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Motion along a cycloid A particle moves in the \(x y\) -plane in such a way that its position at time \(t\) is \begin{equation} \mathbf{r}(t)=(t-\sin t) \mathbf{i}+(1-\cos t) \mathbf{j} \end{equation} \(\begin{array}{l}{\text { a. Graph } \mathbf{r}(t) . \text { The resulting curve is a cycloid. }} \\ {\text { b. Find the maximum and minimum values of }|\mathbf{v}| \text { and }|\mathbf{a}| \text { . }} \\ {\text { (Hint: Find the extreme values of }|\mathbf{v}|^{2} \text { and }|\mathbf{a}|^{2} \text { first and take }} \\ {\text { square roots later.) }}\end{array}\)

Short Answer

Expert verified
The graph is a cycloid. The maximum speed is 2 and the acceleration is constant at 1.

Step by step solution

01

Express the function

The position vector given for the particle, as a function of time \( t \), is \( \mathbf{r}(t) = (t - \sin t) \mathbf{i} + (1 - \cos t) \mathbf{j} \). Each component describes how the position changes with time in the \( x \) and \( y \) directions.
02

Graph the Cycloid

To graph the cycloid, plot \( x(t) = t - \sin t \) and \( y(t) = 1 - \cos t \) over a range of \( t \) values. Generally, a cycloid starts at the origin, rises to a maximum point, and forms a series of arches as \( t \) increases.
03

Find the Velocity Vector

The velocity vector \( \mathbf{v}(t) \) is the derivative of \( \mathbf{r}(t) \) with respect to \( t \): \[ \mathbf{v}(t) = \left(1 - \cos t\right) \mathbf{i} + \sin t \mathbf{j}. \]
04

Compute the Speed

The speed \( |\mathbf{v}(t)| \) is the magnitude of the velocity vector:\[ |\mathbf{v}(t)| = \sqrt{(1 - \cos t)^2 + (\sin t)^2}. \]Simplifying gives:\[ |\mathbf{v}(t)| = \sqrt{2(1 - \cos t)} = 2\sin(\frac{t}{2}). \]
05

Derivative for Speed Extremes

To find the extremes of \( |\mathbf{v}(t)|^2 = 4\sin^2(\frac{t}{2}) \), note that it ranges from \(0\) to \(4\) as \(\sin(\frac{t}{2})\) varies from \(0\) to \(1\).
06

Find the Acceleration Vector

The acceleration vector \( \mathbf{a}(t) \) is the derivative of \( \mathbf{v}(t) \):\[ \mathbf{a}(t) = \sin t \mathbf{i} + \cos t \mathbf{j}. \]
07

Compute the Magnitude of Acceleration

The magnitude of acceleration \( |\mathbf{a}(t)| \) is:\[ |\mathbf{a}(t)| = \sqrt{(\sin t)^2 + (\cos t)^2} = 1. \]Since the magnitude of acceleration is constant, it has no maximum or minimum other than 1.

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Key Concepts

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

Velocity Vector
When discussing the motion along a cycloid, understanding the velocity vector is key. The velocity vector, represented as \(\mathbf{v}(t)\), gives the rate of change of the position vector \(\mathbf{r}(t)\) with respect to time \(t\). This vector shows how fast and in which direction the point is moving at any point in time.
To find \(\mathbf{v}(t)\), we must derive the position vector, \(\mathbf{r}(t) = (t - \sin t) \mathbf{i} + (1 - \cos t) \mathbf{j}\). Taking the derivative with respect to \(t\) yields \(\mathbf{v}(t) = (1 - \cos t) \mathbf{i} + \sin t \mathbf{j}\). This means the velocity depends on sine and cosine functions, capturing the periodic nature of the cycloid motion.

Another important aspect is the speed of the object, which is determined by the magnitude of the velocity vector. For our cycloid, the speed \(|\mathbf{v}(t)| = \sqrt{(1 - \cos t)^2 + (\sin t)^2}\), which simplifies to \(|\mathbf{v}(t)| = 2\sin(\frac{t}{2})\). The speed varies from 0 to 2, illustrating the changing rate of motion along the cycloid path.
Acceleration Vector
In kinematics, the acceleration vector highlights the change in velocity as a function of time. For our cycloid example, the acceleration vector \(\mathbf{a}(t)\) is found by deriving the velocity vector \(\mathbf{v}(t)\). Thus, \(\mathbf{a}(t) = \sin t \mathbf{i} + \cos t \mathbf{j}\).
The acceleration vector reflects how the velocity is adjusted in response to the cycloid path. In this scenario, the expressions involve sine and cosine, indicating circular motion influences. Observing the magnitude \(|\mathbf{a}(t)|\), we derive it as \(\sqrt{(\sin t)^2 + (\cos t)^2} = 1\), which is constant. This illustrates that the acceleration, although always present, maintains a steady value throughout the motion.
This consistency in magnitude makes the motion easier to predict and analyze, as changes in direction occur smoothly without abrupt accelerations or decelerations.
Graphing Parametric Equations
Graphing the motion described by the parametric equations like a cycloid requires understanding the behavior of each component. Here, the equations \(x(t) = t - \sin t\) and \(y(t) = 1 - \cos t\) define the path as a series of loops or arches as \(t\) progresses.

