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Chapter 14: Problem 40

In Exercises v and a are given at a certain instant of time. Find \(a_{T}, a_{N}, \mathbf{T},\) and \(\mathbf{N}\) at this instant. $$. \mathbf{v}=\mathbf{i}+2 \mathbf{j}, \quad \mathbf{a}=3 \mathbf{i}$$

Short Answer

Expert verified
\(a_T = \frac{3}{\sqrt{5}}, a_N = \frac{6}{\sqrt{5}}, \mathbf{T} = \frac{\mathbf{i} + 2\mathbf{j}}{\sqrt{5}}, \mathbf{N} = \frac{-2\mathbf{i} + \mathbf{j}}{\sqrt{5}}\).

Step by step solution

01

Understand the Problem

We need to find the tangential (\(a_T\)) and normal (\(a_N\)) components of acceleration, and the unit tangent (\(\mathbf{T}\)) and unit normal (\(\mathbf{N}\)) vectors at a given instant. We are provided with the velocity vector \(\mathbf{v} = \mathbf{i} + 2\mathbf{j}\) and the acceleration vector \(\mathbf{a} = 3\mathbf{i}\).
02

Compute the Magnitude of v

First, calculate the magnitude of the velocity vector \(\mathbf{v}\):\[\|\mathbf{v}\| = \sqrt{1^2 + 2^2} = \sqrt{5}\]
03

Find the Unit Tangent Vector T

The unit tangent vector \(\mathbf{T}\) is given by \(\mathbf{T} = \frac{\mathbf{v}}{\|\mathbf{v}\|}\). Substitute the values:\[\mathbf{T} = \frac{\mathbf{i} + 2 \mathbf{j}}{\sqrt{5}}\]
04

Compute a_T, the Tangential Component of Acceleration

The tangential component of acceleration \(a_T\) is the dot product of the acceleration vector \(\mathbf{a}\) and the unit tangent vector \(\mathbf{T}\):\[a_T = \mathbf{a} \cdot \mathbf{T} = (3\mathbf{i}) \cdot \left(\frac{\mathbf{i} + 2 \mathbf{j}}{\sqrt{5}}\right) = \frac{3}{\sqrt{5}}\]
05

Compute a_N using a and a_T

The normal component of acceleration \(a_N\) can be found using the relation \(\|\mathbf{a}\| = \sqrt{a_T^2 + a_N^2}\). First, find \(\|\mathbf{a}\|\):\[\|\mathbf{a}\| = \sqrt{3^2} = 3\]Then solve for \(a_N\):\[3 = \sqrt{\left(\frac{3}{\sqrt{5}}\right)^2 + a_N^2}\]\[3 = \sqrt{\frac{9}{5} + a_N^2}\]\[9 = \frac{9}{5} + a_N^2\]\[a_N^2 = 9 - \frac{9}{5} = \frac{36}{5}\]\[a_N = \frac{6}{\sqrt{5}}\]
06

Find the Unit Normal Vector N

The normal vector \(\mathbf{N}\) is perpendicular to \(\mathbf{T}\). Since \(\mathbf{T} = \frac{\mathbf{i} + 2\mathbf{j}}{\sqrt{5}}\), one such perpendicular vector can be \(\mathbf{N} = \frac{-2\mathbf{i} + \mathbf{j}}{\sqrt{5}}\) using a standard rotation in the plane.

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Key Concepts

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

Tangential Component
The tangential component of acceleration, denoted as \(a_T\), describes how much of the acceleration is directed along the tangent to the path of a moving object.
It provides insight into the change in speed of the object. To determine \(a_T\), we calculate the dot product of the acceleration vector \(\mathbf{a}\) with the unit tangent vector \(\mathbf{T}\).
This yields the component of acceleration in the direction of the velocity.
  • The formula is \(a_T = \mathbf{a} \cdot \mathbf{T}\).
  • In our example, with \(\mathbf{a} = 3\mathbf{i}\) and \(\mathbf{T} = \frac{\mathbf{i} + 2 \mathbf{j}}{\sqrt{5}}\), we find \(a_T = \frac{3}{\sqrt{5}}\).
The value \(\frac{3}{\sqrt{5}}\) signifies that part of the acceleration vector is contributing to a change in the speed of the object along its path.
Normal Component
The normal component of acceleration, \(a_N\), is the part that acts perpendicular to the path, contributing to the change in direction of the motion.
This is essential for understanding how sharply or smoothly a path curves. To find \(a_N\), use the Pythagorean theorem in the context of vectors. The total acceleration can be resolved into its tangential and normal parts:
\(\|\mathbf{a}\| = \sqrt{a_T^2 + a_N^2}\).
  • With our example鈥檚 \(\|\mathbf{a}\| = 3\) and \(a_T = \frac{3}{\sqrt{5}}\), solve for \(a_N\).
  • This results in \(a_N = \frac{6}{\sqrt{5}}\).
This calculation shows how much the acceleration is pulling the object away from its tangent line, thus changing its direction.
Unit Tangent Vector
The unit tangent vector, \(\mathbf{T}\), is a normalized vector that points in the direction of motion.
It provides a meaningful direction along the path without scaling by the speed. To find \(\mathbf{T}\), divide the velocity vector \(\mathbf{v}\) by its magnitude \(\|\mathbf{v}\|\), thereby ensuring its length is one:
  • For \(\mathbf{v} = \mathbf{i} + 2\mathbf{j}\), first calculate \(\|\mathbf{v}\| = \sqrt{5}\).
  • Then, the unit tangent vector is \(\mathbf{T} = \frac{\mathbf{i} + 2 \mathbf{j}}{\sqrt{5}}\).
This vector is highly important for visualizing the direction of motion at any given point on a path.
Unit Normal Vector
The unit normal vector, \(\mathbf{N}\), is orthogonal to the unit tangent vector \(\mathbf{T}\) and signifies the direction towards which the path is bending.
It is crucial for seeing how a path deviates from its straight-line trajectory.To determine \(\mathbf{N}\), ensure it is perpendicular and of unit length. For a planar movement:
  • If \(\mathbf{T} = \frac{\mathbf{i} + 2 \mathbf{j}}{\sqrt{5}}\), a perpendicular vector that achieves unit length is \(\mathbf{N} = \frac{-2\mathbf{i} + \mathbf{j}}{\sqrt{5}}\).
This choice ensures that \(\mathbf{T} \cdot \mathbf{N} = 0\), confirming orthogonality.
The unit normal vector helps in analyzing the curvature of the path.

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