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In unit vector notation, find the torque about a point at coordinates \((2.0 \mathrm{~m}, 1.0 \mathrm{~m}, 3.0 \mathrm{~m})\) on a particle located at \((3.0 \mathrm{~m}, 1.0 \mathrm{~m}\), \(2.0 \mathrm{~m})\) when a force \(\vec{F}=(1.0 \mathrm{~N}) \hat{\mathrm{i}}-(3.0 \mathrm{~N}) \hat{\mathrm{k}}\) acts on the particle.

Short Answer

Expert verified
The torque is \(-3 \hat{\mathrm{i}} \).

Step by step solution

01

Define the position vector

First, determine the position vector \( \vec{r} \) from the point \( (2.0\, \mathrm{m}, 1.0\, \mathrm{m}, 3.0\, \mathrm{m}) \) to the particle at \( (3.0\, \mathrm{m}, 1.0\, \mathrm{m}, 2.0\, \mathrm{m}) \). This is done by subtracting the coordinates of the point from the coordinates of the particle: \( \vec{r} = (3.0 - 2.0)\hat{\mathrm{i}} + (1.0 - 1.0)\hat{\mathrm{j}} + (2.0 - 3.0)\hat{\mathrm{k}} = 1.0\hat{\mathrm{i}} + 0\hat{\mathrm{j}} - 1.0\hat{\mathrm{k}} \).
02

Express the force vector

The force vector \( \vec{F} \) is given as \( \vec{F} = 1.0 \mathrm{~N} \hat{\mathrm{i}} - 3.0 \mathrm{~N} \hat{\mathrm{k}} \).
03

Calculate the torque

Torque \( \vec{\tau} \) is given by the cross product of the position vector \( \vec{r} \) and the force vector \( \vec{F} \):\[\vec{\tau} = \vec{r} \times \vec{F} = \begin{vmatrix}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \1 & 0 & -1 \1 & 0 & -3 \ \end{vmatrix}\]Expanding this determinant, we get:\[\vec{\tau} = \hat{\mathrm{i}}(0 - 3) - \hat{\mathrm{j}}(-1 - (-1)) + \hat{\mathrm{k}}(0 - 0)\]So, \(\vec{\tau} = -3 \hat{\mathrm{i}} + 0 \hat{\mathrm{j}} + 0 \hat{\mathrm{k}} = -3 \hat{\mathrm{i}} \).
04

State the result

Thus, the torque in unit vector notation is \(-3 \hat{\mathrm{i}} \).

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Key Concepts

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

Cross Product
The cross product is a crucial mathematical operation often used in physics to find quantities like torque. It operates on two vectors in three-dimensional space and results in a new vector that is perpendicular to both input vectors.
Think of it as helping you find the twist or rotation effect when a force is applied. \(\vec{C} = \vec{A} \times \vec{B}\)
The resulting vector \(\vec{C}\) has a direction given by the right-hand rule:
  • Point your index finger in the direction of the first vector \(\vec{A}\).
  • Bend your middle finger in the direction of the second vector \(\vec{B}\).
  • Your thumb then points in the direction of the cross product \(\vec{C}\).
The magnitude, or size, of the cross product depends on the sine of the angle between the two original vectors:
\(|\vec{C}| = |\vec{A}| |\vec{B}| \sin(\theta)\)
In physics problems like torque calculation, this operation helps you determine how effectively a force can cause an object to rotate around a pivot point.
Unit Vector Notation
Unit vector notation is a convenient way to express vectors in a coordinate system using unit vectors \((\hat{i}, \hat{j}, \hat{k})\). These unit vectors each have a magnitude of one and point in the direction of the coordinate axes.
It's like using specific building blocks to describe exactly where something is or how something moves in space.
In physics, it's essential because it simplifies the process of working out results and helps visualize directionality of vectors.
  • \(\hat{i}\) points in the x-direction.
  • \(\hat{j}\) points in the y-direction.
  • \(\hat{k}\) points in the z-direction.
For instance, a vector \(\vec{A}\) could be written as \(\vec{A} = a_{x} \hat{i} + a_{y} \hat{j} + a_{z} \hat{k}\), where \(a_{x}\), \(a_{y}\), and \(a_{z}\) are the vector's projections along the respective axes.
This notation is particularly useful in torque calculations, as it allows easy visualization and manipulation of the position and force vectors in the operation.
Position Vector
The position vector \(\vec{r}\) is a vector that extends from a reference point to a particle or object in space. It essentially acts as the locational address of the particle in the coordinate system.
By subtracting the coordinates of a point from the coordinates of the particle, you can determine the position vector. This is particularly vital in torque calculations because it helps specify the arm's length.
  • Used for defining where the particle is in relation to a specific location.
  • Crucial for finding subtleties of rotational movement.
For example, if a particle is at \((3.0 \, \mathrm{m}, 1.0 \, \mathrm{m}, 2.0 \, \mathrm{m})\) and a reference point is at \((2.0 \, \mathrm{m}, 1.0 \, \mathrm{m}, 3.0 \, \mathrm{m})\), the position vector is \(\vec{r} = 1.0 \hat{i} + 0 \hat{j} - 1.0 \hat{k}\).
Understanding this helps achieve accuracy in physical phenomena computation, like calculating torques where knowing the pivot point to the point of force application is important.
Force Vector
A force vector \(\vec{F}\) represents the force applied to an object, depicted with both magnitude and direction. In essence, it is a representation of pushing or pulling action applied to an object in a coordinate space.
Think of it as an instruction for how and in which way something is being pushed or pulled.
You define this force using unit vectors just as other physical vectors.
  • Magnitude, which tells how much strength is applied.
  • Direction through unit vector notation.
Expressed as \(\vec{F} = F_x \hat{i} + F_y \hat{j} + F_z \hat{k}\), where \(F_x, F_y,\) and \(F_z\) are the force components along the x, y, and z-axes respectively.
For example, in a scenario where \(\vec{F} = 1.0 \, \mathrm{N} \hat{i} - 3.0 \, \mathrm{N} \hat{k}\), the force is acting in the positive x-direction and negative z-direction.
This strong component in physics forms the basis for evaluating how forces influence movements, particularly in calculating effects like torque.

