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 ωC , i.e., ω₀ = 0.90ωC, determine the ball’s cm velocity vCM when it starts to roll without slipping.
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16. Angular Momentum
Conservation of Angular Momentum
Problem 40b
Textbook Question
A small block on a frictionless, horizontal surface has a mass of 0.0250 kg. It is attached to a massless cord passing through a hole in the surface (Fig. E10.40). The block is originally revolving at a distance of 0.300 m from the hole with an angular speed of 2.85 rad/s. The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.150 m. Model the block as a particle. What is the new angular speed?


1
Identify the principle of conservation of angular momentum, which states that if no external torque acts on a system, the angular momentum of the system remains constant.
Write the expression for angular momentum L of the block: L = I * ω, where I is the moment of inertia and ω is the angular speed.
For a point mass m at a distance r from the axis of rotation, the moment of inertia I is given by I = m * r^2.
Set up the equation for conservation of angular momentum: m * r1^2 * ω1 = m * r2^2 * ω2, where r1 and ω1 are the initial radius and angular speed, and r2 and ω2 are the final radius and angular speed.
Solve for the new angular speed ω2: ω2 = (r1^2 * ω1) / r2^2. Substitute the given values: r1 = 0.300 m, ω1 = 2.85 rad/s, and r2 = 0.150 m to find ω2.

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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
In a closed system with no external torques, the angular momentum remains constant. For a particle moving in a circle, angular momentum L is given by L = Iω, where I is the moment of inertia and ω is the angular speed. When the radius changes, the moment of inertia changes, but the product Iω remains constant, allowing us to solve for the new angular speed.
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Conservation of Angular Momentum
Moment of Inertia for a Point Mass
The moment of inertia (I) for a point mass revolving around an axis is calculated as I = mr², where m is the mass and r is the radius of the circle. This concept is crucial for understanding how changes in the radius affect the angular momentum and, consequently, the angular speed of the block.
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Moment of Inertia & Mass Distribution
Relationship Between Radius and Angular Speed
As the radius of rotation decreases, the angular speed must increase to conserve angular momentum, assuming no external torques. This inverse relationship is derived from the conservation principle, where the initial and final angular momenta (I₁ω₁ = I₂ω₂) are equal, allowing us to calculate the new angular speed when the radius changes.
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