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Chapter 13: Problem 22

Show that a moving particle will move in a straight line if the normal component of its acceleration is zero.

Short Answer

Expert verified
A particle moves in a straight line if the normal component of its acceleration is zero.

Step by step solution

01

Understand the Components of Acceleration

The acceleration of a moving particle can be broken down into two components: the tangential component, which is parallel to the direction of motion, and the normal component, which is perpendicular to the direction of motion.
02

Express the Total Acceleration

The total acceleration \( \mathbf{a} \) can be expressed as the vector sum of the tangential component \( \mathbf{a}_t \) and the normal component \( \mathbf{a}_n \), i.e., \( \mathbf{a} = \mathbf{a}_t + \mathbf{a}_n \).
03

Define the Normal Component

The normal component of acceleration \( \mathbf{a}_n \) is responsible for changing the direction of the velocity. If \( \mathbf{a}_n = 0 \), it means there is no force acting perpendicular to the direction of motion, which implies that the direction remains unchanged.
04

Resultant Motion Analysis

If \( \mathbf{a}_n = 0 \), the entire acceleration is along the tangential direction. In such a case, changes can only occur in the speed of the particle, not in its direction. Hence, the particle will continue moving in a straight line, following its initial velocity path.

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Key Concepts

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

Tangential Component of Acceleration
The tangential component of acceleration plays a crucial role in understanding how an object moves in its path. This component, denoted as \( \mathbf{a}_t \), is aligned parallel to the direction of the object's velocity. Therefore, it influences the speed of the object directly. Changes in speed, whether increasing or decreasing, are due to the tangential component.
Imagine a car speeding down a straight highway. The tangential component of its acceleration would be responsible if the car were to speed up or slow down. It doesn't, however, cause any change in the direction of the car. The tangential component is purely about the rate of change in speed.
  • If \( \mathbf{a}_t > 0 \), the object gains speed that is, it accelerates.
  • If \( \mathbf{a}_t < 0 \), the object loses speed, and it decelerates.
  • If \( \mathbf{a}_t = 0 \), the object moves at a constant speed.
The tangential component is essential for determining how an object's velocity changes as it moves along a path, especially when considering how speed alterations impact overall motion.
Understanding Straight Line Motion
Straight line motion is essential in physics as it simplifies the analysis of a moving object's trajectory. This type of motion occurs when an object's path does not curve; it follows a direct line from one point to another. The simplest form of motion, it occurs when the normal component of the acceleration is zero.
For straight line motion, it's crucial to note the importance of the normal component's absence. When the normal component is zero, no perpendicular force acts to alter the course of the object. This results in the object continually following its initial path.
  • Initial velocity dictates the direction, and since there's no normal force, direction stays constant.
  • The tangential component may still affect speed, but not direction.
Straight line motion is foundational in physics because it allows for straightforward predictions about an object's path and behavior, simplifying calculations and assumptions about movement in more complex systems.
Vector Sum of Acceleration
Acceleration as a vector involves both magnitude and direction, which means it can be broken down into components. The total acceleration \( \mathbf{a} \) is the vector sum of both the tangential and normal components: \( \mathbf{a} = \mathbf{a}_t + \mathbf{a}_n \). Understanding this vector sum is essential in analyzing an object's motion.
The tangential component \( \mathbf{a}_t \) alters the speed, directly contributing to how fast an object moves. On the other hand, the normal component \( \mathbf{a}_n \) is responsible for changes in direction. Together, they determine the complete trajectory of a moving object.
  • The vector sum shows cumulative effects on motion from both components.
  • Any movement will combine changes in speed and direction unless constrained to a straight line.
In scenarios where the normal component is zero, the vector sum aligns with just the tangential component. Therefore, only speed, not direction, is affected, underscoring why objects with zero normal acceleration move in straight lines.

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Most popular questions from this chapter

Find the velocity and acceleration vectors in terms of \(\mathbf{u}_{r}\) and \(\mathbf{u}_{\theta} .\) \begin{equation}r=a(1+\sin t) \quad \text { and } \quad \theta=1-e^{-t}\end{equation}

You will use a CAS to explore the osculating circle at a point \(P\) on a plane curve where \(\kappa \neq 0 .\) Use a CAS to perform the following steps: a. Plot the plane curve given in parametric or function form over the specified interval to see what it looks like. b. Calculate the curvature \(\kappa\) of the curve at the given value \(t_{0}\) using the appropriate formula from Exercise 5 or \(6 .\) Use the parametrization \(x=t\) and \(y=f(t)\) if the curve is given as a function \(y=f(x) .\) c. Find the unit normal vector \(N\) at \(t_{0}\) . Notice that the signs of the components of \(N\) depend on whether the unit tangent vector \(T\) is turning clockwise or counterclockwise at \(t=t_{0}\) . (See Exercise 7\()\) d. If \(\mathbf{C}=a \mathbf{i}+b \mathbf{j}\) is the vector from the origin to the center \((a, b)\) of the osculating circle, find the center \(\mathbf{C}\) from the vector equation $$ \mathbf{C}=\mathbf{r}\left(t_{0}\right)+\frac{1}{\kappa\left(t_{0}\right)} \mathbf{N}\left(t_{0}\right) $$ The point \(P\left(x_{0}, y_{0}\right)\) on the curve is given by the position vector \(\mathbf{r}\left(t_{0}\right) .\) e. Plot implicitly the equation \((x-a)^{2}+(y-b)^{2}=1 / \kappa^{2}\) of the osculating circle. Then plot the curve and osculating circle together. You may need to experiment with the size of the viewing window, but be sure the axes are equally scaled. $$ \mathbf{r}(t)=\left(\cos ^{3} t\right) \mathbf{i}+\left(\sin ^{3} t\right) \mathbf{j}, \quad 0 \leq t \leq 2 \pi, \quad t_{0}=\pi / 4 $$

Find an equation for the circle of curvature of the curve \(\mathbf{r}(t)=t \mathbf{i}+(\sin t) \mathbf{j}\) at the point \((\pi / 2,1) .\) (The curve parametrizes the graph of \(y=\sin x\) in the \(x y\) -plane.)

Use a CAS to perform the following steps. \begin{equation} \begin{array}{l}{\text { a. Plot the space curve traced out by the position vector } \mathbf{r} \text { . }} \\ {\text { b. Find the components of the velocity vector } d \mathbf{r} / d t \text { . }} \\ {\text { c. Evaluate } d \mathbf{r} / d t \text { at the given point } t_{0} \text { and determine the equation of }} \\ {\text { the tangent line to the curve at } \mathbf{r}\left(t_{0}\right) .} \\ {\text { d. Plot the tangent line together with the curve over the given interval. }}\end{array} \end{equation} \begin{equation} \begin{array}{l}{\mathbf{r}(t)=\left(\ln \left(t^{2}+2\right) \mathbf{i}+\left(\tan ^{-1} 3 t\right) \mathbf{j}+\sqrt{t^{2}+1} \mathbf{k}\right.} \\ {-3 \leq t \leq 5, \quad t_{0}=3}\end{array} \end{equation}

Motion along a parabola A particle moves along the top of the parabola \(y^{2}=2 x\) from left to right at a constant speed of 5 units per second. Find the velocity of the particle as it moves through the point \((2,2) .\)
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