without slipping.

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Related Practice

Textbook Question

A toy gyroscope consists of a 170-g disk with a radius of 5.5 cm mounted at the center of a thin axle 21 cm long (Fig. 11–42). The gyroscope spins at 45 rev/s. One end of its axle rests on a stand and the other end precesses horizontally about the stand. How long does it take the gyroscope to precess once around?

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Textbook Question

A merry-go-round with a moment of inertia equal to 860 kg·m² and a radius of 3.0 m rotates with negligible friction at 1.70 rad/s. A child initially standing still next to the merry-go-round jumps onto the edge of the platform straight toward the axis of rotation causing the platform to slow to 1.25 rad/s. What is her mass?

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

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.

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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 ω_C such that, if ω₀=ω_C, kinetic friction will bring the ball to a complete (as opposed to momentary) stop.

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Textbook Question

Suppose the solid wheel of Fig. 11–42 has a mass of 260 g and rotates at 85 rad/s; it has radius 6.0 cm and is mounted at the center of a horizontal thin axle 25 cm long. At what rate does the axle precess?

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