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

Giancoli Douglas 5th edition

Ch. 11 - Angular Momentum; General Rotation

Problem 39

Chapter 11, Problem 39

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?

Verified step by step guidance

Step 1: Recall the formula for angular momentum. The angular momentum $ \mathbf{L} $ of a particle about the origin is given by the cross product of its position vector $ \mathbf{r} $ and its linear momentum $ \mathbf{p} $: $ \mathbf{L} = \mathbf{r} \times \mathbf{p} $.

Step 2: Write the position vector $ \mathbf{r} $ and velocity vector $ \mathbf{v} $ in component form. Here, $ \mathbf{r} = (1.0, 2.0, 3.0) \, \text{m} $ and $ \mathbf{v} = (-5.0, 2.8, -3.1) \, \text{m/s} $.

Step 3: Calculate the linear momentum $ \mathbf{p} $ using the formula $ \mathbf{p} = m \mathbf{v} $, where $ m $ is the mass of the particle. Multiply each component of $ \mathbf{v} $ by the mass $ m = 4.3 $ kg to get $ \mathbf{p} = (4.3 \times -5.0, 4.3 \times 2.8, 4.3 \times -3.1) \, \text{kg·m/s} $.

Step 4: Compute the cross product $ \mathbf{L} = \mathbf{r} \times \mathbf{p} $. Use the determinant method for the cross product: $ \mathbf{L} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1.0 & 2.0 & 3.0 \\ (4.3 \times -5.0) & (4.3 \times 2.8) & (4.3 \times -3.1) \end{vmatrix} $, where $ \mathbf{i}, \mathbf{j}, \mathbf{k} $ are the unit vectors in the x, y, and z directions.

Step 5: Expand the determinant to find the components of $ \mathbf{L} $. The x-component is $ L_x = (2.0)(4.3 \times -3.1) - (3.0)(4.3 \times 2.8) $, the y-component is $ L_y = -[(1.0)(4.3 \times -3.1) - (3.0)(4.3 \times -5.0)] $, and the z-component is $ L_z = (1.0)(4.3 \times 2.8) - (2.0)(4.3 \times -5.0) $. Simplify each term to express $ \mathbf{L} $ in component form.

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Key Concepts

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

Angular Momentum

Angular momentum is a vector quantity that represents the rotational motion of an object. It is calculated as the cross product of the position vector and the linear momentum vector. For a particle, angular momentum (L) can be expressed as L = r × p, where r is the position vector from the origin to the particle, and p is the linear momentum, given by p = mv, with m being mass and v being velocity.

Cross Product

The cross product is a mathematical operation that takes two vectors and produces a third vector that is perpendicular to the plane formed by the original vectors. In the context of angular momentum, the cross product of the position vector and the momentum vector determines the direction and magnitude of the angular momentum. The magnitude of the cross product is given by |A × B| = |A||B|sin(θ), where θ is the angle between the two vectors.

Position and Velocity Vectors

Position and velocity vectors are fundamental concepts in physics that describe the location and motion of an object in space. The position vector indicates the object's location relative to a reference point, typically the origin, while the velocity vector describes the object's speed and direction of motion. In this problem, the position vector is (1.0, 2.0, 3.0) m, and the velocity vector is (-5.0, +2.8, -3.1) m/s, both of which are essential for calculating angular momentum.

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Related Practice

Textbook Question

Two identical particles have equal but opposite momenta, $\vec{p}$ and $- \vec{p}$ , but they are not traveling along the same line. Show that the total angular momentum of this system does not depend on the choice of origin.

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. 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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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

Calculate the angular momentum of a particle of mass m moving with constant velocity υ for two cases (see Fig. 11–34): about O′.

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Textbook Question

Calculate the angular momentum of a particle of mass m moving with constant velocity υ for two cases (see Fig. 11–34): about origin O.

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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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