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Chapter 11: Problem 28

A \(2.0 \mathrm{~kg}\) particle-like object moves in a plane with velocity components \(v_{x}=30 \mathrm{~m} / \mathrm{s}\) and \(v_{y}=60 \mathrm{~m} / \mathrm{s}\) as it passes through the point with \((x, y)\) coordinates of \((3.0,-4.0) \mathrm{m}\). Just then, in unitvector notation, what is its angular momentum relative to (a) the origin and (h) the noint located $$ (-2.0,-2.0) \mathrm{m} ? $$

Short Answer

Expert verified
(a) From the origin: \( \vec{L} = 1200\hat{k} \text{ kg m}^2/\text{s} \). (b) From point \((-2.0, -2.0)\): \( \vec{L} = 1440\hat{k} \text{ kg m}^2/\text{s} \).

Step by step solution

01

Calculate Position Vector

The position vector \( \vec{r} \) from the origin to the point \((3.0, -4.0) \text{ m} \) is \( \vec{r} = 3.0\hat{i} - 4.0\hat{j} \). For the point located at \((-2.0, -2.0) \text{ m}\), the position vector is \( \vec{r}' = 5.0\hat{i} - 2.0\hat{j} \), since \(3.0 - (-2.0) = 5.0\) and \(-4.0 - (-2.0) = -2.0\).
02

Calculate Velocity Vector

The velocity vector \( \vec{v} \) is given as \( v_x = 30 \text{ m/s} \) and \( v_y = 60 \text{ m/s} \). Therefore, the velocity vector is \( \vec{v} = 30\hat{i} + 60\hat{j} \).
03

Calculate Angular Momentum about the Origin

The angular momentum \( \vec{L} \) about a point is given by the cross product \( \vec{L} = \vec{r} \times m\vec{v} \). For the origin, \( \vec{r} = 3.0\hat{i} - 4.0\hat{j} \) and \( m = 2.0 \text{ kg} \). So, \( \vec{L} = (3.0\hat{i} - 4.0\hat{j}) \times 2(30\hat{i} + 60\hat{j}) \).
04

Calculate the Cross Product

Compute the cross product: \((3.0\hat{i} - 4.0\hat{j}) \times (60\hat{i} + 120\hat{j}) = (3)(120)\hat{k} - (-4)(60)\hat{k} \), which results \( 360\hat{k} + 240\hat{k} = 600\hat{k} \). Therefore, \( \vec{L} = 2(600\hat{k}) = 1200\hat{k} \text{ kg m}^2/\text{s} \).
05

Calculate Angular Momentum about the Point \((-2.0, -2.0)\)

The position vector relative to the point \((-2.0, -2.0)\) is \( \vec{r}' = 5.0\hat{i} - 2.0\hat{j} \). Again, use \( \vec{L} = \vec{r}' \times m\vec{v} \). So, \( \vec{L} = (5.0\hat{i} - 2.0\hat{j}) \times 2(30\hat{i} + 60\hat{j}) \).
06

Compute the Cross Product for the Point \((-2.0, -2.0)\)

Compute the cross product: \((5.0\hat{i} - 2.0\hat{j}) \times (60\hat{i} + 120\hat{j}) = (5)(120)\hat{k} - (-2)(60)\hat{k} \), leading to \( 600\hat{k} + 120\hat{k} = 720\hat{k} \). Therefore, \( \vec{L} = 2(720\hat{k}) = 1440\hat{k} \text{ kg m}^2/\text{s} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Position Vector
In physics, a position vector is a vector that extends from a reference point, typically the origin of a coordinate system, to a specific point in space. This vector is instrumental in describing the location of a particle or object.

For example, in the given exercise, the position vector to the coordinates
  • From the origin to the point a
    • (3.0, -4.0) m is given by
      • \( \vec{r} = 3.0 \hat{i} - 4.0 \hat{j} \)

  • Similarly, for the point
    • (-2.0, -2.0) m, the position vector becomes
      • \( \vec{r}' = 5.0 \hat{i} - 2.0 \hat{j} \)
These vectors are expressed in unit vector notation, employing \( \hat{i} \) and \( \hat{j} \) to represent the x-axis and y-axis components respectively. Calculating position vectors helps in determining where the particle is located relative to a chosen reference point. They form the base in further vector calculations, including velocity and angular momentum.
Velocity Vector
A velocity vector defines the speed and direction of an object's motion in space. It includes both magnitude and direction, making it a vector quantity, which distinctly separates it from speed, a scalar quantity.

In this exercise, the velocity vector of the particle is given by:
  • Velocity components:
    • \( v_x = 30 \text{ m/s} \)
    • \( v_y = 60 \text{ m/s} \)

  • Combined, they describe a velocity vector:
    • \( \vec{v} = 30 \hat{i} + 60 \hat{j} \)
Understanding the velocity vector is crucial as it defines how the position of the particle changes over time. In this context, it's a step towards determining how angular momentum is formulated, demonstrating not just the magnitude, but also the trajectory of the object's movement.
Cross Product
The cross product is a mathematical operation that provides a vector which is perpendicular to two input vectors in three-dimensional space. The cross product is pivotal when calculating quantities like torque and angular momentum in physics.

