/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 The position of a moving particl... [FREE SOLUTION] | 91影视

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The position of a moving particle is given as a function of time \(t\) to be $$\mathbf{r}(t)=\hat{\mathbf{x}} b \cos (\omega t)+\hat{\mathbf{y}} c \sin (\omega t)$$ where \(b, c,\) and \(\omega\) are constants. Describe the particle's orbit.

Short Answer

Expert verified
The particle follows an elliptical path centered at the origin with axes lengths determined by \( b \) and \( c \).

Step by step solution

01

Identify the Position Function Components

The position function provided is \( \mathbf{r}(t)=\hat{\mathbf{x}} b \cos (\omega t)+\hat{\mathbf{y}} c \sin (\omega t) \). This indicates that the position of the particle in the \( x \)-direction at any time \( t \) is \( b \cos(\omega t) \) and in the \( y \)-direction is \( c \sin(\omega t) \).
02

Eliminate Time Parameter

To find the shape of the orbit, we need to eliminate the time variable \( t \) by expressing both \( x \) and \( y \) in terms of each other. From the position function, set \( x = b \cos(\omega t) \) and \( y = c \sin(\omega t) \).
03

Use Trigonometric Identity

Using the Pythagorean identity \( \cos^2(\theta) + \sin^2(\theta) = 1 \), solve for \( \cos(\omega t) \) and \( \sin(\omega t) \). Substitute these back into the identity: \( \left(\frac{x}{b}\right)^2 + \left(\frac{y}{c}\right)^2 = 1 \).
04

Recognize the Standard Equation of an Ellipse

The equation \( \left(\frac{x}{b}\right)^2 + \left(\frac{y}{c}\right)^2 = 1 \) is the standard form of an ellipse centered at the origin with semi-major axis \( b \) and semi-minor axis \( c \). Thus, the particle's motion traces out an elliptical path.

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Key Concepts

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

Position Function
In the world of physics, a position function provides essential information about the location of a particle at any given point in time. The position function in the exercise is \[\mathbf{r}(t)=\hat{\mathbf{x}} b \cos (\omega t)+\hat{\mathbf{y}} c \sin(\omega t)\]Here, the particle's position is captured in two parts:
  • The term \(\hat{\mathbf{x}} b \cos (\omega t)\) describes the particle's movement in the \(x\)-direction
  • The term \(\hat{\mathbf{y}} c \sin (\omega t)\) denotes its movement in the \(y\)-direction
The constants \(b\) and \(c\) represent the maximum reach or amplitude of the particle's movement in the respective directions. Meanwhile, \(\omega\) is the angular frequency, which tells us how quickly the particle completes a cycle of its motion.
This position function combines trigonometric components to define precisely where the particle is as time \(t\) progresses. It is an example of parametric equations, where two separate equations describe a singular movement of a particle in space.
Ellipse
An ellipse is a geometric shape that resembles an elongated circle. The equation \(\left(\frac{x}{b}\right)^2 + \left(\frac{y}{c}\right)^2 = 1\) as derived in the exercise is a standard representation of an ellipse.
  • Centered at the origin.
  • With a semi-major axis of length \(b\) in the x-direction
  • And a semi-minor axis of length \(c\) in the y-direction
The significant property of an ellipse is that for any point on it, the sum of the distances from two fixed points (foci) is constant.
This property helps us understand the particle's path in the exercise. By eliminating the parameter \( t \) from the position function, we can visualize the particle's motion more broadly as a distinct, recognizable shape.
The particle's path is clearly illustrated as it travels, demonstrating how mathematics and geometry intersect to provide a clear picture of motion.
Trigonometric Identities
Trigonometric identities are mathematical equations that relate trigonometric functions to one another. In this exercise, they are crucial for understanding the relationship between the components of the position function.A powerful identity used is the Pythagorean identity:\[\cos^2(\theta) + \sin^2(\theta) = 1\]This identity allows us to transform our position functions from involving the explicit parameter \( t \) to a statement about \( x \) and \( y \), converting the problem into a geometric interpretation.
For instance, when we have:
  • \(x = b \cos(\omega t)\)
  • \(y = c \sin(\omega t)\)
We can use the Pythagorean identity to say:\[\left(\frac{x}{b}\right)^2 + \left(\frac{y}{c}\right)^2 = \cos^2(\omega t) + \sin^2(\omega t) = 1\]Thus, the trigonometric identities serve as fundamental building blocks, helping us derive a comprehensive picture of the path a particle takes, such as in this motion along an ellipse.

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Most popular questions from this chapter

You lay a rectangular board on the horizontal floor and then tilt the board about one edge until it slopes at angle \(\theta\) with the horizontal. Choose your origin at one of the two corners that touch the floor, the \(x\) axis pointing along the bottom edge of the board, the \(y\) axis pointing up the slope, and the \(z\) axis normal to the board. You now kick a frictionless puck that is resting at \(O\) so that it slides across the board with initial velocity \(\left(v_{\mathrm{ox}}, v_{\mathrm{oy}}, 0\right) .\) Write down Newton's second law using the given coordinates and then find how long the puck takes to return to the floor level and how far it is from \(O\) when it does so.

By evaluating their dot product, find the values of the scalar \(s\) for which the two vectors \(\mathbf{b}=\hat{\mathbf{x}}+s \hat{\mathbf{y}}\) and \(\mathbf{c}=\hat{\mathbf{x}}-s \hat{\mathbf{y}}\) are orthogonal. (Remember that two vectors are orthogonal if and only if their dot product is zero.) Explain your answers with a sketch.

In elementary trigonometry, you probably learned the law of cosines for a triangle of sides \(a, b,\) and \(c,\) that \(c^{2}=a^{2}+b^{2}-2 a b \cos \theta,\) where \(\theta\) is the angle between the sides \(a\) and \(b\). Show that the law of cosines is an immediate consequence of the identity \((\mathbf{a}+\mathbf{b})^{2}=a^{2}+b^{2}+2 \mathbf{a} \cdot \mathbf{b}\)

Let \(\mathbf{u}\) be an arbitrary fixed unit vector and show that any vector \(\mathbf{b}\) satisfies $$b^{2}=(\mathbf{u} \cdot \mathbf{b})^{2}+(\mathbf{u} \times \mathbf{b})^{2}$$ Explain this result in words, with the help of a picture.

[Computer] The differential equation (1.51) for the skateboard of Example 1.2 cannot be solved in terms of elementary functions, but is easily solved numerically. (a) If you have access to software, such as Mathematica, Maple, or Matlab, that can solve differential equations numerically, solve the differential equation for the case that the board is released from \(\phi_{\mathrm{o}}=20\) degrees, using the values \(R=5 \mathrm{m}\) and \(g=9.8 \mathrm{m} / \mathrm{s}^{2} .\) Make a plot of \(\phi\) against time for two or three periods. ( \(\mathbf{b}\) ) On the same picture, plot the approximate solution (1.57) with the same \(\phi_{\mathrm{o}}=20^{\circ}\) Comment on your two graphs. Note: If you haven't used the numerical solver before, you will need to learn the necessary syntax. For example, in Mathematica you will need to learn the syntax for "NDSolve鈥漚nd how to plot the solution that it provides. This takes a bit of time, but is something that is very well worth learning.

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