Problem 114 - Fundamentals of Physics

Question

The position vector \(\vec{r}\) of a particle moving in the \(x y\) plane is \(\vec{r}=2 \hat{i}+2 \sin [(\pi / 4 \mathrm{rad} / \mathrm{s}) t] \hat{\mathrm{j}}, \quad\) with \(\vec{r}\) in meters and \(t\) in seconds. (a) Calculate the \(x\) and \(y\) components of the particle's position at \(t=0,1.0,2.0,3.0\), and \(4.0 \mathrm{~s}\) and sketch the particle's path in the \(x y\) plane for the interval \(0 \leq t \leq\) \(4.0 \mathrm{~s}\). (b) Calculate the components of the particle's velocity at \(t=1.0,2.0\), and \(3.0 \mathrm{~s}\). Show that the velocity is tangent to the path of the particle and in the direction the particle is moving at each time by drawing the velocity vectors on the plot of the particle's path in part (a). (c) Calculate the components of the particle's acceleration at \(t=1.0,2.0\), and \(3.0 \mathrm{~s}\).

Step-by-Step Solution

Answer

The particle oscillates in the \( y \)-direction; velocities are tangent; and acceleration is sinusoidal.

Step 1: Extract Position Components

The position vector is given by \( \vec{r} = 2 \hat{i} + 2 \sin\left(\frac{\pi}{4} t\right) \hat{j} \). For any time \( t \), the components of the position are \( x(t) = 2 \) and \( y(t) = 2 \sin\left(\frac{\pi}{4} t\right) \).

Step 2: Evaluate Position at Given Times

Calculate \( x \) and \( y \) for \( t = 0, 1.0, 2.0, 3.0, 4.0 \) s. - **At \( t = 0 \):** \( x = 2 \), \( y = 2 \sin(0) = 0 \)- **At \( t = 1.0 \):** \( x = 2 \), \( y = 2 \sin\left(\frac{\pi}{4}\right) = \sqrt{2} \approx 1.414 \)- **At \( t = 2.0 \):** \( x = 2 \), \( y = 2 \sin\left(\frac{\pi}{2}\right) = 2 \)- **At \( t = 3.0 \):** \( x = 2 \), \( y = 2 \sin\left(\frac{3\pi}{4}\right) = \sqrt{2} \approx 1.414 \)- **At \( t = 4.0 \):** \( x = 2 \), \( y = 2 \sin(\pi) = 0 \)

Step 3: Sketch the Path

The particle's path is a vertical oscillation along the line \( x = 2 \) between \( y = 0 \) and \( y = 2 \). Sketch this motion on the \( xy \) plane.

Step 4: Differentiate Position to Find Velocity

The velocity \( \vec{v} \) is the derivative of \( \vec{r} \) with respect to time: \( \vec{v} = \frac{d}{dt}[2 \hat{i} + 2 \sin(\frac{\pi}{4}t) \hat{j}] = 0 \hat{i} + 2 \cos(\frac{\pi}{4}t) \cdot \frac{\pi}{4} \hat{j} = \frac{\pi}{2} \cos(\frac{\pi}{4}t) \hat{j} \). Thus, \( v_x = 0 \) and \( v_y = \frac{\pi}{2} \cos(\frac{\pi}{4}t) \).

Step 5: Evaluate Velocity Components

Calculate the velocity components at \( t = 1.0, 2.0, 3.0 \) s.- **At \( t = 1.0 \):** \( v_x = 0 \), \( v_y = \frac{\pi}{2} \cos\left(\frac{\pi}{4}\right) = \frac{\pi \sqrt{2}}{4} \approx 1.11 \)- **At \( t = 2.0 \):** \( v_x = 0 \), \( v_y = \frac{\pi}{2} \cos\left(\frac{\pi}{2}\right) = 0 \)- **At \( t = 3.0 \):** \( v_x = 0 \), \( v_y = \frac{\pi}{2} \cos\left(\frac{3\pi}{4}\right) = -\frac{\pi \sqrt{2}}{4} \approx -1.11 \)

Step 6: Velocity Vectors on Path

Draw the velocity vectors on the path. At \( t = 1.0 \) and \( 3.0 \), the velocities are in positive and negative \( y \)-directions, respectively. At \( t = 2.0 \), the velocity is zero, indicating a momentary stop at the maximum displacement.

