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Chapter 9: Problem 9

What is the torque about the origin on a particle located at \(x=1.5 \mathrm{~m}, \quad y=-2.0 \mathrm{~m}, z=1.6 \mathrm{~m}\) due to a force \(\overrightarrow{\mathbf{F}}=\) \((3.5 \mathrm{~N}) \hat{\mathbf{i}}-(2.4 \mathrm{~N}) \hat{\mathbf{j}}+(4.3 \mathrm{~N}) \hat{\mathbf{k}} ?\) Express your result in unit vector notation.

Short Answer

Expert verified
The torque of the force on the particle about the origin, expressed in unit vector notation, is \( \tau = \overrightarrow{r}\times\overrightarrow{F} = -8.6\hat{i} +5.45 \hat{j} -11 \hat{k} \, Nm \).

Step by step solution

01

Identify the Position and Force Vectors

First, identify the position and force vectors from the exercise. The position vector \(\overrightarrow{r}\) from the origin to the particle in unit vector notation is \(1.5\hat{i} - 2.0\hat{j} + 1.6\hat{k}\) and the force vector \(\overrightarrow{F}\) is \((3.5 N)\hat{i} - (2.4 N)\hat{j} + (4.3 N)\hat{k}\).
02

Compute the cross product

Calculate the cross product of the position vector and the force vector, that is \(\overrightarrow{r}\times\overrightarrow{F}\). This is done by determining the determinant of the following matrix:\[\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \1.5 & -2.0 & 1.6 \3.5 & -2.4 & 4.3\end{vmatrix}.\] Expanding the determinant along the first row, we get:\[\overrightarrow{r}\times\overrightarrow{F} = (-2.0*4.3+1.6*-2.4)i - (1.5*4.3-1.6*3.5)j + (1.5*-2.4+3.5*-2.0)k.\] Calculating gives: \(-8.6\hat{i} +5.45 \hat{j} -11 \hat{k} \, Nm\).
03

Express Result in Unit Vector Notation

The final step is to express the result in unit vector notation. So, the torque of the force on the particle is \( \tau = \overrightarrow{r}\times\overrightarrow{F} = -8.6\hat{i} +5.45 \hat{j} -11 \hat{k} \, Nm \).

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Key Concepts

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

Torque
Torque is a concept in physics that refers to the rotational effect of a force applied to a point on a body. It's like the rotational equivalent of linear force. Here鈥檚 how it works:
  • Understanding Torque: Torque (\( \tau \)) is the product of force and the distance from the point of rotation, which determines how much the force acts to rotate an object around an axis.
Torque is calculated using the formula:\[\tau = \overrightarrow{r} \times \overrightarrow{F}\]Where \( \overrightarrow{r} \) is the position vector from the axis of rotation to the point where the force is applied, and \( \overrightarrow{F} \) is the force vector.
  • Significance of Direction: The direction of the torque vector is perpendicular to the plane formed by the position and force vectors due to the nature of the cross product.
This concept is essential when analyzing forces in systems that involve rotation, such as gears or levers.
Vector notation
Vectors are fundamental in physics as they allow us to express quantities that have both a magnitude and a direction, such as force or velocity.
  • Expressing Vectors: A vector is typically represented in unit vector notation using i, j, and k. For example, the position vector \( \overrightarrow{r} = 1.5\hat{i} - 2.0\hat{j} + 1.6\hat{k} \) is written to show the displacement in each orthogonal direction.
Vectors have various operations like addition and scalar multiplication, but the cross product is especially useful when dealing with problems like rotating forces or torque.
  • Importance: Understanding how to represent and manipulate vectors is crucial for solving physics problems involving multiple dimensions.
By using vector notation, we can succinctly describe scenarios in three-dimensional space.
Cross product
The cross product is a key mathematical tool in vector calculus, particularly when dealing with three dimensions. It allows us to find a vector that is perpendicular to two given vectors, which is highly useful in physics.
  • Definition: The cross product of two vectors, say \( \overrightarrow{A} \) and \( \overrightarrow{B} \), is denoted by \( \overrightarrow{A} \times \overrightarrow{B} \) and results in a vector perpendicular to both \( \overrightarrow{A} \) and \( \overrightarrow{B} \).
The magnitude of the cross product is equal to the area of the parallelogram formed by the two vectors:\[|\overrightarrow{A} \times \overrightarrow{B}| = |\overrightarrow{A}| \, |\overrightarrow{B}| \, \sin(\theta)\]
  • Computation: Using the determinant of a matrix formed by vectors, as shown in the solution for torque calculation, helps to compute the cross product effectively.
    • Understanding the cross product is important as it often appears in physics, especially in angular momentum and torque calculations.
    Unit vectors
    Unit vectors are a fundamental concept in vector mechanics and are particularly important in simplifying the analysis of vectors in three-dimensional space.
    • Explanation: A unit vector is a vector with a magnitude of one. It's used to indicate direction without affecting magnitude.
    For example, in three-dimensional space, the standard unit vectors are: \( \hat{i} \), \( \hat{j} \), and \( \hat{k} \). Each represents the x, y, and z axes, respectively.
    • Utility in Vector Notation: By expressing vectors in terms of unit vectors, we can easily understand the directional components of physical quantities like forces or momentum.
    Knowing how to work with unit vectors is crucial for decomposing forces and performing operations, such as the cross product, in problems involving spatial dimensions.