In the given problem, the angular momentum calculation involves computing a cross product between the position vector \( \vec{r} \) and the momentum \( m\vec{v} \).
  • For the origin:
    • \( \vec{L} = (3.0\hat{i} - 4.0\hat{j}) \times 2(30\hat{i} + 60\hat{j}) \)
    • This yields \( 600\hat{k} \) as the result of \((360 + 240)\hat{k}\), reflecting the angular momentum at the origin.

  • For the point
    • (-2.0, -2.0):
      • \( \vec{L} = (5.0\hat{i} - 2.0\hat{j}) \times 2(30\hat{i} + 60\hat{j}) \)
      • Gives \( 720\hat{k} \) as \((600 + 120)\hat{k}\), denoting a different angular momentum value at this reference point.
The result is a vector along the z-axis (\( \hat{k} \)), showing the direction of rotation perpendicularly out of the plane formed by the position and velocity vectors.
Unit Vector Notation
Unit vector notation is a way to express vectors using standard directional components, denoted as \( \hat{i} \), \( \hat{j} \), and \( \hat{k} \). These represent unit vectors in the directions of the x-axis, y-axis, and z-axis respectively.

This notation helps in simplifying vector equations and visualizing them in 3D space.
  • In our exercise, positions and velocities are expressed in unit vector form:
    • \( 3.0\hat{i} - 4.0\hat{j} \) for position
    • \( 30\hat{i} + 60\hat{j} \) for velocity.

  • When vectors are multiplied, such as in the cross product, the unit vectors follow specific rules to ensure accurate mathematical calculations.
The simplicity of unit vector notation makes it easier to add, subtract, and compute products between vectors, thus playing a crucial role in solving physics problems involving vector quantities.

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Most popular questions from this chapter

In a long jump, an athlete leaves the ground with an initial angular momentum that tends to rotate her body forward, threatening to ruin her landing. To counter this tendency, she rotates her outstretched arms to "take up" the angular momentum (Fig. 11-18). In \(0.700 \mathrm{~s}\), one arm sweeps through \(0.500\) rev and the other arm sweeps through \(1.000\) rev. Treat each arm as a thin rod of mass \(4.0 \mathrm{~kg}\) and length \(0.60 \mathrm{~m}\), rotating around one end. In the athlete's reference frame, what is the magnitude of the total angular momentum of the arms around the common rotation axis through the shoulders?

In a playground, there is a small merry-go-round of radius \(1.20 \mathrm{~m}\) and mass \(180 \mathrm{~kg}\). Its radius of gyration (see Problem 79 of Chapter 10 ) is \(91.0 \mathrm{~cm}\). A child of mass \(44.0 \mathrm{~kg}\) runs at a speed of \(3.00 \mathrm{~m} / \mathrm{s}\) along a path that is tangent to the rim of the initially stationary merry-go-round and then jumps on. Neglect friction between the bearings and the shaft of the merry-go-round. Calculate (a) the rotational inertia of the merry-go-round about its axis of rotation, (b) the magnitude of the angular momentum of the running child about the axis of rotation of the merry-go-round, and (c) the angular speed of the merry-go- round and child after the child has jumped onto the merry-go-round.

A car travels at \(80 \mathrm{~km} / \mathrm{h}\) on a level road in the positive direction of an \(x\) axis. Each tire has a diameter of \(66 \mathrm{~cm}\). Relative to a woman riding in the car and in unit-vector notation, what are the velocity \(\vec{v}\) at the (a) center, (b) top, and (c) bottom of the tire and the magnitude \(a\) of the acceleration at the (d) center, (e) top, and (f) bottom of each tire? Relative to a hitchhiker sitting next to the road and in unit-vector notation, what are the velocity \(\vec{v}\) at the \((\mathrm{g})\) center, (h) top, and (i) bottom of the tire and the magnitude \(a\) of the acceleration at the (j) center, (k) top, and (1) bottom of each tire?

Force \(\vec{F}=(2.0 \mathrm{~N}) \hat{\mathrm{i}}-(3.0 \mathrm{~N}) \hat{\mathrm{k}}\) acts on a pebble with posi- tion vector \(\vec{r}=(0.50 \mathrm{~m}) \hat{\mathrm{j}}-(2.0 \mathrm{~m}) \hat{\mathrm{k}}\) relative to the origin. In unitvector notation, what is the resulting torque on the pebble about (a) the origin and (b) the point \((2.0 \mathrm{~m}, 0,-3.0 \mathrm{~m})\) ?

A wheel of radius \(0.250 \mathrm{~m}\), which is moving initially at \(43.0\) \(\mathrm{m} / \mathrm{s}\), rolls to a stop in \(225 \mathrm{~m}\). Calculate the magnitudes of \((\mathrm{a})\) its lin- ear acceleration and (b) its angular acceleration. (c) The wheel's rotational inertia is \(0.155 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about its central axis. Calculate the magnitude of the torque about the central axis due to friction on the wheel.
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