Step 7: Differentiate Velocity to Find Acceleration

The acceleration \( \vec{a} \) is the derivative of velocity:\( \vec{a} = \frac{d}{dt}[\frac{\pi}{2} \cos(\frac{\pi}{4}t) \hat{j}] = -\frac{\pi}{2} \cdot \frac{\pi}{4} \sin(\frac{\pi}{4}t) \hat{j} = -\frac{\pi^2}{8} \sin(\frac{\pi}{4}t) \hat{j} \).Thus, \( a_x = 0 \) and \( a_y = -\frac{\pi^2}{8} \sin(\frac{\pi}{4}t) \).

Step 8: Evaluate Acceleration Components

Calculate the acceleration components at \( t = 1.0, 2.0, 3.0 \) s.- **At \( t = 1.0 \):** \( a_x = 0 \), \( a_y = -\frac{\pi^2}{8} \sin\left(\frac{\pi}{4}\right) = -\frac{\pi^2 \sqrt{2}}{16} \approx -0.87 \)- **At \( t = 2.0 \):** \( a_x = 0 \), \( a_y = -\frac{\pi^2}{8} \sin\left(\frac{\pi}{2}\right) = -\frac{\pi^2}{8} \approx -1.23 \)- **At \( t = 3.0 \):** \( a_x = 0 \), \( a_y = -\frac{\pi^2}{8} \sin\left(\frac{3\pi}{4}\right) = -\frac{\pi^2 \sqrt{2}}{16} \approx -0.87 \)

Key Concepts

Position Vector

A position vector is a crucial concept in describing the location of a particle in space. For a particle moving in the \(xy\) plane, like in our exercise, we use the notation \( \vec{r} = x(t) \hat{i} + y(t) \hat{j} \). This vector represents where the particle is at any given time \( t \).

• The \( x(t) \) component tells us the particle's location along the x-axis.
• The \( y(t) \) component tells us the particle's location along the y-axis.

In our example, the position vector \( \vec{r} = 2 \hat{i} + 2 \sin \left( \frac{\pi}{4} t \right) \hat{j} \) describes a simple oscillation along the y-axis. This means as time passes, this particle oscillates up and down vertically, while remaining at the same horizontal position \( x = 2 \). Reviewing the position at specific times gives insights into the path the particle traces over time.

Velocity Calculation

Velocity is all about how fast something is moving and in which direction. For particles, it's a vector just like position. In this case, we find velocity by differentiating the position vector with respect to time.To calculate the velocity, take the derivative of the position vector \( \vec{r} \). The result is \( \vec{v} = 0 \hat{i} + \frac{\pi}{2} \cos \left( \frac{\pi}{4}t \right) \hat{j} \). This shows us:

• \( v_x = 0 \): There is no movement left or right since the x-component is constant.
• \( v_y = \frac{\pi}{2} \cos \left( \frac{\pi}{4}t \right) \): The motion involves only the vertical direction (y-axis).

The calculation of the velocity at specific times, like \( 1.0, 2.0, \) and \( 3.0 \) seconds, shows how the particle's speed and direction vary with its vertical oscillation. The cosine factor influences the speed and direction, indicating whether the particle is moving upwards or downwards and how swiftly it's doing so.

Acceleration Components

Acceleration is how quickly a particle's velocity changes, either in speed, direction, or both. We find acceleration by differentiating the velocity vector.For the given problem, when we differentiate \( \vec{v} = \frac{\pi}{2} \cos \left( \frac{\pi}{4}t \right) \hat{j} \), we find the acceleration vector \( \vec{a} = 0 \hat{i} - \frac{\pi^2}{8} \sin \left( \frac{\pi}{4}t \right) \hat{j} \). This means:

• \( a_x = 0 \): No acceleration in the x-direction.
• \( a_y = -\frac{\pi^2}{8} \sin \left(\frac{\pi}{4}t \right) \): Acceleration only occurs in the vertical direction.

This vertical acceleration is dependent on the sine function, which fluctuates with time. It shows periodic changes as the particle accelerates upwards or downwards, reversing directions at specific points. Calculating the acceleration at various times reveals these dynamics.

XY Plane

The \(xy\) plane is a flat, two-dimensional surface where we can easily represent positions, paths, and movement of particles. Imagine it as a drawing paper with two axes: horizontal \( x \) and vertical \( y \).In our exercise, the path of the particle is sketched in the \(xy\) plane, which helps visualize how the particle moves over time. Here's how the \(xy\) plane is utilized:

• **Vertical Movement**: The particle oscillates up and down along a straight vertical line at \( x = 2 \).
• **Path Sketching**: By plotting points for different \( t \) values, you create an oscillation pattern.

This plane is essential for understanding how changes in time affect the motion of particles. By observing paths on the \(xy\) plane, we gain insights about directions, velocities, and accelerations visually, making it easier to grasp complex motion concepts.