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The coordinates of an object moving in the \(x y\) plane vary with time according to the equations \(x=-(5.00 \mathrm{m}) \sin (\omega t)\) and \(y=(4.00 \mathrm{m})-(5.00 \mathrm{m}) \cos (\omega t),\) where \(\omega\) is a constant and \(t\) is in seconds. (a) Determine the components of velocity and components of acceleration at \(t=0 .\) (b) Write ex- pressions for the position vector, the velocity vector, and the acceleration vector at any time \(t>0 .\) (c) Describe the path of the object in an \(x y\) plot.

Short Answer

Expert verified
Velocity at t=0 is \( \mathbf{v}(0) = 0\mathbf{i} + 5.00 \mathrm{m}\omega\mathbf{j} \), and acceleration at t=0 is \( \mathbf{a}(0) = 5.00\,\mathrm{m}\omega^2\mathbf{i} - 0\mathbf{j} \). The position, velocity, and acceleration vectors for any time t are \( \mathbf{r}(t) = -(5.00\,\mathrm{m}) \sin(\omega t)\mathbf{i} + ((4.00\,\mathrm{m})-(5.00\,\mathrm{m}) \cos(\omega t))\mathbf{j} \), \( \mathbf{v}(t) = 5.00\,\mathrm{m}\omega \cos(\omega t)\mathbf{i} + 5.00\,\mathrm{m}\omega \sin(\omega t)\mathbf{j} \), and \( \mathbf{a}(t) = (5.00\,\mathrm{m}\omega^2 \sin(\omega t))\mathbf{i} - (5.00\,\mathrm{m}\omega^2 \cos(\omega t))\mathbf{j} \) respectively. The path of the motion is a clockwise circle with radius 5 m centered at (0, 4.00 m).

Step by step solution

01

Find the Components of Velocity at t=0

The velocity components are found by taking the first derivative of the position functions with respect to time. For the x-component: \( v_x(t) = \frac{dx}{dt} = -(-5.00\,\mathrm{m}) \cdot \omega \cos(\omega t) \). For the y-component: \( v_y(t) = \frac{dy}{dt} = 0 - (-5.00\,\mathrm{m}) \cdot \omega \sin(\omega t) \). Evaluate these at \( t = 0 \) to find \( v_x(0) \) and \( v_y(0) \).
02

Find the Components of Acceleration at t=0

The acceleration components are given by the second derivative of the position functions with respect to time. For the x-component: \( a_x(t) = \frac{d^2x}{dt^2} = -\omega^2 \cdot -(5.00\,\mathrm{m}) \sin(\omega t) \). For the y-component: \( a_y(t) = \frac{d^2y}{dt^2} = -\omega^2 \cdot -(5.00\,\mathrm{m}) \cos(\omega t) \). Evaluate these at \( t = 0 \) to obtain \( a_x(0) \) and \( a_y(0) \).
03

Write Expressions for the Position Vector

The position vector \( \mathbf{r}(t) \) is given in vector form combining the x and y position equations: \( \mathbf{r}(t) = x(t)\mathbf{i} + y(t)\mathbf{j} = -(5.00\,\mathrm{m}) \sin(\omega t)\mathbf{i} + ((4.00\,\mathrm{m})-(5.00\,\mathrm{m}) \cos(\omega t))\mathbf{j} \).
04

Write Expressions for the Velocity Vector

Based on the first derivatives found in Step 1, the velocity vector \( \mathbf{v}(t) \) can be expressed as: \( \mathbf{v}(t) = v_x(t)\mathbf{i} + v_y(t)\mathbf{j} = 5.00\,\mathrm{m}\omega \cos(\omega t)\mathbf{i} + 5.00\,\mathrm{m}\omega \sin(\omega t)\mathbf{j} \).
05

Write Expressions for the Acceleration Vector

Using second derivatives found in Step 2, the acceleration vector \( \mathbf{a}(t) \) is: \( \mathbf{a}(t) = a_x(t)\mathbf{i} + a_y(t)\mathbf{j} = (5.00\,\mathrm{m}\omega^2 \sin(\omega t))\mathbf{i} - (5.00\,\mathrm{m}\omega^2 \cos(\omega t))\mathbf{j} \).
06

Describe the Path of the Object

The equations for \(x\) and \(y\) suggest a circular motion with a radius of 5.00 m, centered at (0, 4.00 m) in the xy-plane, since \(x\) and \(y\) are sinusoidal functions out of phase by 90 degrees. The negative sign in the \(x\) position function indicates the motion starts at the max x-value and moves clockwise.