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Most popular questions from this chapter

Nonuniform ball. In Fig. 11-29, a ball of mass \(M\) and radius \(R\) rolls smoothly from rest down a ramp and onto a circular loop of radius \(0.48 \mathrm{~m}\). The initial height of the ball is \(h=0.36 \mathrm{~m}\). At the loop bottom, the magnitude of the normal force on the ball is \(2.00 \mathrm{Mg}\). The ball consists of an outer spherical shell (of a certain uniform density) that is glued to a central sphere (of a different uniform density). The rotational inertia of the ball can be expressed in the general form \(I=\beta M R^{2}\), but \(\beta\) is not \(0.4\) as it is for a ball of uniform density. Determine \(\beta\).

Figure 11-45 shows an overhead view of a ring that can rotate about its center like a merry-goround. Its outer radius \(R_{2}\) is \(0.800 \mathrm{~m}\), its inner radius \(R_{1}\) is \(R_{2} / 2.00\), its mass \(M\) is \(6.00 \mathrm{~kg}\), and the mass of the crossbars at its center is negligible. It initially rotates at an angular speed of \(8.00 \mathrm{rad} / \mathrm{s}\) with a cat of mass \(m=M / 4.00\) on its outer edge, at radius \(R_{2}\). By how much does the cat increase the kinetic energy of the cat-ring system if the cat crawls to the inner edge, at radius \(R_{1}\) ?

A top spins at \(25 \mathrm{rev} / \mathrm{s}\) about an axis that makes an angle of \(30^{\circ}\) with the vertical. The mass of the top is \(0.50 \mathrm{~kg}\), its rotational inertia about its central axis is \(5.0 \times 10^{-4} \mathrm{~kg} \cdot \mathrm{m}^{2}\), and its center of mass is \(4.2 \mathrm{~cm}\) from the pivot point. If the spin is clockwise from an overhead view, what are the (a) precession rate and (b) direction of the precession as viewed from overhead?

A track is mounted on a large wheel that is free to turn with negligible friction about a vertical axis (Fig. 11-42). A toy train of mass \(m\) is placed on the track and, with the system initially at rest, the train's electrical power is turned on. The train reaches speed \(0.15 \mathrm{~m} / \mathrm{s}\) with respect to the track. What is the wheel's angular speed if its mass is \(1.1 m\) and its radius is \(0.43 \mathrm{~m}\) ? (Treat it as a hoop, and neglect the mass of the spokes and hub.)

A disk with a rotational inertia of \(7.00 \mathrm{~kg} \cdot \mathrm{m}^{2}\) rotates like a merry-go-round while undergoing a variable torque given by \(\tau=(5.00+2.00 t) \mathrm{N} \cdot \mathrm{m}\). At time \(t=1.00 \mathrm{~s}\), its angular momentum is \(5.00 \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\). What is its angular momentum at \(t=5.00 \mathrm{~s}\) ?

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