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    Most popular questions from this chapter

    In the act of jumping off a diving board, a diver changed his angular velocity from zero to \(6.20 \mathrm{rad} / \mathrm{s}\) in \(220 \mathrm{~ms}\). The diver's rotational inertia is \(12.0 \mathrm{~kg} \cdot \mathrm{m}^{2} .(a)\) Find the angular acceleration during the jump. (b) What external torque acted on the diver during the jump?

    Vectors \(\overrightarrow{\mathbf{a}}\) and \(\overrightarrow{\mathbf{b}}\) lie in the \(x y\) plane. The angle between \(\overrightarrow{\mathbf{a}}\) and \(\overrightarrow{\mathbf{b}}\) is \(\phi\), which is less than \(90^{\circ}\). Let \(\overrightarrow{\mathbf{c}}=\overrightarrow{\mathbf{a}} \times(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{a}})\). Find the magnitude of \(\overrightarrow{\mathbf{c}}\) and the angle between \(\overrightarrow{\mathbf{b}}\) and \(\overrightarrow{\mathbf{c}}\).

    Calculate the rotational inertia of a meter stick, with mass \(0.56 \mathrm{~kg}\), about an axis perpendicular to the stick and located at the 20 -cm mark.

    Two vectors \(\overrightarrow{\mathbf{r}}\) and \(\overrightarrow{\mathbf{s}}\) lie in the \(x y\) plane. Their magnitudes are \(r=4.5\) units and \(s=7.3\) units. Their directions are, respectively, \(320^{\circ}\) and \(85^{\circ}\) measured counterclockwise from the positive \(x\) axis. Find the magnitude and the direction of \(\overrightarrow{\mathbf{r}} \times \overrightarrow{\mathbf{s}}\)

    A yo-yo (see Sample Problem \(9-13\) ) has a rotational inertia of \(950 \mathrm{~g} \cdot \mathrm{cm}^{2}\) and a mass of \(120 \mathrm{~g}\). Its axle radius is \(3.20 \mathrm{~mm}\) and its string is \(134 \mathrm{~cm}\) long. The yo-yo rolls from rest down to the end of the string. ( \(a\) ) What is its acceleration? (b) How long does it take to reach the end of the string? \((c)\) If the yo-yo "sleeps" at the bottom of the string in pure rotary motion, what is its angular speed, in rev/s? (d) Repeat \((c)\), but this time assume that the yo-yo was thrown down with an initial speed of \(1.30 \mathrm{~m} / \mathrm{s}\)
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