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Key Concepts

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

Kinematics in Two Dimensions
Kinematics is the branch of classical mechanics that describes the motion of objects without considering the forces that cause the motion. In two dimensions, this involves motion along the xy-plane, where an object's position is given by coordinates \(x, y\). To fully understand two-dimensional kinematics, one must be familiar with vector quantities like displacement, velocity, and acceleration, and how they vary with time.

When dealing with problems like the movement of an object in the xy-plane, we translate the object's motion into mathematical expressions. In this case, parametric equations represent the x and y coordinates of the object as separate functions of time, which allow us to analyze the motion along each axis independently. For more comprehensive analysis, these parametric equations are often differentiated or integrated to find the associated velocity and acceleration vectors.
Velocity Vector
The velocity of an object in motion is a vector quantity that describes both the speed and direction of the object's movement. In two-dimensional kinematics, the velocity vector has both an x-component \(v_x\) and a y-component \(v_y\). These components can be calculated by taking the time derivative of the position functions.

For example, in the given exercise, the velocity components at any time t can be written as \( v_x(t) = 5.00\,\mathrm{m}\omega \cos(\omega t) \) and \( v_y(t) = 5.00\,\mathrm{m}\omega \sin(\omega t) \) by deriving the x and y position functions. At \( t = 0 \) specifically, the values of \( v_x(0) \) and \( v_y(0) \) give us the initial velocity vector. It's important to recognize that while the magnitude of the velocity can give us speed, the vector nature of velocity is essential for understanding the direction of motion in the plane.
Acceleration Vector
Acceleration is another vector quantity that represents the rate of change of velocity. In the context of two-dimensional motion, this can be separated into components as well鈥擻( a_x \) for the x-direction and \( a_y \) for the y-direction. These components are found by taking the second derivative of the position functions with respect to time, or equivalently, the first derivative of the velocity functions.

For instance, the object moving in the xy-plane in our exercise has acceleration components \( a_x(t) \) and \( a_y(t) \) given by the second derivatives of the parametric equations for x and y. At time \( t = 0 \) specifically, \( a_x(0) \) and \( a_y(0) \) provide the initial acceleration vector, which is crucial for understanding how the object's velocity is changing at the start of our observation.
Parametric Equations
Parametric equations are a pair (or set) of equations where the coordinates are expressed as functions of a common parameter, typically time (t). This approach is especially useful in two-dimensional kinematics, as it allows for a more straightforward analysis of the motion along each axis.

As demonstrated in our problem, the object's x and y coordinates are represented by separate functions linked by the parameter \( \omega t \). These equations encapsulate information about the position of the object as it moves through time. Utilizing parametric equations, we can easily calculate the derivative to find velocity and acceleration vectors, as well as deal with more complex paths like circular or projectile motion.
Sinusoidal Motion
Sinusoidal motion is characterized by the sinusoidal (sine or cosine) waveforms in the position functions of an object. This type of motion is periodic and can be found in systems that oscillate back and forth, such as pendulums, springs, or in this exercise, circular motion. When the position functions of the object involve both sine and cosine terms offset by a phase of 90 degrees, as they do here, the resultant motion is circular.

Specifically, the exercise given outlines motion starting at the maximum x-value and moving clockwise, as indicated by the negative sign in the \(x\) position function. The presence of sinusoidal functions in parametric equations often simplifies computation of velocity and acceleration, since the derivatives of sine and cosine functions are well-understood and lead to motion which is readily describable in terms of circular paths or harmonic oscillators.

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