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Chapter 12: Problem 45

Use the given information to find the normal scalar component of acceleration at time \(t=1\) $$ \mathbf{a}(1)=\mathbf{i}+2 \mathbf{j}-2 \mathbf{k} ; \quad a_{T}(1)=3 $$

Short Answer

Expert verified
The normal scalar component of acceleration at \( t=1 \) is 0.

Step by step solution

01

Understanding the problem

We are given the acceleration vector \( \mathbf{a}(1)=\mathbf{i}+2\mathbf{j}-2\mathbf{k} \) and the tangential component of acceleration \( a_T(1)=3 \). We need to find the normal scalar component of acceleration at \( t=1 \).
02

Formula for normal component of acceleration

The normal component of acceleration \( a_N \) is given by the formula \( a_N = \sqrt{\| \mathbf{a} \|^2 - a_T^2} \), where \( \| \mathbf{a} \| \) is the magnitude of the acceleration vector and \( a_T \) is the tangential component of acceleration.
03

Find the magnitude of the acceleration vector \( \| \mathbf{a} \| \)

Calculate the magnitude of the vector \( \mathbf{a}(1)=\mathbf{i}+2\mathbf{j}-2\mathbf{k} \). The magnitude is given by \( \| \mathbf{a} \| = \sqrt{1^2 + 2^2 + (-2)^2} \).
04

Compute the magnitude \( \| \mathbf{a} \| \)

Calculate \( \| \mathbf{a} \| = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \).
05

Substitute values in the formula for \( a_N \)

Substitute \( \| \mathbf{a} \| = 3 \) and \( a_T = 3 \) into the formula \( a_N = \sqrt{\| \mathbf{a} \|^2 - a_T^2} \) to find \( a_N \).
06

Calculate normal component of acceleration \( a_N \)

Substitute and simplify: \( a_N = \sqrt{3^2 - 3^2} = \sqrt{9 - 9} = \sqrt{0} = 0 \).

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Key Concepts

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

Acceleration Vector
When we talk about an acceleration vector, we're diving into a concept that describes how the velocity of an object changes over time, considering both the speed and direction. In our case, the acceleration vector is given at a specific time: \( \mathbf{a}(1) = \mathbf{i} + 2\mathbf{j} - 2\mathbf{k} \). This is a representation in three-dimensional space, where:
  • \( \mathbf{i} \) represents the change in the x-direction,
  • \( 2\mathbf{j} \) indicates the change in the y-direction,
  • \( -2\mathbf{k} \) represents the change in the z-direction.
Each component signifies how the object's velocity in that particular direction is accelerating or decelerating. The primary purpose of an acceleration vector is to convey not just how fast, but also in what direction these changes in motion are taking place.
Tangential Component
The tangential component of acceleration refers to the part of acceleration that is directly related to changing the speed of an object along its path. It's one component of the entire acceleration vector, and it tells us how quickly the speed of the object is increasing or decreasing.

In our example, we have the tangential acceleration at \( t=1 \) given by \( a_T(1) = 3 \). This positive value indicates that the object is speeding up in the direction it is currently moving. The tangential component is crucial because it's directly related to the rate at which an object covers distance, and it helps identify changes in speed along the trajectory without considering directional changes.
Magnitude of a Vector
Understanding the magnitude of a vector is essential when working with vectors like acceleration. The magnitude essentially measures the length or size of the vector, ignoring its direction, and is calculated with the help of Pythagoras鈥� theorem.

For the acceleration vector \( \mathbf{a}(1) = \mathbf{i} + 2\mathbf{j} - 2\mathbf{k} \), the magnitude \( \| \mathbf{a} \| \) is found using:\[ \| \mathbf{a} \| = \sqrt{1^2 + 2^2 + (-2)^2} \]This simplifies to:\[ \| \mathbf{a} \| = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]

This value of 3 tells us about the overall strength of the acceleration at \( t=1 \), without concern for which direction it's pointing. Understanding the magnitude helps in evaluating vectors separately from their directions, useful in complex physics problems where multiple vector components interact.

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Most popular questions from this chapter

Consider the various forces that a passenger in a car would sense while traveling over the crest of a hill or around a curve. Relate these sensations to the tangential and normal vector components of the acceleration vector for the car's motion. Discuss how speeding up or slowing down (e.g., doubling or halving the car's speed) affects these components.

Assume that \(s\) is an arc length parameter for a smooth vector-valued function \(\mathbf{r}(s)\) in 3 -space and that \(d \mathbf{T} / d s\) and \(d \mathbf{N} / d s\) exist at each point on the curve. (This implies that \(d \mathbf{B} / d s\) exists as well, since \(\mathbf{B}=\mathbf{T} \times \mathbf{N} .)\) (a) Show that \(d \mathbf{B} / d s\) is perpendicular to \(\mathbf{B}(s) .\) (b) Show that \(d \mathbf{B} / d s\) is perpendicular to \(\mathbf{T}(s)\). [Hint: Use the fact that \(\mathbf{B}(s)\) is perpendicular to both \(\mathbf{T}(s)\) and \(\mathbf{N}(s),\) and differentiate \(\mathbf{B} \cdot \mathbf{T} \text { with respect to } s .]\) (c) Use the results in parts (a) and (b) to show that \(d \mathbf{B} / d s\) is a scalar multiple of \(\mathbf{N}(s) .\) The negative of this scalar is called the torsion of \(\mathbf{r}(s)\) and is denoted by \(\tau(s) .\) Thus, $$\frac{d \mathbf{B}}{d s}=-\tau(s) \mathbf{N}(s)$$ (d) Show that \(\tau(s)=0\) for all \(s\) if the graph of \(\mathbf{r}(s)\) lies in a plane. [Note: For reasons that we cannot discuss here, the torsion is related to the "twisting" properties of the curve, and \(\tau(s)\) is regarded as a numerical measure of the tendency for the curve to twist out of the osculating plane. \(]\)

Let \(\mathbf{r}(t)=\ln t \mathbf{i}+2 t \mathbf{j}+t^{2} \mathbf{k} .\) Find $$ \begin{array}{llll}{\text { (a) }\left\|\mathbf{r}^{\prime}(t)\right\|} & {\text { (b) } \frac{d s}{d t}} & {\text { (c) } \int_{1}^{3}\left\|\mathbf{r}^{\prime}(t)\right\| d t} & {}\end{array} $$

A shell, fired from ground level at an elevation angle of \(45^{\circ},\) hits the ground \(24,500 \mathrm{m}\) away. Calculate the muzzle speed of the shell.

Find an arc length parametrization of the cycloid $$ \begin{array}{l}{x=a t-a \sin t} \\ {y=a-a \cos t}\end{array} \quad(0 \leq t \leq 2 \pi) $$ with (0,0) as the reference point.
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