- Start by varying \(t\) over a range, typically from 0 onwards, observing how the position \((x, y)\) evolves in the plane.
- The unique shape, the cycloid, emerges due to the combined sinusoidal effects in both components. At \(t = 0\), the curve starts at the origin \((0,0)\).

Each loop of the cycloid illustrates a complete cycle of motion where the particle rolls along the path, its height determined by the cosine terms and distance by sine. The cycloid is not just a mathematical curiosity but a practical path studied for optimizing motion in physics and engineering, recognized for its efficient traversal and minimal energy paths.

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Most popular questions from this chapter

Colliding marbles The accompanying figure shows an experiment with two marbles. Marble \(A\) was launched toward marble \(B\) with launch angle \(\alpha\) and initial speed \(v_{0 .}\) At the same instant, marble \(B\) was released to fall from rest at \(R \tan \alpha\) units directly above a spot \(R\) units downrange from \(A .\) The marbles were found to collide regardless of the value of \(v_{0}\) . Was this mere coincidence, or must this happen? Give reasons for your answer.

Use a CAS to perform the following steps in Exercises \(35-38\) \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)=(\sin t-t \cos t) \mathbf{i}+(\cos t+t \sin t) \mathbf{j}+t^{2} \mathbf{k}} \\ {0 \leq t \leq 6 \pi, \quad t_{0}=3 \pi / 2}\end{array} \end{equation}

In Exercises \(11-14,\) find the arc length parameter along the curve from the point where \(t=0\) by evaluating the integral $$ s=\int_{0}^{t}|\mathbf{v}(\tau)| d \tau $$ from Equation \((3) .\) Then find the length of the indicated portion of the curve. $$ \mathbf{r}(t)=(1+2 t) \mathbf{i}+(1+3 t) \mathbf{j}+(6-6 t) \mathbf{k}, \quad-1 \leq t \leq 0 $$

As mentioned in the text, the tangent line to a smooth curve \(\mathbf{r}(t)=f(t) \mathbf{i}+g(t) \mathbf{j}+h(t) \mathbf{k}\) at \(t=t_{0}\) is the line that passes through the point \(\left(f\left(t_{0}\right), g\left(t_{0}\right), h\left(t_{0}\right)\right)\) parallel to \(\mathbf{v}\left(t_{0}\right),\) the curve's velocity vector at \(t_{0}\) Find parametric equations for the line that is tangent to the given curve at the given parameter value \(t=t_{0}\) . \begin{equation} \mathbf{r}(t)=\ln t \mathbf{i}+\frac{t-1}{t+2} \mathbf{j}+t \ln t \mathbf{k}, \quad t_{0}=1 \end{equation}

Ellipse $$ \begin{array}{c}{\text { a. Show that the curve } \mathbf{r}(t)=(\cos t) \mathbf{i}+(\sin t) \mathbf{j}+(1-\cos t) \mathbf{k}} \\ {0 \leq t \leq 2 \pi, \text { is an ellipse by showing that it is the intersection }} \\\ {\text { of a right circular cylinder and a plane. Find equations }} \\\ {\text { for the cylinder and plane. }} \\ {\text { b. Sketch the ellipse on the cylinder. Add to your sketch the unit }} \\ {\text { tangent vectors at } t=0, \pi / 2, \pi, \text { and } 3 \pi / 2 \text { . }}\\\\{\text { c. Show that the acceleration vector always lies parallel to the }} \\ {\text { plane (orthogonal to a vector normal to the plane). Thus, if }} \\ {\text { you draw the acceleration as a vector attached to the ellipse, it }} \\ {\text { will lie in the plane of the ellipse. Add the acceleration vectors }} \\\ {\text { for } t=0, \pi / 2, \pi, \text { and } 3 \pi / 2 \text { to your sketch. }}\\\\{\text { d. Write an integral for the length of the ellipse. Do not try to }} \\ {\text { evaluate the integral; it is nonelementary. }} \\\ {\text { e. Numerical integrator Estimate the length of the ellipse to }} \\\ {\text { two decimal places. }}\end{array} $$

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